Hasiak ksiazka

rnz

.11 G)2rnrn3321G)

mOI>z5cn

ENGINEERING MECHANICS

DYNAM ICSTHIRTEENTH EDITION

12Kinematics of a Particie 3

Chapter Object ives 31 2 . 1 I n t r o d u c t i o n 312.2 Rectilinear Kinematics: Continuous Motion 512.3 Rectilinear Kinematics: Erratic Motion 19

12.4 General Curvilinear Motion 3212.5 Curvilinear Motion: Rectangular Components 3412 .6 Mot ion o f a P ro jec t i l e 3912.7 Curvilinear Motion: Normal and Tangential Components 5312.8 Curvilinear Motion: Cylindrical

Components 6712.9 Absolute Dependent Motion Analysis of Two Particles 8112.10 Relative-Motion of Two Particles Using Translating

Axes 87

13Kinetics of aParticie: Force and Acceleration 107

Chapter Object ives 10713 .1 Newton ' s Second Law o f Mo t i on 107 13 .2 The Equa t i on o f Mo t i on 11013.3 Equation of Motion for a System of Particles 11213.4 Equations of Motion: Rectangular Coordinates 11413.5 Equat ions of Mot ion: Normal and

Tangential Coordinates 131 13.6 Equations of Motion: Cyl indrical

Coord inates 144*13.7 Central-Force Motion and Space Mechanics 155

XVIIICONTENTS

14Kinetics of a Particie: Work and Energy 169

Chapter Objectives 16914.1 The Work of a Force 16914.2 Principle of Work and Energy 17414.3 Principle of Work and Energy for a System

of Particles 17614.4 Power and Efficiency 19214.5 Conservative Forces and Potential

Energy 20114.6 Conservation of Energy 205

15Kinetics of aParticie: Impulseand Momentum 221

Chapter Objectives 22115.1 Principle of Linear Impulse and

Momentum 22115.2 Principle of Linear Impulse and Momentum

for a System of Particies 22815.3 Conservation of Linear Momentum for a

System of Particies 23615.4 Impact 24815.5 Angular Momentum 26215.6 Relation Between Moment of a Force and

Angular Momentum 26315.7 Principle of Angular Impulse and

Momentum 26615.8 Steady Flow of a Fluid Stream 277*15.9 Propulsion with Variabie Mass 282

Review1.Kinematics and Kinetics of a Particie 298

CONTENTSX I X

1bPlanar Kinematics of a Rigid Body 311

Cha pter Objectives 31116.1 Planar Rigid-Body Motion 31116.2 Translation 31316.3 Rotation about a Fixed Axis 31416.4 Absolute Motion Analysis 32916.5 Relative-Motion Analysis: Velocity 33716.6 Instantaneous Center of Zero Velocity 35116.7 Relative-Motion Analysis:

Acceleration 36316.8 Relative-Motion Analysis using Rotating

Axes 377

17Planar Kinetics of a Rigid Body: Force and Acceleration 395

Chapter Objectives 39517.1 Mass Moment of Inertia 39517.2 Planar Kinetic Equations of Motion 409 17.3 Equations of Motion: Translation 412 17.4 Equations of Motion: Rotation about a

Fixed Axis 42517.5 Equations of Motion: General Piane

Motion 440

XXCONTENTS

18Planar Kinetics of a Rigid Body: Work and Energy

455Chapter Objectives 455

18.1Kinetic Energy 45518.2 The Work of a Force 45818.3 The Work of a Couple Moment 460 18.4 Principle of Work and Energy 462 18.5 Conservation of Energy 477

19Planar Kinetics of a Rigid Body: Impulse and Momentum 495

Chapter Objectives 49519.1 Linear and Angular Momentum 495 19.2 Principle of Impulse and Momentum 501 19.3 Conservation of Momentum 517 *19.4 Eccentric Impact 521

Review2.Planar Kinematics and Kinetics of a Rigid

Body 534

PREFACEX X I

20Three-Dimensional Kinematics of a Rigid Body 549

Chapter Objectives 54920.1 Rotation About a Fixed Point 549

20.2 The Time Derivative of a Vector Measured from Either a Fixed or Translating-Rotating System 552

20.3 General Motion 55720.4 Relative-Motion Analysis Using

Translating and Rotating Axes 566

21Three-Dimensional Kinetics of a Rigid Body 579

Chapter Objectives 579*21.1 Moments and Products of Inertia 57921.2 Angular Momentum 58921.3 Kinetic Energy 592*21.4 Equations of Motion 600*21.5 Gyroscopic Motion 61421.6 Torque-Free Motion 620

XXIIPREFACE

22Vibrations 631

Chapter Objectives 631*22.1 Undamped Free Vibration 631 *22.2 Energy Methods 645*22.3 Undamped Forced Vibration 651 *22.4 Viscous Damped Free Vibration 655 *22.5 Viscous Damped Forced Vibration 658 *22.6 Electrical Circuit Analogs 661

AppendixA.Mathematical Expressions 670B.Vector Analysis 672C.The Chain Rule 677

Fundamental Problems Partial Solutions and Answers 680

Answers to Selected Problems 701

Index 721

CREDITS

Chaptcr opening images are credited as follows:

Chapter 12, O Bano12007 I Dreamstime.com

Chapter 13, 0 Skvoor I Dreamstime.com

Chapter 14, O Alan Haynes.com / Alamy

Chapter 15, 0 Charles William Lupica /

Alamy Chapter 16, Faraways/Shutterstock.

com Chapter 17, O Toshil2 I Dreamstime.

com Chapter 18, 0 Oleksiy Maksymenko /

Alamy Chapter 19, iofoto/Shutterstock.com

Chapter 20, O Juice Images / Alamy Chapter

21, 0 imagebroker / Alamy Chapter 22, 0 Joe

Belanger / Alamy Cover O Nick M. Do /

iStockphoto.com

Other images provided by the author.

XXI I I

http://Dreamstime.com


http://Haynes.com

http://Shutterstock.com





http://iStockphoto.com

Chapter 1 2ddmiffi.

Although each of these boats is rather large, from a distance their motion canbe analyzed as if each were a particie.

Kinematics of a Particie

CHAPTER OBJECTIVES•To introduce the concepts of position, displacement, velocity,

and acceleration.

•To study particie motion along a straight line and represent this motion graphically.

•To investigate particie motion along a curved path using different coordinate systems.

•To present an analysis of dependent motion of twe particies.

•To examine the principles of relative motion of twe particles using transiating axes.

12.1 Introduction

Mechanice is a branch of the physical sciences that is concerned with the state of rest or motion of bodies subjected to the action of forces. Engineering mechanics is divided into twe areas of study, namely, statics and dynamics. Statics is concerned with the equilibrium of a body that is either at rest or moves with constant velocity. Here we will consider dynamics, which deals with the accelerated motion of a body.The subject of dynamics will be presented in twe parts: kinematics, which trcats only the geometrie aspects of thc motion, and kinetics, which is thc analysis of the forces causing the motion. To develop these principles, the dynamics of a particie will be discussed first, followed by topics in rigid-body dynamics in twa and then three dimensions.

4 C H A P T E R 1 2 K I N E M A T I C S O F A P A R T I C L E

Historically, the principles of dynamics developed when it was12 possible to make an accurate measurement of time. Galileo Galilei

(1564-1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642-1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by Euler, D'Alembert, Lagrange, and others.

There are many problems in engineering whose solutions require application of the principles of dynamics. Typically the structural design of any vehicle, such as an automobile or airplane, requires consideration of the motion to which it is subjected. This is also true for many mechanical devices, such as motors, pumps, movable tools, industrial manipulators, and machinery. Furthermore, predictions of the motions of artificial satellites, projectiles, and spacecraft are based on the theory of dynamics. With further advances in technology, there will be an even greater need for knowing how to apply the principles of this subject.

Problem Solving. Dynamics is considered to be more involved than statics since both the forces applied to a body and its motion musi be taken into account. Also, many applications require using calculus, rather than just algebra and trigonometry. In any case, the most effective way of learning the principles of dynamics is to solve problems. To be successful at this, it is necessary to present the work in a logical and orderly manner as suggested by the following sequence of steps:

1. Read the problem carefully and try to correlate the actual physical situation with the theory you have studied.

2. Draw any necessary diagrams and tabulate the problem data.3. Establish a coordinate system and apply the relevant principles,

generally in mathematical form.

4. Solve the necessary equations algebraically as far as practical; then, use a consistent set of units and complete the solution numerically. Report the answer with no more significant figures than the accuracy of the given data.

5. Study the answer using technical judgment and common sense to determine whether or not it seems reasonable.

6. Once the solution has been completed, review the problem. Try to think of other ways of obtaining the same solution.

In applying this general procedure, do the work as neatly as possible. Being neat generally stimulates elear and orderly thinking, and vice versa.

12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION5

12.2 Rectilinear Kinematics: Continuous Motion

We will begin our study of dynamics by discussing the kinematics of a particie that moves along a rectilinear or straight-line path. Recall that a particie has a mass but negligible size and shape. Therefore we must limit application to those objects that have dimensions that are of no consequence in the analysis of the motion. In most problems, we will be interested in bodies of finite size, such as rockets, projectiles, or vehicies. Each of these objects can be considered as a particie, as long as the motion is characterized by the motion of its mass center and any rotation of the body is neglected.

Rectilinear Kinematics. The kinematics of a particie is characterized by specifying, at any given instant, the particie's position, veiocity, and acceieration.

Position. The straight-line path of a particie will be defined using a single coordinate axis s, Fig. 12-1a. The origin O on the path is a fixed point, and from this point the position coordinate s is used to specify the location of the particie at any given instant. The magnitude of s is the distance from O to the particie, usually measured in meters (m) or feet (ft), and the sense of direction is defined by the algebraic sign on s. Although the choice is arbitrary, in this case s is positive since the coordinate axis is positive to the right of the origin. Likewise, it is negative if the particie is located to the left of O. Realize that position is a vector quantity since it has both magnitude and direction. Here, however, it is being represented by the algebraic scalar s since the direction always remains along the coordinate axis.

SO

s --I

Position

(a )

Displacement. The displacement of the particie is defined as the change in its position. For example, if the particie moves from one point to another, Fig. 12-1b, the displacement is

s

As1

As = s ' — s

In this case As is positive since the particies finał position is to the right of its initial position, i.e., s' > s. Likewise, if the final position were to the left of its initial position, As would be negative.

The displacement of a particie is also a vecłor quantity, and it shouid be distinguished from the distance the particie travels. Specifically, the distance traveled is a positive scalar that represents the total length of path over which the particie travels.

Displacement

(c )

Fig. 12-1


Velocity. If the particie moves through a displacement As during the12 time interval At, the average velocity of the particie during this time

interval is

Asvavg = At

If we take smaller and smaller values of At, the magnitude of As becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as v = li m (As/At), or

—■0

= d sd —d t

(12-1)

v

Since At or dt is always positive, the sign used to define the sense of the velocity is the same as that of As or ds. For example, if the particie is moving to the right, Fig. 12-1c, the velocity is positive; whereas if it is moving to the left, the velocity is negative. (This is emphasized here by the arrow written at the left of Eq. 12-1.) The magnitude of the velocity is known as the speed, and it is generally expressed in units of m/s or ft/s.

Occasionally, the term "average speed" is used. The average speed is always a positive scalar and is defined as the total distance traveled by a particie, sT, divided by the elapsed time At; i.e.,

s

— IO

V e l o c i t y

( c )

ST( V s p ) „ = —

g A l

For example, the particie in Fig. 12-1d travels along the path of length sTin time At, so its average speed is (vsp)„,g = sT/At, but its average velocity is vavg = -Asi At.

I— os-1

P' PO I . .

TAverage velocity and

Average speed

( d )

Fig. 12-1 (cont.)

s

J


Acceleration. Provided the velocity of the particie is known at two points, the average acceleration of the particie during the time interval At is defined as

A vaav g t

Here &v represents the difference in the velocity during the time interval At, i.e., Av = v' - v, Fig. 12-1e.

The instantaneous acceleration at time t is a vector that is found by taking smaller and smaller vaiues of At and corresponding smaller and smaller values of Av, so that a = lim (A v/At), or

Acceleration

( e )

(d v

a =d t

(12-2)

a

oł

— O O s

v v'

Substituting Eq. 12-1 into this result, we can also write

( a = d2s(112

Both the average and instantaneous acceleration can be either positive or negative. In particular, when the particie is slowing down, or its speed is decreasing, the particie is said to be decelerating. In this case, v' in Fig. 12-1f is less than y, and so Av = v' - u will be negative. Consequently, a will also be negative, and therefore it will act to the len, in the opposite sense to v. Also, notice that if the particie is originally at rest, then it can have an acceleration if a moment later it has a velocity v'; and, if the velocity is constant, then the acceleration is zero since Av = v- v= O. Units commonly used to express the magnitude of acceleration are m/s2 or ft/s2.

Finally, an important differential relation involving the dispiacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between Eqs. 12-1 and 12-2, which gives

a

P P 's

OV

Deceleration

(f)

Fig. 12-1 (cont.)

a ds = v dv (12-3)

Although we have now produced three important kinematic equations, realize that the above equation is not independent of Eqs. 12-1 and 12-2.


Constant Acceleration, a ac. When the acceleration is12 constant, each of the three kinematic equations = dv/dt, v = ds/dt,

and ac ds = v dv can be integrated to obtain formulas that relate ac, v, s, and t.

Velocity as a Function of Time. Integrate a, = dvIdt,assuming that initially v = vo when t = O.

vd v = a , d t

o

-±› )v = vo + act

Constant Acceleration (12-4)

Position asa Function of Time. Integrate v = ds/dt = vo + art, assuming that initially s = so when t = O.

ds = (vo + ant) dtso o

(s = so + vot +14,ł2Constant Acceleration (12-5)

Velocity as a Function of Position. Either solve for t in Eq.12-4 and substitute into Eq.12-5,or Integrate v dv = acds, assuming that initially v = vo at s = so.

s

J v d v = a c d s O o s o

(V2 = 2a,(s — so)Constant Acceleration

(12-6)

The algebraic signs of so, vo, and ac, used in the above three equations, are determined from the positive direction of the s axis as indicated by the arrow written at the left of each equation. Remember that these equations are useful only when the acceleration is constant and when t = 0, s = so, v = vo. A typical example of constant accelerated motion occurs when a body falls freely toward the earth. If air resistance is neglected and the distance of fall is short, then the downward acceleration of the body when it is close to the earth is constant and approximately 9.81 m/s2 or 32.2 ft/s2.The proof of this is given in Example 13.2.


rtant Points

• Dynamics is concerned with bodies that have accelerated motion.

• Kinematics is a study of the geometry of the motion.

• Kinetics is a study of the forces that cause the motion.

• Rectilinear kinematics refers to straight-line motion.

• Speed refers to the magnitude of velocity.

• Average speed is the total distance traveled divided by the total time. This is different from the average velocity, which is the displacement divided by the time.

• A particie that is slowing down is decelerating.

• A particie can have an acceleration and yet have zero velocity.

• The relationship a ds = v dv is derived from a = dv/dt and v = ds/dt, by eliminating dt.

During the time this rocket undergoes rectilinear motion, its altitude as a function of time can be measured and expressed as s = s(t). Its velocity can then be found using v = ds/dt, and its acceleration can be determined from a = dv/dt.

Procedure for Analysis

Coordinate System.• Establish a position coordinate s along the path and specify its fixed origin and positive direction.

• Since motion is along a straight line, the vector quantities position, velocity, and acceleration can be represented as algebraic scalars. For analytical work the sense of s, v, and a is then defined by their algebraic signs.

• The positive sense for each of these scalars can be indicated by an arrow shown alongside each kinematic equation as it is applied.

Kinematic Equat ions.

• If a relation is known between any two of the four variables a, v, s, and t, then a third variable can be obtained by using one of the kinematic equations, a = dv I dt, v = ds/dt or a ds = v dv, since each equation relates all three variables.*

• Whenever integration is performed, it is important that the position and velocity be known at a given instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits of integration if a definite integral is used.

• Remember that Eqs. 12-4 through 12-6 have only limited use. These equations apply only when the acceleration is constant and the initial conditions are s = so and v = yo when t = O.

*Some standard differentiation and integration formulas are given in Appendix A.

1 O CHAPTER 12 KINEMATICS OF A PARTICLE

EXAMPLE

12.1

The car on the left in the photo and in Fig. 12-2 moves in a straight line such that for a short time its velocity is defined by v= (312 + 2t) ft/s, where t is in seconds. Determine its position and acceleration when t = 3 s. When t = 0, s = 0.

s - I 9 , V

Fig. 12-2

SOLUTION

Coordi n ate System. The position coordinate extends from the fixed origin O to the car, positive to the right.

Position. Since v = f (t), the car's position can be determined from = dsldt, since this equation relates v, s, and t. Noting that s = O when t =0, we have*

dv= s —dr = (312 + 21)

S T

ds — f (3t2 + 2t)dtf o — o

iis = r3 + r'

os = t3 + t2

When t = 3 s,

s = (3)3 + (3)2 = 36 ft Ans.

Acceleration. Since v = f (t), the acceleration is determined from a =dv I dt, since this equation relates a, v, and t.

a = —dv

= —d (312

+ 21)

dt

= 6t + 2

When t = 3 s,

a = 6(3) + 2 = 20 ft/s2 --) Ans.

NOTE: The formulas for constant acceleration cannot be used to solve this problem, because the acceleration is a function of time.

*The same result can be obtained by evaluating a constant of integration C rather than using definite Iimits on the integral. For example, integrating ds = (312 + 21)dt yields s = t3 + 12 + C. Using the condition that at r = O. s = 0, then C = 0.

1 2


EXAMPLE

A smali projectile is fired vertically downward into a fluid medium with an initial velocity of 60 m/s. Due to the drag resistance of the fluid the projectile experiences a deceleration of a = (-0.4v3) m/ s2, where v is in m/s. Determine the projectile's velocity and position 4 s after it is fired.

SOLUTIONCoordinate System. Since the motion is downward, the position

coordinate is positive downward, with origin located at 0, Fig. 12-3.Velocity. Here a = ,j(v) and so we must determine the velocity as a

function of time using a = dv Idt, since this equation relates v, a, and t. (Why not use v = vo + a ct?) Separating the variables and integrating, with vc, = 60 m/s when t = 0, yields

dv(+ 1) a = — = -0.43

dtFig. 12-3

vdv

— = d t f 6 0m / s - 0 . 4v 3 jo

( 1 -0.4 -2.).)2

1 E 1

= t - 0601

—08 (60)2 J= t1/2

= {[(60)2 0.8I] m/s

Here the positive root is taken, since the projectile will continue to move downward. When t = 4 s,

v = 0.559 m/s1 Ans.

Position. Knowing v = f (t), we can obtain the projectile's position from v = ds I dt, since this equation relates s, v, and t. Using the initial condition s = 0, when t = 0, we have

(+

When t = 4 s,

v = ds —d

ds =

s _

1s = —

0.4

=t

f

0.8

1

dt

to

60

(60 — + 0.8t )2

1 [— + 0.81]a (60)2

f l + 0.8ti1/2

I (6o)2I/2

÷ 0.81] —(60)-

s = 4.43m

12 CHAPTER 12 KINEMATICS OF A PARTICLE

Fig. 12-4

During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails. Determine the maximum height S B reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s2 due to gravity. Neglect the effect of air resistance.

SOLUTION

Coordinate System. The origin O for the position coordinate s is taken at ground level with positive upward, Fig. 12-4.

Maximum Height. Since the rocket is traveling upward, vA = +75 m/s when t = 0. At the maximum height s = ss the velocity vB = 0. For the entire motion, the acceleration is a. = -9.81 m/s'` (negative since it acts in the opposite sense to positive velocity or positive displacement). Since is constant the rocket's position may be related to its velocity at the two points A and B on the path by using Eq. 12-6, namely,

(+?) v2B = 7A + 2ac(sa - SA)

O = (75 m/s)2 + 2(-9.81 m/s2)(4 - 40 m)ss = 327 m A

Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12-6 between points B and C, Fig. 12-4.

(+T) = v2s + 2at(sc - SB)

= O 2(-9.81 m/s2)(0 - 327 m)

= -80.1 m/s = 80.1 m/s Ans.

The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12-6 may also be applied between points A and C, i.e.,

(+f) = v2A + 2ae(sc SA)

= (75 m/s)2 + 2(-9.81 m/s2)(0 - 40 m)= -80.1 m/s = 80.1 m/s Ans.

NOTE: It should be realized that the rocket is subjected to a deceleration from A to B of 9.81 m/s2, and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at B (v B = 0) the acceleration at B is still 9.81 m/s2 downward!

12


EXAMPLE

12.4

A metallic particie is subjected to the influence of a magnetic field as it travels downward through a fluid that extends from plate A to plate B, Fig. 12-5. If the particie is released from rest at the midpoint C, s = 100 mm, and the acceleration is a = (4s) m/s2, where s is in meters, determine the velocity of the particie when it reaches plate B, s = 200 mm, and the time it takes to travel from C to B.

SOLUTIONCoordinate System. As shown in Fig. 12-5,s is positive downward, measured from plate A.

Velocity. Since a = .f(s), the velocity as a function of position can be obtained by using v dv = a ds. Realizing that v = O at s = 0.1 m, we have

v dv = a dsv s

f o v d v = f 4 s d s0.1 in

1 v 2 1 = s 2 '2 l o 2 1 0 . 1 , „

v = 2(s2 — 0.01)1/2 m/s (1)

At s = 200 mm = 0.2 m,

vił = 0.346 m/s = 346 mm/s 1 Ans.

The positive root is chosen since the particie is traveling downward, i.e., in the +s direction.Time. The time for the particie to travel from C to B can be obtained using v = ds/dt and Eq. 1, where s = 0.1 m when t = 0. From Appendix A,

(+ ds = v dt

= 2(52 — 0.01)1/2dt[s ds — d tJ , a ( s 2 — 0 . 0 1 ) 1 / 2 o

I n ( N i s 2 — 0 . 0 1 + s ) S = 2 t

0.1 lf

oln(N7s2 — 0.01 + s) + 2.303 = 2t

200 mm

Fig. 12-5

A

At s = 0.2 m,

t —2 — 0.658 s Ans.

NOTE: The formulas for constant acceleration cannot be used here because the acceleration changes with position, i.e., a = 4s.

In( \/(0.2)2 — 0.01 + 0.2) + 2.303


12

t (s)

(1 s, —3 m/s)

(b)

v = 3t2 —

(2 s, O)

In order to determine the distance traveled in 3.5 s, it is necessary to investigate the path of motion. If we consider a graph of the velocity function, Fig. 12-6b, then it reveals that for O < t < 2 s the velocity is negative, which means the particie is traveling to the lep, and for t > 2 s the velocity is positive, and hence the particie is traveling to the right. Also, note that v = O at t = 2 s. The particle's position when t = 0, t = 2 s, and t = 3.5 s can be determined from Eq. 1. This yields

Sit=0 = O Slt=2s = 4.0m s t=3.5, = 6.125mFig. 12-6

EXAMPLE

A particie moves along a horizontal path with a velocity of v = (3/2 -

6t) m/s, where t is the time in seconds. If it is initially located at the origin 0, determine the distance traveled in 3.5 s, and the particle's average velocity and average speed during the time interval.

SOLUTIONs = —4.0 m s = 6.125 m Coordinate System. Here positive motion is to the right, measured

from the origin 0, Fig. 12-6a.ODistance Traveled. Since v = f (t), the position as a function of time

t = 2 s t = O s t = 3.5 s may be found by integrating v = ds/dt with t = 0, s = 0.

(a) (;)ds = v dt= (3t2 - 6t) dt

jos ds = (312 - 61) dt

os = (t3 - 3t2) m

The path is shown in Fig. 12-6a. Hence, the distance traveled in 3.5 s is

sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m Ans

Velocity. The displaeement from t = O to t = 3.5 s isAs = s I t=3.5, — Sit=0 = 6.125 m - O = 6 125 m

and so the average velocity is

As 6.125 m v a " A t 3 . 5 s - O 1 . 7 5 m i s — >

The average speed is defined in terms of the distance traveled sT. This positive scalar is

(v ) = sT 14.125 m 4.04 m/ssP avg At 3.5 s - O

NOTE: In this problem, the acceleration is a = dv I dt = (6r - 6) m/s2, which is not constant.

(1)

Ans.

12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTiON 15

It is highly suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equations of kinematics must be used and how they are applied in order to determine the unknowns. Then before solving any of the problems, try and solve some of the Fundamental Problems given below.The solutions and answers to all these problems are given in the back of the book. Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills.

• FUNDAMENTAL PROBLEMS

F12-1. Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 10 m/s in 15 s, determine the constant deceleration of the car.

F12-5. The position of the particie is given by s = (212 - 8r +6) m, where t is in seconds. Determine the time when the velocity of the particie is zero, and the total distance traveied by the particie when t = 3 s._4~

—Tł>

FU-5

s

F 1 2 - 1

F12-2. A bali is thrown vertically upward with a speed of 15 m/s. Determine the time of flight when it retums to its original position.

F12-6. A particie travels aiong a straight line with an acceleration of a = (10 - 0.2s) m/s2, where s is measured in meters. Determine the velocity of the particie when s= 10 m if = 5 m/s at s = O.

• s

S

FU-6

F 1 2 - 2

F12-3. A particie travels aiong a straight line with a velocity of v = (4t - 31 .2) m/s, where t is in seconds. Determine the position of the particie when t = 4 s. s= O when t = O.

F12-7. A particie moves aiong a straight line such that its acceleration is a = (4/2 - 2) m/s2, where t is in seconds. When t = 0, the particie is located 2 m to the left of the angin, and when r = 2 s, it is 20 m to the left of the angin. Determine the position of the particie when t = 4 s.

F12-4. A particie travels aiong a straight line with a speedv = (0.513 - m/s, where t is in seconds. Determine theacceleration of the particie when t = 2 s.

F12-8. A particie travels along a straight line with a velocity of v = (20 - 0.05s2) m/s, where s is in meters. Determine the acceleration of the particie at s = 15 m.

1 6 CHAPTER 1 2 KINEMATICS OF A PARTICLE

1 2 ■ P R O B L E M S

12-1. A baseball is thrown downward from a 50-ft tower with an initial speed of 18 ft/s. Determine the speed at which it hits the ground and the time of travel.

12-2. When a tram is traveling along a straight track at 2 m/s, it begins to accelerate at a = (60 v-;) m/s2, where v is in m/s. Determine its velocity v and the position 3 s after the acceleration.

Prob. 12 -2

12-3. From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed of 80.7 ft/s (55 mi/h) when it hits the ground? Each floor is 12 ft highcr than the one below it. (Note: You may want to remember this when traveling 55 mi/h.)

*12-4. Traveling with an initial speed of 70 km/h, a car accelerates at 6000 km/h2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

12-5. A bus starts from rest with a constant acceleration of 1 m/s2. Determine the time required for it to attain a speed of 25 m/s and the distance traveled.

12-6. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

12-7. A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine his acceleration if it is constant. Also, how long does it take to reach the speed of 30 km/h?

■12-8. A particie moves along a straight line with an acceleration of a = 51(15'13 + ss12) m/s2, where s is in meters. Determine the particle's velocity when s = 2 m, if it starts from rest when s = 1 m. Use a numerical method to evaluate the integral.

12-9. If it takes 3 s for a bali to strike the ground when it is released from rest, determine the height in meters of the building from which it was released. Also, what is the velocity of the bali when it strikes the ground?

12-10. The position of a particie along a straight line is given by s = (1.5t3 - 13.5,2 + 22.50 ft, where t is in seconds. Determine the position of the particie when t= 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

12-11. If a particie has an initial velocity of yo = 12 ft/s to the right, at so = 0, determine its position when t = 10 s, if a = 2 ft/s2 to the left.

*12-12. Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the mad. The car can accelerate at 1.5 m/s2 and decelerate at 2 m/s2.

12-13. Tests reveal that a norma) driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 2 ft/s2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don't drive!

Prob. 12-13

12.2 RECTIUNEAR KINEMATICS: CONTINUOUS MOTION17

12-14. A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 0.6 ft/s2, decelerate at 0.3 ft/s2, and reach a maximum speed of 8 ft/s, determine the shortest time to make the lift, starting from rest and ending at rest.

12-15. A train starts from rest at station A and accelerates at 0.5 m/s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m /s2 until it is brought to rest at station B. Determine the distance between the stations.

*12-16. A particie travels along a straight line such that in 2 s it moves from an initial position sA = +0.5 m to a position SB = —1.5 m. Then in another 4 s it moves from .59 to sc = +2.5 m. Determine the particle's average velocity and average speed during the 6-s time interval.

12-17. The acceleration of a particie as it moves along a straight line is given by a = (2: — 1) m/s2, where t is in seconds. If s = 1 m and v = 2 m/s when t = 0, determine the particle's velocity and position when t = 6 s. Also, determine the total distance the particie travels during this time period.

12-18. A freight train travels at v = 60(1 — e-1) ft/s, where t is the elapsed time in seconds. Determine the distance traveled in three seconds, and the acceleration at this time.

*12-20. The velocity of a particie traveling along a straight line is v = (3t2 — 6t) ft/s, where t is in seconds. If s = 4 ft when t = 0, determine the position of the particie when t = 4 s. What is the total distance traveled during the time interval t = 0 to i = 4 s? Also, what is the acceleration when t = 2 s?

12-21. If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a = 9.81[1 — V2 (10 —)1 m/s2, where v is in m/s and the positive direction is downward. H the body is released from rest at a very high alatude, determine (a) the velocity when t = 5 s, and (b) the body's terminal or maximum attainable velocity (as t —> oo ).

12-22. The position of a particie on a straight line is givenby s = (t 3 — 912 15t) ft, where t is in seconds. Determinethe position of the particie when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

12-23. Two partides A and B start from rest at the origins = O and move along a straight line such that aA = (6t — 3) ft/s2 and aB = (12/2 — 8) ft/s2, where t is in seconds. Determine the distance between them whent = 4 s and the total distance each has traveled in t = 4 s.

*12-24. A particie is moving along a straight line such that its velocity is defined as v = (-4s2) m/s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time.

12-25. A sphere is fired downwards finto a medium with an initial speed of 27 m/s. If it experiences a deceleration of a= (-6t) m/s2, where t is in seconds, determine the distance traveled before it stops.

12-26. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vp, respectively. If 8 maintains its constant speed, while A begins to decelerate at aA, determine the distance d between the cars at the instant A stops.

Prob. 12-18A B

12-19. A particie travels to the right along a straight line with a velocity v = [5/(4 + s)] m/s, where s is in meters. Determine its position when i = 6 s if s = 5 m when t = 0.

4 4 1 1 § k l e i g ~ . _

Prob. 12-26

1 8 CHAPTER 1 2 K I N E M A T I C S O F A P A R T I C L E

12-27. A particie is moving along a straight line such that 12 when it is at the origin it has a velocity of 4 m/s. If it begins to decelerate at the rate of a = (-1.5v 1/2) m/s2, where v is in m/s,

determine the distance it travels before it stops.

12-34. A boy throws a bali straight up from the top of a 12-m high tower. If the bali falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the bali when it strikes the ground, and the time of flight.

*12-28. A particie travels to the right along a straight line with a velocity v = [5/(4 + s)] m/s, where s is in meters. Determine its deceleration when s = 2 m.

12-29. A particie moves along a straight line with an acceleration a = 2 v% m/s2, where vis in m/s. If s= 0, v =4 m/s when t = 0, determine the time for the particie to achieve a velocity of 20 m/s. Also, find the displacement of particie when t= 2 s.

12-30. As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m/s and then 10 m/s. Determine the train's velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.

12-31. The acceleration of a particie along a straight line is defined by a = (21 — 9) m/s2, where r is in seconds. At t =0, s = 1m and v = 10 m/s. When t = 9 s, determine (a) the particle's position, (b) the total distance traveled, and (c) the velocity.

*12-32. The acceleration of a particie traveling along a1

straight line is a =— 4 m/s2, where s is in meters. If v = 0, s

=1 m when t= 0, determine the particle's velocity at s =2 m.

12-33. At t = O bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

12-35. When a particie falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf If this variation of the acceleration can be expressed asa = (g/v2f)(v2f — v2), determine the time needed for the velocity to become v = vf/2. Initially the particie falls from rest.

*12-36. A particie is moving with a velocity of vo when s = O and t = 0. If it is subjected to a deceleration of a = —kv3, where k is a constant, determine its velocity and position as functions of time.

12-37. As a body is projected to a high altitude above the earth's surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formuła a = —go[R2/(R + y)2], where go is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If go = 9.81 m/s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth's surface so that it does not fali back to the earth. Hint: This requires that v = O as y —) a.

12-38. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12-37), derive an equation that relates the velocity of a freely falling particie to its altitude. Assume that the particie is released from rest at an altitude yo from the earth's surface. With what velocity does the particie strike the earth if it is released from rest at an altitude yo = 500 km? Use the numerical data in Prob. 12-37.

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION19

12.3 Rectilinear Kinematics: Erratic Motion

When a particie has erratic or changing motion then its position, velocity, and acceleration cannot be described by a single continuous mathematical function along the entire path. Instead, a series of functions will be required to specify the motion at different intervals. For this reason, it is convenient to represent the motion as a graph. If a graph of the motion that relates any two of the variables s,v, a, t can be drawn, then this graph can be used to construct subsequent graphs relating two other variables since the variables are related by the differential relationships v = ds/dt, a = dv I dt, or a ds = v dv. Several situations occur frequently.

The s-t, v-t, and a-t Gra phs. To construct the v-t graph given the s-t graph, Fig. 12-7a, the equation v = ds/dt should be used, since it relates the variables s and t to v. This equation states that

ds— = v dt

slope of= velocity

s-t graph

For example, by measuring the slope on the s-t graph when t = tl, the velocity is vl , which is plotted in Fig. 12-7b. The v-t graph can be constructed by plotting this and other values at each instant.

The a-t graph can be constructed from the v-t graph in a similar manner, Fig. 12-8, since

d v = a d tslope of

= accelerationv-t graph

Examples of various measurements are shown in Fig. 12-8a and plotted in Fig. 12-8b.

If the s-t curve for each interval of motion can be expressed by a mathematical function s = s(t), then the equation of the v-t graph for the same interval can be obtained by differentiating this function with respect to time since v = ds/dt. Likewise, the equation of the a-t graph for the same interval can be determined by differentiating v = v(t) since a =dv I dt. Since differentiation reduces a polynomial of degree n to that of degree n - 1, then if the s-t graph is parabolic (a second-degree curve), the v-t graph will be a stoping line (a first-degree curve), and the a-t graph will be a constant or a horizontal line (a zero-degree curve).

_ ds'vo drit = '2-7112

_ds _ dsi7114 U3dti4

,41111(

2 (a)

5 3

O

( b )

Fig. 12-7

U

1 2

(a)

(b)

Fig. 12-8


tli

— av = a ch r

1 2

t

(a)

If the a—t graph is given, Fig. 12-9a, the v—t graph may be constructed using a = dv/dt, written as

v = J a d t

change in _ area undervelocity a—t graph

Hence, to construct the v—t graph, we begin with the particle's initial velocity vo and then add to this smali increments of area (Av) determined from the a—t graph. In this manner successive points, v, = vo + A v, etc., for the v—t graph are determined, Fig. 12-9b. Notice that an algebraic addition of the area increments of the a—t graph is necessary, since areas lying above the t axis correspond to an increase in v ("positive" area), whereas those lying below the axis indicate a decrease in v ("negative" area).

Similarly, if the v—t graph is given,Fig.12-10a, it is possible to determine the s—t graph using v = ds I dt, written as

(b)

Fig. 12-9

t

& s = v d t

displacement =area underv—t graph

( a )

(b)

Fig. 12-10

In the same manner as stated above, we begin with the particle's initial position so and add (algebraically) to this smali arca increments As determined from the v—t graph, Fig. 12-10b.

If segments of the a—t graph can be described by a series of equations, then each of these equations can be integrated to yield equations describing the corresponding segments of the v—t graph. In a similar manner, the s—t graph can be obtained by integrating the equations which describe the segments of the v—t graph. As a result, if the a—t graph is linear (a first-degree curve), integration will yield a v—t graph that is parabolic (a second-degree curve) and an s—t graph that is cubic (third-degree curve).

1 2.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 2 1

The v-s and a-s Graphs. If the a-s graph can be constructed, then points on the v-s graph can be determined by using v dv = a ds. Integrating this equation between the limits v = vc, at s = so and v = v, at s = si , we have,

a

[s,joa ds = ł (v

z 1 2 — Y0 2 )

s

( a )4 ( v ; - v 8 ) =

rs I a d s

arca under a-s graph

UI

U p

Therefore, if the red area in Fig. 12-11a is determined, and the initial velocity vo at so = O is known, then v1 = (2 f: Ja ds + vW72, Fig. 12-11b. Successive points on the v-s graph can be constructed in this manner.

If the v-s graph is known, the acceleration a at any position s can be determined using a ds = v dv, written as

v( dvd s

velocity timesacceleration = slope of

v-s graph

a =

(b )

Fig. 12-11

Vo

I . - S . - . I

(a)

z

a = 'v(dvIds)

s'

(b)

Fig. 12-12

Thus, at any point (s, v) in Fig. 12-12a, the slope dv/ds of the v-s graph is measured.Then with v and dv/ds known, the value of a can be calculated, Fig. 12-12b.

The v-s graph can also be constructed from the a-s graph, or vice versa, by approximating the known graph in various intervals with mathematical functions, v = f(s) or a = g(s), and then using a ds = v dv to obtain the other graph.


A bicycle moves along a straight road such that its position is described by the graph shown in Fig. 12-13a. Construct the v-t and a-t graphs for O 5 t 5-30 s.

s (ft)

500

1 0 0

( a )

a (ft/s2)

SOLUTIONv-t Graph. Since v = ds I dt, the v-t graph can be determined by differentiating the equations defining the s-t graph,Fig.12-13a.We have

dsO 5- t < 10 s; s = (t2) ft v = — = (20 ft ł s

dtds

10 s < t 5- 30 s; s = (20t - 100) ft v = —dt = 20 ft/s

The results are plotted in Fig. 12-13b. We can also obtain specific values of v by measuring the slope of the s-t graph at a given instant. For example, at t = 20 s, the slope of the s-t graph is determined from the straight line from 10 s to 30 s, i.e.,

A s 5 0 0 f t - 1 0 0 f tt = 20 s; v - - - 20 ft/s0130 s - 10 s

a-t Graph. Since a = dv dt, the a-t graph can be determined by differentiating the equations defining the lines of the v-t graph. This yields

O S t < 10S; v = (2t) ft/s

10 < t 30 s; V = 20 ft/s

The results are plotted in Fig. 12-13c.

NOTE: Show that a = 2 ft/s2 when t = 5 s by measuring the slope of the v-t graph.

y (ft/s)

= 2/ v = 2020

10 t (s)

(b)

I0 30 t (s)

( c )

Fig. 12-13

dva= — = 2 ft/s2

dr

a= —dv =0dt

1 2

12.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 23

EXAMPLE

The car in Fig. 12-14a starts from rest and travels along a straight track such that it accelerates at 10 m/s2 for 10 s, and then decelerates at 2 m/s2. Draw the v—t and s—t graphs and determine the time t' needed to stop the car. How far has the car traveled?

SOLUTION

v—t Graph. Since dv = a dt, the v—t graph is determined by integrating the straight-line segments of the a—t graph. Using the initial condition v = O when t = 0, we have

0 t < 10 s; a = (10) ml s2;

When t = 10 s, v = 10(10) = 100 m/s. Using this as the initial condition for the next time period, we have

10 s < t t'; a = (-2) mjs2; f dv = f —2 dt, v = (-2t + 120) m/s

mo m/s ur sWhen t = t' we require v = 0. This yields, Fig. 12-14b,

t' = 60 s Ans.

A more direct solution for t' is possible by realizing that the area under the a—t graph is equal to the change in the car's velocity. We require Av = O = Al + A2, Fig. 12-14a. Thus

O = 10 m/s2(10 s) + (-2 m/s2)(ts — 10 s)

t' = 60 s Ans.

s—t Graph. Since ds = v dt, integrating the equations of the v—t graph yields the corresponding equations of the s—t graph. Using the initial condition s = O when t = 0, we have

0-ty 10 s; v = (101) ml s; ds = 10t dt, s = (5t2) mo o

When t = 10 s, s = 5(10)2 = 500 m. Using this initial condition,s t

10 s t 60 s; v = (-2t + 120) m/s; f ds = (-2t + 120)dt

500m 10ss — 500 = —t2 + 120t — [—(10)2 + 120(

10)1 s = (-12 + 1201 — 600) m When t' =60 s, the position is

s = —(60)2- + 120(60) — 600 = 3000 m Ans. 500

The s—t graph is shown in Fig. 12-14c.

NOTE: A direct solution for s is possible when t' = 60 s, since the triangular area under the v—t graph would yield the displacement As = s — O from t = O to t' = 60 s. Hence,

As = :12(60 s)(100 m/s) = 3000 m Atu

dv = 10 dt, v = 10tof.

3000

s(m)

( c )

Fig. 12-14

4 (m/s)- ~ 1 1 1 1 1 1 1 . 1 1

10

Ai

t (s )

—2

10

A 2

(a

)

( b )


EXAMPLE

12.8

AJL

200 400

( a )

50

W

y (ft/s)

s (ft)

a (ft /S2)/ s 2 )

a = 0.04s + 2

The v-s graph describing the motion of a motorcycle is shown in Fig. 12-15a. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 400 ft.

SOLUTION

a-s Graph. Since the equations for segments of the v-s graph are given, the a-s graph can be determined using a ds = v dv.

0 s < 200 ft; v = (0.2s + 10) ft/sdv

a = v— = (02s + 10)±(0.2s + 10) = 0.04s + 2ds ds

200 ft < s s 400 ft; v = 50 ft/s

a = v—dv = (50)—d(50)ds = O

d s

The results are plotted in Fig. 12-156.

Time. The time can be obtained using the v-s graph and v = ds I dt, because this equation relates v, s, and t. For the first segment of motion, s = O when t = 0, so

a = O

(

b) Fig.

12-15

200 400

10

2

O s < 200 ft; v = (0.2s + 10) ft/s;ds

dt = =v 0.2s + 10

d s dst =L lo. 0.2s + 10t = (5 In(0.2s + 10) - 5 ln 10) s

s (ft)

At s = 200 ft, t = 51140.2(200) + 10] - 5 In 10 = 8.05 s. Therefore, using these initial conditions for the second segment of motion,

d s d s200 ft < s v 400 ft; v = 50 ft/s;dt = — = —

50

1 dt = is —ds• 8.05 s 2c0 .50

s s ..,rt _ 8 .05 = 50 _ 4 ; t = 50+ 4..u.3) s

Therefore, at s = 400 ft,

t = —400 + 4.05 = 12.0 s Ans.50

NOTE: The graphical results can be checked in part by calculating slopes. For example, at s = 0, a = v(dv/ds) = 10(50 - 10)/200 = 2 m/s2. Also, the results can be checked in part by inspection. The v-s graph indicates the initial increase in velocity (acceleration) followed by constant velocity (a = 0).

12


■ FUNDAMENTAL PROBLEMS

F12-9. The particie travels along a straight track such that F12-12. The sports car travels along a straight road suchits posilion is described by the s-t graph. Construct the v-t that its position is described by the graph. Construct the v-tgraph for the same time interval. and a-t graphs for the time interval O t t 10 s.

s (m) s (m)

411101»_

t (s)I0

t (s)6 8 10 F12-12

F12-9 F12-13. The dragster starts from rest and has anF12-10. A van travels along a straight road with a velocity acceleration described by the graph. Construct the v-tdescribed by the graph. Construct the s-t and a-t graphs graph for the time interval O t s t', where t' is the timeduring the same period. Take s = O when t = 0. for the car to come to rest.

108

O

s = 108

= 0.5 t3

2 2 5

s = 30t 7

7 5s = 3

O5

v (ft/s)

t ( s )5

a (m/s2)

20-

t '

t (s)

O

10

80

v= —4t + 80

F12-10F12-11. A bicycle travels along a straight road where its velocity is described by the v-s graph. Construct the a-s graph for the same time interval.

v (m/s)

10

40

F12-11

F12-13F12-14. The dragster starts from rest and has a velocity described by the graph. Construct the s-1 graph during the time interval O s t s 15 s. Also, determine the total distance traveled during this time interval.

v (mis)

F12-14

150

s (m)

C H A P T E R 1 2 K I N E M A T I C S O F A P A R T I C I E

PROBLEMS

2 6

12 ■12-39. A freight tram starts from rest and travels with a constant acceleration of 0.5 ft/s2. After a time ł it maintains a constant speed so that when t = 160 s it has traveled 2000 ft. Determine the time t' and draw the v—t graph for the motion.

*12-40. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car starts from rest, determine the distance s' the car travels until it stops. Construct the v—s graph for Osss s'.

(ft/s2)

12-43. If the position of a particie is defined by s = [2 sin (7r15), + 41 m, where t is in seconds, construct the s—t, v—t, and a—t graphs for O s_ t s 10 s.

*12-44. An airplane starts from rest, travels 5000 ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. It then climbs in a straight line with a uniform acceleration of 3 ft/s2 until it reaches a constant speed of 220 mi/h. Draw the s—t, v-1, and a—t graphs that describe the motion.

12-45. The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft/s2 and then decelerate at 2 ft/s2. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a—t, v—t, and s—t graphs for the motion.

1000

6

4

l'rob. 12-40

s (ft)40 f t

Prob. 12-45

12-46. The velocity of a car is plotted as shown.Determine the total distance the car moves until it stops (t = 80 s). Construct the a—t graph.

v (m/s)

40 80

Prob. 12-46

10

t (s)

12-41. A tram starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the tram decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v—t graph and determine the maximum speed of the tram.

12-42. A particie starts from s = O and travels along a straight line with a velocity v = (t2 — 4t + 3) m/s, where t is in seconds. Construct the v—t and a—t graphs for the time interval 0s t s 4 s.

2712.3 RECTILINEAR KINEMATICS: ERRATIC MOTION

12-47. The v-s graph for a go-cart traveling on a straight road is shown. Determine the acceleration of the go-cart at s = 50 m and s = 150 m. Draw the a-s graph.

12-50. The v-t graph of a car while traveling along a road is shown. Draw the s-t and a-t graphs for the motion.

v (m/s)

(m)100 200

Prob. 12-47

2 0

20 30t (s)

*12-48. The v-t graph for a particie moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m/s2. If the plates are spaced 200 mm apart, determine the maximum velocity vmax and the time t' for the particie to travel from one plate to the other. Also draw the s-t graph. When t = t'/2 the particie is at s = 100 mm.

12-49. The v-t graph for a particie moving through an eiectric field from one plate to another has the shape shown in the figure, where = 0.2 s and v. = 10 m/s. Draw the s-t and a-t graphs for the particie. When t = t'/2 the particie is at s = 0.5 m.

Prob. 12-50

12-51. The a-t graph of the bullet tram is shown. If the trafie starts from rest, determine the elapsed time t' before it again comes to rest. What is the total distance traveied duńng this time interval? Construct the v-t and s-t graphs.

S m a x " I

Probs. 12-48/49

a (m /s2)

t (s)

Prob. 12-51


*12-52. The snowmobile moves along a straight course 12 according to the v—t graph. Construct the s—t and a—t graphs for the same 50-s time interval. When t = 0, s = 0.

12-54. The dragster starts from rest and has an acceleration described by the graph. Determine the time t' for it to stop. Also, what is its maximum speed? Construct the v—t and s-1 graphs for the time interval O 5 t t'.

v (m/s)

12

3 0

Prob. 12-52

a (ft/s2)

80

t (s)

Prob. 12-54

t ( s )

12-53. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the v—t and s—t graphs which describe the two-stage motion of the missile for O t 5 20 s.

12-55. A race car starting from rest travels along a straight road and for 10 s has the acceleration shown. Construct the v—t graph that describes the motion and find the distance traveled in 10 s.

B A

a = t

a (m/s2)

25

18

t (s)

a (111 IS2)

r (s)15 20

Prob. 12-53

6 10

Prob. 12-55

1 2.3 RECTILINEAR KINEMATICS: ERRATIC MOTION 29

*12-56. The v—t graph for the motion of a car as if moves along a straight road is shown. Draw the a—t graph and determine the maximum acceleration during the 30-s time interval.The car starts from rest at s = 0.

12-57. The v—t graph for the motion of a car as it moves along a straight road is shown. Draw the s—t graph and determine the average speed and the distance traveled for the 30-s time interval. The car starts from rest at s = 0.

12-59. An airplane lands on the straight runway, originally traveling at 110 ft/s when s = 0. If it is subjected to the decelerations shown, determine the time t' needed to stop the piane and construct the s—t graph for the motion.

v (ft/s)

60= t + 30

10 30

Probs. 12-56/57

t (s)

a (ft/s2)

20 t't ( s )

Prob. 12-59

12-58. The jet-powered boat starts from rest at s = O and travels along a straight line with the speed described by the graph. Construct the s—t and a—t graph for the time interval O -s t s 50 s.

*12-60. A car travels along a straight road with the speed shown by the v—t graph. Plot the a—t graph.

12-61. A car travels along a straight road with the speed shown by the v—t graph. Determine the total distance the car travels until it stops when t = 48 s. Also plot the s—t graph.

25 50

Prob. 12-58 Probs. 12-60/61


y = 20 ł - 50

12-66. A two-stage rocket is fired vertically from rest at s = O with an acceleration as shown. After 30 s the first stage A burns out and the second stage B ignites. Plot the v-t graph which describes the motion of the second stagefor O t 60s .

12-67. A two-stage rocket is fired vertically from rest at s .= O with an acceleration as shown. After 30 s the first stage A burns out and the second stage B ignites. Plot the s-t graph which describes the motion of the second stage for Osts 60 s.

12-62. A motorcyclist travels along a straight road with 12 the velocity described by the graph. Construct the s-t and a-t graphs.

v (ft/s)

150

5 0 a (m 152) AB

Probs. 12-66/67

30 60 t (s)

a = 0.01r

15

9

t ( s )1 0

Prob. 12-62

12-63. The speed of a train during the first minute hasbeen recorded as follows:

t (s) O 20 40 60

v (m/s) O 16 21 24

Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

*12-64. A man riding upward in a freight elevator accidentally drops a package off the elevator when it is 100 ft from the ground. If the elevator maintains a constant upward speed of 4 ft/s, determine how high the elevator is from the ground the instant the package hits the ground. Draw the v-t curve for the package during the time it is in motion. Assume that the package was released with the same upward speed as the elevator.

12-65. 'No cars start from rest side by side and travel along a straight road. Car A accelerates at 4 m/s'- for 10 s and then maintains a constant speed. Car B accelerates at 5 mis zuntil reaching a constant speed of 25 m/s and then maintains this speed. Construct the a-1, v-t, and s-t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s?

*12-68. The a-s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v-s graph. At s = O, v = 0.

a (m/s2)

Prob. 12-68

= 2 t'

100 350s (ft)

Prob. 12-71


12-69. The v-s graph for the car is given for the first 500 ft of its motion. Construct the a-s graph for Osss 500 ft. How long does it take to travel the 500-ft distance? The car starts at s = O when t = O.

v (ft /s)

60

= 0 .1s y

I0

12-71. The v-s graph of a cyclist traveling along a straight road is shown. Construct the a-s graph.

v (ft/s)

151— v

= 0.1s + 5

5

s (ft)

12-70. The boat travels along a straight line with the speed described by the graph. Construct the s-t and a-s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = O when t = 0.

"12-72. The a-s graph for a boat moving along a straight path is given. If the boat starts at s = O when v = 0, determine its speed when it is at s = 75 ft, and 125 ft, respectively. Use a numerical method to evaluate v at s =125 ft.

500

Prob. 12-69

a (f /s2)

100s (ft)10(1100

v (m/s)

80

20

Prob. 12-70 Prob. 12-72

.....~1~1115.1•- -.1-1

32 CHAPTER 12 KINEMATICS OF A P A R T I C L E

12 12.4 General Curvilinear Motion

s

Position Path

( a )

Displacement

(b)

Curvihnear motion occurs when a particie moves along a curved path. Since this path is often described in three dimensions, vector analysis will be used to formulate the particle's position, velocity, and acceleration.* In this section the general aspects of curvilinear motion are discussed, and in subsequent sections we will consider three types of coordinate systems often used to analyze this motion.

Position. Consider a particie located at a point on a space curve defined by the path function s(t), Fig. 12-16a. The position of the particie, measured from a fixed point 0, will be designated by the position vector r = r(t). Notice that both the magnitude and direction of this vector will change as the particie moves along the curve.

Displacement. Suppose that during a smali time interval At the particie moves a distance As along the curve to a new position, defined by r' = r + Ar, Fig. 12-16b.The displacement Ar represents the change in the particle's position and is determined by vector subtraction; i.e., Ar= r' - r.

Ve I ocity. During the time At, the average velocity of the particie is

ArVavg =AI

The instantaneous velocity is determined from this equation by letting At -› 0, and consequently the direction of Ar approaches the tangent to the curve. Hence, v = Alim (Ar/At) or

t—>0

d rv

d t(12-7)

s -

Velocity

(c)

Fig. 12-16

Since dr will be tangent to the curve, the direction of v is also tangent to the curve, Fig. 12-16c. The magnitude of v, which is called the speed, is obtained by realizing that the length of the straight line segment Ar in Fig. 12-16b approaches the arc length As as At -> 0, we havev = 1;—> lim (Ar I At) = Ar—> lim (s 1 At), or

0

d sd t (12-8)

Thus, the speed can be obtained by differentiating the path function s with respect to time.

*A summary of some of the important concepts of vector analysis is given in Appendix B.

12.4 GENERAL CURVILINEAR MOTION33

Acceleration. If the particie has a velocity v at time t and a velocity v' = v + Av at t + At, Fig. 12-16d, then the average acceleration of the particie during the time interval At is

Ava a v g =

At ( d )

where = v' - v. To study this time rate of change, the two velocityvectors in Fig. 12-16d are plotted in Fig. 12-16e such that their tails are located at the fixed point O' and their arrowheads touch points on a curve.This curve is called a hodograph, and when constructed, it describes the locus of points for the arrowhead of the velocity vector in the same manner as the path s describes the locus of points for the arrowhead of the position vector, Fig. 12-16a.

To obtain the instantaneous acceleration, let -› O in the aboveequation. In the limit Av will approach the tangent to the hodograph, and so a = lim (Av/4, or (e)

a = d vd ł

(12-9) Hodograph

f v a O '

Substituting Eq. 12-7 into this result, we can also write

a = der

dt2

By definition of the derivative, a acts tangent to the hodograph, Fig. 12-16f, and, in generał it is not tangent to the path of motion, Fig. 12-16g. To clarify this point, realize that Ov and consequently a must account for the change made in both the magnitude and direction of the velocity v as the particie moves from one point to the next along the path, Fig. 12-16d. However, in order for the particie to follow any curved path, the directional change always "swings" the velocity vector toward the "inside" or "concave side" of the path, and therefore a cannot remain tangent to the path. In summary, v is always tangent to the path and a is always tangent to the hodograph.

Acceleration path

(g)

Fig. 12-16


12 12.5 Curvilinear Motion: RectangularComponents

Occasionally the motion of a particie can best be described along a path that can be expressed in terms of its x, y, z coordinates.

z

kr = xi + yj + zk

z

y

Y

P o s i t i o n

( a )

Position. If the particie is at point (x, y, z) on the curved path s shown in Fig. 12-17a, then its location is defined by the position vector

r = xi + yj + zk

(12-10)

When the particie moves, the x, y, z components of r will be functions of time; i.e., x = x(t), y = y(t), z = z(t), so that r = r(t).

At any instant the magnitude of r is defined from Eq. B-3 in Appendix B as

r= VX2 + y2 + Z2

And the direction of r is specified by the unit vector u, = r/r.

Velocity. The first time derivative of r yields the velocity of the particie. Hence,

d r d d dv = v.,i + vj + vzic V = = dt dł (xi) ł dt —(xj) ł dt(zk)

Y

/ When taking this derivative, it is necessary to account for changes in both

■the magnitude and direction of each of the vector's components. For example, the derivative of the i component of r is

—d(xi) = (-Lxi + x—di

dt dt dłFig. 12-17 The second term on the right side is zero, provided the x, y, z reference

frame is ftxed, and therefore the direction (and the magnitude) of i does not change with time. Differentiation of the j and k components may becarried out in a similar manner, which yields the finał result,

v = ( L r = v i + v j + v , kx Y

dt

where

V e l o c i t y

( b )

= k v y = v , = ż (12-12)

12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 35

The "dot" notation ż, j‚, ż represents the first time derivatives of x = x(t), Y = y(t), z = z(t), respectively.

The velocity has a magnitude that is found from

= VV + V2 + V2-x y

and a direction that is specified by the unit vector uv = v/v. As discussed in Sec. 12-4, this direction is always tangent to the path, as shown in Fig. 12-17b.

Acceleration. The acceleration of the particie is obtained by taking the first time derivative of Eq. 12-11 (or the second time derivative of Eq. 12-10). We have

a = + aj + azIc

a= dv — = axi + aYj a-k dt

(12-13)Acceleration

(c)

where

a , = = a y

= = 3 ; = = ti

(12-14)

Here (sx, ay, a, represent, respectively, the first time derivatives of vx = vx(t), vy = vy(t), v, = vz(t), or the second time derivatives of the functions x = x(t), y = y(t), z = z(t).

The acceleration has a magnitude

a , v 'a x2 a2 v

and a direction specified by the unit vector up = a/a. Since a represents the time rate of change in both the magnitude and direction of the velocity, in general a will not be tangent to the path, Fig. 12-17c.

36 CHAPTER 1 2 KINEMATICS OF A PARTICLE

12

Important Points

• Curvilinear motion can cause changes in both the magnitude and direction of the position, velocity, and acceleration vectors.

• The velocity vector is always directed tangent to the path.

• In general, the acceleration vector is not tangent to the path, but rather, it is tangent to the hodograph.

• If the motion is described using rectangular coordinates, then the components along each of the axes do not change direction, only their magnitude and sense (algebraic sign) will change.

• By considering the component motions, the change in magnitude and direction of the particle's position and velocity are automatically taken into account.


Coordinate System.

• A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y, z components.

Kinematic Quantities.

• Since rectilinear motion occurs along each coordinate axis, the motion along each axis is found using v = dsldt and a = dvIdt: or in cases where the motion is not expressed as a function of time, the equation a ds = v dv can be used.

• In two dimensions, the equation of the path y = f(x) can be used to relate the x and y components of velocity and acceleration by applying the chain rule of calculus. A review of this concept is given in Appendix C.

• Once the x, y, z components of v and a have been determined, the magnitudes of these vectors are found from the Pythagorean theorem, Eq. B-3, and their coordinate direction angles from the components of their unit vectors, Eqs. B-4 and B-5.

12.5 CURVILINEAR MOTION: RECTANGULAR COMPONENTS 37

At any instant the horizontal position of the weather balloon in Fig. 12-18a is defined by x = (8t) ft, where t is in seconds. If the equation of the path is y = x2/10, determine the magnitude and direction of the velocity and the acceleration when t = 2 s.

SOLUTION

Velocity. The velocity component in the x direction is

dv, = = —(81) = 8 ft/s

di

To find the relationship between the velocity components we will use the choin rule of calculus. When t = 2s, x = 8 (2) = 16 ft, Fig. 12-18a, and so

= y = 411 (x2/10) = 2.x.k/10 = 2(16)(8)/10 = 25.6 ft/s ł

When t = 2 s, the magnitude of velocity is therefore

v = N./(8 ft/s)2 + (25.6 ft/s)2 = 26.8 f tl s Ans.

The direction is tangent to the path, Fig. 12-18b, whereV. 25.6

O v = tan - l= = tan - I — 72.6 °

v, 8Ans.

Acceleration. The relationship between the acceleration components is determined using the choin rule. (See Appendix C.) We have

a, = v,. = 7f(8) = O

. Y dt d .aY = v = —(2xx/10) = 2(k)V10 + 2,4)/10

= 2(8)2/10 + 2(16)(0)/10 = 12.8 ft/s2 ł

Thus,a = V(0)2 + (12.8)2 = 12.8 ft/s2 Ans.

The direction of a, as shown in Fig. 12-18c, is

x

(a)

y = 26.8 ft „,s

O, = 72.6°

B --a—

(b)

a = 12.8 ft

= 9t3P

B( c )

12.8= = 90°

I d " O Ans. Fig. 12.-18

NOTE: It is also possible to obtain y,. and ay by first expressing y = f (t) = (8t)-/ 10 = 6.4t2 and then taking successive time derivatives.

38 CHAPTER 12 KINEMATICS OF A PARTICIE

EXAMPLEFor a short time, the path of the piane in Fig. 12-19a is described by y = (0.001x'-) m. If the piane is rising with a constant upward velocity of 10 m/s, determine the magnitudes of the velocity and acceleration of the piane when it reaches an altitude of y = 100 m.

SOLUTIONWhen y = 100 m, then 100 = 0.001x2 or x = 316.2 m. Also, due to constant velocity v,, = 10 m/s, so

100m = (10 m/s) t t = 10sy = vvt.,

Velocity. Using the chain rule (see Appendix C) to find the relationship between the velocity components, we have

y = 0.001x2

v). = = (0.001X) = (0.002x)X = 0.002x vdt (1)

Thus

źly = 0.001x2

10 m/s = 0.002(316.2 m)(vx) v., = 15.81 m/s

100 m I

x

The magnitude of the velocity is therefore

(a) v = = V(15.81 m/s)2 + (10 m/s)2 = 18.7 m/s Ans.

Acceleration. Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components.

ay = vy = (0.002X)X + 0.002x(..) = 0.002(y? +

xa,) When x = 316.2 m, vx = 15.81 m/s, by = ay = O,

V, O = 0.002[(15.81 m/s)2 + 316.2 m(a„)]a a, = -0.791 m/s2

1110 ni

(b)

Fig. 12-19

x The magnitude of the plane's acceleration is therefore

a = Vax2 + ay = V"(-0.791 m/s2)2 + (0 m/s2)2

= 0.791 m/s2 Ans.

These results are shown in Fig. 12-196.

12

12.6 MOTION OF A PROJECTILE 39

12.6 Motion of a Projectile

The free-flight motion of a projectile is often studied in terms of its rectangular components. To illustrate the kinematic analysis, consider a projectile Iaunched at point (x0, yo), with an initial velocity of v0, having components (v0)., and (v0)y, Fig. 12-20. When air resistance is neglected, the only force acting on the projectile is its weight, which causes the projectile to have a constant downward acceleration of approximately

= g = 9.81 m/s2 or g = 32.2 ft/s2.*

Fig. 12-20

H ori zonta I Motion. Since ax = 0, application of the constant acceleration equations, 12-4 to 12-6, yields

v = vo + ad' v, = (v0),

x = xo + vot + x = x0 + (v0).,t

112 = /17j + 2a,(x — xo); vx = (v0)„

The first and last equations indicate that the horizontal component of velocity always remains constant during the motion.

Vertical Motion. Since the positive y axis is directed upward, then ay

= —g. Applying Eqs. 12-4 to 12-6, we get

vy = (2,0)y — gt

Y = Yo + (vo)); — g g2 vy= (1.10)y — 2g(y - Yo)

Recall that the last equation can be formulated on the basis of eliminating the time t from the first two equations, and therefore only two of the above three equations are independent of one anołher.

*This assumes that the earth's gravitational field does not vary with altilude.

X

v = vo + act;Y = Yo yol 4a,/.2;

y 2 = 2 , 8

Each picture in this sequence is taken after the same time interval. The red bali falls from rest, whereas the yellow bali is given a horizontal velocity when released. Both balls accelerate downward at the same rate, and so they remain at the same elevation al any instant. This acceleration causes the difference in elevation between the balls to increase between successive photos. Also, note the horizontal distance between successive photos of the yellow bali is constant since the velocity in the horizontal direction remains constant.

4O CHAPTER 12 KINEMATIOS OF A PARTICLE

To summarize, problems involving the motion of a projectile can have12at most three unknowns since only three independent equations can be

written; that is, one equation in the horizontal direction and two in the vertical direction. Once vx and vy are obtained, the resultant velocity v, which is always tangent to the path, can be determined by the vector sum as shown in Fig. 12-20.


Coordinate System.

• Establish the fixed x, y coordinate axes and sketch the trajectory of the particie. Between any two points on the path specify the given problem data and identify the three unknowns. In alt cases the acceleration of gravity acts downward and equals 9.81 m/s2

or 32.2 ft/s2. The particle's initial and final velocities should be represented in terms of their x and y components.

• Remember that positive and negative position, velocity, and acceleration components always act in accordance with their associated coordinate directions.

Kinematic Equations.

• Depending upon the known data and what is to be determined, a choice should be made as to which three of the following four equations should be applied between the two points on the path to obtain the most direct solution to the problem.

Horizontal Motion.

• The velocity in the horizontal or x direction is constant, i.c.. yx = (1,0),,,, and

x=xo+(yo )x t

Vertical Motion.

• In the vertical or y direction only two of the following three equations can be used for solution.

vy = (vo)y + act

Y = yo + (va)y t + 4a,r1

2Vy = (2,43)3, 2a , ( — Yo )

For example, if the particle's final velocity vy is not needed, thenthe first and third of these equations will not be useful.

Gravel falling off the end of this conveyor bełt follows a path that can be predicted using the equations of constant acceleration. In this way the location of the accumulated pile can be determined. Rectangular coordinates are used for the analysis since the acceleration is only in the vertical direction.

12.6 MOTION OF A PROJECTILE41

EXAMPLE

12.11

A sack slides off the ramp, shown in Fig. 12-21, with a horizontal velocity of 12 m/s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where sacks begin to pile up.

SOLUTION

Coordinate System. The origin of coordinates is established at the beginning of the path, point A, Fig. 12-21. The initial velocity of a sack has components (vA), = 12 m/s and (vA), = O. Also, between points A and B the acceleration isay = —9.81 m/s2.Since (vB)x = (vA), = 12 m/s, the three unknowns are (vB)y, R, and the time of flight tAB. Here we do not need to determine (vB)y.

Vertical Motion. The vertical distance from A to B is known, and therefore we can obtain a direct solution for tAB by using the equation

Ya = YA (VAVAII acaB

—6 m = O + O + m/s2)t,21B

tAB = 1.11 s Ans.

Horizontal Motion. Since tAB has been calculated, R is determined as follows:

xs = (vA),JABR = O + 12 m/s (1.11 s)

R = 13.3 mAns.

NOTE: The calculation for tAB also indicates that if a sack were released from rest at A, it would take the same amount of time to strike the floor at C, Fig. 12-21.

x12 m/s

a = g

Fig. 12-21


12 EXAMPLE

12.12

The chipping machine is designed to eject wood chips at yo = 25 ft/s as shown in Fig. 12-22. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if at this instant they land on the pile 20 ft from the tube.

Fig. 12-22

SOLUTION

Coordinate System. When the motion is analyzed between points O and A, the three unknowns are the height h, time of flight toA, and vertical component of velocity (vA)y. [Note that (vA)x = (%),.] With the origin of coordinates at 0, Fig. 12-22, the initial velocity of a chip has components of

(vo), = (25 cos 30°) ft/s = 21.65 ft/s (v

o)y = (25 sin 30°) ft/s = 12.5 ft/4

Also, = (1)0)x = 21.65 ft/s and ay = -32.2 ft/s2. Since we donot need to determine (vA)y, we haveHorizontal Motion.

XA = X0 ÷ (V0)„10A20 ft = O + (21.65 ft/s)t0A

toA = 0.9238 s

Vertical Motion. Relating toA to the initial and final elevations of a chip, we have

(+I) YA = Yo (vo)ytoA źekt2oA(h - 4 ft) = O + (12.5 ft/s)(0.9238 s) + ft/s2)(0.9238 s)2

h = 1.81 ft Ans.

NOTE: We can determine (vA)y by using (vA)y = (vo)y + actoA.


The track for this racing event was designed so that riders jump off the slope at 30°, from a height of 1 m. During a race it was observed that the rider shown in Fig. 12-23a remained in mid air for 1.5 s. Determine the speed at which he was traveling off the ramp, the horizontal distance he travels before striking the ground, and the maximum height he attains. Neglect the size of the bike and rider.

( a )

SOLUTION

Coordinate System. As shown in Fig. 12-23b, the origin of the coordinates is established at A. Between the end points of the path AB the three unknowns are the initial speed vA range R, and the vertical component of velocity (v8)y.

Vertical Motion. Since the time of flight and the vertical distance between the ends of the path are known, we can determine vA(+T) YB = YA (vA)yrAB "laci,b?

-1 m = O + vA sin 30°(1.5 s) + '1(-9.81 m/s2)(1.5 s)2

vA = 13.38 m/s = 13.4 m/s Ans.

Horizontal Motion. The range R can now be determined.

±›) XB = XA (VAIAB

R = O + 13.38 cos 30° m/s (1.5 s)= 17.4 m Ans.

In order to find the maximum height h we will consider the path AC, Fig. 12-23b. Here the three unknowns are the time of flight tAc, the horizontal distance from A to C, and the height h. At the maximum height (vc,)y = 0, and since vA is known, we can determine h direcdy without considering tAc using the following equation.

(vc)y = (vA)y 2ac[Yc YA]

02 = (13.38 sin 30° m/s)2 + 2(-9.81 m/s2)[(h - 1 m) -

h = 3.28 m Ans.

NOTE: Show that the bike will strike the ground at B with a velocity having components of

(y/3)y = 8.02 m/s4,

(v8), = 11.6 m/s —> ,

Y

B

Fig. 12-23

4 4 C H A P T E R 1 2 K I N E M A T I C S O F A P A R T I C L E

FUNDAMENTAL PROBLEMS

F12-15. If the x and y components of a particle's velocity are vx = (321) m/s and v,. = 8 m/s, determine the equation of the path y = j(x). x = O and y = O when t = 0.

F12-16. A particie is traveling along the straight path. If its position along the x axis is = (8t) m, where t is in seconds, determine its speed when t = 2 s.

F12-18. A particie travels along a straight-Iine path y = 0.5x. If the x component of the particle's velocity is v., = (2t2) m/s, where t is in seconds, determine the magnitude of the particle's velocity and acceleration when t = 4 s.

3 m

x

F12-18

F12-19. A particie is traveling along the parabolic path y =0.25x2. If x = (2t2) m, where t is in seconds, determine the magnitude of the particle's velocity and acceleration when r = 2 s.

7= y 0.25.v2

x

F I 2 - 1 9

F12-20. The box slides down the siope described by the equation y = (0.05x2) m, where x is in meters. If the box has x components of velocity and acceieration of vx = —3 m/s and a., = —1.5 m/s2 at x = 5 m,determine the y components of the velocity and the acceieration of the box at this instant.

Y

2 = : . . . , c 2

x

F12-20

V4 m

F12-16

F12-17. A particie is constrained to travel along the path. If x = (44) m, where I is in seconds, determine the magnitude of the particle's velocity and acceieration when t = 0.5 s.

F12-17

12.6 MOTION OF A PROJECTILE45

F12-21. The bali is kicked from point A with the initial F12-25. A bali is thrown from A. If it is required to elearvelocity vA = 10 m/s. Determine the maximum height h it the wali at B, determine the minimum magnitude of itsreaches. initial velocity vA.

F12-22. The bali is kicked from point A with the initial velocity vA = 10 m/s. Determine the range R, and the speed when the bali strikes the ground.

x B - - = 10 m sB

F12-21/22

F12-23. Determine the speed at which the basketball at A musi be thrown at the angle of 30° so that it makes it to the basket at B.

v B

/x

1 .Ś m

10 m

F12-23

12 f t -

F12-25

F12-26. A projectile is fired with an initial velocity of vA = 150 m/s off the roof of the building. Determine the range R where it stńkes the ground at B.

3m

F12-24. Water is sprayed at an angle of 909 from the slope at 20 m/s. Determine the range R.

F12-24

vA = 150 m/s

F12-26

Y


1 2

12-73. The position of a particie is defined by r = {5(cos 201 + 4(sin 2t)J } m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particie when t = I s. Also, prove that the path of the particie is elliptical.

12-74. The velocity of a particie is v = { 31+ (6 — 2t)j } m/s, where t is in seconds. If r = O when t = 0, determine the displacement of the particie during the time interval t= Istot= 3 s.

12-75. A particie, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = {6ti + 12t2k} ft/s2. Determine the particie's position (x, y, z) at t = I s.

*12-76. The velocity of a particie is given by v =-16t2i +4t3j + (5t + 2)k m/s, where t is in seconds. If the

particie is at the origin when i = O. determine the magnitude of the particie's acceleration when t = 2 s. Also, what is the x, y, z coordinate position of the particie at this instant?

12-77. The car travels from A to B, and then from B to C, as shown in the figure. Determine the magnitude of the displacement of the car and the distance traveled.

12-78. A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed?

12-79. A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A, B, and C. If it takes 3 s to go from A to B, and then 5 s to go from B to C, determine the average acceleration between points A and B and between points A and C.

qab, v,, = 20 niA

Prob. 12-79

*12-80. A particie travels along the curve from A to B in 2 s. It takes 4 s for it to go from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D.

2km

AMIO

3 tiu„

'C

Prob. 12-77 Prob. 12-80


12-85. A particie travels along the curve from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C.

12-81. The position of a crate sliding down a ramp is given by x = (0.25t3) m, y = (1.512) m, z = (6 - 0.75/5/2) m, where t is in seconds. Determine the magnitude of the crate's velocity and acceleration when t = 2 s.

12-82. A rocket is fired from rent at x = O and travels along a parabolic trajectory described by y21 = [120(103)x] m. If the x component of acceleration is ax = (412) m/s2, wheret is in seconds, determine the magnitude of the rocket's velocity and acceleration when t = 10 s.

12-83. The particie travels along the path defined by the parabola y = 0.5x2. If the component of velocity along the x axis is r), = (5t) ft/s, where t is in seconds, determine the particle's distance from the origin O and the magnitude of its acceleration when t = 1 s. When t= 0, x = 0, y = 0.

1 y = 0.5x2

0

x

Prob. 12-83

• C

Prob. 12-85

R

20 m

A

12-86. When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path (y - 40)2 = 160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at v, = 180 m/s, determine the magnitudes of the rocket's velocity and acceleration when it reaches an altitude of 80 m.

*12-84. The motorcycle travels with constant speed yoalong the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve.

Y

- 4o), = 160x

y = c sin (-y-, x)

c_lx

x

Prob. 12-84 Prob. 12-86


12-87. Pegs A and B are restricted to move in the elliptical 12 slots due to the motion of the slotted Iink. If the link moves with

a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m.

y

Prob. 12-87

*12-88. The van travels over the hill described by y = (-1.5(10-3)x2 + 15)ft. If it has a constant speed of 75 ft/s, determine the x and y components of the van's velocity and acceleration when x = 50 ft.

Y

15 ft Y = (-1.5 (10') x2 + 15) ft

x

100 ft

Prob. 12-88

1249. It is observed that the time for the bali to strike the ground at B is 2.5 s. Determine the speed y4 and angle 04 at which the bali was thrown.

12-90. Determine the minimum initial velocity vo and the corresponding angle 00 at which the bali must be kicked in order for it to just cross over the 3-m high fence.

ni

Prob. 12-90

12-91. During a race the dirt bike was observed to leap up off the smali hill at A at an angle of 60° with the horizontal. If the point of landing is 20 ft away, determine the approximate speed at which the bike was traveling just before it Ieft the ground. Neglect the size of the bike for the calculation.

•S.

A/ 6°' \

1.2m

50 m Prob. 12-89 Prob. 12-91


*12-92. The girl always throws the toys at an angle of 30° from point A as shown. Determine the time between throws so that both toys strike the edges of the pool B and C at the same instant.With what speed musi she throw each toy?

12-95. A projectile is given a velocity vo at an angle above the horizontal. Determine the distance d to where 12 it strikes the sloped ground. The acceleration due to gravity is g.

*12-96. A projectile is given a velocity vo. Determine the angle ib at which it should be Iaunched so that d is a maximum. The acceleration due to gravity is g.

Prob. 12-92 Probs. 12-95/96

12-93. The player kicks a football with an initial speed of vo

= 90 ft/s. Determine the time the bali is in the air and the angle d of the kick.

12-94. From a videotape, it was observed that a player kicked a football 126 ft during a measured time of 3.6 seconds. Determine the initial speed of the bali and the angle O at which it was kicked.

12-97. Determine the minimum height on the wali to which the firefighter can project water from the hole, so that the water strikes the wali horizontally.The speed of thewater at the nozzle is = 48 ft/s.

•12-98. Determine the smallest angle O. measured above the horizontal, that the bose should be directed so that the water stream strikes the bottom of the wali at B. The speed of the water at the nozzle is vc = 48 ft/s.

Prol)... 12-97/98Probs. 12-93/94

1212-99. Measurements of a shot recorded on a videotape during a basketball game are shown. The bali passed through the hoop even though it bardy cleared the hands of the player B who attempted to block it. Neglecting the size of the bali, determine the magnitude vA of its initial velocity and the height h of the bali when it passes over player B.

SO ft


30°

7 ft

25 ft

Prob. 12-99

*12-100. It is observed that the skier leaves the ramp A at an angle BA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the time of flight 1,4R.

12-101. It is observed that the skier leaves the ramp A at an angle BA = 25° with the horizontal. If he strikes the ground at B, determine his initial speed vA and the speed at which he strikes the ground.

Probs. 12-100/101

12-102. A golf bali is struck with a velocity of 80 ft/s as shown. Determine the distance d to where it will land.

Prob. 12-102

12-103. The bali is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the bali strikes the slope. Also, determine the speed at which the bali hits the ground.

Prob. 12-103

*12-104. The projectile is launched with a velocity vo. Determine the range R, the maximum height h attained, and the time of flight. Express the results in terms of the angle d and ry0. The acceleration due to gravity is g.

V■1

. = x

Prob. 12-104

5 ft —

R

I0 ft


12-105. Determine the horizontal velocity vA of a tennis bali at A so that it just clears the net at B. Also, find the distance s where the bali strikes the ground.

*12-108. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart 12 is thrown with a speed of 10 m/s, determine the angles Bcand OD at which they should be thrown and the time between each throw. Note that the first dart must be thrown at OC (> O

D), then the second dart is thrown at O.

B-r-

3 ft

i— s 2 1 f t - - • 1

Prob. 12-105

7.5 ft1 5m

O c C

BA

Prob. 12-108

12-106. The bali at A is kicked with a speed v,, = 8 ft/s and at an angle 0A = 30°. Determine the point (x, y) where it strikes the ground. Assume the ground has the shape of a parabola as shown.

12-107. The bali at A is kicked such that BA = 30°. If it strikes the ground at B having coordinates x = 15 ft, y = -9 ft, determine the speed at which it is kicked and the speed at which it strikes the ground.

12-109. A boy throws a bali at O in the air with a speed rioat an angle O. If he then throws another bali with the same speed ty0 at an angle 02 < Bt, determine the time between the throws so that the balls collide in midair at B.

Prob. 12-109P r4). 12-106/107

x

52CHAPTER 12 K1NEMATICS OF A PARTICLE

12-110. Smali packages traveling on the conveyor bełt fali12 off finto a 1-m-long loading car. If the conveyor is running at a

constant speed of vc = 2 m/s, determine the smallest and largest distance R at which the end A of the car may be placed from the conveyor so that the packages enter the car.

*12-112. The baseball player A hits the baseball at vA = 40 ft/s and 0, = 60° from the horizontal. When the bali is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the bali was hit.

d15 ft

Prob. 12-110

VA = 40 0 iS

Prob. 12-112

12-111. The fireman wishes to direct the flow of water from his hose to the fire at B. Determine two possible angles 01 and 02 at which this can be done. Water flows from the hose at vA = 80 ft/s.

12-113. The man stands 60 ft from the wali and throws a bali at it with a speed vo = 50 ft/s. Determine the angle O at which he should release the bali so that it strikes the wali at the highest point possible. What is this height? The room has a ceiling height of 20 ft.

35 ft

Prob. 12-111 Prob. 12-113

■7. .• i FA L'IrW 1•■.̀ 1 F■TM FA i ~a ∎T:

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS5 3

12.7 Curvilinear Motion: Normal and Tangential Components

When the path along which a particie travels is known, then it is often convenient to describe the motion using n and t coordinate axes which act norma] and tangent to the path, respectively, and at the instant considered have their origin located at the particie.

Planar Motion. Consider the particie shown in Fig. 12-24a, which moves in a piane along a fixed curve, such that at a given instant it is at position s, measured from point O. We will now consider a coordinate system that has its origin on the curve, and at the instant considered this origin happens to coincide with the location of the particie. The t axis is tangent to the curve at the point and is positive in the direction of increasing s.We will designate this positive direction with the unit vector u,. A unique choice for the normal axis can be macie by noting that geometrically the curve is constructed from a series of differential arc segments ds, Fig. 12-24b. Each segment ds is formed from the arc of an associated circle having a raditts of curvature p (rho) and center of curvature D'. The normal axis n is perpendicular to the t axis with its positive sense directed toward the center of curvature O', Fig. 12-24a. This positive direction, which is always on the concave side of the curve, will be designated by the unit vector u.. The piane which contains the n and t axes is referred to as the embracing or osculating piane, and in this case it is fixed in the piane of motion.*

Velocity. Since the particie moves,s is a function of time. As indicated in Sec. 12.4, the particle's velocity v has a direction that is always tangent to the path, Fig. 12-24c, and a magnitude that is determined by taking the time derivative of the path function s = s(t), i.e., v = ds/dt (Eq. 12-8). Hence

V =(12-15)

where

Position(a)

o'ds

Radius of curvature

(b)

Velocity(c)v = s (12-16)

Fig. 12-24

The osculating piane may also be defined as the piane which has che greatest contact with the curve at a point. It is the limiting position of a piane contacting both the point and the arc segment ds. As noted above, the osculating piane is always coincident with a piane curve; however, each point on a three-dimensional curve has a unique osculating piane.

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 55

To better understand these results, consider the following two special cases of motion.

1. If the particie moves along a straight line, then p -) cc and from Eq. 12-20, a„ = 0. Thus a = a, = b, and we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of the velocity.

2. If the particie moves along a curve with a constant speed, then a, = v = O and a = a„ = v2/p. Therefore, the normal component of acceleration represents the time rate of change in the direction ofthe velocity. Since always acts towards the center of curvature,this component is sometimes referred to as the centripetal (or center seeking) acceleration.

As a result of these interpretations, a particie moving along the curved path in Fig. 12-25 will have accelerations directed as shown.

r" at

Change inmagnitude of velocity

Fig. 12-25

Three-Dimensional Motion. If the particie moves along a space curve, Fig. 12-26, then at a given instant the t axis is uniquely specified; however, an infinite number of straight lines can be constructed normal to the tangent axis. As in the case of planar motion, we will choose the positive n axis directed toward the path's center of curvature O'. This axis is referred to as the principal normal to the curve. With the n and t axes so defined, Eqs. 12-15 through 12-21 can be used to determine v and a. Since u, and u„ are always perpendicular to one another and lie in the osculating piane, for spatial motion a third unit vector, ub, defines the binormal axis b which is perpendicular to u, and u„, Fig. 12-26.

Since the three unit vectors are related to one another by the vector cross product, e.g., ub = u, x u,„ Fig. 12-26, it may be possible to use this relation to establish the direction of one of the axes, if the directions of the other two are known. For example, no motion occurs in the ubdirection, and if this direction and u, are known, then u„ can be determined, where in this case u„ = ub x u„ Fig. 12-26. Remember, though, that un is always on the concave side of the curve.

As the boy swings upward with a velocity y, his motion can be analyzed using n—r coordinates. As he rises, the magnitude of his velocity (speed) is decreasing, and so a, will be negative. The rale at which the direction of his velocity changes is a„, which is always positive, that is, towards the center of rotation.

8 = a,

Increasing speed

a

Change in direction of velocity

an

b osculating piane• O' n

Ull

Fig. 12-26


12Acceleration. The acceleration of the particie is the time rate of change of the velocity. Thus,

a = v = vu, + vu, (12-17)

In order to determine the time derivative note that as the particiemoves along the arc ds in time dt, u, preserves its magnitude of unity; however, its direction changes, and becomes , Fig. 12-24d. As shown in Fig. 12-24e, we require ui = u, + du,. Here du, stretches between the arrowheads of u, and which lie on an infinitesimal arc of radius u, = 1. Hence, du, has a magnitude of du, = (1) dO, and its direction is defined by un. Consequently, du, = Mu,,, and therefore the time derivative becomes

= Bu„. Since ds = pdO, Fig. 12-24d, then B = Słp, and therefore

( d )

" f t

4

( e )

o '

a,Acceleration

(f)

Fig. 12-24 (mut)

= Óti„ = = u„P P

Substituting into Eq. 12-17, a can be written as the sum of its two components,

a = a,u, + anun (12-18)

where

a ,= t i or a,ds = v dv (12-19)

and

v2 a„ = — (12-20) P

These two mutually perpendicular components are shown in Fig. 12-24f. Therefore, the magnitude of acceleration is the positive value of

a = (12-21)


12


Coordinate System.

• Provided the park of the particie is known, we can establish a set of n and t coordinates having a fixed origin, which is coincident with the particie at the instant considered.

• The positive tangent axis acts in the direction of motion and the positive normal axis is directed toward the path's center of curvature.

Velocity.

• The particle's velocity is always tangent to the path.• The magnitude of velocity is found from the time derivative of the

path function.

v = s

Tangential Acceleration.

• The tangential component of acceleration is the result of the time rate of change in the magnitude of velocity. This component acts in the positive s direction if the particle's speed is increasing or in the opposite direction if the speed is decreasing.

• The relations between a,, v, t and s are the same as for rectilinear motion, namely,

a , =v a i d s = v dv

• I f a , i s c o n s t a n t , a , = ( a , ) , . , t h e a b o v e e q u a t i o n s , w h e n i n t e g r a t e d , y i e l d s = s o

+ v o r + i ( a , ) , 1 2

v = vp + (a ,),tv2 = vó + 2(a,),(s — so)

Normal Acceleration.

• The normal component of acceleration is the result of the time rate of change in the direction of the velocity. This component is always directed toward the center of curvature of the path, i.e., along the positive n axis.

• The magnitude of this component is determined from V2

a„ = —P

• If the path is expressed as y = j(x), the radius of curvature p at any point on the path is determined from the equation

[1 + (dy/dX")2]3/2

P —I d2y I dx= I

The derivation of this result is given in any standard calculus text.

Motorists traveling along Ibis cloverleaf interchange expericnce a normal acceleration due to the change in direction of their velocity. A tangential component of acceleration occurs when the cars speed is increased or decreased.


When the skier reaches point A along the parabolic path in Fig. 12-27a, he has a speed of 6 m/s which is increasing at 2 m/s2. Determine the direction of his velocity and the direction and magnitude of his acceleration at this instant. Neglect the size of the skier in the calculation.

SOLUTION

Coordinate System. Although the path has been expressed in terms of its x and y coordinates, we can still establish the origin of the n, t axes at the fixed point A on the path and determine the components of v and a along these axes, Fig. 12-27a.

Velocity. By definition, the velocity is always directed tangent tothe path. Since y = 2+0 x2, dyldx = then at x = 10 m, dyldx = I.Hence, at A, v makes an angle of 9 = tan-II = 45° with the x axis, Fig. 12-276. Therefore,

vA = 6 m/s 45° Y Ans.

The acceleration is determined from a = vu, + (v2 p)u„ However, it is first necessary to determine the radius of curvature of the path at A (10 m, 5 m). Since d2y1dx2 = Wio, then

[ + (dy/dx)2]3/2 [I + (*) x.)2 3ł

P - I d2Y/6-1 hi(31

The acceleration becomes

= 28.28 mx = 1 0 m

Y 2)"2

v2aa = but + —un

P(6 m/s)2

= 2u, +28.28 m u"

- {2u, + 1.273u„}m/s2

As shown in Fig. 12-276,

a = 17(2 m/s2)2 + (1.273 m/s2)2 = 2.37 m/s2

2 ct• = tan-I1.273 - 57.5°

Thus, 45° + 90° + 57.5° - 180° = 12.5° so that,

a = 2.37 m/ s2 12.5° Y Ans.

NOTE: By using n, t coordinates, we were able to readily solve thisproblem through the use of Eq. 12-18, since it accounts for the separate changes in the magnitude and direction of v.

I - 1 0 m

( a )

(b)Fig. 12-27

n\ 1.273 m/s2

90°

s2

45°

58 CHAPTER 1 2 KINEMATICS OF A PARTICIE

EXAMPLE 12.15

A race car C travels around the horizontal circular track that has a radius of 300 ft, Fig. 12-28. If the car increases its speed at a constant race of 7 ft/s2, starting from rest, determine the time needed for it to reach an acceleration of 8 ft/s2. What is its speed at this instant?

— —Fig. 12-28

SOLUTION

Coordinate System. The origin of the n and t axes is coincident withthe car at the instant considered. The t axis is in the direction of motion, and the positive n axis is directed toward the center of thecircle. This coordinate system is selected since the path is known.

Acceleration. The magnitude of

acceleration can be related to its components using a = 1./a;2 + a;2,. Here a, = 7 ft/s2. Since a„ = v2/p, the velocity as a function of time must be determined first.

v = yo + (a,),1

v = O + 71

Thusv 2 ( 7 0 2

a„ = — p 300 = = 0.163/2 ft/s2

The time needed for the acceleration to reach 8 ft/s2 is therefore a

= Va;2 + a,2,

8 ft/s2 = V(7 ft/s2)2 + (0.163/2)2

Solving for the positive value of t yields

0.163t2- = V(8 ft/s2)2 — (7 ft/s2)2

t = 4.87 s Ans

Velocity. The speed at time t = 4.87 s isv = 71 = 7(4.87) = 34.1 ft/s Ans.

NOTE: Remember the velocity will always be tangent to the path. whereas the acceleration will be directed within the curvature of the path.

12

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS59

EXAMPLE

1 2 . 1 6

( d )

. 1 1 " \ t w t •

7,

B

( b )

5.242 m/s drr

B 1.138 m/s2

(c)Fig. 12-29

O

n

The boxes in Fig. 12-29a travel along the industrial conveyor. If a box as in Fig. 12-29b starts from rent at A and increases its speed such that a, = (0.2t) m/s2, where t is in seconds, determine the magnitude of its acceleration when it arrives at point B.

SOLUTIONCoordinate System. The position of the box at any instant is defined from the fixed point A using the position or path coordinate s, Fig. 12-29b. The acceleration is to be determined at B, so the origin of the n, t axes is at this point.

Acceleration. To determine the acceleration components a, = and a„ =v2/p, it is first necessary to formulate v and i) so that they may be evaluated at B. Since vA = O when t = 0, then

a, = 0.2t (1)

j o u f

dv = 0.2to

= 0.1/2 (2)

The time needed for the box to reach point B can be determined by realizing that the position of B is s8 = 3 + 21r(2)/4 = 6.142 m, Fig. 12-29b, and since sA = O when t = O we have

dsv = —dt = 0.1t2

f06.142 m ft,ds = 0.1t2dt

o6.142 m = 0.03334 ta

= 5.690s

Substituting into Eqs. 1 and 2 yields(a8), = it8 = 0.2(5.690) = 1.138 m/s2

v9 = 0.1(5.69)2 = 3.238 m/s

At B, p8 = 2 m, so that

avi(3.238 m/s)2= = — 5.242 m/s2

2 m

The magnitude of a8, Fig. 12-29c, is therefore

( IB = V(1.138 !nisz)'- + (5.242 misy = 5.36 m/s2 Ans.

6 0 C H A P T E R 1 2 K I N E M A T I C S O F A P A R T I C L E

1 2 ■ F U N D A M E N T A L P R O B L E M S

F12-27. The boat is traveling along the circular path with a speed of v = (0.062512) m/s, where t is in seconds. Determine the magnitude of its acceleration when t = 10 s.

40 m

F12-27

F12-28. The car is traveling along the road with a speed of v = (2 s) m/s, where s is in meters. Determine the magnitude of its acceleration when s = 10 m.

F12-28

F12-29. If the car decelerates uniformly along the curved road from 25 m/s at A to 15 m/s at C, determine the acceleration of the car at B.

F12-30. When x = 10 ft, the crate has a speed of 20 ft/s which is increasing at 6 ft/s2. Determine the direction of the crate's velocity and the magnitude of the crate's acceleration at this instant.

F12-30

F12-31. If the motorcycle has a deceleration of a, = -(0.001s) m/s2 and its speed at position A is 25 m/s, determine the magnitude of its acceleration when it passes point B.

F12-31

F12-32. The car travels up the hill with a speed of v = (0.2s) m/s, where s is in meters, measured from A. Determine the magnitude of its acceleration when it is at point s = 50 m, where p = 500 m.

F12-32

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTSb1• PROBLEMS

12-114. A car is traveling along a circular curve that has a radius of 50 m. If its speed is 16 m/s and is increasing uniformly at 8 m/52, determine the magnitude of its acceleration at this instant.

12-115. Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed 7.5 m/s2 while rounding a track having a radius of curvature of 200 m.

*12116. A car moves along a circular track of radius 250 ft such that its speed for a short period of timeO t 4 s, is v = 3(t + t 2) ft/s, where t is in seconds.Determine the magnitude of its acceleration when t = 3 s. How far has it traveled in t = 3 s?

12-117. A car travels along a horizontal circular curved road that has a radius of 600 m. If the speed is uniformly increased at a rate of 2000 km/h2,deterrnine the magnitude of the acceleration at the instant the speed of the car is 60 km/h.

12-118. The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m/s. For a short distance from s= 0, its speed is increased by i, = (0.05s) m/s-, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

= (0.05s) m/s2

y= 4m/s

Prob. 12-118

12-119. The automobile is originally at rest at s = 0. If its speed is increased by i, = (0.05/2) ft/s2, where t is in seconds, determine the magnitudes of its velocity and acceleration when t = 18 s.

*12-120. The automobile is originally at rest s = 0. If it then starts to increase its speed at i) = (0.05t 2) ft/s2, where t is in seconds, determine the magnitudes of its velocity and acceleration at s = 550 ft.

Probs. 12-119/120

12-121. When the roller coaster is at B, it has a speed of 25 m/s, which is increasing at a, = 3 m/s2. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle it makes with the x axis.

12-122. If the roller coaster starts from wsi at A and its speed increases at a, = (6 - 0.06s)m/s2,determine the magnitude of its acceleration when it reaches B where ss = 40 m.

Probs. 12-121/122


12-123. The speedboat travels at a constant speed of 12 15 m/s while making a turn on a circular curve from A to B. If it takes

45 s to make the turn, determine the magnitude of the boat's acceleration during the tum.

Prob. 12-123

12-124. The car travels along the circular path such that its speed is increased by a, = (0.54 m/s'-, where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s = 18 m starting from rest. Neglect the size of the car.

12-125. The car passes point A with a speed of 25 m/s after which its speed is defined by v = (25 — 0.15s) m/s. Determine the magnitude of the car's acceleration when it reaches point B, where s = 51.5 m and x = 50 m.

12-126. If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of a, = 0.5 m/s', determine the magnitude of the car's acceleration when s = 100 m and x = 0 .

Y Y =16 — k5x2

a sA

x

Probs. 12-125/126

12-127. A tram is traveling with a constant speed of 14 m/s along the curved path. Determine the magnitude of the acceleration of the front of the tram, B, at the instant it reaches point A (y = 0).

Prob. 12-124

x (m)

Prob. 12-127

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTSó3

*12-128. When a car starts to round a curved road with the radius of curvature of 600 ft, it is traveling at 75 ft/s. If the car's speed begins to decrease at a rate of

= (-0.06t2) ft /s2, determine the magnitude of the acceleration of the car when it has traveled a distance of s = 700 ft.

Prob. 12-128

12-129. When the motorcyclist is at A, he increases his speed along the vertical circular path at the rate of

= (0.3t) ft/s2, where t is in seconds. If he starts from rest at A, determine the magnitudes of his velocity and acceleration when he reaches B.

12-130. When the motorcyclist is at A, he increases his speed along the vertical circular path at the rate of

= (0.04s) ft/s2 where s is in ft. If he starts at vA = 2 ft/s where s = O at A, determine the magnitude of his velocity when he reaches B. Also, what is his initial acceleration?

12-131. At a given instant the train engine at E has a speed of 20 m/s and an acceleration of 14 m/s2 acting in the 12 direction shown. Determine the rate of increase in the train's speed and the radius of curvature p of the path.

*12-132. Car B turns such that its speed is increased by (a,)9= (0.5e) m/s2, where t is in seconds. If the car starts from rest when O = 0°, determine the magnitudes of its velocity and acceleration when the arm AD rotates O = 30°. Neglect the size of the car.

12-133. Car B turns such that its speed is increased by (ar)9 = (0.5es) m/s2, where t is in seconds. If the car starts from rest when O = 0°, determine the magnitudes of its velocity and acceleration when t = 2 s. Neglect the size of the car.

Probs. 12-132/133

300 ft 60°

300 ft

Probs. 12-129/130

-M. tint/t/stip20 m/s

//:1:*

a= 14 m/s2

Prob. 12-131


12-134. A boat is traveling along a circular curve having a 12 radius of 100 ft. If its speed at t = O is 15 ft/s and is

increasing at v = (0.8t) ft/s2, determine the magnitude of its acceleration at the instant t = 5 s.

12-135. A boat is traveling along a circular path having a radius of 20 m. Determine the magnitude of the boat's acceleration when the speed is v = 5 m/s and the rate of increase in the speed is v = 2 m/s2.

*•12-136. Starting from rest, a bicyclist travels around a horizontal circular path, p = 10 m, at a speed of v = (0.09(2 + 0.1t) m/s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

12-137. A particie travels around a circular path having a radius of 50 m. If it is initially traveling with a speed of 10 m/s and its speed then increases at a rate of v = (0.05 v) m/s2

, determine the magnitude of the particle's acceleraton four seconds later.

12-138. When the bicycle passes point A, it has a speed of 6 m/s, which is increasing at the rate of v = 0.5 m/s2. Determine the magnitude of its acceleration when it is at point A.

y= 12 In (ii)

12-139. The motorcycle is traveling at a constant speed of 60 km/h. Determine the magnitude of its acceleration when it is at point A.

*12-140. The jet piane travels along the vertical parabolic path. When it is at point A it has a speed of 200 m/s, which is increasing at the rate of 0.8 m/s2. Determine the magnitude of acceleration of the piane when it is at point A.

Y

I--- 5 km - - I

Prob. 12-140

12-141. The bali is ejected horizontally from the tube with a speed of 8 m/s. Find the equation of the path, y = f(x), and then find the ball's velocity and the normal and tangential components of acceleration when t = 0.25 s.

Prob. 12-138

I0 km

v

vA = 8 m/sA-dt"------- x

o

~ 1 1 1 1 1 ~ - -

Prob. 12-139 Prob. 12-141

Ą ( Y 2 =


12-142. A toboggan is traveling down along a curve which 12-145. If the speed of the crate at A is 15 ft/s, which iscan be approximated by the parabola y = 0.01x2. increasing at a rate i, = 3 ft/s2, determine the magnitude ofDetermine the magnitude of its acceleration when it reaches the acceleration of the crate at this instant.point A, where its speed is vA = 10 m/s, and it is increasing at the rate of vA = 3 m/s2.

12-146. The race car has an initial speed vA = 15 m/s at A. If it increases its speed along the circular track at the rate at = (0.4s) m/s2, where s is in meters, determine the time needed for the car to travel 20 m. Take p = 150 m.Prob. 12-142

Prob. 12-145

12-143. A particie P moves along the curve y = (x2 - 4) m with a constant speed of 5 m/s. Determine the point on the curve where the maximum magnitude of acceleration occurs and compute its value.

*12-144. The Ferris wheel turns such that the speed of the passengers is increased by V = (4t) ft/s2, where t is in seconds. If the wheel starts from rest when O = 0°, determine the magnitudes of the velocity and acceleration of the passengers when the wheel turns O = 30°.

Prob. 12-1-14

Prob. 12-146

12-147. A boy sits on a merry-go-round so that he is always located at r = 8 ft from the center of rotation. The merry-go-round is originally at rest, and then due to rotation the boy's speed is increased at 2 ft/s2. Determine the time needed for his acceleration to become 4 ft/s2.

*12-148. A particie travels along the path y = a + bx + cx2, where a, b, c aro constants. If the speed of the particie is constant, u = vo, determine the x and y components of velocity and the norma) component of acceleration when x = 0.


12-149. The two particles A and B start at the origin O and12 travel in opposite directions along the circular path at

constant speeds vA = 0.7 m/s and vi? = 1.5 m/s, respectively. Determine in t = 2 s, (a) the displacement along the path of each particie, (b) the position vector to each particie, and (c) the shortest distance between the particles.

12-150. The two particles A and B start at the origin O and travel in opposite directions along the circular path at constant speeds vA = 0.7 m/s and v9 = 1.5 m/s, respectively. Determine the time when they collide and the magnitude of the acceleration of B just before this happens.

'12-153. A go-cart moves along a circular track of radius 100 ft such that its speed for a short period of time, 0 s t s 4 s, is v = 60(1 - e-'2) ft/s. Determine the magnitude of its acceleration when t = 2 s. How far has it traveled in t = 2 s? Use a numerical method to evaluate the integral.

•12-154. The bali is kicked with an initial speed vA = 8 m/s at an angle BA = 40° with the horizontal. Find the equation of the path, y = fix), and then determine the ball's velocity and the normal and tangential components of its acceleration when t = 0.25 s.

12-151. The position of a particie traveling along a curved path is s = (3t3 - 442 + 4) m, where t is in seconds. When t =2 s, the particie is at a position on the path where the radius of curvature is 25 m. Determine the magnitude of the particle's acceleration at this instant.

*12-152. If the speed of the box at point A on the track is 30 ft/s which is increasing at the rate of v = 5 ft/s2,determine the magnitude of the acceleration of the box at this instant.

Y

x

Prob. 12-154

12-155. The race car travels around the circular track with a speed of 16 m/s. When it reaches point A it increases its speed at a, = (3yl/4) m/s', where v is in m/s. Determine the magnitudes of the velocity and acceleration of the car when it reaches point B. Also, how much time is required for it to travel from A to B?

Prob. 12-155

x- v4 = 0.7 m/s

Probs. 12-149/150

y = 0.004x2 +10

50 ft

Prob. 12-152

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS67

*12-156. A particie P travels along an elliptical spiral path such that its position vector r is defined by r = {2 cos(0.1t)i + 1.5 sin(0.10j + (20k } m, where t is in seconds and the argumenty for the sine and cosine are given in radians.When t = 8 s, determine the coordinate direction angles a, 13, and y, which the binormal axis to the osculating piane makes with the x, y, and z axes. Hint:• Solve for the velocity vp and acceleration ap of the particie in terms of their i, j, k components. The binormal is parallel to Vp x ap. Why?

12-157. The motion of a particie is defined by the equations x = (2t + t2) m and y = (t2) m, where t is in 12 seconds. Determine the normal and tangential components of the particle's velocity and acceleration when t = 2 s.

12-158. The motorcycle travels along the elliptical track at a constant speed v. Determine the greatest magnitude of the acceleration if a > b.

Prob. 12-156 Prob. 12-158

1 2.8 Curvilinear Motion: Cylindrical Components

Sometimes the motion of the particie is constrained on a path that is best described using cylindrical coordinates. If motion is restricted to the piane, then polar coordinates are used.

Poler Coordinates. We can specify the location of the particie shown in Fig. 12-30a using a radial coordinate r, which extends outward from the fixed origin O to the particie, and a transverse coordinate O, which is the counterclockwise angle between a fixed reference line and the r axis. The angle is generali)/ measured in degrees or radiany, where1 rad = 180°//t. The positive directions of the r and O coordinates are Odefined by the unit vectors u, and uo, respectively. Here ur is in the direction of increasing r when 8 is held fixed, and uf, is in a direction of increasing O when r is held fixed. Note that these directions are perpendicular to one another.

r

Position

(a)

Fig. 12-30

6H CHAPTER 12 KINEMATICS OF A PARTICLE

1 2

r

Position

( a )

u,

Position. At any instant the position of the particie, Fig. 12-30a, is defined by the position vector

r = ru, (12-22)

Velocity. The instantaneous velocity v is obtained by taking the time derivative of r. Using a dot to represent the time derivative, we have

v = = r u , +

To evaluate n„ notice that u, only changes its direction with respect to time, since by definition the magnitude of this vector is always one unit. Hence, during the time At, a change Ar will not cause a change in thedirection of ur; however, a change AO will cause ur to become whereu', = u,. + Au,, Fig. 12-306. The time change in u, is then Aur. For smaliangles AO this vector has a magnitude Au, 1(A0) and acts in the uedirection.Therefore, Au, Aeue, and so

A u r ( A OU, = lim.0 At —) ueAt— ,t--*0

ur = eUe (12-23)

oV e l o c i t y

( c )

Fig. 12-30 (cont.)

Substituting into the above equation, the velocity can be written in component form as

V = 21,11". VoU0 (12-24)

where

v, = r(12-25)

Ve =

These components are shown graphically in Fig. 12-30c. The radial component v, is a measure of the rate of increase or decrease in the length of the radial coordinate, i.e., whereas the transverse component vo

can be interpreted as the rate of motion along the circumference of a circie having a radius r. In particular, the term B - de/dt is called the angolar velocity, since it indicates the time rate of change of the angle O. Common units used for this measurement are rad/s.

Since v, and vo are mutually perpendicular, the magnitude of velocity or speed is simply the positive vaiue of

V = N7(r)2 + (r0)2 (12-26)

and the direction of v is, of course, tangent to the path, Fig. 12-30c.

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS b9

Acceleration. Taking the time derivatives of Eq. 12-24, using Eqs. 12-25, we obtain the particle's instantaneous acceleration,

a = V = rur + ;lir+ ifiuo + rdue + rÓtle

To evaluate tio, it is necessary only to find the change in the direction of uosince its magnitude is always unity. During the time At, a change Ar will not change the direction of uo, however, a change AO will cause uo to become ue, where ue = uo + Auo, Fig. 12-30d. The time change in uo isthus pue. For smali angles this vector has a magnitude Auo I (AO) andacts in the -u, direction; i.e., Auo = -AOur. Thus,

= =

Auo pelu0 — u —)

.sri-o At At->m0 ptu,

U0 = -Our (12-27)

(d )

Substituting this result and Eq. 12-23 into the above equation for a, we can write the acceleration in component form as

a = arur + aouol (12-28)

where

a r = - a o = +

(12-29)

The term 9 = d2O/dt2 = d/dr(dO/dt) is called the angular acceleration since it measures the change made in the angular velocity during an instant of time. Units for this measurement are rad/s2.

Since a, and ao are always perpendicular, the magnitude of acceleration is simply the positive value of

O

Aceduration

( f )

a =- 162)2 + (r.ti + 21"-Ó)2 (12-30)

The direction is determined from the vector addition of its two components. In general, a will not be tangent to the path, Fig. 12-30e.

O

7OCHAPTER 12 KINEMATICS OF A PARTICLE

Cylindrical Coordinates. If the particie moves along a space curve as shown in Fig. 12-31, then its location may be specified by the three cylindrical coordinates, r, 0, z. The z coordinate is identical to that used for rectangular coordinates. Since the unit vector defining its direction, u-, is constant, the time derivatives of this vector are zero, and therefore the position, velocity, and acceleration of the particie can be written in terms of its cylindrical coordinates as follows:

rp = /ur + zu,

v = rur + rÓue + żu,(12-31)

a = — + + 2i.Ó)ne + (12-32)

Time Derivatives. The above equations require that we obtain thetime derivatives Ó, and d in order to evaluate the r and O componentsof v and a.'INvo types of problems generally occur:

1. If the polar coordinates are specified as time parametric equations, r = r(t) and B = 0(t), then the time derivatives can be found directly.

2. If the time-parametric equations are not given, then the path r = AO) must be known. Using the chain rule of calculus we can then find the relation between i- and Ó, and between 7- and d. Application of the chain rule, along with some examples, is explained in Appendix C.


Coordinate System.

• Polar coordinates are a suitable choice for solving problems when data regarding the angular motion of the radial coordinate r is given to describe the particle's motion. Also, some paths of motion can conveniently be described in terms of these coordinates.

• To use polar coordinates, the origin is established at a fixed point. and the radial line r is directed to the particie.

• The transverse coordinate B is measured from a fixed reference line to the radial line.

Velocity and Acceleration.

• Once r and the four time derivatives Ó, and d have beenevaluated at the instant considered, their values can be substituted into Eqs. 12-25 and 12-29 to obtain the radial and transverse components of v and a.

• If it is necessary to take the time derivatives of r = f(0), then the chain rule of calculus must be used. See Appendix C.

• Motion in three dimensions requires a simple extension of the above procedure to include ż and ż.

12

The spiral motion of this girl can be followed by using cylindrical components. Here the radial coordinate r is constant, the transverse coordinate O will increase with time as the girl rotates about the vertical, and her altitude z will decrease with time.

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 71

EXAMPLE

The amusement park ride shown in Fig. 12-32a consists of a chair that is rotating in a horizontal circular path of radius r such that the arm OB has an angular velocity 9 and angular acceleration B. Determine the radial and transverse components of velocity and acceleration of the passenger. Neglect his size in the calculation.

(a) (b)

Fig. 12-32

SOLUTIONCoordinate System. Since the angular motion of the arm is reported, polar coordinates are chosen for the solution, Fig. 12-32a. Here O is not related to r, since the radius is constant for all B. Velocity and Acceleration. It is first necessary to specify the first and second time derivatives of r and B. Since r is consłara, we have

r = r = O r=0

Thus,

v,. = i- = O Ans.

ve = r6 Ans.

a, = _ ró2 = Ans.ao = + = r6 Ans.

These results are shown in Fig. 12-328.

NOTE: The n, t axes are also shown in Fig. 12-32b, which in this special case of circular motion happen to be callinear with the r and O axes, respectively. Since v = v9 = v, = r9, then by comparison,

( r ) 2

— a , = a „ = = — = r O 2

P rdv d • dr • tiOao = a, = dt dt — = (rO) dt

= — O + r—dt = O + rO


EXAMPLE

1 2 . 1 8

The rod OA in Fig. 12-33a rotates in the horizontal piane such that = (t3) rad. At the same time, the collar B is sliding outward along OA so

that r = (100r2) mm. If in both cases t is in seconds, determine the velocity and acceleration of the collar when t = I s.

( a )

( b )

= 5730

(c)

Fig. 12-

33

SOLUTIONCoordinate System. Since time-parametric equations of the path are given, it is not necessary to relate r to O.

Velocity and Acceleration. Determining the time derivatives and evaluating them when t = 1 s, we have

r = 100t2 = 100mm O = t31= 1 rad = 57.3°

r=1 st=ls

= 200/1 = 200 mm/s B = 3t21= 3 rad/s

1=1 si=ls=2001 = 200 mm/s-

-, d = 6t1 = 6 rad/s2.t= i s t=i s

As shown in Fig. 12-33b,

V = + rÓut,= 200u, + 100(3)ue = {200u„ + 300u0} mm/s The

magnitude of v is

v = 1(200)2 + (300)2 = 361 mm/s Ans.

8 = tan-1(2:—M) = 56.3°

200 8 + 57.3° = 114° Ans.

As shown in Fig. 12-33c,

a = (r - r92)u,. + + 2i-9)uo

= [200 - 100(3)2]u, + [100(6) + 2(200)31uo

as = 1800 mm/52= { -700u, + 1800u0} mm/s2

The magnitude of a is

a` = 14-700)2 + (1800)2 = 1930 mm/s2 Ans.

NOTE: The velocity is tangent to the path; however, the acceleration is directed within the curvature of the path, as expected.

1 2

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 73

1 2 . 1 9

The searchlight in Fig. 12-34a casts a spot of light along the face of a wali that is located 100 m from the searchlight. Determine the magnitudes of the velocity and acceleration at which the spot appears to travel across the wali at the instant 8 = 45°. The searchlight rotates at a constant rate of Ó = 4 rad/s.

SOLUTIONCoordinate System. Polar coordinates will be used to solve this problem since the angular rate of the searchlight is given. To find the necessary time derivatives it is first necessary to relate r to O. From Fig. 12-34a,

r = 100/cos B = 100 sec 0

Velocity and Acceleration. Using the chain rule of calculus, noting that d(sec 0) = sec 0 tan 8 da, and d(tan 0) = sec2 B da, we have

= 100(sec O tan 0)9r = 100(sec 0 tan 0)9(tan 0)0 + 100 sec 0(sec2 0)0(Ó)

+ 100 sec O tan 0(d)= 100 sec O tan20 (Ó)2 + 100 sec30 (Ó)2 + 100(sec O tan 0)

Since 9 = 4 rad/s = constant, then 9 = 0, and the above equations, when O = 45°, become

r = 100 sec 45° = 141.4= 400 sec 45° tan 45° = 565.7= 1600 (sec 45° tang- 45° + sec3 45°) = 6788.2

As shown in Fig. 12-34b,

v = + rfiuo

= 565.7u, + 141.4(4)%

= { 5 6 5 . 7 u , + 5 6 5 . 7 % } m / s v = = N7(565.7)2 + (565.7)2

= 800 m/s Ans.

As shown in Fig. 12-34c,

a = — r02)u, + (rd + 2i-Ó)uo= [6788.2 — 141.4(4)2]11r + [141.4(0) + 2(565.7)41% = {4525.5u„ + 4525.5u0} m/s2

a = 14,2 + a20- = 144525.5)2 + (4525.5)2

= 6400 m/s2 Ans.

NOTE: It is also possible to find a without having to calculate r (or ar). As shown in Fig. 12-34d, since ao = 4525.5 m/s2, then by vector resolution, a = 4525.5/cos 45° = 6400 m/s2.

100 m

( b )

100 ni

( c )

(d)

Fig. 12-34


EXAMPLE

2 .20

r = 0.5 (1 - cos o) ft

( a )

r

(b )

Fig. 12-35

Due to the rotation of the forked rod, the bali in Fig. 12-35a travels around the slotted path, a portion of which is in the shape of a cardioid, r = 0.5(1 - cos e) ft, where O is in radians. If the ball's velocity is v = 4 ft/s and its acceleration is a = 30 ft/s2 at the instant O= 180°, determine the angular velocity Ó and angular acceleration

of the fork.

SOLUTION

Coordinate System. This path is most unusual, and mathematically it is best expressed using polar coordinates, as done here, rather than rectangular coordinates. Also, since Ó and d must be determined, then r, O coordinates are an obvious choice.

Velocity and Acceleration. The time derivatives of r and O can be determined using the chain rule.

r = 0.5(1 - cos O)

r = 0.5(sin 0)9

= 0.5(cos 61)Ó(9) + 0.5(sin

Evaluating these results at O = 180°, we haver = 1 ft r = r=-0.562

Since v = 4 ft/s, using Eq. 12-26 to determine Ó yields

v = N4)2 + (rÓ)2

4 = 17(0)2 + (16)2

Ó = 4 rad/s

In a similar manner, B can be found using Eq. 12-30.

a = 1/('r - r2)2 + + 2i.Ó)2

30 = 14-0.5(4)2 - 1(4)212 + 11i5i + 2(3)(4)]*2 (

30)2 = (-24)2 +

B = 18 rad/s2 Ans.

Vectors a and v are shown in Fig. 12-35b.

NOTE: At this location, the O and t (tangential) axes will coincide. The +n (normal) axis is directed to the right, opposite to +r.

12



F12-33. The car has a speed of 55 ft/s. Determine the angular velocity 6 of the radial line OA at this instant.

F12-36. Peg P is driven by the forked link OA along the path described by r = , where r is in meters. When 0 = rad, the link has an angular velocity and angular acceleration of

= 2 rad/s and (i = 4 rad/s2. Determine the radial and transverse components of the peg's acceleration at this instant.

r=eB

O

F12-33

F12-34. The platform is rotating about the vertical axis such that at any instant its angular position is B = (4t3'2) rad, where t is in seconds. A bali rolls outward along the radial groove so that its position is r = (0.1t3) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the bali when t = 1.5 s.

F12-34

F12-35. Peg P is driven by the fork link OA along the curved path described by r = (20) ft. At the instant

= /r/4 rad, the angular velocity and angular acceleration of the link are B = 3 rad/s and B = I rad/s2. Determine the magnitude of the peg's acceleration at this instant.

O u,

F12-36 F12-37. The collars are pin connected at B and are free to move along rod OA and the curved guide OC having the shape of a cardioid, r = [0.2(1 + cos d)] m. At 0 = 30°, the angular velocity of OA is B = 3 rad/s. Determine the magnitude of the velocity of the collars at this point.

A

= 3 rad/s F12-37F12-38. At the instant B = 45°, the athiete is running with a constant speed of 2 m/s. Determine the angular velocity at which the camera must turn in order to follow the motion.

r = (30 esc 0) m

F12-35 F12-38

7 6 CHAPTER 12 KINEMATICS O F A P A R T I C L E

12 • PROBLEMS

12-159. A particie is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by 0 = cos 2r, where O is in radians and t is in seconds. Determine the magnitude of the acceleration of the particie when O = 30°.

*12-160. A particie travels around a limgon, defined by the equation r = b — a cos 0, where a and b are constants. Determine the particle's radial and transverse components of velocity and acceieration as a function of O and its time derivatives.

12-161. If a particle's position is described by the polar coordinates r = 4(1 + sin t) m and O = (2e') rad, where t is in seconds and the argument for the sine is in radians, determine the radial and tangential components of the particie's velocity and acceleration when t = 2 s.

12-162. An airpiane is flying in a straight line with a velocity of 200 mi/h and an acceleration of 3 mi/h2. If the propeller has a diameter of 6 ft and is rotating at a constant angular rate of 120 rad/s, determine the magnitudes of veiocity and acceleration of a particie iocated on the tip of the propeller.

12-163. A car is traveling along the circuiar curve of radius r = 300 ft. At the instant shown, its anguiar rate of rotation is B = 0.4 rad/s, which is increasing at the rate of d = 0.2 rad/s2. Determine the magnitudes of the car's velocity and acceieration at this instant.

Prob. 12-163

*12-164. A radar gun at O rotates with the angular velocity of d = 0.1 rad/s and angular acceleration of B = 0.025 rad/s2

, at the instant O = 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200 m. Determine the magnitudes of velocity and acceleration of the car at this instant.

OProb. 12-164

12-165. If a particie moves along a path such that r = (2 cos t) ft and 0 = (t/2) rad, where t is in seconds, plot the path r = f(0) and determine the particie's radial and transverse components of velocity and acceleration.

12-166. If a particie's position is described by the polar coordinates r = (2 sin 20) m and O = (4t) rad, where t is in seconds, determine the radial and tangential components of its velocity and acceieration when t = 1 s.

12-167. The car travels along the circular curve having a radius r = 400 ft. At the instant shown, its anguiar rate of rotation is B = 0.025 rad/s, which is decreasing at the rate

= —0.008 rad/s2. Determine the radial and transverse components of the car's velocity and acceieration at this instant and sketch these components on the curve.

*12-168. The car travels along the circuiar curve of radius r = 400 ft with a constant speed of v = 30 ft/s. Determine the angular rate of rotation 9 of the radial line r and the magnitude of the car's acceleration.

Probs. 12-167/168


12-169. The time race of change of acceleration is referred to as the jerk, which is often used as a means of measuring passenger discomfort. Calculate this vector, Ś. in terms of its cylindrical components, using Eq. 12-32.

12-170. A particie is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by O = sin 31, where O and the argument for the sine are in radians, and t is in seconds. Determine the acceleration of the particie at O = 30°. The particie starts from rest at O = 0°.

12-171. The slotted link is pinned at 0, and as a result of the constant angular velocity B = 3 rad/s it drives the peg P fora short distance along the spiral guide r = (0.4 0) m, where O is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant O = ir/3 rad.

12-172. Solye Prob. 12-171 if the slotted link has an angular acceleration 8 = 8 rad/s2 when B = 3 rad/s at O= ir/3 rad.

12-173. The slotted link is pinned at O, and as a result of the constant angular velocity B = 3 rad/s it drives the peg P for a short distance along the spiral guide r = (0.4 0) m, where O is in radians. Determine the velocity and acceleration of the particie at the instant it leaves the slot in the link, i.e., when r = 0.5 m.

Probs. 12-171/172/173

12-174. A particie moves in the x -y piane such that its position is defined by r = (2ii + 4t2j} ft, where t is in seconds. Determine the radial and transverse components of the particie's velocity and acceleration when t = 2 s.

12-175. A particie P moves along the spiral path r = (1010) ft, where O is in radians. If it maintains a constant 12 speed of v = 20 ft/s, determine vr and vo as functions of O and evaluate each at O = 1 rad.

v. 1 4 ■ .

Prob. 12-175

*12-176. The driver of the car maintains a constant speed of40 m/s. Determine the angular velocity of the camera tracking the car when O = 15°.

12-177. When O = 15°, the car has a speed of 50 m/s which is increasing at 6 m/s2. Determine the angular velocity of the camera tracking the car at this instant.

Probs. 12-176/177

78 CHAPTER 12 K INEMATICS OF A PARTICLE

12-178. The smali washer slides down the cord OA.12 When it is at the midpoint, its speed is 200 mm/s and its

acceleration is 10 mm/s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical compone n ts.

A

___ 7-7— Y

700 mm

O_

400 3°° min

Prob. 12-178

*12-180. Pin P is constrained to move along the curve defined by the lemniscate r= (4 sin 20) ft. If the slotted arm OA rotates counterclockwise with a constant angular velocity of B = 1.5 rad/s, determine the magnitudes of the velocity and acceleration of peg P when O = 60°.

12-181. Pin P is constrained to move along the curve defined by the lemniscate r = (4 sin 20) ft. If the angular position of the slotted arm OA is defined by O = (31312) rad, determine the magnitudes of the velocity and acceleration of the pin P when d= 60°.

Probs. 12-180/181

12-179. A block moves outward along the slot in the platform with a speed of = (4t) m/s, where t is in seconds. The platform rotates at a constant rate of 6 rad/s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = l s.

12-182. A cameraman standing at A is following the movement of a race car, B, which is traveling around a curved track at a constant speed of 30 m/s. Determine the angular rate d at which the man must turn in order to keep the camera directed on the car at the instant d = 30°.

vB = 30 m/słkik

20 m

2 0 m 2 0 n i

Prob. 12-182

20 m

LA

Prob. 12-179


12-183. 'The slotted arm AB drives pin C through the spiral groove described by the equation r = a0. If the angular velocity is constant at Ó, determine the radial and transverse components of velocity and acceleration of the pin.

*12-184. The slotted arm AB drives pin C through the spiral groove described by the equation r = (1.5 0) ft, where O is in radians. If the arm starts from rest when O = 60° and is driven at an angular velocity of Ó = (4t) rad/s, where t is in seconds, determine the radial and transverse components of velocity and acceleration of the pin C when t = 1 s.

Probs. 12-183/184

12-185. If the slotted arm AB rotates countercloekwise with a constant angular velocity of 9 = 2 rad/s, determine the magnitudes of the velocity and acceleration of peg P at O = 30°. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB.

12-186. The peg is constrained to move in the slots of the fixed bar CD and rotating bar AB. When 0= 30°, the angular velocity and angular acceleration of arm AB are Ó = 2 rad/s and d = 3 rad/s2, respectively. Determine the magnitudes of the velocity and acceleration of the peg Pat this instant.

12-187. If the circular plate rotates clockwise with a constant angular velocity of e = 1.5 rad/s, determine the magnitudes of the velocity and acceleration of the follower rod AB when O = 2/3tr rad.

*12-188. When O = 2/3/r rad, the angular velocity and angular acceleration of the circular plate are 9 = 1.5 rad/s and = 3 rad/s2, respectively. Determine the magnitudes of the velocity and acceleration of the rod AB at this instant.

12-189. The box slides down the helical ramp with a constant speed of v = 2 m/s. Determine the magnitude of its acceleration. The ramp descends a vertical distance of I m for every full revolution.The mean radius of the ramp is r =0.5 m.

12-190. The box slides down the helical ramp such that r = 0.5 m, O = (0.5t3)rad, and z = (2 - 0.2,2) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the box at the instant O = 2tr rad.

r = (10 + 50 0112) mm

Probs. 12-187/188

D

B

4 f t

Probs. 12-185/186 Probs. 12-189/190

8O CHAPTER 12 KINEMATICS OF A PARTICLE

12-191. For a short distance the train travels along a track12 having the shape of a spiral, r = (1000/B) m, where O is in

radians. If it maintains a constant speed v = 20 m/s, determine the radial and transverse components of its velocity when O = (97r/4) rad.

*12-192. For a short distance the train traveis along a track having the shape of a spiral, r = (1000/9) m, where O is in radians. If the angular rate is constant, 8 = 0.2 rad/s, determine the radial and transverse components of its velocity and acceleration when O = (9/r/4) rad.

Probs. 12-191/192

12-193. A particie moves along an Archimedean spiral r = (80) ft, where O is given in radians. If Ó = 4 rad/s (constant), determine the radial and transverse components of the particle's velocity and acceleration at the instant

= /r/2 rad. Sketch the curve and show the components on the curve.

12-194. Solve Prob. 12-193 if the particie has an angular acceleration d = 5 rad/s2 when Ó = 4 rad/s at O = 7r/2 rad.

Probs. U-193/194

12-195. The arm of the robot has a fixed length of r = 3 ft and its grip A moves along the path z = (3 sin 49) ft, where O is in radians. If B = (0.50 rad, where t is in seconds, determine the magnitudes of the grip's velocity and acceleration when t = 3 s.

*12-196. Fora short time the arm of the robot is extending at a constant rate such that r = 1.5 ft/s when r = 3 ft, z = (4t2) ft, and O = 0.51 rad, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the grip A when t = 3 s.

Probs. 12-195/196

12-197. The partial surface of the cam is that of a logarithmic spiral r = (40e "5') mm, where O is in radians. If the cam is rotating at a constant angular rate of

= 4 rad/s, determine the magnitudes of the velocity and acceleration of the follower rod at the instant O = 30°.

12-198. Solve Prob. 12-197, if the cam has an angular acceleration of d = 2 rad/s2 when its angular velocity is Ó = 4 rad/s at O = 30°.

ó= 4 rad/s

Probs. 12-197/198

12.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF Two PARTICLES 8 1

12.9 Absolute Dependent Motion Analysis of Two Particles

In some types of problems the motion of one particie will depend on the corresponding motion of another particie. This dependency commonly occurs if the particles, here represented by blocks, are interconnected by inextensible cords which are wrapped around pulleys. For example, the movement of block A downward along the inclined piane in Fig. 12-36 will cause a corresponding movement of block B up the other incline. We can show this mathematically by first specifying the location of the blocks usingposition coordinates sA and 5.8. Note that each of the coordinate axes is (1) measured from a fixed point (0) or fixed datum line, (2) measured along each inclined piane in the direction of motion of each block, and (3) has a positive sense from the fixed datums to A and to B. If the total cord length is IT, the two position coordinates are related by the equation

sA + /cD + sB = /T

Here łaj is the length of the cord passing over arc CD. Taking the time derivative of this expression, realizing that łaj and IT remain constant, while sA and sB measure the segments of the cord that change in length, we have

d s A d s a+ — = O o r V g = - V Adt dt

The negative sign indicates that when block A has a velocity downward, i.e., in the direction of positive sA, it causes a corresponding upward velocity of block B; i.e., B moves in the negative sB direction.

In a similar manner, time differentiation of the velocities yields the relation between the accelerations, i.e.,

ap = -aA

A more complicated example is shown in Fig. 12-37a. In this case, the position of block A is specified by sA, and the position of the end of the cord from which block B is suspended is defined by sa. As above, we have chosen position coordinates which (1) have their origin at fixed points or datums, (2) are measured in the direction of motion of each block, and (3) from the fixed datums are positive to the right for sA and positive downward for sB. During the motion, the length of the red colored segments of the cord in Fig. 12-37a remains constant. If 1 represents the total length of cord minus these segments, then the position coordinates can be related by the equation

2sa + h + sA = 1

Since 1 and h are constant during the motion, the two time derivatives yield

2vB = 21113 = -aA

Hence, when B moves downward (+4), A moves to the left (-4) with twice the motion.

Fig. 12-36

H2 CHAPTER 12 KINEMATICS OF A PARTICLE

This example can also be worked by defining the position of block B from the center of the bottom pulley (a fixed point), Fig. 12-37b. In this case

2(h — sa) + h + sA = I

Time differentiation yields

2va = vA 2aa = aA

Here the signs are the same. Why?

12

Datum

(b)

Fig. 12-37 (cont.)


The above method of relating the dependent motion of one particie to that of another can be performed using algebraic scalars or position coordinates provided each particie moves along a rectilinear path. When this is the case, only the magnitudes of the velocity and acceleration of the particles will change, not their line of direction.

Position-Coordinate Equation.

• Establish each position coordinate with an origin located at a fixed point or daturo.

• k is not necessary that the origin be the same for each of the coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particie.

• Using geometry or trigonometry, reiate the position coordinates to the total length of the cord, /T, or to that portion of cord, which exchides the segments that do not change length as the particies move —such as arc segments wrapped over pulleys.

• If a problem involves a system of two or more cords wrapped around pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure. Separate equations are written for a fixed length of each cord of the system and the positions of the two particles are then related by these equations (see Examples 12.22 and 12.23).

Time Derivatives.

• 7ivo successive time derivatives of the position-coordinate equations yield the required velocity and acceieration equations which reiate the motions of the particies.

• The signs of the terms in these equations will be consistent with those that specify the positive and negative sense of the position coordinates.

lite motion of the lift on this crane depends upon the motion of the cable connected to the winch which operates it. It is important to be able to relate these motions in order to determine the power requirements of the winch and the (orce in the cable caused by any accelerated motion.

12.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF Two PARTICLES83

EXAMPLE

2._

Determine the speed of block A in Fig. 12-38 if błock B has an upward speed of 6 ft/s.

SOLUTION

Position-Coordinate Equation. There is one cord in this system having segments which change length. Position coordinates sA and ss

will be used since each is measured from a fixed point (C or D) and extends along each block's path of motion. In particular, sB is directed to point E since motion of 13 and E is the same.

The red colored segments of the cord in Fig. 12-38 remain at a constant length and do not have to be considered as the blocks move. The remaining length of cord. I, is also constant and is related to the changing position coordinates sA and ss by the equation

sA + 3s8 = I

Time Derivative. Taking the time derivative yields

vA + 3vs = O

so that when v s = -6 ft/s (upward).

vA = 18 ft/s

D Datum

S

5.1

Fig. 12-38

B


Determine the speed of A in Fig. 12-39 if B has an upward speed of 6 ft/s.

SOLUTION

Position-Coordinate Equation. As shown, the positions of blocks A and B are defined using coordinates sA and sa. Since the system has two cords with segments that change length, it will be necessary to use a third coordinate, sc, in order to relate sA to sa. In other words, the length of one of the cords can be expressed in terms of sA and sc, and the length of the other cord can be expressed in terms of s8 and sc.

The red colored segments of the cords in Fig. 12-39 do not have to be considered in the analysis. Why? For the remaining cord lengths. say II and 12, we have

sA + 2sc = /1 sa + (sB - sc) = /2

Tim e Derivative. Taking the time derivative of these equations yields

VA 2vc = O - vc O

Eliminating vc produces the relationship between the motions of each cylinder.

vA + 4va = O

so that when va = -6 ft/s (upward),

vA = +24 ft/s = 24 ft/s Ans.

S A

sr■

ft/s

Fig. 12-39

12

1 2.9 ABSOLUTE DEPENDENT MOTION ANALYSIS OF Two PARTICLES 85

Determine the speed of block B in Fig. 12-40 if the end of the cord at A is pulled down with a speed of 2 m/s.

Fig. 12-40

SOLUTION

Position-Coordinate Equation. The position of point A is defined by sA

, and the position of block B is specified by SB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fzxed pin at pulley D. Since the system consists of two cords, the coordinates sA

and sB cannot be related directly. Instead, by establishing a third position coordinate, sc, we can now express the length of one of the cords in terms of s8 and sc, and the length of the other cord in terms of sA, SB, and sc.

Excluding the red colored segments of the cords in Fig. 12-40, the remaining constant cord lengths 1, and /2 (along with the hook and link dimensions) can be expressed as

sc + SB = lI(SA SC) + (SB — SC) + sB = 12

Time Derivative. The time derivative of each equation gives

vc + VB = OvA — 2vc + 2v8 = O

Eliminating vc, we obtain

vA + 4vB = O

so that when vA = 2 m/s (downward),

vB = —0.5 m/s = 0.5 m/s I In \. \.

8 6 CHAPTER 12 KINEMATICS OF A PARTICIE

EXAMPLE

12.24

A man at A is hoisting a safe Sas shown in Fig. 12-41 by walking to the right with a constant velocity vA = 0.5 m/s. Determine the velocity and acceleration of the safe when it reaches the elevation of 10 m.The rope is 30 m long and passes over a smali pulley at D.

SOLUTIONPosition-Coordinate Equation. This problem is unlike the previous examples since rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the positions of C and A, are specified by means of the x and y coordinates since they must be measured from a fixed point and directed along the paths of motion of the ends of the rope.The x and y coordinates may be related since the rope has a fixed length / = 30 m, which at all times is equal to the length of segment DA plus CD. Using the Pythagorean theorem to determine /DA , we have1DA = V(15)2 + x2; also, /cr, = 15 — y. Hence,

= 1DA + 1CD

30 = x2 + (15 — y)y= x2— 15 (1)

Time Derivatives. Taking the time derivative, using the chain rule (see Appendix C), where vs = dy/dt and vA = dx/dt, yields

dy [

1

2x dx= =vs

dt2 225+x x dt

Fig. 12-41

v\72 +x'. A (2)

At y = 10 m, x is determined from Eq. 1, i.e., x = 20 m. Hence, from Eq. 2 with vA = 0.5 m/s,

vs = 20 (0 5) = 0.4 m/s = 400mm/s I Ans.

V 2 2 5 , + ( 2 0 ) 2

dr (225 + v

x2)3/211;xeA acceleration is determined by taking the time derivative of Eq. 2. Since vA is constant, then a, = dvAldt = 0, and we haved2y f —x(dx/dt) 1 1 K dx i dvA 225v,2% as — -, f 225 dt )vA + [ x —

1.721- x2] di (225 + X2)312

At x = 20 m, with vA = 0.5 m/s, the acceleration becomes

225(0.5 m/s)2

as — — 0.00360 m/s2 = 3.60 mm/s2 t Aw[225 + (20 m) 2 ] 3 / 2 N O T E : The cons tan t ve loc i ty

a t A causes the o ther end C of the rope to h a v e an

acce le ra t ion s ince v A causes segment DA to change i t s

direction as well as its length.

1 2

8 712.10 RRATivE-MOTiON OF TWO PARTICLES USING TRANSLATING AXES

12.1 0 Relative-Motion of Two Partides Using Translating Axes

Throughout this chapter the absolute motion of a particie has been determined using a single fixed reference frame. There are urany cases, however, where the path of motion for a particie is complicated, so that it may be easier to analyze the motion in parts by using two or more frames of reference. For example, the motion of a particie located at the tip of an airplane propeller, while the piane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particie measured from a reference attached to the airplane.

In this section translating frames of reference will be considered for the analysis.

Position. Consider particles A and B, which move along the arbitrary paths shown in Fig. 12-42. The absolute position of each particie, rA and r8, is measured from the common origin O of the fixed x, y, z reference frame. The origin of a second frame of reference x', y', z' is attached to and moves with particie A. The axes of this frame are only permitted w translate relative to the fixed frame. The position of B measured relative to A is denoted by the relative position vector Using vector addition, the three vectors shown in Fig. 12-42 can be related by the equation

rh. = rA + rB/A (12-33)

Velocity. An equation that relates the velocities of the particles is determined by taking the time derivative of the above equation; i.e.,

Translating observer

r B /A

Fixed observer

Fig. 12-42

VB = VA VBIA (12-34)

Here y8 = *BI cit and vA = drA/dt refer to absolute velocities, since they are observed from the fixed frame; whereas the relative velocity vs/A = dr81A1 dł is observed from the translating frame. It is important to note that since the x', y', axes translate, the components of rB/A will not change direction and therefore the time derivative of these components will only have to account for the change in their magnitudes. Equation 12-34 therefore states that the velocity of B is equal to the velocity of A plus (vectorially) the velocity of "B with respect to A," as measured by the translating observer fixed in the x', y', z' reference frame.

8 8 CHAPTER 12 K INEMATICS OF A PARTICLE

12Acceleration. The time derivative of Eq. 12-34 yields a similar vector relation between the absolute and relative acceleration of particles A and B.

ae = aA aBlA (12-35)

Here apvil is the acceleration of B as seen by the observer located at A and translating with the x', y', z' reference frame.*


• When applying the relative velocity and acceleration equations, it is first necessary to specify the particie A that is the origin for the translating x', y', z' axel. Usually this point has a known velocity or acceleration.

• Since vector addition forms a triangle, there can be at most Iwo unknowns, represented by the magnitudes and/ or directions of the vector quantities.

• These unknowns can be solved for either graphically, using trigonometry (law of sines, law of cosines), or by resolving each of the three vectors into rectangular or Cartesian components, thereby generating a set of scalar equations.

The pilots of these jet planes flying close to one another must be aware of their relative positions and velocities at all times in order to avoid a collision.

* An easy way to remember the setup of these equations is to noce the "cancellation" of the subscript A between the Iwo terms,e.g., ap = a4 + apm.

12.10 RELATIVE-MOTION OF Two PARTICLES USING TRANSLATING AXES 89

I

A train travels at a constant speed of 60 mi/h and crosses over a road as shown in Fig. 12-43a. If the automobile A is traveling at 45 mi/h along the road, determine the magnitude and direction of the velocity of the train relative to the automobile.

SOLUTION I

Vector Analysis. The relative velocity vT/A is measured from the translating x', y' axes attached to the automobile, Fig. 12-43a. It is determined from vT = vA + vT/A. Since vT and vA are known in both magnitude and direction, the unknowns become the x and y components of vT/A. Using the x, y axes in Fig. 12-43a, we have

vT = vA + vT/A

601 = (45 cos 45°i + 45 sin 45°j) + vT/A

„7-/A = { 28.21 — 31.8j } mi/h

The magnitude of vT/A is thus

vT/A = V(28.2)2 + (-31.8)2 = 42.5 mi/h Ans.

From the direction of each component, Fig. 12-43b, the direction ofV T I A

tan B 31.8 o ( v T / A x 2 8 . 2

0 = 48.5° c

Note that the vector addition shown in Fig. 12-436 indicates the correct sense for vT/A . This figure anticipates the answer and can be used to check it.

SOLUTION II

Scalar Analysis. The unknown components of vT/A can also be determined by applying a scalar analysis. We will assume these components act in the positive x and y directions. Thus,

V T = V A V T / A[ 60 —3./hl [

= 4 45a mi/h + [ (vT/A)i + f (VT/A5 °

Resolving each vector into its x and y components yields

60 = 45 cos 45° + (vr/A),, + O(+I) O = 45 sin 45° + O + (OT/A)ySolving, we obtain the previous results,

(v77,i)x = 28.2 mi/h = 28.2 mi/h (vT/

A), = —31.8 mi/h = 31.8 mi/h 1,

A ns .

T ,40.„.,5,

Lx'

= 45 mi/h

(a)

28.2 mi,/h

31.8 mi/h v r

( b )

vA = 45 mi/h

vT = 60 mi/h

( c )

Fig. 12-43


EXAMPLE

12.26Piane A in Fig. 12-44a is flying along a straight-line path, whereas piane B is flying along a circular path having a radius of curvature of p8 = 400 km. Determine the velocity and acceleration of B as measured by the pilot of A.

SOLUTION

Velocity. The origin of the x and y axes are located at an arbitrary fixed point. Since the motion relative to piane A is to be determined, the translating frame of reference x', y' is attached to it, Fig. 12-44a. Applying the relative-velocity equation in scalar form since the velocity vectors of both planes are parallel at the instant shown, we have( a )

( + I ) Vg = VA VBIA

600 km/h = 700 km/h + VgIAVgłA = —100 km/h 1 /h = 100 km/h I Ans.

The vector addition is shown in Fig. 12-44b.

Acceleration. Piane B has both tangential and normal components of acceleration since it is flying along a curved path. From Eq. 12-20, the magnitude of the normal component is

vi? (600 km/h)2

(aa)n = = - 900 km/h2

P 400 kmApplying the relative-acceleration equation gives

V B I A

vA = 700 km/h vs = 600 km/h

( b )

a8 = aA + agpi

900i - 100j = 50j + aNA

Thus,

aBIA = {900i - 150j } km/h2

From Fig. 12-44c, the magnitude and direction of aBIA are therefore

aD/A = 912 km/h2 B = tan —150 = 9.46° Ans.900

NOTE: The solution to this problem was possible using a translating frame of reference, since the pilot in piane A is "translating." Observation of the motion of piane A with respect to the pilot of piane B, however, must be obtained using a rotating set of axes attached to piane B. (This assumes, of course, that the pilot of B is fixed in the rotating frame, so he does not turn his eyes to follow the motion of A.) The analysis for this case is given in Example 16.21.

900 km/h2

;0 a '

150 km 112 aBIA

( c )

Fig. 12-44

12

12.10 RELATIVE-MOTION OF Twa PARTICLES USING TRANSLATING AXES91

At the instant shown in Fig. 12-45a, cars A and B are traveling with speeds of 18 m/s and 12 m/s, respectively. Also at this instant, A has a decrease in speed of 2 rn/s2, and B has an increase in speed of 3 m/s2. Determine the velocity and acceleration of B with respect to A.

SOLUTION

Velocity. The fixed x, y axes are established at an arbitrary point on the ground and the translating x', y' axes are attached to car A, Fig. 12-45a. Why? The relative velocity is determined from v9 = vA + v8/A . What are the two unknowns? Using a Cartesian vector analysis, we have

V B = V A V B I A

- I 2j = (-18 cos 601 - 18 sin 60°j) + vg/A

VBIA = {91 + 3.588j} m/s

Thus,vB/A = V(9)2 + (3.588)2 = 9.69 m/s Ans.

Noting that vB/A has +i and +j components, Fig. 12-45b, its direction is

(vB/A)y 3.588tang -

(va/A)x 9= 21.7° a Ans.

Acceleration. Car B has both tangential and normal components of acceleration. Why? The magnitude of the normal component is uit(12 m/s)2

(aB)„ = - P 100 m - 1.440 m/s2

Applying the equation for relative acceleration yields

= a/4 ł aBIA

(-1.4401 - 3j) = (2 cos 60°i + 2 sin 60°j) + a s / a

•3 m /s2

•

p= 100m4 iU 4 12 m

Is B

( a 1

3.588 m/s

i n S

aB/A = { -2.4401 - 4.732j } m/s2

Here aB has -i and -j components.Thus, from Fig. 12-45c, 2.440

Ans.

,Ans.

m/s2

m/s24.732

aBiA = V(2.440)2 + (4.732)2

(aa/A)y 4.732

= 5.32 m/s2

d

tan =(aB/A)x 2.440

(/) = 62.7° NOTE: Is it possible to obtain the relative acceleration of aA/B using this (c)

method? Refer to the comment made at the end of Example 12.26. Fig. 12-45

1 2

92CHAPTER 12 KINEMATICS OF A PARTICLE


F12-39. Determine the velocity of block D if end A of the F12-42. Determine the velocity of błock A if end F of therope is pulled down with a speed of vA = 3 m/s. rope is pulled down with a speed of i F = 3 m/s.

F12-39

F12-40. Determine the velocity of block A if end B of the rope is pulled down willi a speed of 6 m/s.

F12-40

F12-41. Determine the velocity of block A if end B of the rope is pulled down with a speed of 1.5 m/s.

F12-41

1 vF = 3 m/s F12-42

F12-43. Determine the velocity of car A if point P on the cable has a speed of 4 m/s when the motor M winds the cable in.

F12-43

F12-44. Determine the velocity of cylinder B if cylinder A moves downward with a speed of vA = 4 ft/s.

= 4 ft/s

F12-44

12.10 RELATIVE-MOTION OF Two PARTICLES USING TRANSLATING AXES93

F12-45. Car A is traveling with a constant speed of 80 km/h due north, while car B is traveling with a constant speed of 100 km/h due east. Determine the velocity of car B relative to car A.

F12-47. The boats A and B travel with constant speeds of vA = 15 m/s and vB = 10 m/s when they leave the pier at 12 O at the same time. Determine the distance between them when r = 4 s.

F12-45

J

= 15 m/s30° A

F12-47

F12-46. 'Two planes A and B are traveling with the constant velocities shown. Determine the magnitude and direction of the velocity of piane B relative to piane A.

F12-48. At the instant shown, cars A and B are traveling at the speeds shown. If B is accelerating at 1200 km/h2 while A maintains a constant speed, determine the velocity and acceleration of A with respect to B.

112-46 F12-48

9 4 CHAPTER 12 KINEMATICS O F A P A R T I C L E

1 2 ■ P R O B L E M S

12-199. If the end of the cable at A is pulled down with a speed of 2 m/s, determine the speed at which block B rises.

Prob. 12-199

*12-200. The motor at C pulls in the cable with an acceleration ac = (3t2) m/s2, where t is in seconds. The motor at D draws in its cable at aa = 5 m/s2. If both motors start at the same instant from rest when d = 3 m, determine (a) the time needed from d = 0, and (b) the velocities of blocks A and B when this occurs.

D

Prob. 12-200

12-201. The crate is being lifted up the inclined piane using the motor M and the rope and pulley arrangement shown. Determine the speed at which the cable must be taken up by the motor in order to move the crate up the piane with a constant speed of 4 ft/s.

Prob. I 2-2o 1

12-202. Determine the time needed for the load at B to attain a speed of 8 m/s, starting from rest, if the cable is drawn into the motor with an acceleration of 0.2 m/s2. A

Prob. 12-202

12-203. Determine the displacement of the log if the truck at C pulls the cable 4 ft to the right.

Prob. 12-203

*12-204. Determine the speed of cylinder A, if the rope is drawn toward the motor Mat a constant rate of 10 m/s.

12-205. If the rope is drawn toward the motor M at a speed of vy = (5t3/2) m/s, where t is in seconds, determine the speed of cylinder A when t = 1 s.

Prohs. 12-204/205

d = 3 m

12.10 RELATIVE-MOTION OF Twa PARTICLES USING TRANSLATING AXES95

12-206. If the hydraulic cylinder H draws in rod BC at 2 ft/s, determine the speed of slider A.

A

H

Prob. 12-206

12-207. If block A is moving downward with a speed of 4 ft/s while C is moving up at 2 ft/s, determine the velociły of block B.

*12-208. If block A is moving downward at 6 ft/s while block C is moving down at 18 ft/s, determine the speed of block B.

Probs. 12-207/208

12-209. Determine the displacement of block B if A is pulled down 4 ft.

Prob. 12-2119

12-210. The pulley arrangement shown is designed for hoisting materials. If BC remains fixed while the plunger P is pushed downward with a speed of 4 ft/s, determine the speed of the load at A.

Prob. 12-210

12-211. Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m/s.

Prob. 12-211

*12-212. The cylinder Cis being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drutu with a speed of 2 m/s, determine the velocity of the cylinder.

Prob. 12-212


12-213. The man pulls the boy up to the tree limb C by12 walking backward at a constant speed of 1.5 m/s.

Deterrnine the speed at which the boy is being lifted at the instant 4 -= 4 m. Neglect the size of the limb. When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long.

12-214. The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when xA = O and moves backward with a constant acceleration aA = 0.2 m/s2, determine the speed of the boy at the instant ys = 4 in. Neglect the size of the limb. When xA = 0, y9 = 8 m, so t ha t A and B are coincident, i.e., the rope is 16 m long.

Probs. 12-213/214

12-215. The roller at A is moving upward with a velocity of vA = 3 ft/s and has an acceleration of aA = 4 ft/s2 when sA =4 ft. Determine the velocity and acceleration of block B at this instant.

*12-216. The girl at C stan ds near the edge of the pier and pulls in the rope horizonłally at a constant speed of 6 ft/s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft.

6 ft/s xc-1C

81 t

x8—.1

Prob. 12-216

12-217. Tice crate Cis being lifted by moving the roller at A downward with a constant speed of vA = 2 m/s along the guide. Deterrnine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hini: Relate the coordinates xc and xA using the problem geometry, then take the first and second time derivatives.

-riProb. 12-217

12-218. The man can row the boat in stili water with a speed of 5 m/s. If the river is flowing at 2 m/s, determine the speed of the boat and the angle O he must direct the boat so that it travels from A to B.

5 m/s

5 0 m

2 5 7 1 4 '

Prob. 12-215 Prob. 12-218

3 ft = 4 ft

12.10 RELATIVE-MOTION OF Two PARTICLES USING TRANSLATING AXES97

12-219. Vertical motion of the load is produced by movement of the piston at A on the boom. Determine the distance the piston or pulley at C must move to the left in order to lift the load 2 ft. The cable is attached at B, passes over the pulley at C, then D, E, F, and again around E, and is attached at G.

Prob. 12-219

*12-220. If block B is moving down with a velocity y8 and has an acceleration aB, determine the velocity and acceleration of block A in terms of the parameters shown.

12-221. Collars A and B are connected to the cord that passes over the smali pulley at C. When A is located at D, B is 24 ft to the left of D. If A moves at a constant speed of 2 ft/s to the right, determine the speed of B when A is 4 ft to the right of D.

1 0 i t

/ 6 / 1 = ' , 1

D2 i i

Prob. 12-221

12-222. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km/h and 12 v8 = 500 km/h such that the angle between their straightline courses is O = 75°, determine the velocity of piane B with respect to piane A.

• V A

'r=(=r4 B

ł. 4 _ : v B

Prob. 12-222

12-223. At the instant shown, cars A and B are traveling at speeds of 55 mi/h and 40 mi/h, respectively. If B is increasing its speed by 1200 mi/h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.5 mi.

*12-224. At the instant shown, car A travels along the straight portion of the road with a speed of 25 m/s. At this same instant car B travels along the circular portion of the road with a speed of 15 m/s. Determine the velocity of car B relative to car A.

Prob. 12-224

SA

Prob. 12-220

A 3 0 °

vA = 55 mi/h

Prob. 12-223

98 CHAPTER 1 2 KINEMATICS OF A PARTICLE

12-225. An aircraft carrier is traveling forward with a12 velocity of 50 km/h. At the instant shown, the piane at A has

just taken off and has attained a forward horizontal air speed of 200 km/h, measured from stili water. If the piane at B is traveling along the runway of the carrier at 175 km/h in the direction shown, determine the velocity of A with respect to B.

*12-228. At the instant shown, the bicyclist at A is traveling at 7 m/s around the curve on the race track while increasing his speed at 0.5 m/s2. The bicyclist at B is traveling at 8.5 m/s along the straight-a-way and increasing his speed at 0.7 m /s2. Determine the relative velocity and relative acceleration of A with respect to B at this instant.

Prob. 12-225Prob. 12-228

12-226. A car is traveling north along a straight road at 50 km/h. An instrument in the car indicates that the wind is coming from the east. If the car's speed is 80 km/h, the instrument indicates that the wind is coming from the north-east. Determine the speed and direction of the wind.

12-227. Two boats leave the shore at the same time and travel in the directions shown. If trA = 20 ft/s and v8 = 15 ft/s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 800 ft apart?

Prob. 12-227

12-229. Cars A and B are traveling around the circular race track. At the instant shown, A has a speed of 90 ft/s and is increasing its speed at the rate of 15 ft/s2, whereas B has a speed of 105 ft/s and is decreasing its speed at 25 ft/s2. Determine the relative velocity and relative acceleration of car A with respect to car B at this instant.

Prob. 12-229

B

,30.,

Q

.■40'

12.10 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES99

12-230. The two cyclists A and B travel at the same constant speed v. Determine the speed of A with respect to B if A travels along the circular track, while B travels along the diameter of the circle.

Prob. 12-230

12-231. At the instant shown, cars A and B travel at speeds of 70 mi/h and 50 mi/h, respectively. If B is increasing its speed by 1100 mi/h2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.7 mi.

*1..232. At the instant shown, cars A and B travel at speeds of 70 mi/h and 50 mi/h, respectively. If B is decreasing its speed at 1400 mi/h2 while A is increasing its speed at 800 mi/h2, determine the acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.7 mi.

12-233. A passenger in an automobile obseryes that raindrops make an angle of 30° with the horizontal as the 12 auto travels forward with a speed of 60 km/h. Compute the terminal (constant) velocity vr of the rain if it is assumed to fali vertically.

12-234. A man can swim at 4 ft/s in stili water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. If the river flows with a velocity of 2 ft/s, determine the speed of the man and the time needed to make the crossing. Note: While in the water he musi not direct himself toward point B to reach this point. Why?

60 km/h

Prob. 12-233

, - 70 mi

Probs. 12-231/232 Prob. 12-234

100CHAPTER 12 KINEMATICS OF A PARTICLE

12-235. The ship travels at a constant speed of v, = 20 m/s12 and the wind is blowing at a speed of = 10 m/s, as

shown. Determine the magnitude and direction of the horizontal component of velocity of the smoke coming from the smoke stack as it appears to a passenger on the ship.

12-238. Al a given instant the football player at A throws a football C with a velocity of 20 m/s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football.

20 m/s

a j t B

Prob. 12-238Prob. 12-235

*12-236. Car A travels along a straight road at a speed of 25 m/s while accelerating at 1.5 m/s2. At this same instant car C is traveling along the straight road with a speed of 30 m/s while decelerating at 3 m/s2. Determine the velocity and acceleration of car A relative to car C.

12-237. Car B is traveling along the curved road with a speed of 15 m/s while decreasing its speed at 2 m/s2. At this same instant car Cis traveling along the straight road with a speed of 30 m/s while decelerating at 3 m/s2. Determine the velocity and acceleration of car B relative to car C.

12-239. Both boats A and B leave the shore at O at the same time. If A travels at vA and B travels at vB, write a general expression to determine the velocity of A with respect to B.

Probs. 12-236/237 Prob. 12-239

12,10 R ELATiVE- M OT ON pF TVVO PARTICLES USING TRANSLATING AXFS 1 0 1

■ CONCEPTUAL PROBLEMS

P12-1. If you measured the ti me it take for the construction elevaior l0 go from A to R. then B to C. and [hen C to D. and you also know the distanee between each of the points. how could you determine the average velocity and average aceeleration of the elevator as ii ascends from A to D? Use numerica] values to explain how this can be dane.

P1.2-3. The basketball was thrown al an angle measured front the horizontal to the man's ontstretched arm. If thc basket is 3 m from the ground. make appropriate measurements in the photo and determine if the ball located as shown will pass through the basket.

P 1 2 - 3

P12-4. The pilot teils you the wingspan of her płane and her constant airspeed. How would you determine the aeceleration of the piane at thc moment shown? Use numerteal values and take any necessary measurements from the photo.

P 1 2 - 1

P1.2-2. If the sprinklcr at A is 1 ni from the ground, then seale the necessary measurements from the photo to determine the approximate velocily of the waler jet as it flows from the nozzle of the sprinkler.

1'12-2 P 1 2 - 4

102 CHAPTER 12 KiNIJMATiCS OF A PmcricLE

CHAPT1 2

A sW, gI

ST

( )av. =As

dra dsa= v = .

dra ds = v dv

V = v c +

= so + vor + ;a c?

v2 -- vo2 + 2u,(.5 — go)

Rectilinear Kinematics

Rectdinear kinernatics refers to rnotion dong a straigh line, A position coordinate s specifies the location of the particie on the lnie, and the displacemen t As is the ehange in this position.

The average velocity is a vector quantity, defined as the displacement divided by the linie interv al.

The average speed is a scalar, and is the total distance traveled divided by the time of travd.

The time, position. veocity. and acceleration are related by three differential equations.

If the acceleration is known to be constant. then the differential equations relating tirne, posi lion, velocity, and acceleration can be Integrated.

I— As --łC S

I 'ST

Graphical &dation§

If the motion is erratic, then it can be described by a graph. If one of these graphs is given, then the othcrs can be established using the differenlial relations between a. y, s. and t.

dv=

git,

ds

v = d i .

a ds= v dv

CHAYMR REVIFW1 0 3

Curvilinear Motion, x, y, zCurvilinear motion along the pł.Uh can be resolved into rectilinear motion &mg the x, y, z axes.The equation of the path is used to relaie ihe motion along each axis.

Y

Projectile MalimFree-flight motion of a projectile follows parabolic path. It has a constant velocity in the horizontal direction, and a constant dovvnward acceleration of g = 9.81 m/s2 or 32.2 f« in the vertical direction. Any two of the three equations for constant accełeration apply in the vertieak direction, and in the horizontal direction only one equation applies.

(+1)= (V0)3, + ac't

(+t)Y = Ye(vo).t + 'I ac-12

(±1)y = (yo);>.2a,tv — Yo)

)-r =+

-

v , = k R x = 2 1

= y a „ —

y z = Ż s

r = xi - _vj • zk

Z

104CH APTER 12 KINFMATICS OF A PARTILLE

Curvilinear Motion n, tIf normal and tangenłial axel are used for the anałysis, then v is always in the positive t direction,

The acceleratio-n bas twa components, The tangential component, a1, accounts for the change in the magnitude of the veloeity: a slowing down is in thenegative direction, and a speeding up isa, = vardrdy = v dvin the positive t direction. The norma' component a, accounłs for the changei n t h e d i r e c t i o n o f t h e v e l o c i t y . T h i s =

Qqcomponent is always in the positive n P

direction.

Curvilinear Motion r, CIf the path of motion is expressed in polar coordinates, then the veiocity and accelerałion components can be related to the time derivatives af r and H.

To appiy the time-derivative equations. it is necessary to determine r, ;-, r, i), fł at the instant considered. 1f the path r = fi e) 15 giVen, hen the choin rule af calculus musi be used to obtain time derivatives, (See Appendix C.)

Once the data are substituted into the equations, then the algebraic sign of the resuits will indicate the direction of the components of v ar a along each axis.

v r = r

=

r Ó

= — r 6 2

2H".1

Velocity

OAcceleration

12

C HA RTE R R EVI EW 105

Absolute Dependent -Muflon ofTwo Partieles

The dependent motion of blocks that are suspended from pulleys and cables can be related by the geometry of the system. This is done by first establishing position coordinates, measured from a fixed origin to cach hlock, Each coordinate must be directed atong the line of metion of a block.

Using geometry and/or trigonometry, the coordinates arc thcn rclatcd to the cakle length -in order to fonnulate position coordinate equation.

The first time derivative of this equation gives a retadonsitip between the velocities of the blocks. and a second time derivative gives the relation between their accelerations.

2s s + + sA = i

2 r = V , I

_as = — —aA

r g = r q +

= VB/A

aB = aA aa/ATranslating observer

rw,t

F i x e dobserve r

O

x

Y

Relative-Motion Analysis Using Ttanslating Ases

Tf two particles A and R undergo independent mations, [hen these mołions can be rciated to their rclative motion using a łranslałing set of axes altached to one of the particles (A).

For planar motion, end] vector equation produces two scalar equations. one in the x, and the other in the y dixection. For solution, the vectors can be expressed in Cartesian form, or the x and y sealar components can be written directly,

IChapter

A car driving aiong this road will be subjected to forces that create botki norma' and tangential accelerations. In this chapter we will study how these forces are

related to the accełerations they create.

Kinetics of a Particie:Farne and Acceleration

CHAPTER OBJECTIVES

■To state Newton's Second Law of Motion and to define mass and weig ht.

■To analyze the accelerated motion of a particie usin g the equation of motion with different coordinate systems.

■To investigate central-forte motion and apply it to problems in space mechanics.

13,1 Newton's Second Law of Motion

Kinetics is a branch of dynamics that deals with the relationship between the change, in motion ❑f a body and the forces that cause this change, The basis for kinctics is Newton's seeond law, which states that when an unbalanced furce acts on a particie, the particie will accekrałe in thc direction of the forte with a magnitude that is proportional to the forte.

This law can be verified experimentally by applying a known unbalanced furce F to a particie, and then measuring the acceleration a. Since the forte and acceleration are directly proportional, the eonstant of proportionality, m, may be determined from the ratio in = Fla. This positive scalar m is calied the mass (.>f the particie, Being constant during any acceleration, m provides a guan titative nneasure of the resistance of thc particie to a change in its velocity, that is its incrtia.

108 CHAPTER 13 KINETICS OP A PARTICLE FORCE A n In ACCELERATION

If thc mass of the particie is m, Newton's sccond law of motion may be writtcn in mathematical form as

F = m a

The above cquation, which is referred to as the equarłem of rnottort.iS one of the most important formulations in mechanics.* As previousiy state-d, its validity is bascd soldy on experimentai evidence. In 1905, however, Albert Einstein developcd the theory of reiativity and placed limitations on the use of Newton's second law for describing generai particie =non. Through experiments it was proven that time is not an absolute quantity as assumed by Newton; and as a resulł, the equation of motion fails to predict the exact behavior of a particie, especially when the particles speed approaches the speed of light (0,3 Grn/s), Developrnents of the theory of quantum mechanice by Erwin SchrOdinger and others inz.ficate furthcr that conclusions drawn from using this cquation arc also invalid when particles arc the size cif an atom and move close to one another. For thc most part, however, these requircments regarding particie speed and size arc not encountered in engineering problems, so their effects will not be considered in this book.

Newtonis Law of Gravitational Attraction. Shortly after formulating his three laws of motion, Newton postulated a law governing the mutual attraction between any twa particles. In rnathemafical form this law can be expressed as

mln],F = G (13-1)

whcre

F = G

tni, m2

r=

forte of attracton between the two particlesuniversal constan l ofgravitation;according to experimentalevidence G = 66.73(10-12) m3/(kg • s2)mass of cach of the two particlesdistancc bctwecn the centers of the two particles

*Since ne is constani, we can afro write F = d(mv)fdr, where mv is ihe particle's linear inomentunh Here the unhalanced forte acting on the particie is propodional to the tirne rata of change of the pardcle's linear momentum.

13.1 NEWTON'S SECOND LAW OF MOTION 109

In the case of a particie Iocated at or near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the particie. This fotce is termed the "weight" and, for aur purpose, it will be the ordy gravitational force considered.

From Eq. 13-1, we can deveiop a general expression for finding theweight W of a particie having a mass trel = Let rn2 = M, be the massof the ea.rth and r the distance between the earth's center and the particie. Then,if g = GM,/ r2, wc have

W =-

By eornparison with F = ma, we term g the acceleration due to graviły. For most engineering ea leulations g is measured at a point on the surface of the earth at sca leveI. and at a latitude of 45°, which is considered the "standard location." Herc the values g ,= 9.8 i rnis2 = 32.2 ft/s2 will be used for calculations.

In the S1 system the mass of the body is specificd in kdograms, and the weight must be calculated using the above equatian, Fig. 13-1a. Thus.

m (kg)

W = mg (N) ( g =- 9,81 mjs2) (13-2)a.= g (0-02)

J'

As a result, a body cif mass 1 kg has a weight of 9.81 N: a 2-kg body weighs 19.62 N; and so on.

In the FPS system the weight of the body is specified in pounds. The mass is measured irt slugs, a term derived from "stuggish" which refers to the body's inertia. It musi be calculated, Fig. 13-lb, using

S I s y s t e m

( a )

JEm = g ( dug )

m '= —w (slug} (g -= 32.2 ft/s2) g (13-3)

42 = g (ftis2)

Therefore, a bady weighing 32.2 lb has a mass of 1 sług; a 64.4-1b body has a mass of 2 slugs: and so on.

W (lb)

FPS system

(b)

Fig. 13-1

1 1 OCHAPTER 13 KiNETICS OP A PARTICLE: FORCE A nIn ACCELERATION

a

13.2 The Equation of Motion

( a )

When more Chan one furce acts on a particie, the resultant forte isdetermined by a vector surnmation of all the forces; i.e., FR F. For thismore generai rwe, the equation of motion may be written as

= m a (13-4)

(b)

Kine tle diagram

Fig, 13-2(c)

ma

v

a

CiPalb 'Elf particlei

inertial frame of referenceO x

Fig. 13-3

I Free-body diagram

To illustrate application of this equation, considcr the particie shown in Fig. 13-2a, which bas a mass in and is subjected to thc action of Iwoforces, F, and F1_ can graphically aceount for the magnitude anddirection of each tarce acting un the particie by drawing the particle's free-body diagram, Fig. -13-2b. Since the resuitara of these forcesproduces the vector ma, its magnitude and direction can be represented graphically on the kinełic diagram, shown in Fig. 13-2c.* The equal sign written between the diagrams symbolizes the graphicai equivalency between the free-body diagram and the kinetic diagram; i,e., F = ma.t In particular, nate that if FR = F = 0, then the accelerałion is also zero, so that the particie will either remain at resr or move along a straight-line path with consrard ve‚ociły, Such 41"C the canditianS af 5-fatiC CQUilibriUnT, NeWW1-

1'S firSf law of motion.

Inertial Reference Franie. When applying the equation of motion, it is important that the acccleration of the particie be measured with respect to a reference frame that is eirher frxed or translares witki a constans velocity. In this way, the observer will not accelerate and measurements of the particle's acceleration will be the same from any reference cif this type. Such a frame of reference is commonly known as a Nawianiam or inertial reference frame. Fig. 13-3.

When studying the rno6ons uf rockets and satellites, it is justifiable to considcr the incrtial reference frame as fixed to thc stars, whcreas dynamics problems cancerned with mations on or near thc surface of the earth may be sobred by using an inertial frame which is assumed fixed to the earth. Even though the earth both rotates about its own axis and revolves about the sun, the accelerations created by these rotations are relatively smali and so they can be neglected for most applications.

Re cal I the free-body diagram considers the pa/1i cle to be -Free of its suł•roundingsnpports and shows all the forces acting on the particie. The kinetic diagram pertains tu the particles ~lian as caused by the forces.The equation of motion can also be rewritten in the form S F — ma - O. The vector -

ma is referred to as the ineriia lorce vector. if il is treated in the same way as a "forte vector; then the state of "equilibrium" created is referred to as dynam& equilibrium, This method of application. which will not be uscd in this Text. is often referred to as the DAkrnhen prindpk, named after the French rnathernatician Jean lę Rond d'Alernbert_

132 THE ECUATION OF MOTiON 111

We are alf familiar with the sensation one feels when sitting in a car Chat is subjected to a forward acceeration. Often people think this is caused by a "force" which acts on them and tends to push them back in their seats; however, this is not the case. Instead, this sensation occurs due to their inertia or the resistance of their mass to a change in velocity.

Consider the passenger who is strapped to the seat of a rocket sied. Provided the sied is at rest or is moving with constant velocity, then no fotce is exerted ors his back as shown on his free-body diagram.

At rent or constant velocity

When Che Chrust of the rocket engine causes the sied to accelerate, then the seat upon which he is sitting exerts a force F on him which pushes him forward with the sied. In the pilota, notice that the inertia of his head resists this change in maiian (acceleration), and so his head rnoves back against the seat and his face, which is nonrigid, tends to clistort backward.

N,

Acculeration

Upon deccleration the force of the scatbelt F' tends to puli his body to a stop, but his head leaves contact with the back of the seat and his face distorts forward, again due to his incrtia or tcndency to eon tinue to move forward. No force is pulling him forward, although this is the sensation he receives.

1 1 2 CHARTER 13 KINETIC5 OF A PARTICLE: FORCE AND ACCELERATION

1 3.3 Equation of Motion for a System of Particles

The equation of motion will now be extended ta inciude a system of particies isolated within att enclosed region in space, as shown in Fig. 13-4a, In particular, there is no restriction in the way the particles are connected, so the foilowing analysis applies equally well to the motion of a solid, liquid, ar gas system.

At the instant considered, the arbitrary i-th particie, having a mass mi, is subjected to a system of internal forces and a resultant external forte.The łrnernal forte, represented symbolicaliy as is the resultant °f allthe forces the other particles exert on the ith particie. The resultant

Pytce represents, for example, the effeet of gravitational,electrical, magnetic, or contact forces hetween the ith particie and adjacent bodies ar particles not included within the system.

The free-body and kinetic diagrams for the ith particie are shown in Fig. 13-4b. Appiying the equation of motion,

= ma; F, + pnia;

When the equation of motion is applied to each of the other particies of the system, sirnilar equations will resuit. And, if all thesc equations are added together vectorially, we obtain

iF, + Xf, = im,aj

Inertial coordinate system

(b)

nee-bodyKineticdiagramdiagram

Fig. 13-4

1 1 313.3 EWA-MON er MOTION FOR A SYSTEM er PARTICLES

The summation of the internal forces, if carried aut, will equal zero, since internal forces between any twa particles occur in equal but opposite collinear pairs. Consequently, only the sum of the external forces will rernain, and therefore the equation of rnotion. written for the system of particles, becomes

= (13-5)

If ru is a position vector which locates the center of mass G of the particles, Fig. 13-4a, then by definition of the center of mass, ‚nik = where in =is the total mass of all the particles. DiffereMiating thisequation twice with respect to time, assuming that no mass is entering or leaving the system, yields

mac = /ni ja i

Substituting this result into Eq. 13-5, we obtain

( 1 3 - 6 )

Hence, the sum of the external forces acting nn the system of particles is equal to the total mass of the particles times the acceleration of its center of rnass G. Since in reality all particles musi have a finite sine to possess mass, Eq. 13-6 justifies application of [lic equation of mo Lian to a body that is representecl as a single particie.

• The equation of motion is based on experimen tai evidence and is valid only when applied within an inertial frame cif reference,

• The equation of motion states that the tmbalaneed lorce on a particie causes it to accelerme.

• An inertial frame of reference does not rotate, rather its axes either translate with constant velocity or are at rest.

• Mass is a property of matter that provides a quantitative measure of its resistance ta a change in velocity. It is an absolute quantity and so it does not change from one location to another.

• Weight is a furce that is causcd by the carth's gravltation. It is not absolute; rathcr it depends on the altitudc of the mass from the earth's surface.

11 4 CHARTER 13 K INET ICS OP A PART ICLE : FORCE A n I n A C C E L E R A T I O N

13.4 Equations of Motion: Rectangular Coordinates

When a particie movesrelative to an inertial x, y, z fratrie of reference, the forces acting on the particie, as well as its acceleration. can be expressecl in terms of their i. j, k components, Fig. 13-5. A pplying the equation of motion, we have

= ma; 5F_,1 5F,j + 5F,k = m(ca,i + a,k)

For this equation to be satisfied, the respective i, j, k components on the Ieft side must equal the eorresponding components on the right side. Consequently, we may write the folinwing three scalar equations;

(13-7)

IF,_ = ma,

ZFy may F.

= ma,

Fig. 13-5

In particular, if the particie is constrained to move only in the x—y piane, then the first two of these equations are used to specify the motion.

ocedure for Analysis

The equations of motion a-re used to solve problems which require a reiationship between the forces acting on a particie and the accelerated motion they cause.

Free-Body Diagram.

• Select the inertial coordinate system. Most often, rectangular nr x, y, z coordinates are chosen to analyze probIems for which the particie has reetiiirgear morion.

• fince the coordinates are established, draw the particle's freebody diagram. Drawing this diagram is very importarzt siatce it provides a graphical representation that accounłs for all the forces (5F) which act on the particie. and thereby makes it possibIe to resolve these forccs into their x, y, z components.

• The direction and sense of the particle's acceleration u should also be e_stablished. If the sense is unknown, for ma.thematical convenience assurne that the sense of each acceleration component acts in the same directiort as its positive inertial coordinate, axis.

• The acceleration may be represented as the ma vector on the kinetic diagram.*

• Identify the unknowns in the problem.

is a convention in ibis texi always to use the kirietic diagram as a graphical Md when developing the proofs and theory. The particle's acceleration ur its components wili be silowi as blue eolured vcctorfi ncar the frec.body diagram irl file :.fcarnples.

1 1 51 3.4 EQUATIONS. OF MOTION: RECTANGULAR COORDINATES

Equations of Motion.

• If the forces can be resolved directly from the free-body diagram, apply the equations of motion in their scalar component form.

• If the geometry of the problem appears complicated, which often occurs in three dimensions, Cartesian vector analysis can be used for the solution.

• Frietion. If a moving particie contacts a rough surface, it urny be nccessary to usc the frictional equation. which rciates the frictional and normal forces F,' and N acting at the surface of contact by using the coefficient of kinetic friction, i.e., F f = 14N. Remernber that Fialways acts on the free-body diagram such that it opposes the motion of the particie relative to the surface it contacts, If the particie is on Lhe verge of rei ative motion, then the coefficient of static friction shoult.1 be used,

• Spring. If thc particie is connectcd to an elanie spring having negligible mass, the spring furce F, can be related to thedeformation of the spring by the equation = ks. Here k isthe spring's stiffness rneasured as a forte per unit ]ength, and s is the stretch or compression defined as the difference between the deformed length f and the undeforrned length i.e., s = i — I.

Kinematics.

• If the velocity or position cif the particie is to be found, it will be necessary to apply the necessary kinematic equations once the particie's acceleration is determined from F = ma.

• 1f aeceleration is a function of time. use a = dv ł dr and y -= ds I dt which, when inłegrated, yield the particles velocity and position, respectively.

• Ifacceieration is a function cif displacement, integrate a ds = v dv to obtain the velocity as a function of position.

• If acceierałion is consran,,use v = vo + a,r,sa + voi += 2a,(s so) to deterrnine the velocity or position of the

particie.

• 1f the problem involvcs the dependent motion of several particies, use thc method outlined in Sec. 12,9 to [dale their accelcrations. In all cases, make sura the positive inertial coordinate directions used for writing the kinematic equations ara the same as those used for writing the equations of rnotion; otherwise, simultaneous solution of the equations will result in errors.

• 1f the solution for an unknown vectur component yields a negative scalar, it indicates that the component acts in the direction oppositc to that which was assurn cd.

1 1 6 CHARTER 13 KINETIC5 OF A PARTICLE: FORCE AND ACCELERATION

EXAMPLE

( a )

a

The 50-kg crate shown in Fig. 13-6a rests on a horizonłal surface farwhich the coefficient of kinetic friction is0.3. If the crate issubjected to a 400-N towing fotce as shown, determine the velocity of thc crate in 3 s starting from rent.

SOLUTIONUsing the equations of motion, we can rełate the crates acceleration to the furce causing the motion. The crates velocity can tlien be determinecl using kinernatics.Free-Body Diagram. The weight of the crate is W = =50 kg (9.81 rn/s.2) = 490.5 N. As shown in Fig. 13-6b. the frictional furce bas a magnitude F = I.LAN c and acts to the left, since it upposes the motion of the crate. The acceleration a is assumed to act horizontally. in the positive x direction. There are two unknowns, narnely Nr- and a.

Equations of Motion. Using the data shown on the free-body diagram, we have

F,. = max; 400 cos 30° — 0,3Nc = 50a (1)+ F = ma,; Nc — 490.5 + 400 sin 31:r = G (2)

x490.5 NSoIving Eq. 2 for Nc, substituting the result into Eq. 1, and solving

400 Nfor a yieldsNe = 290.5 N a = 5.185 n-1/s2'

Kinematies. Notice that the acceleration is cnpsirtnr, since the applied forte P is constant. Since the initial velocity is zero, the velocity cif the crate in 3 s is

= vo + a,: = O + 5.185(3) 15,6 rn/sNe

(c) Fig. 13—.6 490.5 N

400 N

30°

1— F — 0.31Vc

{c )

NOTE: We can also use the alternafive procedure ()f drawing the crate's free-body and kinetic diagramy, Fig. 13-6c, Kiur to applying the equatkuis of motion.

13.4 EGUATioNs Gr NA GTION: RECTANGULAR COORDNATES 1 1 7

EXAMPLE

1 1 1 E f f i a " f

A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 mis, Fig. 13-7a. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected; and (h) atmospheric resistance is measured as FD = (0.012,2) N, where z, is the speed of the projectile at any instant, measured in m/s.

SOLUTIONIn both cases the known forte on the projectile can be rclated to its acceleration using the equation of motion. Kinerriatics can then be used to relate the projedile's acceleration to its position.

Part (a) Free-Body Diagram. As shown in Fig. 13-7b. the projectile's weight is W = mg = 10(9.81) = 98.1 N. We will assume the unknown acceleration a acts upward in the positive z direction.

Equation of Motion.= ma,: —98.1 = 10a, a = —9.81 m/s2

The rcsult indicatcs that the projectile, like every object having freeflight motion near the earth's surface, is subjectcd to a constant downward acceleration of 9.81 m/s2.

Kinematics. Initially, z = O and vo = 50 m/s, and at the maximum height z = h, y = O. Since the acceleration is L.onsiant, then

(+T) = + 2 a,(z — z())O = (50)2 + 2(-9_81)(h — 0)

h = 127 m Av.v.

Part (b) Free-Body Diagram. Since the forte FD (0.017?) N tendsto retard the upward motion of the projectile, it acts downward as shown on the frce-body diagram, Fig. 13-7c.

Equation of Motion.+ S F, = ma,; —0.0i v2 — 98.1. =- lQa, a = —(0.0017,2 ± 9.8D

Kinematics. He-re the acceleration is not constani since FD depends on the velocty. Since a = fiv), we can relate a to position using

(+ ) a = v dv: —(0.00 + 9.81) dz = v clv

Separating the variables and integrating, realizing that initially zo = 0, vo = 50 m/s (positive upward). and at ti = h, v = 0, we have

11 v dv dz = — —500 In(v2 9810)

50 0.001v= + 9.81

= 114 m Ans.

NUM: The answer indieates a lower elevation than that obtained in part (

a) due to atmospheric resistance or drag.

11

50 mjs

S

( a )

T 9 8 . 1 N

(b)

111 Fn

98.1 N

(e)

Fig. 13-7

1 1 8 CHAPTER 13 KINETfCS OF A PARTICLE FORCE AKH:› AccFLERATioN

(b)

Fig. 13-8

The haggage truck A shown in the photo has a weight of 900 lb and tows a 550-lb cart B and a 325-1b cart C. For a short time the driving frictional force developed at the Mice Is of the truck is FA = (40r) lb, where ł is in seconds. If the truck starts from rest, determine its speed in 2 seconds. Also, what is the horizontal force acting on the coupling between the truck and cal-t B at this instant? Neglect the size of the truck and carts.

9 0 0 l b

N B

( a )

SOLUTION

Free-Body Diagram. As shown in Fig. 13-8a, i t is the frictionaldriving force that gives both the truck and carts an acceleration. Herewe have considered all three vehicles as a single system.

Equation of Motion. Only motion in the horizontal direction has tobe considered.L !F, = ina,;

40t= (904 ± 550 ± 325)a32,2

a = 0.7256t

Kin emat ics. Since the acceleration k a function of time, the velocityof the truck is obtained using a = dvi& with the initial condition thatvo = O at t = 0, We have

2. 2.0.7256t dt; = 0.3628i2 = L45 ft/s

f> v - a ,

Free-Body Diagram. In order to determine the lorce betwecn the truck and cart B, we will consider a frce-body diagram of the truck so that we can "expose" the coupling force T as external to the free-body diagram. Fig. 13-81)

Equation of Motion. When I = 2 s, then!F, = 40(2) - T = 9°0) [0.7256(2)]

32.2

T = 39.41b-1u 1.

NOTE; Try and obtain this same result by considering a free-body diagram

cif carts B and C as a single system.

1,0

, )001b

13.4 EnUATION$ QF KAOTION: R rcraniouLAR CooRENINATE$ 1 19

EXAMPLE

A smooth 2-kg collar C, shown in Fig. 13-9a. is attached ta a spring having a stiffness k = 3 Nim and an unstretehed length of 0.75 m. If the collar is released from rest at A determine its acceleration and the normal furce of the rod on the collar at the instant y = 1 m.

SOLUTION

Free-Bady Diagram. The free-body diagram of the collar when it is located at the a rbi trary position y is shown in Fig, 13-9b. Furthermore, thc collar is assumed to be accelerating so that "a" acts downward in the positive y direetion. There ara kiur unknowns. nameły,Nc, F,., a, and O.


+ = ma,; —Nc + F„ cos 9 = O (1)

4- 1. F. = may; 19.62 — F„ sin O = 2a (2)

From Eq. 2 it fis seen that the accełeration depends on the magnitude and direction of the spring fOTCC. Solution for Arc and a is possible once F, and O ara known.

The magnitude of the spring fotce is a function of the stretch s of the spring; i.e.. F, = ks. Herc thc unstrctched length is AR = 0.75 m,

Fig. 13-9a; therefore, s = CR — AR = ik.2 (0,75)2 — 0,75. Sincek T 3 Nim, [bert

F,. = ks = {V)? + (0.75)2 — 0.75) (3)

From Fig. 13-9a, the angle O is related to y by trigonometry.

tan O =0.75

Substituting y .= 1 ni into Eqs. 3 and 4 yields F, = 1.50 N and O = 53.1°. Substituting thesc results into Eqs. 1 and 2, we obtain

Nc = 0.900 N

= 9.21 infs2 Ans

NOTE: This is not a case of constant accelcration, since the spring furce changes butli its magnitude and direction as the collar moves downward.

y

( u l

v19.62 N

F,4 - 4 4

nr c-(b)

Fig. 13-9

20 CHARTER 13 KINETics OF A PARTICLE: FORCE AND ACCELERATION

E X A M P L E 13.5

Bloek A ,

( I )981 - 2T -= 100aA

(2) 196.2 - T = 20aB

Equations 01

Motion. +1 F, = Inav;

Block B,

+ F,, = ma,,;

(3) = - a B

S A

gsł

196 2 NsM

(d)

Fig. 13-10

The 100-kg block A shown in Fig, 13-10a is released from rest. If the masses of the pulleys and the cord are negIected,determine the velocity cif the 20-kg błock F3 in 2 s.

SOLUTIONFree-Body Diagrarns. Since the mass of the pulleys is negiected, then For pulley C,►na = O and we can apply F, = O as shown in Fig. 13-10b. The free-body diagrarns for blocks A and B are shown in Fig. 13-10c and d, respeetively, Notice that for A to remain stationary T = 490.5 N, whcreas for B to remain static T = 196.2 N. E-lence A will move down while B mores up.Although this is the case, we will assume both blocks accelerate downward, in the direction of +.sA and +sB. The three unknowrts are T, aA, and aB.

Kinenlatics. The necessary third equation is obtained by relating 0, to aB using a dependent motion analysis, dscussed in Sec, 12.9, The coordinates 4. and 4, in Fig, 13-10a measure the positions of A and H from the fixed datum.It is scen that

2sA =

wherc i is constant and represents the total vertical length of Differentiating this expression twiee with respect to tiule yields

Noticc that when writing Eqs. 1 to 3, the posinve direction was always assinned downward. It is nery important to be consistent in this assumption since we are seeking a simultaneous solution of equations. The results are

T = 327.0 N

aA = 3.27 m/s:-'

aB -= -6.54 mis2

Hence whenblock A accelerates downward,block B accelerate5 upward m expeeted. Since as is constant, the velocity of błock B in 2 s is łhus

= a B t= O + (-6,54)(2) = -131 m/s

The negative sign indicates that błock Bis moving upward.

2 T ( b )

T

12113.4 EQUATION5 OF MOTiQN: RFCTANGULAR COORDiNATEs•FUNDAMENTAL PROBLEMS

F13-1. The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance

= 6 m in 3 s, starting from rest. Determine the tension developed in the cable. The coefficient of kinetic friction between the crate and the piane is p.k = 0.3.

F13-1

F13-4. The 2-Mg car is king towed by a wind. If the winch exerts a force of T = 100(s + I) N on the cable, where s is the displacĘ,,mcnt of the car in meter& determine the speed of the car when s -= IQ rn, starting from rest. Neglect rrslling resistance of the car.

F13-4

F13-5. The spring has a stiffness k = 200 Nim mid is unsłretched when the 25-kg block is at A. Dctermine the acceleration of the block when s = 0,4 m. The contaet surface betwcen the block and the piane is smooth,

F13-2F13-5

F13-1 A spring of stiffness k = 500 N/in is moun led against the 10-kg błock. If the block is subjected to the force of F = 500 N, determine its velocity at s = 0,5 m, When s -= 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.

F13-6. Block B rests upon a smooth surface, If the coefficients of stalic and kinetic friction hetween A and 13are = 0.4 and p. = 0.3, respectively, determine theacceleration of each block if P = fa lb.

F13-2. If motor M exerts a force of = (10r2 + 100} N on the c,able, where t is in seconds, determine the velocity of the 25-kg crate when r = 4 s. The coefficients of static and kinetic friction between the crate and the piane are = 0,3and pet = 0.25, respectively. The crate is initial), at rest.

201hp A- 1 >

B 50 lb

F13-6

F 500 N

F13-3

122 CHAPTER 13 KINETICS OF A PARTICIE: FORCE AND ACCELERATION

PROBLEMS

13-1. The 6-Ib particie is subjected to the action of i ts weight and forces F, = 121 + 6j — 2tiz} Ib, F, =

r2 — 4t,1 — lk } Ib. and Fs = —211} ib, where r is in seconds. Determirae the distance the bali is from the origin 2 s after being released from rest.

Proh. 13-1

13-2, The 10-lb błock bas an mitial veiocity of 10 ft/s on the smooth piane. If a force F = (2.5h ib, where t is in seconds, acts on the blvck for 3 s, dełermine the final velocity of the block and the distance the block travels during this time.

u= 10fti's

F = (2.5r)]!

• • • • • • • • • •

Prob. 13-2

13-3. 1f the coefficienł of kinetic friction between the 50-kg crate and the ground is µk "= 0.3, deterrnine the distance the crate travels and its velocily when r = 3 s. The crate starts from rest, and P =- 200 N.

If the 50-kg erate starts from rest and achieveswiodły of4 m/s when it travels a distanee cif 5 m to theright, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between theerate and the ground is --- 0.3.

P

Prohs. 13-3/4

13-5. The water-park ride conssts of an 800-lb sied which slidcs from rest down the incllne and Ulen juto the pooi. If the frictional resistance on the incline is F, = 30 ib,and in the pool for a short distance = 80 lb, determinehow fast the sled is traveling when s 5 ft.

Prob. 13-5

13-6. If P = 401) N and the coefficient cif kinetic friction between thc 50-kg crate and the inclined piane is rsx = 0.25. delermine the velocity of the erate after it travels 6 m up the piane. The crate starta from rest.

13-7. If the 50-kg crate starts from rest and travels a distance of 6 m up the piane in 4 s, determinc thc magnitude of fotce P acting on the crate. The coefficient of kinetic friction between the crate and the ground is µx = 0.25.

Prohs. 13—if1/7

13.4 EQUATIO1`,15 OF MOTION: RFCTANGULAR COORNNATES123

*13-8, The speed cif the 3500-113 sporits car is plotted over the 30-s dme period. Plot the variation of the traction force F needed to cause the motion_

„t is,

1.3-11. The safe S hasa weight of 200 Ib and is supported by the rope and puliey arrangernent shossm. If the end of the rope is given to a boy B of weight 90 Ib. determine his acceleration if rri the confusion he doesn 't let go of the rope. Neglect the mass ol` the pulieys and rope.

13

+ K F

80

t (s)

3 0

Prob. 13-8

Prob. 13-11

13-9. The trale hasa mass of 80 kg and is being towed by a chain which is always directed at 20' from the horizontal as shown. If the magnitucte of P is increased until the crate begins to siide. determine the crate's initial acceleration if the coefficient of stalic friction is g, = 0.5 and the coefficient of kinetic friction is pk = 0.3.

13-10. The crate has a mass of 80 kg and is being towed by a chain wbici' is aiways directed at 200 from thc horizontal as shown. Determine the crate's acceleration in t = 2 s ifthe cctefficienl of stalic friction is = 0.4, the coefficient ofkinetic friction is pk 0.3. and the towing force isP = (90/2) N. where t is in seconds.

Probs. 110

*13-12. The boy having a weight of 80 lb hangs uniformly from the bar. Determine the force in each of his arms in 1 = 2 s if the bar śs ~ing upward with (a) a constant velocity of 3 ft/s, and (b} a speed of tJ = (4:2) ft/s, wherc is in secands.

Prob. 13-12

124CHAI5TER 13 KIWTfCS OF A f ARTILLE= FORCE AKH:› A CCFLERATicht

13-13. The bullet of mass m is given a velocity due to gas pressure caused by the burning of powder within the ehatriber of the gun. Assuming this pressure creates a formof F= (77-r /t„) on the bullet, determine the velociły ofthe bullet at any instant it isin the barrel.What is the bullets niaximurn velocity? Also. determine the posiłion of the bullet in thc barrel as a funetion of thnc.

13-16. The man pushes on the 60-lb crate with a furce F. The tarce is always directed down at 30 from the hori7ontal as shown. and its magnitude is increased unii the trata begius to slide. Determine the crate's initial acceleration if the coefficient cif static friction is p., = 0.6 and thecoefficient of kinetic friction is = 0.3,

F

Prob. 13-13Prob. 13-16

13-14. The 2-Mg truck is traveling at 15 m /s when the brakes on all its wheels are applied. causing it ta skid for a distance of 10 m before cornirig lo rest. Deierrnine the constan i horizonłaI furce developed in the coupling C, and the frictional furce deveioped between the tires «the truck and the ruad during this tirne.nie total mass of the beat and trailer is I Mg.

13-15. A freight elewator, including its load, hasa mass of 500 kg. It is prevented from rotating by the truck and whee]s mounted along its sides. When r = 2 s. the motor M clraws in the cable with a speed of 6 m/s. ,?-leusured rekuive 10 fihe eievator. If it starts from rest, determine the constant acceleration of file elewator and the tension in the cable. Neglect thc masa of the pullc,,ys, motor, and cables,

13-17. The double inclined piane supports iwo błocks A and B. each having a weight of 10 lb. If the coefficient of

kinetic friction between the, blocks and the piane is = 0.1, deterrnine the acceleration of each block.

Prob. 13-17

13-18. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point whcre it strikes thc ground at C. How long does it take to go from A to C"'

13-19. Solve Prob. 13-18 if the suitcase has an initial velocity down the ramp of v,r = 10 ft/s. and the coefficient of kinetic friction aloug A$ is N.k = 0.2.

Prob. 13-14

Prob. 13-15 Probs. 13-18119

13.4 FgUATIQNS OF MoTioN: R ECTANGULAR COO RDINATES 125

13-20. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time. the forte in the cable 18 F = (3200/2) N, where 1 is in seconds. If the car bas an initial velocity vi = 2 m/s when r = 0. determine its velocity when i = 2 s.

13-21. The 400-kg mine car is hoisted up the inchne using the cable and motor M. For a short time, thc forte in the cabłe is F = (3200t2) N, where t is in seconds. If the car has an initial velocity tri = 2 mis at s = 0 and / = 0, determine the distance it moves up the piane when / = 2 s.

Probs, 13-20/21

13-23. The winding drutu D is drawing in the cable at an accelerated rate of 5 m/s2. Dełermine the cable tension if the suspencled erate lias a mass of 800 kg.

Prob. 13-23

13-22. Delermine the required mass of block A so that when it is reieased from rest it moves the 5-kg block B 0/5 m up along the smooth inclined piane in t = 2 s. Neglect the mass of the pulleys and cords.

Prob. 13-22

13-2,4. If the motor draws in the cable at a race of o = (0.05s-9 m/s. where s is in rneters, determine the

tension developed in thc cable when s = IQ m_ The cratc bas a mass of 20 kg, and the coefficient of kinetic friction between the trale and the ground is f.t,t = 0.2.

13-25. If the motor draws in the cable at a rate of(0.05/?-) m/ s, where t is in seconds. determine the

tension developed in the capie when t = 5 s, The cratc has a mass of 20 kg and the coefficient of kinetic friction between the crate and the ground is z = 0.1

Probs. 13-24/25

126 CHAPTER 13 KINEliCS OF A PART1CLE= FORCE M41:› AccFLERATioN

*13-28. The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originalty

rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of stanc friction

hetween the crate and the rond is = 0.4, and the

coef ficien i of kinetic friction is kk = 0.3.

Prob. 13-28

13-29. The force exerted by the motor on the cable is shown in the graph. Determine the velocity of the 200-1b

crate when t= 25 s, F(Ib)

250113

t ( s )2_5

Prob. 13-29

13-30. The forw of the motor M on the cable is shown in the uaph.Determine the velocity of the 400-kg crate A when t =2 s. F (N)

- F 6254 , 2

2t (s)

2500L

_Ib

13-26. The 2-kg shaft CA passes through a smooth joumal bearing at B. Initially, the spring& which aro coiled loosely around the shaft, arc unstretchcd when no force is applicd to the shaft. In this position s = s' = 250 mm and the shaft is at rest. If a horizontat force of P = 5 kN is applied. determine the speed of the shaft at the instant s = 50 min. s' = 450 mm. The, ends of the springs arc attached to the bearing at i3 and the capy at t:and A.

13-27. The 30-lb crate is being hoisted upward with a const ant acceleration of b ft/s2. If the uniform heam AB has a weight cif 200 lb. determ in e the components of reaction at A. Neglect the size and mass of the pulley al B. Flint. First fluid the tension in the cable. then analyze the forces in the heam using statics.

Prob. 13-27

Prob. 13-26

AProb.13-30

13.4 EMATIONS OF OTION: REcrANG0LAR COORDINATES1 27

13-31. The Lractor is used to lift the 150-kg bad B with the 24-m-long rope, boom, and pulley system. If the łractor travels to the right at a constant speed of 4 m/s. dełerm1ne the tension in the rope when s,ti = 5 m. When

= 0, s8 = 0.

13-32. The tractor is used to lift the [50-kg load B with the 24-m-long rope, boom, and pulley system, H thc tractortravels to the right willi arl acceleration of 3and bas avelocity of 4 mis al the instant sA = 5 rn. deterrnine Lhe tension in the rope at ibis instant.When sti = 0, sB = 0.

s k i

P robs. 13-31/32

13-33. Each of the three plates has a mass of 10 kg. U. the coefficients of stalic and kinetic friclion at each surface of contact are g, = 0.3 and gk = 4.2. respecfively, determine the acccicration of cach plate when the thrcc horizontal forces are applied.

18 N .«C lo IM N

15NA

Prob. 13-33

13-34. Each of the Iwo blocks hasa mass m. The COefficient of kinetic friction at all surfaces of contact is g. If a hortzontal force P moves the 1->ottom błock, detcrmine the accełeration of the bottom block in cach case.

(b)Prob. 13-34

13-35. The conveyor be[t is = ing at 4 mis. If the coefficient of stalic friction between the conveyor and the10-kg package B is= 0.2, determine the shortest liniethe bełt can stop so that the package does not slide on the hcit.

B

Prob. 13-35

*13-36. The 2-lb collar C fits loosely on the smooth shaft. II the spring is unstretched when s = 0 and the collar is given a velocity of 15 flis, determine the velocity of the collar when s = 1 ft.

15 ft ls

Prob. 13-36

128 CHArTER 13 KIWTfCS OF A f ARTICLF= FORCE AKH:› AccFLERATioN

13-37. Cylinder a lias a masa m and is hoisted using the cord and pulley system shown. Detennine the rnagnitude of force F as a function elf the cylinder's walca] position y so that when r is appl4ed the cylinder rises with a constant acceleration as. Negleci the mass of file cord, pulleys, hook aud chamn.

13-39. An electron of masa m is dischmsecl with an inilial horizontal velocity of v0. If it is subjected to two fields of forte for which Fx = F0 and F, = 0.34 where F0 is constant, determine the equation of the path, and the speed of the electron ai any time

Prob. 13-37

Prob. 13-39

13-38. The conveyor belt delivers each 12-kg crate to theramp at A such that the crate's speed is = 2.5 m/s,directed clown aiong the ramp. If file coefficient of kinetic fricfion between each crate and the ramp is µk =- 0.3, determine the speed at wbici, cech crate slides off the ramp at tt. Assume that no tipping occurs, Take H = 30°.

Proh. 13-3N

*13-40. The engine of the van produces a constant driving traetion force F at the wheets as it ascends the slope at a constant velocity v. Determine the acceleration of the van when it passes point A and begins to travel on a ievel ruad. proyicled that it rnaintains the same traction force.

Prob. 13-4E1

1 2 913,4 EQuArioNs OF MOTION: RECTANGULAR COORoINATES

13-41. The 2-kg collar Cis free to slide along the smooth shaft AR. Determine the acceleration of collar Cif (a) the shaft is fixed from moving. (b) eoliar A. which is fixed ta shaft A R. moves downward at «Instant vekwity along the verfical rod. and (c) collar A is subjected Lo a downward acceleration of 2 m/s'. In al] cases, the collar moves in the piane.

13-42. The 2-kg collar Cis free to slide along the smooth shaft A B. Determine the acceleration of collar Cif collar A is subjected to an upward acceleration of 4 mł s2.

Probs. 13-41142

*13-44. When the blocks ara released. determine their acceleration and the lension af the cable. Neglect the mass of the pulley.

I k g

3i3 kg

Prob. 13-14

13-43. The cncfficient af static friction hetween the 200-kg crate and the fiat bed of the truck is p, = 0.3. Determine the shortest linie for the truck to reach a speed of 60 km / h. starting from Test with constant acceleration, sn that the erate does not slip.

13-45. if the force exerted on cable AR by the motor is F= (1001-') Al, where t is in seconds. deterrnine the 50-kg crate's velocity when t= 5 s. The coefficients of static and kinetic ffiction between the erato and the ground ara

= 0.4 and juk = 0.3, respectiveiy. Initially the erate is at Test.

Prob. 13-13 Prob. 13-45

13DCHAI5TER 13 KiWiics OF A f ARTICLE= FORCE AKH:) ACCFLERAriorkt

13-46. Blocks A and B cach have a mass m. Determine the largest horizontal force P which can be applied to R so that A will not morie relative to B. All surfaces are smooth.

13-47. Blocks A and B each have a mass m. Determine the iargcst horizontal force P which can be applied to B so that A will not slip on F3. The coefficienł of static friction between Aand BisNeglect any frietion between i3 and C

13-50. A projectile. of mass m is fired into a liquid at ananglewith an initial velocity vo as shown, If the liquiddevelops a frictional or drag resistance on the projectile which is proportional to its velocity. i.e.. F = kv. where k is a constant,determine the x and y components of its position at any instant. Also, what is the maximum distancex,„, that il travels?

A.4- p

Probs. 13-46/47

13-48. A parachutist having a mass m opens his parachute from an at-rest position at a very high alfitude. If the atmospheric drag resistance is F0 = kv 2. where k is constant. determine his velocity when he has Mien for a time t. What k his velocity when he lands on the grouncl? "Mis velocity is referred to as the wrminal vełocily. which is found by letting the time of fal I r

Prob. 13-48

13-49. The smooth błock B of negligible size has a mass and rests on the horizontal planc.If the board AC pushes on the błock at an angIe O with a constans acceleratkin aa, deterrnine the velocity of the block along ihe board and the distance s the block moves along the board as a function of time i. The błock starts from rest when i = 0, r = D.

Prob. 13-50

13-51. The block A has a mass ni,4 and rests on the pan B, which bas a massina, Both are supported by a spring Ęavilg a stiffness k that is attached to the bottom of the pan and to the ground. Detennine the disłance d Ule pan should be pushed down from the equilibrium position and then released from rest so that separation of the błock will take place from the surface of the pan at the instant the spring becomes unstretched.

Prob. 13-51Prob. 13-49

13.5 EinuATIONS OF Mo-npN: NORMAL AND TANOENTIAL GOOROINATES1 3 1

1 3.5 Equations of Motion: Normaland Tangential Coordinates

When a particie moves along a eurvcd path which is known, the equation of motion for the particie may be written in the tangcntiał, normal, and binormal directions, Fig. 13-11.Note that there is no motion of the particie in the binormai direction, since the particie is constrained to move along the path. We have

2T = ma

łFbui, = ona, + ma„

b[ -Srbaó

2, ...„. ł r i " ' " - . - S F A ,

£F,u,

P

inertiaI coordictaie system

This e-quation is satisfied provided

= ma,

1F„ = ma„

= O

( 1 3 - 8 )

Recall Chat a,(= dvIdr) represents the time rate of change in the magnitude of velocity. So ifF, acts in the direction of motion, the parłieles speed will increase, whereas if it acts in the opposite direction, the particie will slow down. Likewise, a„ (= v2//i) represents the time rate of change in the vełocitys direction. It is caused byF, which ativays acts in the positive n clireciion, i.e., toward the path's center of curvature. From this reason it is often referred to as the cerariperal form

Fig. 13-11

The centrifuge is used ta subject a passenger to a very large normal aeceleration eaused by rapid roiation. Realize that łhis acceieraiiorE is causeci by the urtbnłaneed norma] forek exerted on the passenger by the seat of the eentrifuge.

132CHAPTER 13 KINTłCS OF A PARTICLF: FORCE ANID ACCFLFRATION

When a problem involves the motion of a particie along a known curved pałh. norma and tangential coordinałes should be considered for the analysis since the aceeleration eomponents can be readily formulated, The method for applying the equations of motion, which relate thc forees to the. acceleration, has been outhned in the proce-dure given in Sec, 13,4. Specifieal I y, for ł, n, b coordinates it may be stated as follows:

Free-Body Diagram.

• Establish the inertial t, n, b coordinate system at the particie and <kaw the partcle's free-body diagram.

• The particle's normai aceeleration a„ aiways acts in the positive direction.

• If the tangential aceeierati on a, is unknown, assume it acts in the positive r direction,

• There is no acceleration in the b dircction.• Identity thc unknowns in the problem.


• Apply the equations of motion, Eq. 13-8.

K inernat cs.

• Formulałe the tangential and normal cornponents of aceeleradon; i.e., a, = dv kil arar = v chich and a„ =

• If the path is defined as y = f(x), the radius of eurvature at the point where the particie is located can be obtained from

p =WY 1 clx)21312 ild2Y 1421-

13.5 ECUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES1 3 3

Determine the banking angle O for the race track so that the wheels of the racing cars shown in Fig. 13-12a will not have to depend iwon friction to prevent any car from sliding up ar down the track. Assume the cars have negligib]e size, a mass m, and travel around the curve of radius p with a constant speed v.

( a }

SOLUTIONB efore kw king at the fol lowing solution, give some thought as to why it should be solved using ł, n. b coardinates.

Free-Body Diagram. As shown in Fig. t3-12b, and as stated -in the problem, no frictional force acts on the. car. Here N,- represents the wstałam of the ground on a II four wheels. Since can be calculated. the unknowns ara Mc and O

Equations of Motion. Using the rt, b axes shown,

-t-} I.rT „ „;

+t Fn = 0;

Ar-sit] i3 = -P

cos -= O

(.1.)

( 2 )

W m g

(b)

Fig. 13-12

Eliminating Nc and m from thesc equations by dividing Eq. 1 by Eq. 2, we obtain

V2

tan O = —gP

V2o = t a n i ( —

gP

NOTE: The resul t is independent of the mass of the car, Also, a force summation in the tangential direction is cif no consequence to the solution. 1f it were considered, then ar = dv jdr = 0, sincc the car riwyes with constant speed. A furtki- analysis of this problem is eliscussed in Prob. 21-53.

134 CHARTER 13 KINETICS OF A PARTICLE: FORCE AND ACCELERATION

Moliaii of platform

(1 )

(2 )

(3 )

Equations of Motion.1

2,F = mar,; T = 3()Fr 1

= ma,; 0..1ND = 3a,

= O; ND 29.43 = 0

F =, 0_1 Na

The 3-kg disk D is attached to the end of a cord as shown in Fig.13-13a. The other end of the cord is attached to a hall-and-socket joint located at the center of a platform, If the platform rotates rapidly, and the disk is placed on it and reIcased from Fest as shown, determine the linie it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N. and the coefficient of kinetic friction between the disk and the platform is p,& = 0.1.

SOLUTION

Free-Body Diagram. The frictional fotce has a rn agnitudc F = "..1,

<ND = 0.1% and a sense of dircetion that opposes the reiarive morion of the disk with respect to the platform. It is this forcc that giyes the disk a tangential component of acceleration causing v to increase, thereby causing T to increase unłil it reaches 100 N. The weight of the disk is W = 3(9.81) = 29.43 N. Since a, can be related to v, the unknowns are Nu, a,, and v.

Setting T = 100 Eq. I can be solved for the critical speed y of the diak needed to break the cord. Solving all the equations., we obtain

Ni) = 29.43 N0.981 m/s2

v„ = 5.77 m/

Kinematies. Since a, is ~stara, the tinie needed to break the cord is

v , = V o a r t

5.77 = O + (0.981)1

= 5.89 s A r ;s,

EXAMPLE

b

(b)

Fig. 13-13

1 3 513.5 EQUATIONS. OF MOTION: NORMAL AND TANOENTIAL COORDINATES

2 0 0 f t

Ans.aA=a„=42.2ftfs2I

Design of the ski jump shown in the photo requires knowing the type of forces that wilI be exerted on the skier and her approximate trajectory.lf in this case the jump can be approximated by the parabolashown in Fig. 13-14a. dełermine the normal force on theskierthe instant she arrives at the end of the jump, point A, where her veiacity is 65 ft/s. Also, what is her aceeleration at this point?

SOLUTIONWhy consider using n, f coordinates to solve (flis problem?

Free-Body Diagram. Since dył dx = xi1001,_0 = 0, the slope at A is horizontal. The free-body diagram of the skier when she is at A is shown in Fig. 13-14b. Since the path is curved, there arc twa componcnts of acceleration, a„ and a,. Since a can be calculated, the unknowns are ar and NA.

150 ((65),2)N A - 1 5 Q = 3 2 . 2 p

1500 = — a

32.',

The radius of curvaturc p for the path must be detcrmined at point A(0, -200 ft). Herc y _W x2 - 200, cly ł dx = 7,1-jox, d2y/dx2 = W. so that at x = 0,

+ (dY ic11)2.13/2

P - [1 (0)21312 = 100 ftd2y/c1x2 I x=o Irhm

Substituting this into Eq, 1 and solving for NA, we obtain

= 347 lb Ans,

Kinematits. From Eq, 2,aś = O

Thus.

v2

a, - p(65)2

100 - 42,2 ft/s2

NOTE: App]y the equation of motion in the y direction and show that when the skier is in midair her acceleration is 322 ft/s2.


+1` = ma„;

"SF, = ma,;

•

= 22 i n n

, 0{)

( a )

n

Fig. 13-14

136CHArTER 13 KIlysriCS OF A f ARTICLE= FORCE AKH:› AccFLERATioN

EXAMPLE

13.9

( a )

60(9..81)N

n

(b)

(c)

Fig. 13-

15

The 60-kg skateboarder in Fig. 13-15a eoasts down the circular traek. if he starts from rest when ł1 = il determine the magnitude of the normal reaction the track exerts on him when O = 60°. Neglect his size far the ealeulation.

SOLUTIONFree-Body Diagram. The free-body diagram of the skateboarder when be is at 21n £/ rIntrary pvsirion O is shown in Fig. 13-15h. At = 60° there are three unknowns, N,, a„ and a„ (or v),Equations of Motion.

+7 F„ = ona„ N - [60(9 .81)N] s in = (60 ka a1` 4m

rna,; [60(9.81)N] cos O = (60 kg) a,

a, = 9_81 (ms O

( 1 )

Kinematics. Since a, is expressed in terrns of 0, the equation v dv = a rds must be used to determine the speed of the,skateboarder when B = Using the geometrie relation s = Or,where ds = rd0 = (4 m)dt9, Fig. 13-be, and the initial condition v = O at 0=0°, we have,

v dv = a, ds6(r

av dv = 9.81 cos 0(4 dO)soc.

= 39.24 sin O2 o

v2- O = 39.24(sin 61.r - 0)

V2 = 67.97 m2/s2

Substituting this result and O = int° Eq. (1), yields

N, = 1529.23 N = 1.53 kN

13.5 ECUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES 137

■ FUNDAMENTAL PROBLEMSF13-7. The błock rests at a distance Qf 2 m from the center of the platform. If the coefficient of static frietion between the błock and the piat [orni is p = 4.3, determine the maximum speed which the błock can attain before it begins to slip. Assume the angolar motion of the disk is slowly increasing.

FI3-111. The sports car is traveling along a 30° bankcd road having a radius of curvature of p = 5100 ft. If the coefficient of static friction between the tires and the wad is

= 0.2. determine the inaximurn safe speed so no slipping occurs.Neglect the size of the car.

p - 500 ft

F13-10

F13-11. If the ł0-kg bali bas a velocity of 3 m/s when it is at the position A. along the vertical path. delermine the tension in the cord and the increase in the speed of the bali at

F13-7 Ibis position,

F13-8. Determine the maximum speed that the jeep can travel over the erest of the bill and not lose contact with the road.

p = 250 fl

F 1 3 - 8

113-9. A pilot weighs 150 lb and is traveling at a constant speed of 120 ft/s., Determine the norma' force hG exerts an the seat of the piane when be is upside down at A.'The loop hasa radius of curvalure of 400 ft.

i Ate=q201--

400 ft

F1 3-q

F13-12. The motoreycle bas a mass of 0.5 Mg and a negligible size, ft passes point A traveling with a speed of 15 m/s, which is increasing at a constant ryte of 1.5 m/ Deterrnine the resultant frictional force exerted by the ruad on the tires at this instant,

PA = 20Q ni

F13-12

13$CHARTER 13 KINETICS OF A PARTICLE: FORCE AND ACCELERATION

■ PROBLEMS

13-52. A gid, having a mass of 15 kg, sits motionless relative to the, surface of a horizontal platform at a distance of r= 5 m from the platforrns center. H ihe anguIar motion of the platform is słowiy inereased so that the girls tangential component of acceleration can be neglected, determine the maximum speed which the gir! win have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is pl. =0.2.

Prob. 13-52

13-53. The 2-kg block B and 15-kg cylinder A are connected ta a light cord that passes through a hole in the center of the srnooth labie. If the block is g-iven a speed of v = 10m/s, determine the radiu roi the circular path along which i t travels.

13-54. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth labie. If the block travels along a circular path ol radius r = ł .5 m. determine the speed of the block_

P roby. 13-53/54

13-55. The 5-kg collar A is stiding around a smooth vertical guide rod. At the instant shown, the speed of the Miar is v= 4 rnis..which is increasing at 3 m/s2.Determlne the normal reaction cif the guide rod on the collar, and furce P at this instant.

113-56. Cartons having a mass of 5 kg are required to move along the assembly lina at a constanl speed of S m/s. Determine the smallest radius of curvature, p, for the conveyor sa the cartons do not slip_The coefficients of static and kinetic friction bełween a carton and the conveynr are Fts = 0.7 and j = 0.5, respectively.

8 m

P— 4 m/s

3W

Prob. 13-55

13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDiNATFS139

13-57. The block B. having a mass of 02 kg, is attached to the vertex A of the right cirettlar cone wilg a light cord, The cone is rotating al a constant angular race about the z axis such that the błock attains a speed of 0.5 ni/s. At this speed, determine the, tension in the card and the reaction which the cone exerts on the blach. Neglect the size of the błock and the effect of friction.

Prob. 13-57

13-58. The 2-kg spod S fits laasely on the inclined rod forwhich the coefficicnt af static friction is j. 0.2. If thespool is Iocated 0.25 m from A, deterrnine the minimum constant speed the spod can have so that il does not slip down the rod.

13-59. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is p =-- 0.2. If thc spool is located 0.25 rn from A, deterrnine the maximum constant speed the spool can have so that it does not slip up the rod.

Probs. 13-58159

*13-60. At the instant H = 60°, the boy's center of mass Ghas a downwardspeed = 15 ft/s. Determine the rale afincrease in bis speed and the tension in each cif the twa supporting cords of the swing at Ibis instant, The boy has weight of 60 1K Neglect bis size and the mass of the seat and cords.

1 313-61. At the instant fi = 60°, the boy's center of mass Gis momentarily at resł.Determine his speed and the tensionin each of thc twa supporting cords of the swing when

= 90°. The boy lias a weight of 60 Ib. Neglect los size andthe mass of the seat and cords.

Probs. 13-60161

13-62. The 10-lb suitcase slides down the eurved ramp for which the coeflicient of kinetic friction is A = 0.2, If at ihe instant il reaches point A ił has a speed of 5 fifs, dełermine the narrnal tarce on the suitcase and the rale of increase cif its speed.

Prob. 13-62

1 4 0 C H A R T E R 1 3 K N E T L C S O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

13-61 Thc 150-lb man lies against the cushion for whichthe coefficient of static friction = 0.5, Determine theresullard normał and frictionai forces the cushion exerts an him if. due to rotation abont the z axis. be has a constant speed za = 20 Itp, Neglcct the size of the man. Take O = 60°.

93-64. The I 50-lb man lies against the cushion for which the coefficient of static fricłion is p., = 0.5. If he mtates about the z axis with a constant speed v = 30 ft/s, determine the .smallest angle H of the cushion at which he will begM to slip off.

Z

8 ft

13-66. The man bas a mass of 80 kg and sits 3 m from the center of the rotating platform. Due to the rotation his speed is increased from Test by u = U.4 m/s2. If the cocfficient of static frietion between his clothes andthe platform is = 0.3, determine the time required tocause bim to slip.

Prob. 13-66

Probs. 13-63/64

13-65, Deterrnine the constant speed of the passengers on the amusemenł-park ride if it is observed that the supporting cables are directed at O = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what arc the components of force in che n, !, and h direcflons which the chair exerłs on a 50-kg passenger during the mołion?

Prob. 13-65

13-67. Thc vehicle is designed to combine the feel of a motorcycle with the comfort and safely of an automobile.ff the vehicie is traveling al a constant speed of 80 km/h along a circular curved road of radius 100 m, determine the fili angle H of the vehicie so that only a norma' force froni the seat acts on the driver. Neglect the size of the driver.

Prob. 13-67

13.5 EQUATIONS. OF MOTION: NORMAL AND TANGENTIAL COORDiNATES1 4 1

*13-68. The 0.8-Mg car travels over the hill having the shape of a parabola. Tf the driver main tains a constant speed of 9 m/s, determine both the resultant norma' force and the resultant frictional force that alt the whecls of the car czert on the road at the instant it reaches point A. Neglect the size of the car.

13-69. The 0.8-Mg car travels over the hill having the shape of a parabola. When thc car is at point A, it is traveling at 9 m/s and increasing lis speed at 3 tn/s2.Deterrnine both the resultant normal force and the resultant frictional furce that alt the wheels of the car exert on the ruad at [bis instant. Neglect the size °f the car.

v

2 ° ( 1 T£0 )

A

1 . - 8 ( h n - 1

P r o b s . 1 3 - 6 8 / 6 9

13-70. The package has a weighl of 5 lb and slides down the chute. When it reaches the curved pardon AB, it is łraveling at 1 ft/s (0 = 01, If the chute is smooth, determine the speed of the package when it reaches the intermediate point C(0 = 3T) and when it reaches the horizontal plam (O = 45v). Also. Pind the normal furce on the package at C.

Prob. 13-70

13-71. If thc hall has a mass of 30 kg and a speed= 4 m f s at the instant it is at its lowest point. 6 = 0°.

determine the tension in the cord at this instant. Also, determine the angle O to which the bali swings at the instant it momentarily stopa. Neglect the size of the bali.

*13-72. The ball has a mass of 30 kg and a speed = 4 m/s at the instant it is at its lowest point, O =- 0°. determine the tension in the card and the rato at which the ball's speed is decreasing at the instant O = 20°. Neglect the size of the half.

4 rn

13-73. Deterrnine thc maximum speed at which the car with mass m can pass over the top point A of the vertical curved rond and stili maintain contact wich the wad. If the car maintains this speed. what is the normal reaction the ruad cxerts on the car when it passes the lowest point 13 on the road?

Prob. 13-73

13-74. If the crest of the hill has a radius of curvature p = 200 ft., determine the maximum constant speed at wbici] the car can travel over ii withoui leaving the surface of the road. Neglect the size °f the car in the calcuIałion. The car has a weight of 3500 lb.

p 2 0 0 f t

Prob. 13-74

x

1 4 2 C H A R T E R 13 K I N E T I C S O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

13-75, Bobs A and B cif mass mA and my (mA > my) are connected to an inexłensible Iight string of length I that passes through the smooth ring at C. H hob B moves m a conical penduł urn such that A is suspended a distance of h from C. determine the angre H and the speed of bab B. Neglect the size of bołh bobs.

13-77. The skier starts from rest at A(10 m. O) and descends the smooth slope, which may be approximated by a parabola, If sile has a mass of 52 kg, determine the norma] tarce the ground exerts on the skier at the instant she arrives ai point B. Negleet the size of the skier. Pbrrt, Use the result of Prob. 13-76.

Prob. 13-77

Prob. 13-7513-78. A spring, having an unsłretched length of 2 ft. lias one end attached to the W-lb bali. Determine the angłe N of the spring if the bali hasa speed of 6 ft/s tangent to the horizontal circular path.

13-76. Praw that if the block is rełeascd from rest at point B of a smooth path of arblinvy shape. the speed it aitains when it reaehes point A isecival to ihe speed il aitainswhen it falls freely through a distance Nir2.7h.

Prob. 13-76

—6 in.

Prob. 13-78

13.5 EQUATIONS. OF MOTION: NORMAL AND TAN.GENTIAL COORDINATES 143

1-3-79. The airplane. traveling at a constant speed of 50 m/s, is executing a horizontal turn. If the piane is banked at

= 15°, when the pilot experiences ordy a normal kirce on the seat of the piane, determine the radius of curvature p of the turn. Also, what is the normal forte of the seat on the pilot if he hasa mass of 70 kg.

Prob. 13-79

*13-80. A 5-Mg airplane is flying at a constant speed of 350 km/h along a horizontal circular pach of radiusr 3000m. Dełermine the uplift force L acting on theairplane and the banking angle 9, Neglect the size of the airplane.

13-81. A 5-Mg airplane is flying at a constant speed of 350 km/h along a horizontal circular path. If the hanking angie H = 15°. determine the uplift forte L aefing on the airplane and the radius r ❑t the circular path. Neglect the, size of the airplane.

r

Probs. 13-80/81

13-82. The 800-kg motorbike travels with a constant speed of 80 kirtih up the MI. Determine the normal furce the surface exerts on its wheels when it reaches point A. Neglect its size,

100 m

Prob. 13-82

13-83. The bali lias a mass m and is attached to the cord of Iength I. The cord is tied at the top to a swivel and the hall is given a velocity vr,. Show that the angle B which the cord makes with the vertical as the bałt travels around the circular path must satisfy the equation tan O sio O = v3/gi. Negleet air resistance and the size of the bali.

Prob. 13-83

*13-84. The 5-lb collar slides on the smooth rod, so that when it is at A n hasa speed of 10 ft/s. If the Spring to which it is attached has an unstretched length of 3 ft and a stiffness of k = 10 1h/ft, determine the normal force on the collar and the magnitude of the acceleration of the collar at this instant.

2 f t -

Prob. 13-84

1 3

144 CHAPTER 13 KINETICS OF A PARTICIE: FORCE AND ACCELERATION

1 3.6 Equations of Motion: Cylindrical Coordinates

When ail the forces acting on a particie are resoived into cylindricalP .,'F,111, ~po nents, i.e., a I ong the unit-vector directions ur, up, u4, Fig. 13— 16, the

equation of motion can be expressed as

SF = maO

S,F ru r SFouo + S F = mu ru, + maoup ma ru,

To satisfy this equation, we require

Inertial oaordinate system

= ma r

= mao

= ma,

Fig. 13-16

(13-9)

If the particie is constrained to move only in the r—O piane, then only the flrst two of Eq. 13-9 are used to specify the motion.

Tangential and Normal Forces. The most straightforward type of problem invalving cylindrical coordinates requires thedetermination of the resultant force components whichcause a particie to move with a known acceleration. If, however, the particie's accelerated motion is not completely specified at the given instant, then some in fomation regarding the, directions or magnitudes of the forecs acting on the particie must be known er ealculated in order to soivc Eqs. 13-9. For example, the force P causcs the particie in Fig. 13-17a ta move along a path r = ji0). The twrmal forte N which the pach exerts on the particie is always perpendicsdar w the tangent of the path, whereas the frictional force F always acts ałong the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the raclial coardinate by using the angIe (psi),

13-17b, which is defined between the exrended radial line and the tangent to the curve.

r = (O>

Tangent

r = j(H)

Tangent

(a)Fig. 13—.1.7

r

PO J

(b )

As the car of weight W descends the spiral track, the resultant nuntial foroe which the Jack exerts on the car can be represcnted by its dirce cylindrical components, N. , directed horimntally inward, creates a radia] acceleration —a, l'ticreales a transverse acceleratinnat,. and the difference W —createsan azimathal aceeleratinn —a,.

13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES145

rtan afi =dr Id°

(13-10)

This angle can be obtained by noting that when the particie is displaced a distance ds along the path, Fig. 13-17c, the com piment of displacement in the radia] direction is dr and the component of displaceinent in the transverse direction is r dO. Since these two components are mutually perpendicular, the angle r can be determined from tan = r d 0, I dr, ar

If p is caiculated as a positive quantity, it is measured from the exźended rudikl line to the tangent in a counterclockwise sense nr in the positive direction of 0. If it is negative, it is measured in the opposite direction topositive 0. For example. consider the cardioid r = a(I cos O), shownin Fig. 13-18. Because dr/do = -d sin 0, the]] when O = M°,tan i,/F = a(1 cos 30°)A-a sin 30°) = -3.732, ar = -75°, measuredclockwise, opposite to +0 as shown in the figura.


Cylindrical ar polar coordinates ara a suitable choice for the analysis of a problem for which data regarding the angular motion of the radial line r are given, or in cases where the path can be convenientiy expressed in terms af these coordinates, fince these coordinates have been established, the equations of motion can then be applied in order to relate the forces acting on thc particie to its acceleration components. The method for cloing this has been outlined in the procedure for analysis given in See.13.4.The following is a summary of this procedura.

Free-Body Diagram.

• Establish the r, 0, z inertial coordinate system and draw the particie's free-hndy diagram.•ASSUMC that ar., ap, a- act in the positive directions of r, B. z if they are unknown.• Identify alt thc unknowns in the problem.

Equtions of Motion.• Api:11y the equations of motion, Eq.

13-9. Kinematics.

• Use the methnds of Sec. 12.8 to determine r and the timederivatives r, );', 0, O', and then evaluate the aceelerationcomponents a,. = - r02,ao =a, =

• If any of the acceleration components is computed as a negative quantity, it indicates that it acts -in its negative coordinate direction.• When taking the time derivatives of r = 10), it is very important to use the chain rule of calculus, which is discussed in Appendix C.

r = i(e)Tangent

dr

(c)

Fig. 13-17

Tangen

Fig. 13-18

1 4 ( } C H A R T E R 1 3 K I N E T I C S O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

EXAMPLE [ 1 3 . 1 0

The smooth 0.5-kg double-collar in Fig. 13-19a can freely slide on arm AB and the circular guide rod. If the arm rotates with a ennstant angular veioci ty of 9 = 3 rad/s, determine the force the arm exerts on the collar at the instant O ,= 45°. Motion is M the horizontaI piane.

SOLUTION

1..ee-Rody Diagram. The normal reaction Nc of thc circutar guide rod and the furce F of arm AR act on the collar in the p1ane of morion, Fig. 13-19b. Note that F acts perpendicular to the axis of arm AB, that is, in thc direction of thc O axis, whilc Nc acts perpendicular to the tangent of the circuIar path at O = 45°. The four unknowns ara Nc, F, ar, U.

Equations ol Motion.

= mar: -Nc cos = (0.5 kg) a, (1)

+w\ZFe = F - Nc sin 45° -= (0.5 kg) ao (2)

Kinematies. Using the chain rule (see Appendix C), the first and tangent second finie dcrivatives of r when 6 = 45°, 6 = 3 rad/s, 6 = O, are

r = 0.8 cos O = 0.8 cos 45' = 0.5657 m

= -0.8 sin = -0.8 sin 45°0) = -1.6971 m/s

r= [siri cw; O Ó21

= -0.8[sin 45°(0) + cos 45°(32)] = -5.091 m/s2

We have

a, = - = -5.091 - (0.5657 m)(3 rad/s).- = -10.18 mis:

= rE1 + 2i-6 = (0.5657 m)(0) 21-1.6971 m/s)(3 radfs)

= -10,18 m/s2

Substituting these results in to Eqs. (1) and (2) and solving, we- get

Nc = 7.20 N

F = O A n s.

(1))Fig. 13-19

( a )

13.6 EOUATIONS OF M OTION: CYLINPRICAL COORDiNATES1 4 7

The smooth 2-kg cylinder C in Fig. 13-20a has a pin P through its center which passes through the slot in arm 0A. H the arm is forced to rotate in the veracal piane at a constant rate 9 = 0.5 rad/s, determine the forte that the arm exerts on the peg at the instant O = 60°.

SOLUTIONWhy is it a good idea to use polar coorclinates to solve this problem?

Free-Body Diagram. The frere-body diagram for the cylinder is shown in Fig. 13-20b, The force on the peg, Fp, acts perpendicular to the slot in the ann. As usual. a, and a are assurned to act in the directions of posiiive r and 0, respectiveIy. Identify the kiur unknowns.

Equations of Motinn. Using the data in Fig. 13-20b. we have

= mar; 19.62 sin B - Nc sin B = 2a, (1)\I,F0 =. map; 19.62 cos O + Fp - ,Nrc cos O 2a0 (2)

Kinernatics. From Fig. 13-20a, r can be related to O by the equation

0Ar = - 0.4 csc O

sin

Sffice d(csc 8) = -(csc O cot 8) dO and ri(cot O) = -(csc'` O) dO, then r and the necessary time de,rivatives become

=0.5 r = 13.4 csc O= 0 r = -0.4(csc O cot 0)

= -0.2 csc O cot-0.2(-csc O col 0),(Ó) cot O - 0,2 csc Q(-csc' 0)0 =

0.1 csc 0(cot2 O + csc2 O)

Evaluating these formulas at O = 60°, we get

= 0.5 r = 0.462

= 0 r= --D,133

= 0.192

a r = r - r82 -= 0.192 - 0A62(0,5)2 = 0.0770uy = r6 + = O -I- 2(-0.133)(0.5) = -0.133

Substituting these resuits into Eqs. 1 and 2 with 6 = 60° and soIving yieIds

Nc = 19.4 N Fp = -0.356 N A ny_

The negative siga indicates that Fp acts opposite to the, direction shown in Fig. 13-20b,

( a )

r i

( b )

F i g .

1 3 - 2 0

1 4$ CHAPTER 13 KNyEliCS OF A f ARTICLE= FORCE Ahk:› AccFLERATioN

EXAMPLE

{ a }

( b )

A can C. having a mass of 0.5 kg, moves along a grooved horizontal słot shown in Fig. 13-21a. The slot is in the form of a spiral, which is defined by the equation r = (0,10) rn, where 0 is in radians. if the arm OA rotates with a cunstant rate B = 4 rad/s in the horizontal piane, dcterrnine the force it Gxcrts on the can at the instant O = Tr rad. Ncglect friction and the size of the can,

SOLUT1ON

Fr9e-Body Diagram. The driving furce Fc acts perpendicular to the arni 0A, whereas the normaI forze of the wali of the slot on the can, acts perpendicular to the tangent to the curve at O = 7F rad, Fig. 13-

21b. M usuaI, a r and a5 are assumed to act in the positive direefions of r and 0, respectively. Since the path is specified, the angle

which the extended radial line r rnakes with the tangent,Fig. 13-21c, can be deterrnined from Eq. 13-10. We have r = 0.10, so that

dr/ do = 0.1, and therefore

r 0.10tan rif = = = O

dr/d0 0.1

When O = = tan-17r = 72.3°, so that cl) = 90° - tif = i7.7°, asshown in Fig. 13-21c. Identify the four unknowns in Fig. 13-21b.

Equations of Motion. Using = 17.7° and the data shown in Fig. 13-21b, we have

SF, = mar; cos 17.7° = 0.5a, (1)

i,SF6 = mao; Fe - Ale sin 17.7° =- 13,5a0 (2)

Kinernatics. The time derivatives ot r and O are

= 4 rad/s r = 0.10

B=4 = 0.10 = 0.1(4) = 0.4 mis

= o

At the instant O = zr rad, e=r

a,. = i - = O - 0.1(7r)(4)2 = -5.03 mis"

Lie = rd ł 2k0 = O + 2(0.4)(4) = 3.20 mirs2

Substituting these results finta Eqs_ 1 and 2 and solving yieldsN e = -2 .64 N

F e = 0 ,800 N

What does the negative sign for Ne indicate?

Tangunt

(e)

F. 13-21

Ans .

13.6 EQuATIONS OF MOTION: CYLINORICAL COORDiNATES 149

■ FUNDAMENTAL PROBLEMSE13-13. Determine thc constant angular vdocity fl of the vertical shaft of the amusement ricie if q!, = 45°. Negieet the mass of the eables and the sine of the passengers.

I 5

F13-1.3

F13-14. The 0.2-kg bali is hlown through the smooth vertical circular lube whose shape is defined byr = (0.6 siu 0) m, where H is in radiany. If O = t2) rad.where t is in seconds. deterrnine the magnitude of furce F exerted by the blower on the hall when t = 0.5 s.

F13-13. The 2-Mg car is traveling along thc curvcd road described by r = (50r'-fi) rn, where O is in tacham. If a carnera is located at A and it roiaies with an angular velociły of fJ = 1105 rad /s and an angular acceleration of

0,01 rad/s2 at the instant O = rad, detertnine theresuitant friction forte devełoped bełweentires and themad at this instant.

F13-1.5

113-16. The 0.2-kg pisz P is coristrairied to rnove in the smooth curved siut, which is defined by the lemniscate r = (0.6 cos 20) ni. hs motion is controlied by the rotation of the slotted arm 0A, which has a eonstant dockwise angular velocity of tl = -3 radfs. Determine the furce arm OA .exerts on ihe piu P when 0 = 0°. Motion is in the vertical piane.

1 3

A

f - 1 3-1 1.13-16

' l 50CHARTER 13 KINETICS OF A PARTICLE: FORCE AND ACCELERATION

• PROBLEMS

13-85. The spring-held follower AB bas a weigitt of 0.75 Ib and moves back and forth as its end rolls on the conłoured surface of the carn, where r = 0.2 ft and r = (0.1 siu 20) ft. 1f the carn is rotating at a constant rale of 6 rad/ s. determine the furce at the end .4 of the follower when O = 45°. In thls position the spring is compressed 0.4 ft. Neglect friction at the bearing C

Prob. 13-85

13-86. Detem-Line the magnitude of the resultant furce acting on a 5-kg particie at the instant i =- 2 s, if the particie is movingatong ahorizontal path defmedbytheequations r = + I()) mand H = (1.512 — 6t)rad.where.tis in seconds,13-87. The path of motion of a 5-lb particie in the horizontal piane is described in terms of polar coordinates as r = (21 + 1) ft and O = (0.5/2 — o rad, Mlew I is in seconds. DeterrnMe the rnagnitude of the unbalanced furce acting on the particie when r = 2 s.*13-88. A particie, having a mass of 1.5 kg, rnoves alonga path defined by the equations r = (4 +m.

= (12 + 2) rad, and z— r 3)in. where i is in seconds.Determine the r, H, and z components of forte which the path exerts on the particie when r = 2 s.13-89. Rod DA rotates counterclockwise with a constant angular velocity of ą = 5 radfs. The double coliar B is pin-connected together such that one coIlar slides over the rntating rod and the other slides over the horizonłał curved rod. of which the shape is described by the equation r = 1.5(2 — cos 0) ft. If botki cobars weigh 0.75 lb, determine the normal forte which the curved rod exerts on one collar at the instant O =- 12(r. Neglect friction.

13-90. The boy of niass 40 kg is sliding down the spiral s]ide at a constant speed such that his position. measured front thc top of the ehute, has components r = 1.5 m,O = (0.7.0 rud. and z =m. where t is in seconds.Deterrnine the componen t s of [orce Fr,and F, which theslide exerts on him at the Instant i = 2 s. Neglect the size of the boy.

Prob. 13-90

13-91. The 0.5-lb particie is guided along the circular path using thc slotted arm guide. If the arm has an angolar velocity ri = 4 radl s. and an arigular aceeleration F1 = 8 radN at the instant Fi = 31]-. determine the fotce of the guide on the particie. Motion occurs in the horizontai piane.

Prob. 13-89 Prob. 13-91

15113-6 EQUATIONS QF MOTION: CYLINDRICAL COORDINATES

X13 92. Using a forked rod, a smooth cylinder C having mass of 0.5 kg is forced to more along the verticał siorted pach r = (030) m, where O is in radians. If the angular position of the arm is O = (0.5r 2) rad, where x is in seconds, deterrnine the forte of the rod on the cylinder and the norma] furce of the slot on the cylinder at the instant t = 2 N. The cylinder is in contact with only one edge of the rod and slot at any instant.

Prob. 13-92

13-93. If arm OA rotates with a constant clockwise angular velocity of H = 1.5 rudis. determine the force arm OA exeris on the smooth 4-lb cylinder B when H = 45°.

13-94. The collar hasa mass of 2 kg and travels along the smooth honzontal rod defined by the equitinguIar spiral r = {e °)m, where O is in radians. Determine the tangential furce F and the norma] furce N acting on the collar when O = 900, if the tarce F maintains a constant angular motionO = 2 rad/s. 13

Prob. 13-94

13-95. The bali has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius ro = 0.5 m such that the angular rate of rołation is tje, = I rudfs, If the attached cord ARC is drawn down through the hole at a constant speed of 0.2 mis, determine the tension the cord exerts on the bali at the instant r = 0.25 m. Also, compute the angular velocity of the hall at [bis instant. Neglect the effects of friction between the bali and horizontal piane. 1-fint First show that the equation of motion in the B direchan yields

(1/r1 (d(r'Fl= O. When Integrated.r2? = c, where the constant c is determineci from the problem data.

' • • • • - - i f l • - -

Prob. D-93 Prob. 13-95

152 CHARTER 13 KINETLCS OF A PARTICLE: FORCE AND ACCELERATION

*13-96. The particie lias a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rolalion of the arm 0A. Deterinine the lorce of the rod cm the particie and the norma lorce of the slot on the particie when fl = 30. The rod is rotating with a conslani angular velocity d = 2 rad/s. Assume the particie contacts only one side of the slot at any instant.

13-97, Solve Problem 13-96 if the arm has afl angularacceleration of3 rad/s2 and O = 2 rad/s at Mis instant.Assume the particie contacts only one side cif the slot at any instant.

0.5 m

Probs. 13-96197

13-98. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangukr spiral

=

" (es) ni, where O is in radiany, Determine the tangential lorce F and Ihe normal force N acling im the collar when O = 45°, if the force F niaintains a constant angular motion

= 2 rad/s,

Prob. 13-98

13-99. For a short tirne, the 250-kg miler coaster car is traveling along the spiral track such that its position measured from the top of the track has components r = 8 m, H = (G. II 0.5) rad. and z = (-0.20 ni where t is in seconds.Determine the magniiudes of the componenls of lorce which the track exerts on the car in the r. 0, and z directions at the instan = 2 s. Neglect the size of the car.

Prob. 13-99

'13-100. The 03-lb bali is guided along the verticalcircular pach r2r. cny H using the arm DA. If the armlias an angular velocity H = 0.4 rad/s and an angularacceleration= 0.8 rad/s2 al the Instant O =determine the force of the arm on i he hall. Neglect friction and the size of the bali. Set r, = 0.4ft.

13-101. The bali of mass m is guided along the vertical circular path r = 2r, cna d using the arm DA. If the armbas a constant angular velocitydetermine the angle

45' at which the bali starts to leave the surface of the semicyiinder. Neglect friction and the size of the bali.

Probs. 13-100/101

?3.6 EQUATIONS OF MOTION: CYLINDRICAL COORnINATES153

13-102. Using a forked rod, a srnooth cylinder P, having mass of 0.4 kg, is forced to move along the verncal sloned path r = (0.60) m, where H is in radians.If the cylinder has a COnStan t speed of vc = 2 m/s, determine the forte of the rod and the norrnal forte of the slot on the cylinder at the instant

= 7r rad. Assume the cylinder is in contact with ordy one edge of the rod and slot at any instant. flint To obtain the time derivatives necessary to compute, the cylinder's acceieration components a, and ao, take the first and second linie derivatives of r = 0.60. Then, for further information. ust; Eq. 12-26 to determine fl. l lisa, take the time derivative. of Eq.12-26,nofing that v = 0 to determine

Prob. 13-102

13-103. A ride in an amusement park consists of a eart which fis supported by smali wheeis. lniłialy the cart is traveling in a circular path of radius r4 = 16 ft such that theangolar rate of rotation is= 0.2 rad/s. If the attachedcable OC is drawn inward at a eonstant speed of

= -0.5 flis, determine the tension it exerts on the carl at the instant r = 4 ft. The cart and its passengers have a total weight of 400 lb. Neglect the effects af friction. Hiw: First show that the equation af motion in the O direction yiclds aH = r + 2j-0 = (11/r)d<r" H 1 f dr = O. When integrated, i-2 B = c, where the constant c is determined from the problem data.

Prob. 13-1413

*13-104. The arm is rotating, at a rate of = 5 rad/s when= 2 rad/s2 and fl = 90°, Determine the normal forte it musi

exert on the 0.5-kg particie if the particie is confined to move along the slotted path defined by the horizontal hyperholie spiral r() = 0.2 ni.

Prob. 13-104

13-105. The forked rod is used to move the smooth 2-lb particie around the horizontal pach in the shape of a lima■gon, r = (2 + cos O) ft. If al alt times ii = 0,5 radis, determine the lorce which the rod exerts on the particie at the instant O -= 90°. The Fork and path contact d-te particie on only one side.

13-106. Solve Prob. 13-105 at the instant H = 60°.

13-107. The forked rod is used to move the smooth 2-lb particie around the horizontal path in the shape of a lirnaon, r = (2 -i- cos O) ft. If H = (0.5r2? rad, where t is in seconds, determine the forte which the rod exerts on the particie at the instant r = 1 s,.The fork and path eontact the particie on only one side.

Prohs. 13-10.5/1414/11i7

1 5 4 C H A R T E R 1 3 K I N E T I C S O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

*13-108. The collar, which has a weight of 3 kb, slides along the smooth rod lying in the horizonud piane and having the sbape of a parabola r = 4/(1 — cos O), where H is in radians and r is in kei. If the collars angular rate is constant and equals ll = 4 rad/s, determine the tangential retarding force P needed to cause the motion and the norma force that the collar exerts on the rod al the Instant N = 90w.

13-109. The smooth particie has a mass of SO g. It is attached to an elastic cord extending from O to P and due to the sicitted arrn guide :neves along the horizontal circular path r = (0.8 sin 0) m. If the cord has a stiffness k = 30 Nim and an unstretched Iength of 0.25 m. deterrnine the foru of the guide on the particie when O = 60°. The guide has a constant angular velocity E1 = 5 radis.

13-110. Solve Prob. 13-109 if d - 2 raclis. when = 5 rad./6 and!) = 60°.

Probs. 13-109/110

13-111, A 0,2-kg spool slides down along a smooth rod. If the rod bas a constant angular race of rotation El = 2 rad/s in the vertical piane, show that the equations of matron for the spool arc r — 4r — 9.81 sin O = O and0.8.N, — 1962 cos H = - 0, where N, is the rnagnitudeof the norma] force of the rod on the spool. Using the tinctliods of clifferential equations, it can be shown that Ehe sol ution of the first of these equations is r = Cc-• 2

+ C2ezr — (9.81/8) .sin 2t. If r. r, and O are zero when t = 0, evaivate the conslants CI and C, determine r at the instant O = 5.7/4 rad.

Prob. 13-111

13-112. The pilot of an airplane executes a verfical loop which in part follows the pach of a cardioid. r = 600(1 + cos O) ft. If his speed at A (O = 0°) is a canstanl tip = 80 ft/s, determine the verde& force the seat belt -must exert on him to hold kim to his seat when the piane is upsidc down at A. He weighs 150 lb.

Prob. 13-112

Prob. 13-108

13.7 CENTRAL-FOPCE MOTION AND SPACE MECHANICS 155

*1 3.7 Central-Force Motion and Space Mechanice

If a particie is ruoving pnly under the influence of a force having a line of action which is always directed toward a fixed point, the muflon is cal led central-furce morion. This type of muflon is commonly caused by elcctrostatic and gravitationai forces.

In order to analyze the motion, we wili consider the particie P shown in Fig. 13-22a. which has a mass m and is acted upon only by the central force F. The free-body diagram for the particie is shown in Fig. 13-22b. Using poiar coordinates (r, 0), the equations of rnotion, Eq. 13-9, become

1 .̀1", = mar;-F = r(121

dt2 dr= ma o ;

d26 dr dO)O = ar(r=, + 2— —

dr dr dł

The secund of these equations may be written in the form

I r [d (r2d0dt i)1 o

so that integrating yields,d0

r = hdr (13-12) ( h )

( a )

Fig, 13-22Here h is the constant of integration.

From Fig, 13-22a notice that the shaded arca described by the radius r, as r ~yes through an angle dO, is dA = r2 d6. If the awal velocity is defined as

d A I d 0 h d ł 2 d / 2 (13-13)

then it is seen that the areał velocity for a particie subjected to central-force mation is constanł. In other words. the particie will sweep out equal egments of arca per unit of time as it travels along the path.To obtain the pałh of motam., r -= 1(0), the independent variable t must be eliminated frorri Eqs. 13-11. Using the chain rule of calculus and Eq. 13-12. the tlnie derivatives of Eqs. 13-11 may be replaced by

dr dt

dr dO h drdO dr de

d2r ( dr) d ( A dr)d0 [ d A d r ) i l idii k.r2 dd dO\r2 do i dr L da \,r2 d6

56CHAPTER 13 KINETICS OF A PARTICLE: FORCE AND ACCELERATION

Substituting a ncw dependent variablc (xi) = l /r into the secondequation, we Ytavc

der72 .d =-112C----(182

Also, the squarc of Eq. 13-12 becomes

(do

dr

Substituting thesc two equations into the first of Eqs. 13-11 yields

h 2 e

This salaite is subjected to a central [orce and fis orbiłalrnotion can be closely predicied using rhe equations devetopeci in this secrinn.

O l '

d 2 Z = F d e - (13-14)

This differential equation clefines the pach over which the particie travels when it is subjected to the central furce

For apptication, the force, of gravitational attracłion will be considered. Some common examples of centrał-force systems which depend on gravitation inciudc the motion of the rnoon and artificial satellites about the earth, and the motion of the planets about the sun. As a typical problem in space mechanścs, consider the trajectory of a space satellite ar space vehicle launched into free-flight orbit with an initial velocity vo, Fig. 13-23. It will be assumed that this velocity is initialty para ile! to the tangent at the surface of the earth, as shown in the figure.t Just after the satellite is released into free flight, the only force acting on it is the gravitational furce of the earth. (Gravitational attractions invoIving other bodies such as the moon ar sun will be neglectecI, since for orbits close to the carth their cffect is smali in comparśson with the earths gravitation,) According to Newtons law of gravitation, force F will always act between the mass centers of the earth and the satellite. Fig. 13-23. From Eq. 13-1, this furce of attraction has a magnitude of

F = Gjjn

where Me and ni represent the mass of the earth and the satellite, respectively, G is the gravitational constant, and r is the distance between

in the derivation, F is considered p-ositive when it is directed toward point O. II F is oppositely directed, the right side of Eq. 13-14 should be negative.

tThe case where acts at sanie initial angte 49 to lhe tangent is best described using the conservarton of angular momentum,

Frue-flighttrajectory

r4C1 Satellite

Power-fligh r = ro trajectury

Launching

Fig., 13-23

13_7 CENTRAL-FORCE MOT1ON AND SPACE MECHANIICS 157

the mass centers. To obtain the orbital path, wc set = 1/r in theforcgoing equation and substitutc thc result lota Eq. 13-14, Wc obtain

GM,+ = (13-15)

This sccond-ordcr differcntial equation has constant coefficients and is nonhomogeneous. The soIution is the sum of the compiementary and parłicular solutions given by

GMĘ_ - 1 = C cos - f fs)

rh- (13-16)

x

P

Fig. 13-24

Thi's equation represents the freelli& trajectvey of the satenite. It is the equation of a eonie section expressed in terms of polar coordinates.

A geometrie interpretation of Eq. 13-16 requires knowledge of the equation for a eonie section. As shown in Fig. 13-24, a eonie section is defined as the locus of a point P that rnovcs in such a way that the ratio of its distance to a focus, or fixed point F. to its perpcndicular distance to a fixed line DD calIed the diredrix,is constant.This constant rado will be denoted as e and is called the eccentricity. By clefiniłion

F Pe = A

From Fig. 13-24,

FP = r = KPA) = e[p - r eeps(0 - 4))1

o r

-1 =Icos(0 - da) +1

r p ep

Comparing this equation with Eq. 13-16, ii is scen that the fixed distance from thc focus to the directrix is

c l

(13-17)

And the eccentricity of the eonie section for the trajeetory is

Ch2

(13-18)

e =GM8

1 5 8 CHAPTER 13 KINETICS OF A PARTICLE FORCE AND ACCELERATION

di recirix

P0—(1)

r focus

F

DFig. 13-24 (Repeated)

Provided the pular angle 6 is measured from the x axis (an axis of symmetry since it fis perpendicular to the directrix). the angle is zero, Fig. 13-24, and therefore Eq. 13-16 reduces to

1 =Ccos-1- MG ,—

r h2 (13-19)

The constants h and C are determined from the data obtained for the position and velocity of the satellite at the end of the powerilighł trajectory. For example, if the initial height or distance to the space vehicle is r4, measured from the center of the earfh, and its initial speed is yo at the beginning of its free flight, Fig. 13-25, then the constans h may be obtained from Eq. 13-12. When 6 = rp = O°, the velocity v,) has no radial cornponent; therefore, from Eq. 12-25, yo = ro(dO hit), so that

,d0h = r5—

dl

°r

h = rovo (13-20)

To determine C, use Eq. 13-19 wich o = 0°, r = ro, and substitute Eq. 13-20 for h:

c =G M , ) r o v o

The equation for the free-flight trajectory therefore becuTnes

GM,c o s +

ri-)v5

= GM,

(13-21)

(13-22)

The typu of path łraveted by the satellite is determined from the value °f the eccentricity of the eonie section as given by Eq. 13-18. If

e = 0 frec-flight trajectory is a circlee 1 free-flight trajectory is a parabolae < 1 free-flight trajectory is an ellipsee 1 free-flight trajectory is a hyperbola

(13-23)

13,7 CENTRAL-FORCE MOTION AND SPACE MECHANICE 159

Para boiic Path, Each of these trajectories is shown in Fig, 13-25, From the eurves it is scen that when the satellite follows a parabok path, it is "on thc border" of never returning to its initial starting point. The initial launch velocity, vo, required for the satellite to follow a parabolicpath is called the escape velocity. The speed, can be determined byusing the second of Eqs. 13-23, e = 1, with Eqs. 13-18,13-20, and 13-21. It is left as an exercise to show chat

2Gt14,, ro

(1.3-24)

Circular Orbit. The speed v. required to launch a satellite int° a eirettlar orbit can be found using the first of Eqs. 13-23, e = 0. Since e is related to h and C, Eq. 13-18, C must be zero to satisfy this equation (from Eq, 13-20, h cannot be zero); and therefore, using Eq. 13-21, we have

= G M , r a

(13-25)

l'rovided ro represeuts a minimum height for launching, in which frictional resistance from the atmosphere is negleeted, speeds at launch which arc less than U. will cause thc satellite to reenter the earth's atmosphcre and eithcr bum up or crash. Fig. 13-25,

Hypvrbolic trajectr,rye 14,

Para bobo trajectorye = 1 - _

Elli ptica I trajectcryr-e <1

Yil Crash trajectury Circular

traiectorye=49

Fig. 13-25

1 6a C H A R T E R 13 K I N E T I C 5 O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

aa

Fig. 13-26

Effiptical Orbit. All the trajectories attained by planets and most satellites are elliptical, Fig. 13-26. For a satellite's orbit about the earth. the minimum distance from the orbit to the center of the earth O (which is located at one of the foci of the ellipse) is ri, and can be found usingEq. 13-22 with O = Therefore;

= r o (13-26)

This minimom distance is called the perigee of the orbit. The apogee or maximum distance r„ can be found using Eq, 13-22 with H = I SW,* Thus,

(13-27)r o

r, = (2G.+11,,rom?„) - 1

With reference to Fig. 13-26. the half-length of the major axis of the ellipse is

r + ra-

2(13-28)

Using analytical geometry, it can be shown that the half-length of the minor axis is deterrnined from the equation

b = "virpr„ (13-29)

*Actually, the terminoktgy perigee and apogee pertains only to orbila about the E.arih. H any ()filer heavenly body is located at the focus of an cliii:41[u' orbit, the minimum and maximum distanees are referred w re,speetively as the periapsis and apovpsis ol' Me orbit.

133 CENTRAL-FORCE MOT1ON AND SPACE MECHANI{S1 51

Furthermore, by direct intcgration, the area af an ellipse is

7rA = irab = —(r +

23-30)

The areal velocity has been definecl by Eq. 13-13, dAildł = h12. Integrating yields A -= hT112, where T is the period of time required to makr one orbita] revolutiois. From Eq. 13-30, the period is

T = + r )p .

(13-31)

In addition to prcdicting the orbita] trajectury of earth satellites, the theory developed in this section is valid, to a surprisingly close approximation, at predicting the actual motion of the planets traveling around the sun. In ibis case the mass of the sun, M,, should be substituted for M, when the appropriate formulas are used.

The fact that the planets do indeed follow elliptic orbits about the sun was discovered by the German astronomer Johannes Kepler in the early seventeenth century. His discovery was made hefure Newton had developed the laws of rrkotion and the law of gravitation, and so at the time it provided important proof as to the validity of these laws. Kepler's laws, developed altu 20 years of planetary observation, ale summatized as foliows:

1. Every planet travels in its orbit such that the line joining it to the center of the sun swecps over cqual arcas in cqual intervals of time, whatever the iinc's length.

2. The orbit of evcry planet is an eilipse with the sun placed at one of its foci.

3. The square of the period of any planet is directly proportional to the cube of the major axis of its orbit.

A małhematical statement of the first and second laws is given by Eqs. 13-13 and 13-22, respectively. The third law can be shown from Eq. 13-31 using Eqs. 13-19, 13-28, and 13-29. (See Prob, 13-117.)

162 CHAPTER 13 K INETICS O F A P A R T I C L E : F O R C E A N D A C C E L E R A T I O N

EXAMPLE

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 30 Mm/h acting paralIel to the tangent at the surface of the earth, Fig. 13-27.Assuming that the radius of the earth is 6378 km and that its mass is 5.976(1(i:14) kg, determine (a) the eccentricity of the orbital path, and (b) the velocity of the satellite at apogee.

50 LUT IONPart (a). The eccentricity of the orbit is obtained using Eq. 13-18.The constants h and C are first cletermined from Eq. 13-20 and 13-21, Sinee

rP O = r = 6378 km + 600 km = 6.978(106) in

ó(30 km

Fig. 13-27

Hence,

th cn= rrvo = 6.978(106)(8333.3) = 58.15(109) m2/ s (r- 1 ;

1 rPw,P 11 - 66,73(10-12)[5.976(1024)1}, 25.4(10-9) -1-1

- 6.978(106) 6.978(106)(8333,3)-

30 Mmiii

Ch2 2.54(10-N58.15(109)12

= - 0 . 2 1 5 < 1 G M , 6 6 , 7 3 ( 1 0 - J 2 )1 5 . 9 7 6 ( 1 0 2 4 ) 1

From Eqs. 13-23, observe that the orbit is an ehipse.Part (b). 1f the satellite were launehed at the apogee A shown inFig. 13-27, with a vełocity , thc same orbit would bc maintaincdprovided

h = rPvo = r„vA = 58.15(109) m2/s. Using Eq, 13-27, wre have 6.978(106)

- 21.66.73(10-12)115.976(1024)1 _ 6.978(106)(8333,3)2 58.15(109)

VA - 5382.2 m/s, = 19.4 Mm/h Ao.s_10.804( 1 06)

- 10.804(106) m

NOTE: The farther thc satellite is from thc earth. the sIower it moves, which is to be expectcd since h is constans.

rd = 30 1vIrriih = 8333,3 m/s

rr,r - a 2 G M ,

Thus,

1r

13.7 CENTRAL-FORCE MOTłON AND SPACE MECHANICS 163

• PROBLEMS

In the following problems, except where otherwise indicated, assume that the radius of the earth is 6378 km, the earth's mass is 5.976(1024) kg, the mass of the sun is I .99(I035 kg, and the gravitational constant is G = 66.73(10-12) m3/(kg.s2).

13-113. The earth has an orbit with eccentrici ty e = 0.0821 around the sun. Knowing Chat ehe earths minimum disłanee from the sun is 151.3(106) km, find the speed at which the earth travels when it is at this distance. Determine the ey uation in poTar coordinates which describes the earths orbit about the sun.

13-114. A communications satellite is in a circular orbit above the earth such that it always rernains direcf1y over a point on the earths surface. As a result, the period of the satellite musi equal the rotation of the earth, which is approximately 24 hours. Determine the satellites allitude h above the earth's surface and its orbital speed.

13-115. The speed of a satellite taunched into a circular orbit aboui the earth is given by Ey. 13-25. Determine the speed of a satellite launehed paraliel to the surface of the earth so that it travels in a circular orbit 800 kin from the earth's surface.413-116. The rocket is in circular orbit about the earth at an altitude of h = 4 Mm. Determine the minimum inerement in speed it musi have in order to escape the earths gra vitatkinal

13-117. Prove Kepler's third law cif motion. Flint: Use Eys. 13-19, 13-28, 13-29, and 13-31.

13-118. The satellite is moving in an elliptical orbit with 13 an eccentricity e = 0.25. Determine its speed when it is at its maximum distance A and minimum distance B from the earth.

Prob. 13-118

13-119. The elliptical orbit of a satellite orbiting the earth has an eccentricity of e = 0.45. If the satellite has an altitude Wf 6 Mm at perigee. determine the velocity of the satellite at apogee and the period.•

6 Mm 1 Prob. 13-119

93-120. Determine the constant speed cif satellite S so that it eireles the earth with an orbit of radius r = 15 Mm. Use Eq. 13-1.

Prob. 13-116 Prob. 13-120

1 64CHARTER 13 KINETIC5 OF A PARTICLE: FORCE AND ACCELERATION

13-121. The rocket is in free flight along an elliptical irajectory A A. The planet lias no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure. determine the speed of the rocket when it is at point A.

Prob. 13-121

*L3-124. An elliptica] path of a satellite has an eccentricity e = 0.130. rf it has a speed of 15 Mm/h when it is at perigee, determine its speed when it arrives at apogee, A. Also, how far is it from the earth's surface when it is at A?

Prub. 13-124

13-122. A satellite S travels in a circular orbit around the earth, A rocket is located at the apogee of its elliptical orbit for which e = 0.58. Determine the sudden change in speed that must occur at A so that ihe rocket can enter the satellites orbit while in free flight along the blue elliptical trajectory. When it arrives at B. dełcrmine, the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.

Prob. 13-122

1.3-123. An asteroid is in an elIiptical orbit abouł the sun such that its perihelion distance is 9.30(10v) km. If the eccentricily of the orbit is e = 0.073, determine the aphelion distance of the orbit.

13-125. A satellite is launched with an initial velocity ve = 2500 mi /h parallel to the surface of the earth. Determine the required aliitude (or range cif altiludes) above the earths surface for launching if the free-flight trajectory is to be (a) circular. (h} parabolic, (e) ellipłical, and (d) hyperbolie. Take G = 34.A(10 9)(1h - ft2}/slug2, M, = 409( 1011) slug, the earth's radius re = 3960 mi. and I mi = 5280 ft.

13-126. A probe has a circular orbit around a planet af radius R and mass M.1f the radius of the orbit is nR and the explorer is traveling with a conslant speed vo,determine the angle O at which it lands on the surface of the planet B whenits speed is reduccd towhere k < 1 at point A.

Prob. 13-126

13.7 CENTRAL-FORCE MOT1ON AND SPACE MECHANICS1 b5

13-127. Upon completion of the moon exploration rnission. the command module. which was originally in a circular orbit as shown, is given a boost sa that it eseapes from the moon's gravitational field. Determinc the necessary increase in velocity so that the command module follows a parabolic trajectory. The mass of the moon is 0.01230 Me.

13-130. The satelfite is an elliptical orbit having an eccentricity of e = 0.15. If its velocity al perigee is Vp = 15 Mmih, determine its velocity at apogee A and the period cif the satellite.

15 Mm

*13-128. The rocket is traveling in a free-flight elliptical orbit about the earth such that e = 0.76 and its perigee is 9 Mm as shown. Determine its speed when it is at point B. Also determine the sudden decrease in speed the rocket must experience at A in order to trawi in a circular orbit about the earth.

Prob. 13-128

13-129. A rocket is in circular orbit about the earth at an altitude above the earth's surface of h ,= 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth's gravitational field.

Prob. 13-130

13-131. A rocket is in a free-flight elliptical orbit about the earth such [hal the eccentriciiy of its orbit is e and its perigee is rd. Determine the minimum increment of speed it should have in order to escape the earth's gravitational field when it is at this point along its orbit.

*13-132. The rockei shown is originally in a circular orbit 6 Mm abegsre the surface cif the earth. It is required that it trawl in another circular orbit having an altitude of 14 Mm. To do this. the rocket is given a short pulse of power at A so that it travels in free flight along the grat' elliptical path from the first orbit to the second orbit. Determine the necessary speed it must have at A just after the power pulse, and the lime required to get to the (luter orbit aiong the path AA'. What adjustrnent in speed must be made at A' to maintain the second circular orbit?

Prob. 13-132

Prob. 13-127

' 166 CHAPTER 13 KINETC5 OF A PART IC IE . FORCE AND ACCELERAT ION

CONCEPTUAL PROBLEMS

P13-1. If the hox is released from rest at A. use numerical values to show how you woułd estimate the lime for it to arrive at B. AKo, list the assumptions for your tinalysis.

P 1 3 - 1

P13-2. The tugboat has a known mass and its propeller provides a known maximum thrust. When the tug is fully powered you observe the tirne it takes for the ulg to reach a speed of known value starting from rest. Show how you could determine the mass of the barge. Negleet the drag furce of the water on the tug. Use numerical values to explain your answer.

P 1 3 - 2

P13-3. Determine the smallest speed of each car A and so [bai the passengers do not lose contact with the seat wbite the arms tura at a constant rate. What is the largest normal furce of the seat on each passenger? Use numerical►values to explain your answer.

P13-3

P13-4. Each car is pin connected at ils ends to the urn of the wheel which tunis at a constant speed. Using numerka' values, shaw how to determinc the resultant foree the seat exerts on the passenger located in the top car A. The passengers are seated towards the center of the wheeI. Also. list the assumptions for your analysis.

P:13-4

CHAPTER R PON,' 167

CHAPTER REVIEWKineties

Kinctics is the stucly of the relation between forces and the acceieration they cause, This relatiwi is based on Newton's second law of motion, expressed mathernatically asF -= ma.Before applying the equation of motion, it is important to first draw the particle's free-h ody diagram in order to accounł for alt of the forces that act on the particie. Graphically, this diagram is equal to the kinetic diagram, which shows the result of the forces, that is, the ma vector.

m a

Q

Free-budy.; diagram

Kincticdiagram

Inertial Coardinate Systems

When applying the equation of moltom it is important to measure the acceleration from an inertial coordinate system, This system has axes that do not rotate but are either tixed or translate with a constant velocity. Various types of inertial coordinałe systems can be used to apply F = ma in component form.

a

Paty t.If particie

O i n e r t i a l f r a m e o f r e f e r e n c e

Reclangular x. y, z axes are used to describe = ma,„. = ma,, F, = ma,rectilinear motion along each of the axes.

2.F, -= = maj. 2.Fb = Oaś =- dt) /dł or a, = y dv ł ds

ZF., =—SN = in(rti + 2M)

Z=raż

where p —I c2y ch 21

[1 + (dy/dx)2[3.12

Normal and tangential n, i axes arc often used when the path is known. Recall that an is always directed in the -Hn clirection. it indicates the changc in the velocity direction. Also recall that

is tangent to the path. It intheates the change in the velod ty magnitude.

Cylindrical coordinates are useful when angular motion of the radia) line r is specifiecl or when the path can conventetnly be described with these coordmates.

Central-Farce Mation

When a single furce acts upon a particie, such as during the free-flight irajectory of a satellile in a gravitational field. ihen the motion is referred ło as central-force motion. The orbit depends upon the eccentricity e; and as a result, the trajectory can either be circular, parabolic, elliptical,or hyperbolic,

Chapter 14

As the woman faks, her energy will have te be absorbed by the bungee cord. Thz principles of work and energy can be used to predict the rnotion,

Kinetics of a Particie:Work and Energy

CHAPTER OBJECTIVES

• To develop the principle of work and energy and apply it to solve

problems that involve force, velocity, and displacernent.

• To study problems that invalve power and efficiency.

• To introduce the concept of a conservative force and apply the

theorem of consenration of energy to solve kinetic problems.

14.1 The Work of a Forcefn this chapter, we will analyze motion of a particie using the concepts of work and energy. The resulting equation will be useful for solving problems that invoive forte, veloeity, and dispiacement. Before we do this, however, we must first clefine the work of a force. Specifically, a furce F will do work on a particie Duły when the particie undergoes a ~Mement it2 the direc.ott of the force For example. if the force Fin Fig. 14-1 causes the particie to rnove alang the path s from position r to a new position r',the displacement is then dr r — r. The maguiłude, of dr is ds, thelength cif the differential segment aking the path, If the angle between the tails of dr and Fis B, Fig, 14-1. then We work done by Fis a scaiar quantity, defined by

dU = F ds cos Fig. 14-1

170 CHARTER 14 KINETIC5 OF A PARTICLE: WORK AND ENERGY

Fig. 14-1

By definition of the dat product (see Eq. B-14) this equation can also be writtcn as

dU F • d r

This result may be interpreted in one af twa ways; either as the product af F and the component of displacement ds cos O in the direction of the furce, or as the product of ds and the cumponent of force, F cos 6, in thedirection of dispiacernent. Note that if O° O < 90°, then the forcecomponent and the displacement have the same sense so that the work ispositive,- whcreas if 900 < 1800, these vectors will have oppositesense, and therefore the wark is negarive. Also, dLI -= O if the furce is perpendieular to dispIacement, since cos 900 = 0, or if the furce is appłiecl at a fixeci point, in which case the displacement is zero.

The unit of work in SI units is the joule (J). which is rhe amount of work dane by a one-newton furce when it moves through a distance af one meter in the direction af the force (1 J = 1 N • ni). In the FPS system, work is m easured in units of fout-pounds (ft .1b), which is the work done by a one-pound furce acting through a distance of one foot in the direction of the force.*

Work of a Variable Force. If the particie acted upoili by the fotce F undergoes a finite displacement along iłs path from r, to r-, or s i ta s2, Fig. 14-2a. the work of furce F is determined by integration. Provided F and O can be expressed as a function of position, thcn

r2

= F • dr = F cos O ds ( 1 4 - 1 )

Sometimes. this relation may be obtained by using experimental data to plot a graph af F oos 6 vs. s. Then the aren under this graph bounded by s i

and s, represents the total wark, Fig. 14-2.b.

F c o s

C u s t !

(a) (b)

Fig. 14-2

*By convention, the units rur the murnent uf a foru or torque aro wrinun as lb • ft, tu distinguish thern from łhnse used ło signify work, ft lb.

14.1 THE WORK Or A FORCE 171

s

S

S ,

Fig. 14-3

Work of a Constant Force Moving Along a Straight Line. If the (orce F, has a constant magnitude and acts at a constant angle O front its straight-line path, Fig. 14-3a, then the component of F, in the direction of displacement is always F,. cos O. The work done by F, when the particie is dispłaced from s, to s2 is determined tram Eq. 14-1, in which case

5 2

L = Ft, cas O j ds

a r

1.4~z = F 0(s2 — s (14 - 2)

Herc the wark of F, represents the area of the rectangle in Fig. 14-3b.

Work of a Weight. Consider a particie d weight W, which moves up

alang the path s shown in Fig. 14-4 from position s, to position s,. At

an intermediate point, the dispiacement dr = dxi -1- dyj dzk. SutceW = -Wj, applying Eq. 14-1 we have

Ui -2 -= f F • dr = (-Wj) • (dxi ± dyj dzk)j

r,

= - W dY = W(Y2 yl)J,,

a r

= -WAy (14-3)

Thus, the work is independent of the path and is equal to the magnitude of the particle's weight times its vertical dispIacemen t. In the, case shown in Fig. 14-4 the work is negałive, since Wis downward and Ay is upward. Nołe, however, that if the particie is dispłaced downwarti (— y), the work of the weight is positive. Why? Fig. 14-4

172 CHAPTER 14 KINETrCS OF A PARTICLE= WORK AND ENERGY

Work of a Spring Force. If an dastic spring is elongated a distance ds, Fig. 14-5a, then the work dane by the force that acts on the attached particie is dU = —F",ds = —ks ds. The work is negative since F,acts in the opposite sense to ds. If the particie displaces from s i to thework of F, is then

i s ,UE _2 .= F, ds = : —ks ds

s, i

ul = — — ks; )

(14-4)

This work represents the trapezoidal area under the line F5 = kr, Fig. I.4-5b.

A mistake in sign can be avoiderd when applying this equation if one simply notes the direction of the spring force acting on the particie and compares it with thc sense of direction of dispIacement of thc particieif both are in the same sense, positive work results; if they are opposite to one another, the work is negative.

Unstretehed posili" s = O F,

ds ł

~ M A M ,Force an ParLicle

— ks

s

(a) ( b )

Fig. 14-5

The forces acting on the carl., as ił Ts p ulled a &sumce s up the incline,are shown on its free-body diagram. The constant towing force T does positive work of UT = (T cos 43)s. the weighl does negative work of Uw = —(Wsie 6)5. and the normal force N does no work since there is no displacement of Ibis furce alung its line of actinn.

14.1 THE WORK OF A FORCE 1 7 3

EXAMPLE

The 10-kg block shown in Fig. 14-6a rests on the smooth incline. If the spring is originally s-tretchecl 0.5 m, determine the łotat work done by all the forces acting on the block when a horizontal furce P = 400 N pushes the biock up the piane s = 2 ni.

2 sin m

s = 2 ntInitialposition of spring

= 400

k = 30 Nirri

2 f.,:a.-530° rnSOLUTIONFirst the free-body diagram °f the block is drawn in order to account for al] the forces that act on the block, Fig. 14-6b.

Horizontal Force P. Since this foree is con.stant, the work is determined using Eq. 14-2. The result can be caIculated as the force times the component of displacement in the direction of the furce; i.e.,

Up = 400 N (2 m cos 30°) = 692.8 J

ar the displacement times the component of furce in the direction uf displacement, i.e.,

Spring Force F. In the initial position the spring is stretched = 0.5 ni and in the final position it is stretched sz = 0.5 m + 2 111 = 2.

5 ni. We require the wark to be negative since the farce and displacement are opposite to each uther, 'The work °f is thus

U, = E;(30 Nim)(2.5 m)'1- — 1130 Nim)(0,5 rn)2] = —90 J

Weight W. Since the weight acts in the opposite sense tu its vertical displacement, the work is negative; i.e.,

Uw -= -(98.1 N) (2 m sili 3W) = -98.1 J

Note that it is also possible to consider the component of weight in the direction of displacement; i.e.,

Uw = —(98.1 sin 30° N) (2 m) = —98. I J

Normal Force Na. This tarce does no work since it is Ahvuy+s perpendicular to the displacement.

Total Work. The wark of all the forces when the błock is displaced 2 ni is therefore

UT = 692.8 J — 90 J — 98.1 J = 505 J Ans.

( a )

98_1 N rf

P = 400 N -"11•=

F ,

(b)

Up 400 N cos 30'(2 m) 692.8 J Fig. 14-6

174 CHAPTER 14 KINETC5 OF A PARTICLE: WORK AND ENERGY

14.2 Principle of Wark and Energy

Consider the particie in Fig. 14-7, which is łocated on the path defined relative tv an inertial courdinate system. If the particie, has a mass m and is subjected to a system of external forces represcnted by thc resultantFsthen the equation of motion fur the particie in the tangentialdirection is 2 F, = ma,. Applying the kinematic equation cr = v dv 'ds and integrating both sides, assuming initially Chat the particie has a position s = st and a speed v = , and kater at s = s,, v = vx, we have

S i F, ds ,nap dv

dst

2 FR =

n

Fig. 14-7

S 2

f F, ds = -niv.:', - ,L2mtii (14-5)

, From Fig. 14-7, note Chat "XF, = .k-cos O, and since work is

defined

from Eq. 14-1, the final result can be written as

S U1_2 = z rny - (14-6)

This equation represents the principle of work and energy for the particie. The term on the left is the sum of the work done by ali the forces acting on the particie as the particie moves from point l to point 2.The two termson the right side, which are of the form T = define the particie'sfinal and initial kinerze enrrgy,respectively. Like work, kinetic energy is a scałar and has units of joules (.1.) and ft • lb. However, unlike work, which can be either positive or negative, the kinetic energy is always posWue, regardless of the directiun uf motiun of the particie.

When Eq. 14-6 is appłied,it is often expressed in the form

+ (14-7)

which states that the particle's initial kinetic energy plus the work done by ali the forces acting un the particie as it moves kum its initial to its final position is cqual to thc particic's final kinetic energy.

As notcd kum thc derivation, the principle of work and energy represents an integrated fonu of F, = mar, obtained by using the kinematic equation a, = v dv/ds. As a result. this principle will provide a convenient subsł kation for F, = ma, when solving those types of kinetic problems which involve furce, vełocity, and displacernent since these quantities are involved in Eq, 14-7, FOT application, it is suggested Chat the following procedure be used.

14.2 PRINCIPLE CF WORK AND ENERGY175

Wark (Free-Body Diagram).

• Estabiish the inertial coorclinate system and draw a free-body diagram of the particie in order to account for ali the forces that do work on the particie as -ił =yes along its path_

Prindple of Wark and Energy.•Appty the prineiple of work and energy, T[

•The kinetic energy at the initial and finał points is always positive, since it invalves the speed squa red (T = zrnvg.

• A force does work when it rnoves through a dispiacement in the direction of the fotce.

• Wark is posirive when the forcc component is in the same sense oj' direction as its displacemen t, otherwise it is negative.

• Forces that are functions of clsplacement must be inłegrated tn ohtain the work_ Graphically, the work is equal to the area under the forcc-displacement curve.

•The wark of a weight is the produeł of the weight magnitude and the vertical displacement. UW = ± Wy. Ił is positive when the weight tnoves dawnwards.

•The work of a spring is of the form U,r = -;ks2, where k is the spring stiffness and s is the stretch vr compression of the spring.

Numerical application of this procedure is illustrated in the examples fo3łowing Sec. 14.3.

If an oncoming car strikes these crash barrels, the ears kinetic energy win be transforrned into wark, which cause,s Lhe barrels. and Lo some extent the car, to be deffirmed. By knowMg the amounl of energy thatcan be absorbed by each barrel it is possible to design a crash cushion such as this.

176 C H A P T E R 14 K I N E T C 5 O R A P A R T I C L E : W O R K A N D E N E R G Y

14.3 Principle of Work and Energy for a System of Particles

The principle of work and energy can be extended to include a system of particles isolated within an enclosed region of space as shown in Fig.14-8.Here the arbitrary ith particie, having a mass subjected to a resultantexternal force F; and a resultant internal furce which all the other particles exerł on the ith particie, 1f we apply the principle of work and energy to this and cach of the ather particles in the system, thcn since work and energy are scalar quantitics, the cquations can be summcd algebraically, which gives

£T1 +=(14-8)

In this case, the initial kinetic energy of the system plus the work dane by all the exterrial and internal forccs acting an the system is equal to the final kinetic energy af the system.

If the system represents a translating rigid body, ara series of connected translating bodies, then ali the particles in each body will undergo the same dpliKetn111. Therefore, the work af all the internal forces will occur in equal but opposite collinear pairs and so it will cancel aut. On the other hand, if the body is assumecl to be nonngid, the particles of the body may be displaced along differenł paths, and some uf the energy dugi to force interactions would be given off and lost as heat ar stored in the body if permanem deformations occur. Wc will discuss these effeets briefly at the end of this section and in Sec, 15.4. Throughout this łext, however, the principle of work and energy will be applied to problerns where direct accountability af such energy losses does not have to be considered.

incrtial coordinate system

Fig, 14-8

14.3 PRtNCIRIE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES177

Work of Friction Caused by Sliding. A special class of prob I erns will now be investigated which requires a careful application of Eq, 14-8. These probierns involve cases where a body slides over the surface of another body in the presence of friction. Consider, for example, a block which is translating a distance s over a rough surface as shown in Fig. 14-9a. If the applied furce P just balances the resuhanł frictional force ,Lt.k.N, Fig. 14-9b, then due to equilibrium a constant velocity v is malntained, and one would expect Eq. 14-8 to be applied as fol lows;

, MV - -r r S - gkiVS = n i V -1-

Indeed this equałion is satisfied if P = puki+ however, as one realizes from experience, the shding rnotion will generate heat, a form of energy which seerns not to be accounted for in the work-energy equation. In order to explain this paradox and thereby more closely represent the nature of fr-

iction, we should actually model the biock su that the surfaces of contact are deformable (nonrigid).* Recall that the rough portions at the bottom of the block act as -teeth," and whcn the block słides these teeth deform slighrly and either brcak off ar vibratc as they pull away from "treth" at the contacting surface, Fig. l 4-9c. As a result, frictional forces that ad on the błock at these poinłs are clisplaced slightiy, due to the localized deformations, and ]afer they are replaced by other frictional forces as other points of contact are made. At any instant, the resuhanł F of aII these frictional forces remains essentially constant,i.e., i4N; however, dur to the m uny loealized deforrrtations, the actual displacernent s' of p.kN is not the same as the displaccm cni s of the applied fotce P, Instead, s' will be less than s (s' < s), and therefore the external work done by the resultant frictional furce will be I.L4Nrsi and not AkArs. The remaining amount of work. gki\r(s - s"), manifests itself as an increase in interna, energy, which in fact causes the block's temperature W rise.

In summary then, Eq. 14-8 can be applied to prohlems involving sliding friction: however, it shoułd be fully realized that the work of the resułtant frictional furce is not represented by gkivs; instead. this term represents both the external work of friction (NNs") and interna) work[p4Ms - which is converted into various forms of internat energy,such as heat.t

*See Chapter of Engineering A.fechcrnics: Sunics.I-Set: B. A. Sherwoud and W. H. Bernard, "Wark and Neat Transfer in the- Prsencu of

SlidingAin. J. Phys, 52.1001 (1984).

_ffi-11,- P—11. P

(a)

F = g k N

(b )

178 CHAPTER 14 KINETICS OF A PARTICLE: WORK AND ENERGY

E X A M P L E

20 ft ls The 3500-lb automobile shown in Fig. .14-10a travels down the 10' inclined road at a speed of 20 ft/s, If the driver jama on the brakes. causing bis wheds to lock, determine how far s the tires skid on the rond, The coeffiejent ()f kinetie friction between the wheels and therond is = 0.5.

(a )

SOLUTIONThis problem can be solved using the principIe of work and energy, since it involves force, velocity, and displacement.

3500 lbWork (Free-Body Diagram). As shown in Fig_ 14-10b. the normal' force NA does no work since it never undergoes displacement along. its

FA line of action. The weight, 3500 lb, is displaced s sin 10° and doesiti° ' positive work. Why? The frictional force FA does both external and

( b ) NA interna' work when it undergoes a displacement s. This work is negative

since it is in the opposite sense af direction to the dispIacement. Applying the equation of equilibrium normal to the rond, we have.

Fig. 14-10 +NF„ = 0; — 3500 cos lb = O NA = 3446.8 Ib

Thus.F ▪ —= 0.5 (3446.8 Ib) = 1723.4 Ib

Principle M Work and Energy.

Tj +-=1 ( 3500 lb▪ (20 ft/s)2 -I- 3500 lb(s sin W') — (1723.4 113)s = O2 \,32.2 ft/s-

Solving for s yields

s = 19.5 ft Ans.

NOTE If this problem is solved by using the equation of motion, rwo stopa are involved. First. from the free-body diagram. Fig. 14—1 Gb. the equation of motk' is applied along the incline. This yields

=- ma i;3500 sin 10° lb — 1723.4 Ib — 3500 lb,

32.2 flis-

a —10.3 ft/s'

Then, since a is constant, we have

-Fk?") 9,2 = .2),2) 2a,„(s so);(0)2 = (20 fil/s)2 2(.— 10.3 fti&-2)(5 — 0)

s = 19.5 ft

14.3 PRFNCIPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES1 7 9

14.3

For a shorł time the crane in Fig. 14-11a lifts the 2.50.-Mg beatu with a force of F = (28 + 3s2) kN, Determine the speed of the beatu when it has risen s = 3 m. Also, how much time does it take to attain this height starting from rest?

SOLUTIONWe can solve part of this problem using the principłe of work and energy since it involves force, velocity, and displac.ement. Kinematics must be used to derermine the time. Note that at c = D.F = 28(103)N W = 2.50(103)(9.81)N, so motian will occur.

Work (Free-Body Diagram). As show].) on the free-body diagram, Fig. 14-11b, the lifting force F does positive work, which must be determincd by integration since this force is a variable. Also, the weight is constant and will do negativc work since the displacement is upwards.

4a)

Principles of Work and Energy.

+= T2

O + f(28 + 3s2)(103) ds - (2.50)(103)(9.81)s = (2.50)(103)v2 o28(103)s + (103)s3 - 24.525(143)s = 1.25( I 03)1)2

= (2.78s + 0.8s3)1 (1)

When s = 3 m,

v = 5.47 mis

Kinematics. S incc wc were ab le to express the ve łoc i ty as a fune t ion of d i sp lacement , the t ime can be de te rmined us ing

= ds/d1. In this case,

(2.78s,

+ 0.8s-')7 =ds

=./({1

ds

(2.78s + 0.8,s3)

The integration can be performed nurncrically using a pocket calculator.The result is

= 1/9 s Ans.

NCTE: The accełeration af the beatu can be determined by integratingE. (1) using v ch, d ds, or more directly, by applying the equatianof muflon. I F = ma.

2.50 ( 03)(9 81) N(b)

Fig. 1441

180 CHA.PrEp 14 KINETIC5 OF A PARTICLE: WORK ANO EMERGY

EXAMPLE

The platform P. shown in Fig. 14-12a, has negligib]e mass and is tied down so that the 0.4-m-łong cords keep a l-m-long spring compressed 0.6 rn when nothing is on the platform. H a 2-kg błock is placed on the platform and released from rest after the platform is pushed down 0,1 ni, Fig, l 4-12b. &termine the maximum height h the block rises in the air. measured from the ground.

0.3 m

h

(a) (b)Fig. 14-12

19.62 N

I F ,

( c )

SOLLITION

Work (Free-Body Diagram). Since the block is released from rest and later reaches its maximum height, the initial and final velocities arc zero.The free-body diagram of the block when it is stil) in contact with the platform is shown in Fig. 14-12c. Note that the weight does negative work and the spring force does positive work. Why? In particular, the initial cotn p ress io n in the spring is si = 0.6 ni ÷ 0.1 m = 0.7 m. Due to thc cords. the spring's fina? cornpression is = 0.6 m (after the block Ieaves the platform). The bottom of the block rises from a height of (0.4 m - 0.1 m) = 0.3 m to a fina] height h.

Principle of Work and Energy.

+ 1.U1, = T2

4inVf - (J.2' k -;k - W dy}

Nate that here s[ =- 0.7 m = 0.6 m and so thc werk of the, springas determined from Eq. 14-4 will indeed be positive once the caleuIation is made.Thus.

o + - [4-(200 Nim)(0.6 rri) - ;(2(1t) Nim)(0.7 ni)'

- (19.62 N)[h - (0.3 m)] } = 0

Solvin g yields

0.963 rn Ans.

14.3 PRłNCiPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES1 8 1

14.5

The 40-kg boy in Fig. 14-13a slides down the smooth water slide. If he starta fron-i rest at A, determine his speed when he reaches B and the norma' reaetion the slide exerts un the boy at ths positiun.

( a )

SOLUTION

Work (Free-Body Diagram), As shown on the free-body diagram, Fg.14-13b, there are two forces acting on the boy as he goes down the slide. Note that the normal furce does no work.


+ = T8(1 + (40(9.81)N) (7.5 m) = kg)/)2

8

vR = 12.13 m/s = 12.1 m/s

Equation of Motion. Referring to the free-body diagram of the boy when be is at B, Fig. 14-13c, the normal reaetion NR can now bc obtained by applying the equation af motion along the n axis. Here the radius of eurvature of the path is

1 + (dy )213/2.dx) [1 + (0.15x)23/2

PB —= 6.667 md2Y/621(1.15]

Thus.

n40(9_81) N

Nb

(b)

ni

4.0(9.81) N .-

----..., kk

NR

( c )

((12.131111/ 5)2)NB — 40(9.81) N = 40 kg

6.667 m )

NB = 12753 N = 1.28 kN Ans.

+ F „ = m a p ;Fig. 14-13

1 8 2 CHAPTER 14 KINETICS OF A PARTICLE: WORK AND ENERGY

Da t=

10 kg

( a )

R , R z

Fig. 14-14

Blocks A and B shown in Fig. 14-144 have a mass of 10 kg and 100 kg, respectively.Determine the distance B travels when it is released from test to the point where its spec(' becomes 2 m/s.

SOLUTIONThis problem may be solvcd by considering the blocks separate1y and applying the principle of work and energy to each błock. However, the work of the (unknown) cable tension can be eliminated from the analysis by considering blocks A and B together as a singie system.

Work (Free-Body Diagram). As shawn on the free-body diagram of the system, Fig. 14-14b, the cable furce T and reacłions R1 and R, do no work, since these forces represent the reactions at the supports and consequcntly they do not move while the blocks arc displaced, The wcights both do positivc work if we assume both nnove downward, in the positłve sense of direction af sA and sB.

Principle of Work and Energy. Realizing the blocks are released from rest, we have

£ T i £ L / " _ 2 1 7 " ,

{ 4'nA(vs)7 + rriB(vB)i} {WA ASA + Wg Asu} =

ł3m,,(vA);-. + rnii(vh)

{o + o} + 198.1 N i AsA) + 981 N(A,58)

ł .1(10 kg)(vA)'22 + 4(100 kg)(2 rnis)2}(1)

Kinematics. Using rnethods of kinematics, as discussed in Sec, 12.9, it may be seen from Fig. 14-14a that the tołal length/of all the vertical segments of cable may be expressed in termy of the position coordinates SA and sa as

sA + 4s8 =

Hence, a change in position yields the displacement equationasA + 4 ASE, = O As

Ą = —4 As8

Here we see that a downward displacement of one block produces an upward displacement of the other block. Note that As,4 and Asg musi have the same sign convention in butli Eqs. 1 and 2. Taking the time derivative yiclds

vA = -4vB = -4(2 m/s) = -8 ni/s (2)

Rełaining the negative sign in Eq. 2 and substituting into Eq. 1 yields

As B -= 0.883 m J, A

18314_3 PRNIC1PLE OF WORK ANO ENERGY FOR A SYSTEM OF PARTICLES

■FUNDAMENTAL PROBLEMS

50

20

F14-1. The spring is placed between the wali and the 10-kg block. If the block śs subjected to a force of F 500 N,determine its velocity when s = 0.5 m. When s = 0. the blask is al rest and lhe spring is uncompressed. The conlacl surface is srnooth.

500 N

F14-1

F14-2. If the motor exerts a constant fotce of 3001 on the cable, determine the speed of the 20-kg crate when 11 travels s = 10 m up the piane, starting from rest. The coefficient ofkinetśc friction between the crate and the piane is /../k0.3,

F14-2

F14-3. If the motor exerls a forte of = (600 + 2,s21 N on the cable. determine the speed of the 100-kg cratewhen it rises to x 15 m. The crate is initially at rest onthe groun d.

F14-4. The 1.8-Mg dragster is traveling at 125 m/s when the e..ngine is shut off and the parachute is rcicased_ If the drag force of the parachute can be approximated by the graph. deleTmine the speed of the dragster when h, has traveled 400 m.

s(m)F14-4

F14-5. When s = 0.0 m, the spring is unstretehed and the 10-kg Hack has a speed of 5 m/s down the smoolth piane. Determine the distance s when the block stops.

F14-5

F14-6. The 5-lb collar i5 puIled by a cord that passes around a smalt peg at C_ If the card is subjected to a constanł forte of F = I0 Jb, and the collar is at rest when it is at A, determine its speed when it reaches E. Neglect friction,

4 fl

3 ft

F14-6

t c

CF14-3 F= 101b

F0 (kN)

400

200 N /*I

F= 100 N

3t7 f

84CHAPTER 14 KINMCS OF A PARTICLE: WORK AND ENERGY

• PROBLEMS

14-1. The 20-kg crate is subjected to a furce having constant direction and a magniłude F = lt 0 N. When s = 15 m, the crate is moving to the right with a speed of 8 m/s. Determine its speed when s = 25 m. The coefficienł of kinetic friction between the crate and thc ground is

= 0.25.

Prob. 14-1

14-2, For protection. the barrel harrier is place(' in front of the bridge pier. H the rdation hetween the [orce and deflection of the barrier is F= (90(103)x1r2) lb, where x is in ft, determine the car's maximum penetration in the barrier. The car lias a weight of 4000 lb and it is traveling with a speed of 75 ft/s jus( before ii hits the barder.

F (Ib)

F= 90(10)3 x1t2

(ft)

*-14-4. 'Me 2-kg block is subjected to a furce having a constant direction and a magnitude F = (300/(1 + s)) N, where s is in meters. When s = 4 m, the block is moving to the left with a speed of 8 m/s, Determine its speed when s = 12 m. The coefficient of kinetic friction between the błock and the ground is1.tk = 0.25.

Prob. 14-4

14-5. When a 7-kg projedne is fired from a cannon harrel that lias a length of 2 m, the exp1osive furce exerted on the projedne. while it is in the harrel, varies in the manner shown. Determine the approximate muzzie velocity of the projedne at the instant it leaves the barrel. Negiect the effeets of friction inside the barrel wid assume the harrel is hori7ontal,

F (MN)

0.2 04 06 0.8 1.0 1.2 14 1.6 18 2.0 s (m)

11111111••••••11•EENEE•111111MINIVIE• EEE111r

15

Prob. L4-5

Prob. 14-2

14-3. The crate, w kich hasa mass of 100 kg, is suhjected to the action of the two forces, 1f it is originally at Test, determine the distance it slides in order to attain a speed of 6 m/s, The coefficient of kinetic friction between thc crate and the surface is µk = 0.2-

Prob. 14-3

14-6. The spring in the boy gun has an unstreiched Iength of 100 mm. Ii is cornpressed and locked in the position shown. When thc trigger is pulled, the spring unsłretches 123 mm, and the 20-g hall moves along the harrel. Determine the speed of the bali when it leaves the gun.Negled friction.

50 mmk = 2 kN/m--,-D

150 ‚im) —

Prob. 14-6

14.3 PRUNICIPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES1 85

14-7. As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg błock which rests on the smouth surface and is subjected to a horizontal lorce of 6 N. If observer A is in a fixed france x, determine the final speed of the block if it has an initial speed of 5 m/s and travels 10 m, both directed to the right and measured frant the flxed frame. Compare the result with that obtained by an observer B. attached to the x axis and rnoving at a constant velocity of 2 m/s relative to A. flint The distance the block travels will first have to be cornputed for observer B before applying the principle of work and energy.

Prob. 14-7

If the 50-kg crate is subSected to a forte of P= 200 N. determine its speed when it has traveled 15 m starting from rest.The coefficient of kinetic friction between the crate and the ground is pp, = 4,3,

14-9. If the 50-kg crate starts from rest and anains Ft speed of 6 m/s when it bas traveled a distance of 15 m, determine the lorce P acting on the crate. The coefficient nf kinetic friction between the crate and the groundispk= 0.3.

14-10, The 2-Mg car has a velocity of v, = 100km/h when the driver sees an obstacle in front of the car. If it taks 0/5 s for him to react and lock the brakes, causing the car to skid, determine the distance the car travels before it stops. The coefficient of kinetic triction between the tires and the ruad is dek = 0.25.

14-11. The 2-Mg car has a velocity of v, = 100krn/h when the driver sees an obstacic in front of the car. It takes 0.75 s for him to react and Iock the brakes_causiing the car to skid. If the car stops when it has iraveled a distance of 175 m, determine the coefficient of kinetic friction between the tires and the rond.

= 10-0 km,/h

Probs. 14-10111

*14-12. Design considerations far the bumper B on the 5-Mg train car require use of a nonlinear spring having the load-cleflection characteristics shown in the graph. Select the proper value of k sa that the maximum deflection of che spring is Iimited to 0,2. m when the car. traveling at 4 m/s, strikes the ńgid stop. Neglect the mass of the car wheels.

A•B•2

mis

Srus

x`

6 N

1 0 r n -

F(N)F ks2

s(m)

1 4

8

tPa-

Prob,, 14-819 Prob. 14-12

'1 86 CHAPTER 14 KINETC5 OF A PARTICLE: WORK AND ENERGY

14-13. The 2-lb hrick slides clown a srroołh roof, such that when it is at A it hasa vełocity of 5 ft/s. Deiermine the speed of the brick j ust before it leaves the surface at B, the distance d from the wali to where it strikes the ground, and the speed at which it his the ground,

14-15. The crash cushion for a highway barrier consists of a nest of barrels filled with an impact-abaorbing material. The barrier stopping furce is measured versus the vehicle penetration into the barrier. Determine the distance a car having a weight ot 4000 lb will penetrate the barrier if it is odginany traveling at 55 ft is when it strikes the first barrel.

V

Prob. 14-13

14-14. if the cord is subjected to a constarit kirce of F= 300 N and łhe 15-kg smooth collar starts from rest at A, deterrnine the vekicity of the collar when it reaches point B. Neglect the size of the putley.

Prob. 14-14

*14-16. Determine the wiochy of the 60-1h block A if the twa bbeks aro released from rest and the 40-1b hlnek B moves 2 ft up the incline. The coefficient of kinetic frictionbetween both blocks and the inclined pattes is = 0.10.

Prob. 14-16

Barrier ~pping forte {kip)

9

i t

5 10 15 20 25Vehide penetration (ft)

Prob. 14-15

14.3 PRINCiPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES187

14-17. If the cord is subjected to a constant force of F= 30 lb and the smooth 10-lb collar staris from resi at A. determine its speed when it passes point B. NegIect the size of pulley C.

Prob. 1.4-17

14-18. The twa blocks A and B have weights WA = 60 Iband Wp 10 Ib. H the kinetic eoefficient of friction betweenthe incline and block A is lut = 0.2, determine the speed of A after it rnoves 3 ft down the piane starting from Fest. Neglect the mass of the cord and pulleys.

14-19. H the 10-1h No& passes point A on the smooth truck with a speed of vA = 5 ftis, determine the normat reactiOn on the block when it reaches point B.

Prob. 14-19

*14-20. The steel ingot bas a mass cif 1800 kg. It travels along the conveyor at a speed = 0.5 m/s when it collides with the "nested" spring assembly. Delermine the maximum deflection in each spring needed to stop the motion of the ingot. Take k„, = 5 kN/rn, k = 3 kNim.

14-21. The steel ingot has a mass of 1800 kg. ft travels along the conveyor at a speed v = 0.5 m/s when it collides with the "nested" spring assernbly. if the stiffness cif the outer spring is kA = 5 kNint, determine the required stiffness kB of the inner spring so ihai the motion of the ingot is stopped at the moment the front. C, of the ingot is 0.3 m from the wali.

Prob. 14-18 Probs. 14-20121

188 CHARTER 14 KrniunCS oF A PARTICLE: WORK AND ENERGY

14-22. The 25-1b block has an iriitial speed of tac = 10 ftis when it is midway between springs A wid B. Atter striking spring B. it rebounds and slides across ihe horizontaI piane toward spring A. etc. If the coefficient of kinetic friction between the piane and thc błock is p4, = 0.4, determine the tota' distance traveled by the block before it comes to rest.

14-25. The skier starts frorn rest at A and travels down the ramp. If friction and air resistance can be negłected. determine bis speed ?in when he reaches B. Also, find the distance s to where he strikes the ground at C. if he makes the jump traveIing horizontalty at B. Neglect thc skicr's size. He has a mass of 70 kg.

2 11i ft i k = 60 113/in.

- 10 ft is Rommo

kA = 10 113/in.

A

B

Prob. 14-22

14-23. The tramu car has a mass of 10 Mg and is Traveling at 5 mis when it reaches A. If che rolling resistance is 1/100 of the weight of the car, determine the cornpression of each spring when the car is momentariły brougkl to rest.

300 kNim30 m—= 5 rnfs500

4 0 , 1 0 1 A ~ g i l

Prob. 14-23

*14-24. The 0.5-kg bali is fired up the smooth vertical circular track using the spring ptunger. The ptunger keeps łhe spring compressed 0.08 m when s = 0. Determine how far s it musi be pulled back and released so that the bali will begin to leave the track when N --- 135°,

k = 500N/m

Prob, 14-24

3 [ 1 P

Proh. 14-25

14-26. The catapulting mechanism is used to propel the 10-kg lider A to the right along the smooth track. The propelling action is obtained by drawing the pulley attached ło rod BC rapidly to che lat by =ans of a piston P. If thc piston appties a constant fotce F = 20 kN to rod tqC such that il moves it 0.2 rn, determine the speed attained by the slider if it was originally at rest. Neglect the masa of the pulleys, cable, piston. and rod BC.

Prob. 14-26

14_3 PRINCiPLE OF WORK AND ENERGY FOR A SYSTEM OF PARTICLES189

14-27. Block A bas a weight of 60 Ib and błock B lias a weight of 10 lb. Deiermine the distance A musi descend from rest before il obtains a speed of 8 ft/s.Also.what is the tension in the cord supporting błock A? Negiect the mass of the cord and pulleys.

Prob. 14-27

14-28. The cyclist travels to polni A, pedaling umil be reaches a speed vA = 4 m/s. He then coasts freely up the curved surface. Determine how high he reaches up the surface before he comes to a stop. Also, what are the resultant norma' furce on the surface at this point and his acceleration? The total mass d the bike and man is 75 kg. Negłect friction, the mass of the wheels, and the sine of the bicycle.

14-29. The collar has a mass of 20 kg and slides along the srnooth rod. W° springs are attached to it and the ends of the rod as shown. If each spring has dn unconipressed łength of I m and the collar has a speed of 2 m/s when s = 0. determine thc maximum compression of each spring due to the back-and-forth (oscillating) motion of the collar.

(125 m

Prob. 14-29

14-30. The 30-lb box A is released from rest and slides down along thc smooth ramp and anto the surface of a cart, If the cart isprevented from moving, determlne the distance s from the end of the cart to where the box stops. The coefficient of kinetic friction behveen the carl and the box łs Fik = 0.6.

10 ft4 f t

C

Prob. 14-30

14-31. Nfarbles having a mass of 5 g are dropped from Test at A through the smooth glass Lube and accumulate in the can at C. Determine the placement R of the can from the end of the tube and the speed at which the marbles fali int° the can. Negłeci the size of the can.

• A

2 m

R

Prob. 14-31

3 m

B r

4 in

Prob. 14-28

19d CHAPTER 14 KrNETLC5 ❑F A PARTICLE: WARK AND ENERGY

*14-32. The cyclist travels to point A, pedaling until he reaehes a speed vy = 8 m/s. He then coasts freely up the cnrved surface. Determine the normal furce he exerts on che surface when he reaches point H. The total mass of the bike and man is 75 kg. Neglect frictinn. the mass of the wheels_ and the size of che bicycle.

1.4-34. The spring humper is used to arrest the rnotion of the 4-lb block, which is sliding toward ii at zr = 9 ft/s. As shown_ Ehe spring is confined by the place P and wali using cables so that its length is 1.5 ft. If the stiffness of the spring isk = 501bjfl, cletermine the required unstretched length of the spring so that the place is not displaccd more than 0.2 ft after thc blacik colIides into it. Neglect friction, the mass of che place and spring, and the energy lass between the plate and block during the collision.

1-1.5 ft.

5 ft.

4 in

Prob, 14-32

14-33. The man ai the winclow A wishes to throw the 30-kg sack on the ground. To do this be allnws it to swing from rest at B tu point C, when he releases the cord at O -= 30°. Determine the speed at which it strikes the ground and the distance R.

P rob. 14-34

14-35. The collar lias a mass of 20 kg and is supportecl on the sMOOth rod_ The attached springs arc undcformed when d = 0.5 m. Determine the speed of the collar after (be applied furce F = 100 N causes it to be displaced sn that d = 0.3 m. When d = 0.5 m thc collar is at rest.

R

Prob, 14-33

P rob. 14-35

14_3 PRINCiPLE OF WORK AND ENERGY FOR A SYSTEM er PAR-nem 191

*14-36. 1f the fotce exerted by the motor M on the, cable is 250 N, determine the speed of the 100-kg crate when it is hoisted to s = 3 m. The crate is at rest when s = O.

Prob. 14-36

14-31. rf the track is to be designed so that the passengers of the roller coaster do not experience a FIC)1111Z1 [Me equal to zero ar more than 4 times their weight, determine the limiting heights 11,, and lic so that this does not ()ccm The ratler coaster starts from rest at position A. Neglect friction.

Pc = 20

. 4 w . p R = - iff0.21 5 : n 41 — i r A\ A 4> 0139 ~ w w :

0":4504111W ia.11,1» Einolik~

Prob. 14-37

14-38. The 150-lb skaterpasses point A with a speed of 6 ft/s, determine his speed when he reaches point B and the nonnal [orce exerted on bim by the track at this point. Neglect friction.

Y

x

Prob. 14-38

14-39. The 8-kg cylinder A and 3-kg cylinder B a.re released from rest, Detertnine the speed of A after it bas moved 2 m storting from Test. Neglect the mass of the cord and pulleys.

*14-40. Cylinder A lias a mass of 3 kg and cylinder B has a mass of 8 kg. Determme the speed of A atter it bas moved 2 m storting from rest. Neglect the mass of the cord and pulleys.

Prohs. 14-39/40

14-41. A 2-1b błock rests on the smooth sernicylindrical surface. An elastic card having a stiffness k =- 2113 i fr. is at łached to the błock at Band to the bace of the scmicylindcr al point C. If the block is reIeased Eram rest al A (f-/ = determine the unstreiched length of the cord so the blockbcgins to leave the semicylindcr at the instant = 45°,Neglect the size of the block.

Prob. 14-41

1 9 2 CHAPTEP 14 KNEr ic s oF A P A R T I C L E : W A R K A N O E N E R G Y

14.4 Power and Eff ic iency

Power. The term "power" provides a useful basis for chosing the type of motor or machine which is required to do a certain amount of work in a given time. For example, two pumps may each be abie to empty a reservoir if given enough time; however, the pump having the larger powcr will complete the job sooner.

The power generated by a machine or engine that perforins an amount of work dU within the time interval dt is therefore

The power output of Ibis łocomotive comes from the driving fricłional [orce F dcvełoped atitswheelsillis this tarce that overcorries the frictionai resistance of the cars in tow and is able to lift the weight of the train up the grudo.

"If the work dU is expressed as

P =- dUdr

OT

dUP =

d,

then

dr— F --

dr

(14-9)

(14-10)

dU = F -

dr, _ dr

dr

P = F•vi

Hencc. power is a sca far. where in this formulation v rcprescnts the velociły °f the particie which is acted upon by the force F.

The basic units of power used in the ST and FPS systems aro the watt (W) and harsepower (hp), respectively, These units aro defined as

1 W = I f -= 1 N • tri/s

I hp -= 550 ft • lhis

For conversion between the Iwo systems of units, 1 hp =. 746 W.

Efficiency. The mechartical efficlency of a machine is defined as the ratio of the output of useful power produced by the machine to the input of power supplied to the machine. Hence,

power outpute

power input

14.4 POWER AND EFFICIENCY193

If energy supplied to the machine occurs during the same dme Mrerval at which it is drawn, then the efficiency may also be expressed in terms of the ratio

eriergy QII lpuT1E

Cr1Crgy jupll(14 -12 )

Since machines consist of a senes of m oving parts, frictional forces will always be developcd within the machine, and as a result, cxtra encrgy or power is ncedcd to overcome thcsc forces. Consequently, power output will be less than power input and so the effieiency of a machine is always less than 1.

The power supplied to a body can be cletermined using the folIowing procedure.


• First determine the externał force F acting on the body which causes the motion, This furce is usually developed by a machine or engine placed either within or external to the body.

•If the body is accelerating, it may be necessary to draw its free-body diagram and apply the equation of motion= ma) to

determine F.

•Once F and the velocity y of the particie where F is applied have been found, the power is deten-nined by multipiying the furce magnitude with the component of Yclociły acting in the direction

of F. (i.e., P F•v = Fv cos ‚Y).

• In some problems the power may be found by calculating the work done by F per unit of time (P„,z = d U f 40.

The power requiremenis of fhis erevator depend opon the vertical force F that acts on the elevator and eauses it to move upwards. If the veloei fy of the elevalor is v. 'hen Ehe power ()lapo' is P = F • v.

194 CHARTER 14 KINETIC5 OF A PARTICLE: WARK ANO EMERGY

The man in Fig. 14-15a pushes on the 50-kg erate with a fot-w of F = 150 N. Deter-mine the power supplied by the man when s = 4 s,

The coefficient of kinetie frietion between the floor and the crate is = 0.2. Initially the creałe is at rest.

F= ISON

41.

(a)

F1 =02 N"

(b)

2

Fig. 14-15

SOLUTIONTo determine the power developed by thc man, the velocity of thc 150-N foree must be obtained first.The free-body diagram of the crate is shown in Fig. 14-15b. Applying the equation of motion,

= ma,,; N - (n 150 N - 50(9.81) N =

N= 5805 N

_> Fx = ina,; (1)150 N - 0.2(580.5 N) = (50 kg)a

a = 0.078 m/s2

The veloei ty of thc crate when r = 4 s is therefore

v = vo

= Q+ (0.078 mis2)(4 s) = 0.312 mis

The power supplied to the crate by the man when r = 4 s is therefore

P = F•v = Fxv = )(150 N)(0.3 ł 2 mis)

= 37.4 W Ans,


EXAMPLE

The motor M of the hoist shown in Fig. 14-16a lifts the 75-lb crate C so that the acceleration of point P is 4 ft/s2, Determine the power that must be supplicd to the motor at the instant P has a velocity of 2 ft/s. Neglect the mass of the pulley and cable and tako s = 0.85.

D t u r r i

Datum

SOLUTIONIn order to find the power output of the motor, it is first necessary to determine the tension in the cable since this forte is cleveloped by the motor.

From the free-body diagram. Fig:14-16k we have

751b+ -= ma •Y' -2T + 75 lb -322 fL/5241'

The acccIeration of the cratc can be obtained by u.sing kinematics to reiate it to thc known acceleration of point P, Fig. 14-16a. Using thc methods °f absolute dependent motion, the coordinates sc and sp can be related to a constant portion of cable length 1 which is changing in the vertical and horizontal directions. We have 2sc + sp = I. Taking the second tirne derivative of this equation yields

2ac = -ap (2)Sincc ap = +4 ft/s2, then ac = -(4 ft/s2)/2 =- -2 ft/s2. What docs the negative sign indicatc? Substituting this result lotu Eq. 1 and retaining the negative sign since the acceleration in bołh Eq. 1 and Eq. 2 was considered positive downward, we have

S c

( 3 )

( (b)7 5 l b

-2T + 75 lb = )(-2 ft/s2)32.2 ft/s- Fig 14-16

T = 39..83 lb

The power output, measured in units of horsepower,required to drawthe cable in at a rata of 2 flis is therefore

P ,= T - y = (39.83 lb)(2 ft/s)[ I hp/(550 ft • lb/s)]

= 0.1448 hp

This power output requires that the motor provide a power input of

power input = -s (power output)

= 0.85 (0.1448 hp) = 0.170 hp

NOTE: Since the velocity cif the crate is constantly changing, thepower requirement is instanianeous.

1 9 6 C H A R T E R 1 4 K I N E T I C 5 O F A P A R T I C L E : W A R K A N O E M E R G YB FUNDA1VIENTAL PROBLEMS •F 1 4 - 7 . I f t h e c o n i a c t s u r f a c e b e t w e e n t h e 2 0 - k g N o & a n d

t h e g r o u n d i s s m o o t h , d e t e r m i n e t h e p o w e r o f f o r c e F w h e n =

4 5 . I n i t l a l l y , t h c H o & i s a t r e s t ,

F = 3 0 N

F14-7

F14-8. If F = s') N. where s is in metets. and thec o n t a c t s u r f a c e b e t w e e n t h e b ł o c k a n d t h e g r o u n d i s

s m o o t h . d e t e r m i n e t h e p o w e r o f f o r c e F w h e n s 5 m . W h e n

s = 0 , t h e 2 0 - k g b l o c k i s r n o v i n g a t u = I m / s .

F 1 4 - 1 0 . T h e c o e f f i c i e n t o f k i n e t i c f f l e t i o n b e t w e e n t h e

2 0 - k g b l o c k a n d t h c i n c i i n e d p i a n e i s N . k = 0 . 2 . I f t h e b l o c k

i s t r a v e l i n g u p t h e i n c h n e d p i a n e w i t h a c o n s t a n t v e l o c i t y = 5

r o / s , d e t e r m i n e t h e p o w e r o f f o t c e F .

F14-11 . I f t h e 5 0 - k g l o a d A i s h o i s t e d b y m o t o r M s o t h a t t h e

l o a d h a s a c o n s t a n t v e l o c i t y o f 1 . 5 m / s . d e t e r m i n e t h e p o w e r i n p u t t o t h e m o t o r , w i l i a o p e r a t e s a t a n e f f i c i e n c y e = 0 , 8 ,

F14-8

F t 4 - 9 . I f t h e m o t o r w i n d s i n t h c c a b l c w i t h a c o n s t a n t

s p e e d o f = 3 f t / s , d e t e r m i n e i h e p o w e r s u p p ] i e d t o t h e

m o t o r . T h e l o a d w e i g b s 1 0 0 l b a n d [ b e e f f i c i e n c y o f t h e m o t o r i s e = 0 , 8 . N e g l e c t t h e m a s s o f t h e p u l l e y s .

F14-9

A 1 . 5 m / s

F14-11

F 1 4 - 1 2 . A t t h e i n s t a n t s h o w n . p o i n t P o n t h e c a b l e h a s a

v e l o c i t y v p = 1 2 m / s . w h i c h i s i n c r e a s i n g a t a r a j e o f

a p = 6 m i s 2 . D e t e r m i n e t h e p o w e r i n p u t t o m o t o r M a t t h i s i

n s t a n t i f i t o p e r a t e s w i t h a n e f f i c i e n c y E = 0 . 8 . T h e m a s s o f

b l o c k A i s 5 0 k g .

F14-12

( 1 0 s ) P i


■ PROBLEMS

14-42, The jeep has a weight of 2500 lb and an engine which transmits a power of 100 hp to nil the wheels. Assuming the wheels do not slip on the ground, determine the angle N of the largest inclinc the jeep can climb at a constant speed v = 30 ft/s.

Prob. 14-42

14-43. Determine the power input for a motor necessary to lift 300 Ib at a constant rate of 5 ft/s. The efficiency of the motor is a = 0.65.

*14-44. An automobile having a mass of 2 Mg travels up 7° slope al a constant speed of v = I00 km/h. II mechaffical friction and wind resistance are neglecłed, determine the power developed by the engine if the automobile has an efficiency e -= 0.65.

Prob. 14-44

14-45. The Milkim Airc-raft Co. manafactures a turbojet engine that is placed in a piane having a weight of 13000 lb, If the engine develops a constant thrust of 5200 lb, determine the power outpuł of the piane when it iS just ready to take off with a speed of 600 mi /h,

14-46. To dramalize ihe losy of energy in an automobile, consider a car having a weight of 5000 Ib that is traveling at 35 mi/h. If the car is bmught to a stop, determine how tong a 100-W light bulb must bunt to expend the same amountof energy. (1 mi 5280 ft.)

1447. The escalator steps move with a constant speed of 0.6 m/s, If the steps are 125 mm high and 250 mm in length, determine the power cif a motor needed to lift an average mass of 150 kg per step. There are 32 steps.

If the escalator in Prob. 14-46 is not moving, determine the constant speed at which a man having a mass of 80 kg musi walk up the steps to generate 100 W of power-the same amount that is needed to power a standard light bulb.

Probs. 14-47/48

14-49. The 2-Mg car increases its speed uniformly from rest to 25 m/s in 30 s up the incIined ruad Detennine the maximum power that must be supplied by the engine, which operates with an efficiency of e = 0. 8. Also, find the average power supplied by the engine.

Prob. 14-49

1 4

198 CHAPTER 14 KINETC5 OF A PARTICLE: WORK AND ENERGY

14-50. The erate has a mass of 150 kg and rests on a surface for wbich the coefficients of static and kinefic friction are fi, =- 0.3 and juk =, 0.2, respectively. If the motorM supplies a eable foree of F = (812 21A N, where t is inseconds, determine the power output developed by the motor when r = 5 s,

''14-52. The 50-lb bad is hoisted by the puHey system and motor M. If the motor exerts a constant forte of 30 !b on the cable, determine the power that must be supplied to the motor if the load has Been hoisted s = 10 ft starting from test.The motor bas an efficiency of e = 0.76.

Prob. 14-511

M

Prob. 14-52

14-51. The 50-kg crate is hoisted up the 30' indine by thc policy system and motor M. If thc erate starts from rest and by wistarii acce.leraioi altains a speed of 4 tris atter traveling 8 m along ihe piane, determine the power that must be supplied to the motor at this insłant.Neglect frietion along the piane. The motor has an efficiency of e = 0.74,

14-53. The 10-lb colł ar starts from rest at A and is lifted by applying a ~starli vertical [orce of F = 25 lb lo the cord. If the rod is smooth, determine the power developed by the lorce at the instant H = 60°,

14-54. The 10-lb eollar starts from reSI at A and is lifted with a constant speed of 2 ftis afomy; the smooth rod. Determine the power developed by the [orce F at the instant shown,

Prob. 14-51 Pro bs. 14-53/54

14.4 POWER AND EFFICIENCY1 99

14-55. The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of s = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of = 4 m/s.

14-58. The block hasa mass of 150 kg and rests on a surface for which the coefficient of stalic and kineticfriction = 0.5 and pt.,k = 0.4. respectively. If a kirceF = (60/2) N, where t is in seconds, is applied to the cable, determine the power deveIoped by the force when ,r = 5 s, Flint; First determine the dme needecl for the force to cause molion.

Prob. 14-58

14-59. The rocket sled bas a mass of 4 Mg and travels from rest along the horizon tal track for which the coefficient of kinetic fricłion is Juk = 0.20. If the engine provides a constant thrust T = I50 kN, determine the power output of the engine as a function of time. Neglect the "lass of fuel mass and air resistance.

Prob. 14-55

*14-56. The sports car has a mass of 2.3 Mg, and wbite it is traveling at 28 m/s the driver causes ft to accelerate at 5 rn/s2. If the drag resistance on the car due to the wind is F0= (0.3y7) N, whcre v is the velocity in m/s, determine the power supplied to the engine at this instant, The engine hasa running efficiency of e = (1.68.

14-57. The sports car has a mass of 2.3 Mg and accelerates at 6 m/s2, starting from rest. If the drag resistance on the car due to the wind is Fn = (10v) N, where v is the velocity in m/s, determine the power supplied to the engine when

= 5 s. The engine bas a rttnning efficiency of 60.68.

T

li 11 f l fl l I L 11

Prob. 14-59

*14-60. A loaded truck weighs 16(103) lb and accelerates uniforrnly on a level ruad from 15 ft/N to 30 ftp; during 4 s. If the frictional resistance to motion is 325 Ib_ determine the maximum power that must be delivered to the wheels,Probs. 14-56/57

240 CHARTER 14 KINETC5 OF A PARTICIE. WORK AND ENERGY

14-61. If the jet on the dragsłer supplies a constant thrust of T = 20 kN, determine the power generaied by the jet as a function of time. Neglect drag and rolling resistance, and the lass of fuel. The dragsłcr has a mass af 1 Mg and starts srom test,

''14-64. The 500-kg eIevator starts from rest and travelsupward with a constant acceIeration= 2 mis2.Determine the power output of the motor M when t = 3 s. Neglect the mass af the pulleys and cable.

Prob. 14-61

14-62. An aihkte pushes against an exereise machine with a force that varies with time as shown in the first graph, Also, the velocity of the athlete's arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the wark done in t = 0.3 s.

14-63. An athlete pushes against an exereise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete's arm acting in the same direction as the force varies with Inne as shown in the second graph. Determine the maximum power deveoped during the 0.3-second time period. Prob. 14-64

t (s)

14-65. The 50-lb block rests on the rough surface forwhich the coefficient of kinetic friction is µk 0.2.A forceF = (40 + .52) lb, where .s is in ft, acts on the btock in the direction shown. If the spring is originally unstretched (s = 0) and the block is al Test, determine the power developed by the force the instant the block bas =cclN = l .5 ft_

F

30 = 20 Iblft/ ~ 1

t (s)03

Probs. 14-62,63 Prob. 14-65

14,5 CONSERVATNE FORCES AND POT ENT1AL ENERGY 201

14.5 Conservałive Forces and Potential Energy

Conservative Force. If the work of a force is independent of the path and depends only on the foree's initial and final positions on the pach, then we can classify this force as a

cunseryative foree. Examples of conservative forces are the weight of a particie and the force developed b y a s p r i n g . T h e w a r k d o n e b y t h e w e i g h t

d e p e n d s ordy on t h e verticat displacement 14 t h e w e i g h t , a n d t h e w o r k d o n g i b y a s p r i n g f o r c e d e p e n d s

oniy on the s p r i n g ' s efongation a r compression.in contrast to a conservative force, consider the force of frictian exerted

on a sliding objeet by a fixed surface. The work done by the frictional force depends on the path—the longer the path, the greater the wark. Consequently, frictional forces are nonconservadue. The wark is dissipated from the body in the form of heat.

Energy. Energy is defined as the capacity for doing wark. For

example, if a particie is originally at rest, then the princple af wark andWenergy states that f.11__,2 = T2. In other words, the kinetic energy is equal to the wark that must be done on the particie ta bring it from a state of rest to a speed y. Thus, the kinetic energy is a measure of the particle's capacity to do work. w h i c h i s a s s o c i a t e d w i t h t h e motion a f t h e p a r t i c i e . W h e n e n e r g y c o m e s f r o m

t h e position of the particie. measurcd from a fixed datur]] or reference piane. it is calicd potcntial energy,Thus, potential energy is a measure of the arnount af work a conservative force will do when it nioves from a given position ta the datura. In mechanics, the potential energy created by gravity (weight) and an elastic spring is important.

Gravitational Potential Energy. If a particie is located a distance y above an arbitrarily selected datura, as shown in Fig. 14-17. the particie„s weight W has positive gravitational putenttal energy, Y,, s i n c e W has the capacity af doing positive wark when the particle is moved back down to the daturn. Likewise, if the particie is located a distance y be/ow the daturn, V 5 n e g a t i v e s i n c e t h e w e i g h t d o e s n e g a t i v e w o r k w h e n t h e

particie is 'mycki back up to the daturn. At the daturn= O.

I n g e n e r a l , i f y i s p o s i t i v e u p w a r d , t h e g r a v i t a t i t . s n a l p o t e n t i a l e n e r g y o f

t h e p a r t i e c o f w e i g h t W i s *

11, = + Wy

' V

Datura

—v

Gravitational [menda! unergy

Fig. 14-17

V„ = Wy (14-13)

Here the weight is assumed to be constans. This assumption is smtable for smali differences in elevation dy. ri lhe elevation change is significant, however, a variation of weight with elevation rnnst be taken joto account (ser Prob_ I4—?;2}_

2 0 2CHAPTFR 14 KiNFT[CS OF A PARTICIE; WORK AND ENIRGY

Elastic Potential Energy. When an elastie spring is eiongałed or compressed a distance s from its unsłretched position, elastic potential energy V, can bc stored in the spring. This encrgy is

e.= +4,,c52 (14-14)

Here V, is always positive since, in the deformed position, the foree of the spring has the copacity or "pntential" for always doing positive work on the particie when the spring is returneci to izs unstrłched position, Fig. 14-18.

Unstretehed position,s

k~ ~ » 0 v , , = + I s

- 2 12.

Elastiepotential energy

Fig. 14-1.8

The weighl of the sacks resting on this platform causes potential energy to be stored in the sur porting springs. As each sack is removed, the platform will risc slightly sinee sorne of the potential energy within the springs will be transformect in to an increase In gravitational poiential energy of the remaining sacks. Such a device is useful for rernoving the sacks without having to band over to pick [hem up as the are unloacied.

14,5 CONSERVATIVE FC7RCES AND POTENTiAL f NERGY 2Q3

Potential Function. In the gene,rai case, if a particie is subjected tu both gravitational and elastic forces, the particle's potential energy can bc expressed as a potential function, whicli is the algebraic sum

V — V 8 V e (14-15)

Measuremene of V depends on the location of the particie with respect to a seiected datum in accordance with Eqs. 14-13 and 14-14.

The work dane by a conservative furce in rnoving the particie from one point ta another point is measured by the difference of this function, i.e..

U1_2 = — V2 (14-16)

For example, the potential function for a particie of weight W suspended from a spring can be expressed in terms of its position, s, measured from a datum located at the unstretched length of the spring, Fig. 14-19. We have

V = 11, + V,

= - W s - s 2

If the particie moves frOm S i to a kawci- position 52, then applying Eq. 14-16 it can be scen that the wark of W and F, is

U 1 _ 2 = V j — V 2 = ( — W S I - F k s - - w s +

= si) ks4 - dsi}

D atum

F,

Fig. 1449

2 0 4 CHAPTFR 14 KINFT[CS OF A PARTICLE: WORK AND ENERGY

When the dispłacement along the path is infiuitcsimal, i.e., from point (x, y, z) to (x + dx, y + dy, z + dz), Eq, 14-16 becomes

dU = V(x, y, z) - 1,1(x + dx, y + dy, + dz)

-dif(x, y, z) (14-17)

1f we represcnt both the furce and its displacement as Cartesian vectors, then the work can aiso be expressed as

dU = F • dr = (F„1 + F,j + F,k) (dxi + dy + (kk) F„ dx

+ F„ dy + F. dz

Substituting this result iriło Eq. 14-17 and cxpressing the differential dV(x, y, z) in ter= of its padial derivatives yiełds

F, dx + Fydy + F, dzavdx +v c)v —a dy + — dz)

193' az.

Since changes in x, y, and z are ali independent of one another, this equation is satisfied provided

F = Wxax'

(31/F _

32(14-18)

Th us,

aF = ć)v av - —j - —k

x Uy az

= + + t3z,ik)v

u rF = -VV (14-19)

where V (dcl) represents the veełor operator V = (3/19x)i +

(ałav).i (3/34k.Equation 14-19 relates a furce F to its potential function V and thereby

provides a rnathematical criterion for proving that F is conservative. Forexample, the gravitational potential funet-ion for a weight locatcd adistanee y above a daturn is= wy. To prove that W is eonservative,it isnecessary co show that it satisfies Eq, 14-18 (or Eq, 14-19), in whieh case

aVaF . = —, = -(Wy) = -W

av ay

The negative sign indicates that W acts downward, opposite to posłtive y, which is upward.

14.6 CONSERVATION OF -ENERGY205

14.6 Conservation of Energy

When a particie is acted upon by a system of borh conservative and nonconservative forces, the portion of the work done by the conservaiive forces can be written in terms of the differenee in their potential energics using Eq. 14-16, i.e., (54,7,_2)„„,. = V, - V. Asa resuit, the principle of work and energy can be written as

Tj + V1 + (.U1_.1)„,„rieuns, = T2 + V2 (14-20)

Here (/ U, represents the work of the nonconservative torcesacting on the particie. 1f ordy conservative forces do work then we have

T1 + V1 -= T2 -F V2 (14-21)

This equation is rcferred to as the conservadon of rnechanical energy ar simply the conservation of energy, It states that during the motion the sum a the partick's kinetic and potential energies remains constarn. For this to occur, kinetic energy must be transformed into potential energy, and vice versa. For example, if a bali of weight W is dropped from a height h above the ground (daturn), Fig. 14-20, the potential energy of the hall is maximum before it is dropped, at which time its kinetic energy is zero, The total mechanical energy of the bali in its initial position is thus

E = + V = O + W h =

When the bali has fallen a distance h/2, its speed can be determined byusing za2 = v( 24,ty yo), which yields v = V"2g(h/2) = g1i. Theenergy of the bali at the mid-height position is therefore

E — T2 + 1/2 V-7ig )2 + W(—h) wh

lust before the bali strikes the ground, its potential energy is zero and itsspeed is V2gh. Here, again, the total energy of the bali is

IE = T3 + V3 = V 2gh )2 + O = Wh

2 g

Notc that when the hall comes in contact with the ground, it cleforms sornewhat, and provided the ground is hard enough, the bali will rebound off the sarface, reaching a ncw height h', which will be less than the height h from which it was first released, Neglecting air frietion, the difference in height accounts for an energy lass, Eś = W(h h), which occurs during the collision. Portions of this less produce nose, localizecl deformation of the hall and ground, and beat,

Potentiai Energy (max) Kinetic Energy (zero)

h -rD a t = i

PotentiaI Energy and Kinetic Energy

Potential Energy (zero) Kinetic Energy (mas)

Fig. 14-20

206 CHAPTER 14 KINETrCS OF A PARTICLE: WARK AND EMERGY

System of Pa rti cl e s. H a system of partieles is subjeeted anty to conservaiive forces, then an equation similar to Eq.14-21 can be written for the particles. Applying the ideas of thc preceding discussion, Eq, 14-8 (STO + SL/1_2 = T,) b ccomes

ST 1 + y1 = ST 2 + 5V2 (14-22)

Here, the sum of the system's initial kinetic and potential energies is equal to the sum of the systern's fina! kinetic and potential energies, In other words, T + SV = const,

Procedura for Analysis

The conservation of energy equation can be used to solve problems involving wiodły, displacement, and conservative force svstems.lt is generally easter to apply than the principle of work and energy because this equation requires specifying the partic[e's kinetic and potential energies at only IWO points- along the path. rathcr Chan deterTnining the work when the particie moves through a displacement. For application it is suggcstcd that the folłowing procedure be used.

Potential Energy.•Draw two diagrams showing the particie Iocated at its initial and fina]

points along the path.

•If the particie is subjected to a vertical displacement.establish the fixed horizontal datum from which to measure the particles gravitational potentiaI energy Vg.

•Data pertaining to the elevation y of the particie from the datum and the stretch or compression s of any connecting springs can be determined from the geometry assoeiated with the two diagrams.

•Rwa]] Vg = Wy, whcre y is positive upward from the daturo andncgative downward from thc datura; also for a spring, = 4 ks',

which is always positive.

Conservation of Energy.

• Apply the equation T1 + V1 = T2 + y2,

• When determining the kinetic energy. T =, mv2 remember that the particie's speed v musi be measured front an inertial reference frame.

14,6 CONSERVATiON OF -ENZRGY 207

EXAMPLE

!S°

P.000(9S1) N

(b)

Fig. 14-21

The gantry structure in the photo is used to test the response of an airplane during a crash. As shown in Fig. 14-21a. the piane, having a mass of 8 Mg, is hoisted hack until 8 = 60°, and then the pull-back cable AC is released when the piane is at resł.Deterrnine the speed of the piane just before it crashes finta the ground, 9 = 15°. Also, what fis the maximum tension devełoped in the supporting capie during the motion? Negleet the size of thc airplane and thc cffect of lift caused by the wings during the motion.

(a )

SOLUTION

Since the force of the cable does no work on the piane. it must be obtained using the equation of motion. First, however, we must dełermine the plane's speed at B.

Potential Energy. For convenience, the darnin has heen established at the top of the gantry, Fig. 14-2 I a.


TĄ VA = 7", Vg

0 - 8000 kg (9,81 m /s2)(20 cos 60° m) kg).4

- 8000 kg (9.81 misz)(20 cos 15° m)

v8 = 13.52 m/s = 13.5 m/s A ns,

Equation of Motion. From thc free-body diagram when the piane is at B. Fig. 14-21b, wc have

iF, = ma„;

(1152 m/s)2

20 m

T = 149 kN Ans.

T - (8000(9.81) N) cos 15° = (8000 kg)

208 CHAPTER 14 KINETICS OF A PARTICLE: WORK AND ENERGY

14.10

The ram R shown in Fig. 14-22a has a mass of 100 kg and is reIeased from rest 0.75 m from the top of a spring. A, that has a stiffness kA = 12 kN/m. If a second spring B, having a stiffness kB = 15 kN/m, is "nested" in A, determine the maximum displacernent of

R A needed to stop the downward motion of the ram. The unstretchedlcngth of each spring is indicated in the figurc. Neglecł the mass of the springs.

0.75 m SOLUTION

Potential Energy. We will aminie that the ram compresses both springs at the instant it comes to rest.The dawni is located through the center of gravity of the ram at its initial position,Fig.14-22b.When them

n..4 kinetic energy is reduced to zero (v, = 0), A is compressed a distance0-3 m sA and B compresses sB = sA - 0.1 m.


T I + V I = +

O + O = O + HkAs,24 + 71/kB(sA - 0.1)2 - Wh

O + O = O + {-412 000 Nim).534 + -W5 000 N/m)(sA - 0.1 m)2

- 981 N (0.75 m + SA)}

981 N • Rearranging the terms,

Datum 13 500s-21'- - 2481sA - 660.75 = O

Using the quadratic formula and solving for the positive root, we have

981 N

0.7'msA = 0.331 m Ans.

SĄSince sB = 0.331 m - 0.1 m = 0.231 ni, which is positive, theassumption that both springs are compressed by the ram is correct.

sA — 0.1 m

(b)

Fig. 14-22

NOTE: The second root, sA = -0.148 m, does not represent the physical situation. Since positive s is measured downward, the negative sign indicates that spring A would have to be "extended" by an amount of 0.148 m to stop the ram.

14.6 CONSERVATION OF ENERGY209

1 4 . 1 1

(a)

A smooth 2-kg collar, shown in Fig. 14-23a, fits loosely on the vertical shaft. If the spring is unstretched when the collar is in the position A, determine the speed at which the collar is rnoving when y = ł m, if (a) it is released from Test at A, and (b) it is released at A with an

velocity 2 m/s.

SOLUTION

Part (a) Potential Energy, For convenience, the daturo is estabIished through AB, Fig, 14-23b, When the collar is at C, the gravitational potential energy is —(Ing)y, sinee the collar is below the datuni, and theelastic potential energy is Here scR =- 0.5 m, which representsthe strekh in the spring as shown in the figure.

Conservation of Energy. +

VA = Te +

O + O -= 2rirv + — ingy}

O + O = {;(2 kg)4} + N/m)(0.5 m)2 — 2(9.81) N (1 m)}

vc 4.39 m/s L Ans.

Datu

rn This problem can also be solved by using the equation of motion or the principle of work and energy. Note that for bach o( these methods the variation of the magnitude and direetion cif the spring form must be taken into account (see Example 13.4). Here, however, the above solution is elearly advantageous since the całeulations depend only on data calculated at the initial and finał points of the path.

Part (b) Conservation of Energy. If 12,1 = 2 ni/s, using the data in Fig. 14-23b, we have

TA + Ti + vc

;mn', +Q = inv7c. +{;+wsi$—ingy}

4(2 kg)(2 m/s)' O = 4(2 kg)v2c. f4(3 N/m)(0.5 n-)2

— 2(9.81) N (1 m) }

ve = 4.82 m/s .1, Ans.

NOTE: The kinełic cnergy of the collar depends only on the magnrtrtde of velocity, and thcrefare it is immaterial if the collar is moving up or down at 2 m/s when released at A.

n-02 1 (0.75 1-1-)2 = 1.25 m

sen 125 m — 0.75 = 0.5 m (h)

141-23

1.5 m

B F14-13

21 O CHAPTER 14 KINETrCS OF A PARTICLE: WARK AND ENERGY

F14-13. The 2-kg pendulum bob is released from rest when h is at A. Determine the speed of the bob and the tension in the cord when the bob passes through lis lowest position, B.

A

F14-14. The 2-kg package leaves the eon veyor beli at A with a speed of v,k = I m/s and slides down the smooth ramp. Determinc the re.quired speed of the conveyor beli at B so that the package can be delwered wtthout sEpping on the bell. Also, find the normal reaetion the eurved portion of the ramp exerts on the package at B if p8=-- 2 m,

7,8x

F14-14

F14-15. The 2-kg collar is given a downward vełocity of 4 Mł s when it is at A. If the spring lias an unstretehed length of I m and a sfiffness of k = 30 Nim. deterinfile the veioeity of the collar at s = 1 ni.

2 m

F14--I4. The 5-lb collar is released from rest at A and travels along the frictionless guicle.Determine the speed of the collar when it strikes the stop B. The spring has an unsłretchedlength of 0.5 ft.A

1.5 ft

F14-16

F14-17. The 75-Ib block is released from rest 5 ft ahove the plale. Determine the compression of each spring when the block mornentarily comes to rest after stri king the plate. Negleet the mass of the &te. The springs arc initially unstretehed,

025 fl

5 ft

k = 1000 Ibift = 1000Ih/ft

k' = 1500 ibift F14-.17

F14-18. The 4-kg coilar C has a velocity of v, = 2 m/s when it is at A. f f the guide rod is smooth, deterrnin e the speed of the collar when h is at B. The spring lias an unstretched length of

0.2 m.A r 400 Nim

0.4 m

F14-15 F14-18

14.6 CONSERVATION OF ENERCY 211

■ PROBLEMS

14-66. The girl has a mass of 40 kg and cen ler of mass at G. If she is swinging to a maximum height defined by H = 60°, determine the kirce developed along each of the faun supporting posts such as A.13 at the instant O = 0°, The swing is centrally located hetween ihe posis.

Prob. 14-66

14-67. TWo equaI-Iength springs are "nested" together M order to form a shock absorber. If it is desigued to arrest the motion of a 2-kg mass that is dropped s = 0.5 m above the iop of the springs from an at-rest posiiion, and the maximum compression of the springs is to be 0.2 in. determine the required stiffness of the Miler spring, ka, if the nuter spring bas a stiffness kA = 400 Wrn,

Prob. 14-67

"14-68. The collar has a weight of 8 lb. 1f il is pushed down so as to cornpress the spring 2 ft and (hen released from rest (h =- 0), determine its speed when it is displaced h = 4,5 ft, The spring is not attached to the collar. Neglect frietion,

14-69. The collar has a weigh l of 8 lb. I f it is reieased from rest at a height of h = 2 ft from the top of the uncompressed 14 spring, determine the speed of the callar afier it falls and compresses the spring 03 ft,

P ro lis. 14-68/69

14-70. The 2-kg bali of negligible size is fired from point A with an initial velocity of 10 m/s up the smooth inclined piane. Determine the distance from point C to where it hits the horiciutal surface at D. Also, what is its velocily when it strikes the surface?

Prob. 14-70

212 CHAPTFR 14 KINFT4CS OF A PARTICLE: WORK AND ENERGY

14-71. The ride at an amusernent park consists of a gondola which is lifted to a heighl of 120 ft al A. 1f it is released from rest and faIls along the parabolic track, detcrmine the speed at the Instant y = 20 ft, Also determine the normal reaction of the tracks on the gondola at this instant. The gondola and passenger have a total weight of 5130 lb. Neglect the effects of friction and the mass of the wheels.

Prob. 14-71

414-72. The 2-kg collar is attached to a spring that bas an unstretched lengłh of 3 m. If the collar is drawn to point B and released from rest, determine its speed when it arrives at point A.

14-73. The 2-kg edku- is attached to a spring that has an unstretched length of 2 m. If the collar is drawn to point B and released from rest, determine its speed when it arrives at point A.

14-74. The 0.5-lb bal] is shot from the spring devieeshown. The spring hasa stiffness k = 10 and thefour cords C and place P keep the spring compressed 2 in. when no bad is on thc platc.The płate is pushcd back 3 in_ from its initial position. 1f i( is [hen released from rest, determine the speed of the bali when it travels 30 in. up the smooth piane_

14-75. The 0.5-1b ball is shot from the spring device shown. Determine the smallest stiffness k which is required to shoot the bali a maximum distance of 30 in. up the smooth piane afer the spring is pushed back 3 in. and the bali is released from rest, The foru cords C and plac P keep the spring compressed 2 in. when no load is on the p/ate,

Probs. 14-74/75

*14-76. The Killer coaster car having a mass ni is released from rest at point A. 1f the track is to be designed so that the car does not leave it at 8. determine the required height h. Also, find the speed of the car when it reaches point C. Neglect friction.

3 m

Probs. 14-72/73 Prob. 14-.76

14.6 CONSERVATION OF ENERGY21 3

14-77. A 750-m in-long spring is compressed and confined by the plate P, wbici) can sltde freely along the vertical 600-mm-long rods. The 40-kg block is given speed of v =- 5 mirs when it is h = 2 m above the plate. Detcrmine how far che plate maues downwards when the hlock momentarily stops after striking it. Neglect the mass of the plate.

= S m ł

k = 25 k1,1/in

Proh. 14-77

14-78. The 2-1h No& is given an initial velocity of 20 ftis when it is at A. If the spring has an unstretched length of 2 ft

and a stiffness of k = I00 detevnine the velociły ofthe block when s = I ft.

Prob. 14-78

14-79. The błock lias a weight of 1.5 /h and slides along thc smooth chute AR, It is released front rest at A. which bas coordinates of A (5 f1.0, te ft). Determine the speed at which it slides off at B. which has coordinates of 8(0,8 ft, 0).

Prob. 14-79

*14-80. Each of the two elastic rubber bands of the slingshot has an unstretched length of 200 mm. 1f they are pallad back to the position shown and released from Test, determine the speed cif the 25-g pcilet just after thc rubber bands become unstretched. Negleet the mass of the rubber bands. -Bach rubber band hasa stiffness of k = 50 Nim.

14-81. Fach of the Iwo elastic rubber bands of che shngshoł has an unstretched length of 200 mm. If they are pulled back to the position shown and released front rest, determine the maximum height the 2.5-g pellet will reach if it is fired vertically upward.NegIect the mass of the rubber bands and the ch ange in elevation of the pellet while it is constrained by the rubber bands. Each rubber band has a stiffness k = 50 Nim.

Proh,, 14-80/81

h = 2 m

Mi0 mm

214 CHAPTFR 14 KiNFT4C5 OF A PARTICLE: WORK AND ENFRGY

14-82. If the mass of the earth is M. show that the graviłational potentiał energy of a body of mass rn Iocateddistanoe r from the center of ihe earth is-GM,m/ r.Recall that the gravitational forte acting between the earthand the body is F = r-2), Eq, 13-1, For thecalculation.locate the daturo at r-4 x. Also. prane that Fis a conservative furce.

14-83. A rocket of mass m is fircd vertically from the surface of the earth. i.e.. at r = ri. Assuming that no masa is tost as ił travels upward, deterrnine the wark it m ust do against granity to reach a distance r,. The force of granity is F = GM,m/i2 (Eq.13-1),where M, is the masa of the earth and r t lie distance between the rocket and the center of the earth,

Probs. 14-82183

14-84. The firing rnechanism of a pinball niachine consists of a plunger P having a mass of 015 kg and a spring of stiffness k = 300 Nirn.When s = 0, the spring is compressed 54 mm. If the arm is pulled buk such that s = 100 mm and released, determine the speed of the 0.3-kg pinball B jus before the plunger strikes the stop. i.e., s = 0, Assume alt surfaces of contact to be smooth. The bali tnoves in the horizontal piane. Neglect fricłion. the masa of the spring, and the rolling motion of the bali.

14-85. A 60-kg satellite traveis in free flight along anelliptical orbit such that at A, where= 20 Mm. it bas aspeed vA = 40 Mm/h. What is the speed of the satellite when it reaches point B, where rfj -= - 80 Mm? flint See Prob. 14-82, where M, -= 5.976(1024) kg and G = 66.73(10 12) in5i{kg-s2)-

B

= M m

(

Prob. 14-8514-86. Just for fun, twa 150-1b engineering students A and

▪intend to jump off the bridge from Test using an elastic cord (bungee card) having a stiffness k = 80 lbift. They wish to just reach the surface of the river, when A. attached to the cord, tetr go of B at thc instant they touch thc water.

Deterrnine the proper unstretched lengdi of the cord to do the stunt, and calculale the maximum acceleration of student A and the maximum height he reaches above the water after the rebound. Prom your results, comment on the feasibility of doing this

Prob. 14-86

- 1011"Nfro

Prob. 14-84

=20 Mm

14.6 CONSERV ATION OF E N ERG Y 215

14-87. The 20-lb collar slides along the smooth rod. If the collar is released from rest at A. deterrnine its speed when it passes point B. The spring bas an unstretched length of 3 ft.

Prob. 14-87

*14-88. Two equal-length springs having a stiffness icA = 300 N/rn and ke = 200 Nim are "nested- together in order to form a shock absorber. If a 2-kg block is dropped from an at-rest position 0.6 m abace the top of the springs, determlne their deformation when the błock rnomentarily stopy.

14-89. When the 6-kg box reaches point A it hasa speed of ?JA = 2 m/s. Determine the migle H at which it leaves the smooth circular ramp and the distance s to where it falls t-

iło the cart. Neglect friction.

Prob. 14-89

14-90. The Raptor is an outside loop miler coaster in which riders are belted finto seats resernbling ski-lift chairs. Determine the minimum speed tio at which the cars should toast clown from thc top of the kitl, so that passengers can just rnake the loop without Ieaving contact wiłh thein. seats. Neglect friction, ihe size of the car and passenger, and assume each passenger and car hasa mass m.

14-91. The Raptor fis an outside loop roller coaster in which riders are belted finto seats resembling ski-lift chairs. If the cars travel at yo = 4 mis when they are al the top of the hill, determine their speed when they are at the top of the loop and the reaction of the 70-kg passenger ran bis seat al Ibis instaM.The car bas a mass of 50 kg.Take fi = 12 rn. p = 5 m. Neglect ffiction and the size of the car and passenger.

Prob. 14-88 Probs. 14-90/91

0 , 6 n i

A

x

0.5 ni

1

/c = 250 N /m k = 250NA-19-fiffilf——4.4kk.14

216 CHAPTFR 14 KINFTICS OF A PARTICLE: WORK AND ENFRGY

*14-92. The 75-kg man bungee jumps off the bridge at A with an initial clownward speed of 1.5 m/s. Determine the required unstretched length of the elastic ord lo which he is attached in order that lie stops mornentarily just above thc surface of the wałer. The stiffness of the el astic cord is k = 80 Nim. Neglect the size of the man,

A

Prob. 14-9214-93. The 10-kg sphere C is released from rest whend = and the tension in ihe spring is 100 N. Deterrnine thespeed of Ule sphere at the instant 90°.Negket the massof rod AB and the size of the sphcre.

k = 500N/m

0.3 m

0.15m/

Prob. 14-93

0.4 m

14-94. The double-spring burnper is used to stop the 1500-lb steel billet in the rolling mili. Dctcrminc the maximum displacement of the plute A if the billet strikes the plute with a speed of 8 ft is. Neglect the mass of the springs, roilers and the plates A and B. Take k 1 = 30001h/ft, k, = 45 000 lbifit,

PI = 8 ft/sk a k 2

Prob. 14-94

14-95. The 2-lb box lias a velocity of 5 fłis when it begins to silcie down the smooth inclined surface at A. Determine the point C (x, y) where it strikes the lewar incline.

'14-96. The 2-1b box bas a velocity of 5 ft/s when it begins to shde down the smoołh inclined surface at A. Determine its speed jusl before hitting the surface at C and the time to travel from A to C. The coordinates cif point C are x = 17.66 ft, and y = 8,832 ft.

5

Prohs. 14-95/96

14-97. A pan of ncgligible mass is attached to twa identical springs ot stiffness k = 250 Nim. If a I 0-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical ffisplacernent d. Initially each spring lias a teusion of 50 N,

Prob. 14-97

I m —

14.6 CONSERVATION OF ENERCY 217


P14-1. The sopor coaster k momentarily at rest at A. Determine the approximate norma lorce ił exerts ort the track at B. Also determine its approximate acceleration at this point. Use numerical data, and sake sealed measurements from the photo with a knnwn height at

P 1 4 - 1

P14-2. As the largo ring rotates, the operator can apply a hreaking rnechanism that binda the cars to the ring, which then allows the cars to rotate with the ring. Assuming the passcngcrs arc not heited finfo the cars. determine the smatlest speed of the ring (cars) so that no passenger will fali out.When should the operator release the brake so that the cars can achieye their greatest speed as they slide freely on thc ring? Estimate thc grcatcst normal furce of thc seat on a passenger when ibis speed is reached, Use numerical values to explain your answer.

P14-3. The war= pulls the water bailoon launcher back, stretching each of the four elastic cords. Estimate the maximum height and the maximuin range of a bali placed within the container if it is reieased Erom the position shown. Use numerical values and any necessary measurements from the photo. Assume the unstretched lcngth and stiffnessof each cord is known. 14

P14-3

P14-4. The gir] is momentarily at rest in the position shown. If the unstretehed length and sfiffness of c ach of the iwo elastic cords is known. determine approxirnately how far the girl descends before she again becornes momentariły at rest. Use numerical values and tatce any necessary measurernents from the pilota

P14-2 P14-4

218 CHAPTER 14 K I N E T [ C S O F A P A R T I C L E : W O R K » W E N E R G Y

N

S i

F

F cos

T, + U1_2 = T2

The Prineiple of Work and Energy

f f the equat ion of motion in thetangential direction, F, = ma,, iscombined with the kinematic equation, o, ds = v dv, we obtain the principle of work and cnergy. This equation statcs that the initial kinetic energy T. plus the work done 2 U, is equal to the finał kinctic energy.

CHAPTER REVIEW

Work of a Furce

A force does work when il undergoes a displacement along its line of action. If the force varies with the displacement. tben the work is U = f F cvs O ds.

Graphically, this represents the ara under the F—s diagram.

s

ds

If the force is constant. then for a displacement As in the direction of the lorce, U = F, As. A typical example of this case is the work of a weight, U = —W Ay. Here. Ay is the vertical displacement.

Unstretch ed pusiliori,s= O

The work done by a spring force. F = ks, depends upon the stretch or compression s of the spring.

U = —(ż ks; —

F cos fi

F cos fl

2

F,

Furce on Particie

k

CHAPTER R EVI EVV 219

The principle of work and energy is useful for solving problems that involve force, velocity, and dispiacemen t, For application. the free-body diagram of the particie should be drawn in order to identify the forces that do work.

Power and Efficieney

Power is the time rate of doing work, For apphcation. the force F creating the power and iis velociiy v must be specified.

Efficiency represents the ratio of power output to power input. Due to frictional losses, il is always less than one.

dfiP - dr

P = F • v

power outpui

= power input

Conservaffon °f Eoergy

A conservałive force does work that is independent of its path. Twa examples are the weight of a particie and the spring furce,

Friction is a nonconservative [orce lince the wark depends opon the length of the path, The ionger the path, the more work done.

The work done by a conservative force depends tman its position relative to a daturo. When this work is referenced rrum a Jakim, it is called potentialenergy. For a weight, it is = ± Wy,and for a spring it is V, +4, ks2,

Mechanical energy consists of kinetic energy T and gravitational and dastic potential energies V Aecarding ło the conservation of energy. this sum is constant and has the same value at any position on the path.If oni), graviiational and spring forces cause motion of the particie, then the conservation-ofenergy equabon can be used to solve problems invorving [hese conservative forces, displacernent, and velocity.

- v

WGravi[ational potential energy

Elastic potential energy

Ta + V] = Tz + Vz

k

[VR - + Wyj

Datom

mChapter 15

The design of the bumper cars used for this amusement park ride reguires knowledge of the principles of irnpulse and momentum.

Kinetics of a Particie:Impulse andMomentum

CHARTER OBJECTIVES•To clevelop the principle of linear impuls@ and momentum for a

particie and apply ittn solve problerns that invoive furce, velocity,and time.

▪To study the conservation of lineat momentum for particles.

•To analyze the mechanics of impact.

•To introduce the concept of angular impulse and momentum.•To solve problems involving steady fluid streams and propuision

with variable rnass.

15.1 Principle of Linear Impulse and Momentum

In this sedion we will irdegrate the equation of rnotion with respect to time and thereby obtain the principle of impulse and rnomentum. The resulting equation will be useful for solving problems involving lorce, vełoeity, and firne,

Using kinernaties, the equation of motion for a particie of mass m can be written as

XF == rrr cLdr (15-1)

where a and v are both measured from an inertial frame of reference. Rcarranging the term s and integrating betwecn the lirnits v = vi at f = and v = v, at t = s2, we have

j r , F r i ł mi dv

222 CHAPrER 15 ICINETrCS OF A PARTICLE IMPULSE ANO MOMENTUM

The impulse tool is used to rernove the dent in łhe traiier fender. To do so its end is firsl screwed into a hole drilled in the tender, then the weight is gripped and jerked upwards, Wiking the stopThe impulse developed is transferred along the shaft of the [oni and pitas suddenly on the cleni.

ar . 2 : d i = mv, m— vi

fI12 F

, (15-2)This equation is referred to as the prineple of linear imprdse and mornenuan. From the derivation it can be seen that it is simply a time integration of the equation of motion. It provides a direeł means of obtaining the, particles finał velocity v, after a specified time period when the particies initial vełoci ty is known and the forces acting on the particie arc cither constant or can be cxpressed as functions of timc. By compafison, if v, was determined using the equation of motion_ a łwo-step process would be necessary; i.e., apply F = ma to obtain a, then integrate a = dv/dt to obtain ,,,.

Linear Momentum. Each of the two vectors of the form L = my in Eq. 15-2 is referred to as the pa-rticle's linear momentom. Since m is a positive scalar, the linear-momentum vector has the same direction as v, and iłs magnitude mv ha,s units of mass times velocity, e.g., kg • m/s, or sług • ft/s.

Linear I m pu Ise. The integral I = f F di in Eq. 15-2 is referred to as thc linear impidse. This term is a vector quantity which measurcs the effect of a forte during the time the furce acts. Since time is a positive scalał-, the impuise acts in the same direction as the furce, and its magnitude has units of furce times time, e.g., N - s or 113 • s,*

If the furce is expressed as a function of time, the impulse can be determined by direct evaluation of the integral. In particular.if the orce is constant in both magnitude and direction, the resulting impulse bccomcs

I = fr:'F,dt F,(t2 — t1).

Graphically the magnitude of the impulse can be represented by the shaded arca under the curve of furce versus tirne, Fig. 15-1. A constant furce creates the shaded rectangular arca shown in Fig. 15-2.

f—f= j;ż F(t)dł,--- i = F,(/2 ti)

t2 E 2

Vanable Fotceconstant rurce

Fig, 15-1 Fig. 15-2

Although the units fur impulse and momentum are defined differently.tt can be shown that Eq_ 15-2. is dimensionaliy homogeneous

15.1 PRINCFPLE OF LINEAR IMPULSE AND MOMENTLIM 223

Principia of Linear Impulse and Momentum. For problem solving,Eq. 15-2 will be rewritten in the form

invĘ +F dr = onv, (15-3)I

which states that the initial momentum of thc particie at time plus thc sum of ali thc impuiscs applied to thc particie from r, to is equivalent to the finał momentum of the particie at time t2. These three terms are illustrated graphically on the impuise and momentum diagramy shown in Fig. 15-3.The twa ~meritum diagrams are simply outlinecl shapes of the particie which indicałe the direction and magnitude of the partieies initial and finał mornenta, mv, and mv,. Similar to the free-body diagram, the impuise diagram is an outiined shape of the particie showing ali the impulses that act on the particie when it is located at some intermediate point along its path.

If each d the vectors in Eq. 15-3 is resolved into its x, y, z components, we can write the following three scalar equations of lincar impulse and momentum.

15

Man y types of sparty, such as baseball. require appł ication of the principle of lin ear impulse and momenturn.

g2

M(VaT)1 + j F, . dr = ,rt2(1))2 r,

f r,

m(v.„)[ + i Fydt = ON),, s,

i t ,

m(v,), + 2.. F, dr = in(v,)2r,

(15 -4)

muz

LrrlInitia t omenillm diagram

Impulsel diagram

Fina] momentum diagram

1,1g. 15-3

224 CHAPTER 15 KlNETrCS OF A PARTICLE: IMPULSE AND MOMENTUM


The principle af Iinear impulse and momentum is used to soive problems involving force, time. and vełocity, sincc these terms are involved in the formulation. For application it is suggested that the following procedure be used.*

Free-Body Diagram.

• Establish the x, y, z inertial karne of reference and draw the particle's free-body diagram in order to account for all the forces that produce impuIscs on the particie.

• The direction and sense of the particles initial and finał vetocities should be established.

•If a vector is u-nknown,assurne Chat the, sense of its components is in the direction of the positive inertial coordinate(s).

•As an alternative procedure. draw the impulse and momentum diagrams for the particie as discussed in reference to Fig.15-3.

Principle of impulse and Momentum.

•In accordance with the established coordinate system. apply the principle of linearimpuiseand momentum, mv + jr:i2F dt mv2.

If motion occurs in the piane, the two scalar componentequations can be formulatcd by eithcr resolving the vector componerits of F from the frec-body diagram, or by using thc data on the impulse and momentum diagrams.

• Realize that every force acting on the parficle's free-body diagram will create an impulse, even though some of these forces will do no work,

• Forces that are functions of time must be integrated to obtain the impulse. Graphically, the impulse is equal to the area under the foree—time curve.

As thc wheels of the pitehing machina n:nate_ Ihey apply frictional impulses Io the balL ihereby giving it a Iinear montentum.These impulses are shown on the impulse diagram. Here bodli the ffictional and normał impuises vary with time. By comparison, the weight impulse is constant and is very smali since ihe limf.: 9r ihe bal] is in contaet with the wheels is very smal.

*This procedure will be followed when developing the proors and theorv in ihe lext.

i N d ti F e f r

f F ' d t

frsrW At

15.1 PRVJCIPLE OF LINEAR IMPULSZ AND MOMENTUM 225

Ans.

n(v,), + i Fx dr = m(v.,),• t,

O + 200 N cos 45'(10 s) -= (100 kg)v2

vz = 14.1 m/s

(+t)t,

in(v,.)1 + Fyd/ = m(vy)2

EXAMPLE

The 100-kg crate shown in Fig. 15-4a is originally at rent on the smooth horizontal surface. If a towing force of 200 N. acting at an angle of 45°, is applied for 10 s, determine the final velocity and the norrnal force which the surface exerts on the crate during this time in terval.

SOLUT1ONThis problem can be solvecl using the principle of impulse and momentum since it involves tarce, velocity, and time.

Free-Body Diagram. See Fig. 15-4b. Since all the forces acting are constuni, the impulses are simply the product of the lorce magnitudeand 10 s [I = Fr(y2 Note the alternative procedure of drawingthe erate's impulse and momentum diagramy, Fig, 15-4c

Pri-nciple vf IrnpulNe and Momentum. AppIying 645,15-4 yields

v

+ s) - 981 N(10 s) + 200 N sin 45'00 = O

Nc = MO N Ans

NOTE: Since no motion occurs in the y direction, direct application ofthe equilibrium equation = O gives the same result for Arc. Try tosolne the problem by first applying TFY= mt7x, then v = vo

Nr OO

(c)

Fig. 1.5-4

981 N (10 s)200 N (10 s)

J5`

200 N

(a)

200N

226 C H A R T E R 15 K I N E T [ C S O F A P A R T I C E E : I M P U L S E A N D M O M E N T U M

The 50-lb erate shown in Fig, 15-5a is acted upali by a force having a variable rnagnitude P = (213r) lb, where t is in seconds.Determine the erate's velocity 2 s after P lias been applied. The i nitiaI veloeity is vi = 3 ft/s clown the piane, and the coefficicrit of kinetic friction bctwecn the cratc and the piane is gik = 0.3.

SOLLIT1ON

Free-Body Diagram, Sce Fig, 15-5b, Since the magnitude of furce P = 20/ varics with time, the impulse it creatcs m ust be determincd by integrating over the 2-s time interval.

Principle of Impulse and Momentum. Applying Eqs. 15-4 in the directinn, we have

t ,

nz(v_,)1 + F,dt = sn(v),)27 ]

(a )

2 ,50 lb 50 lb , (3 1't/s) + f 2W dt 0.31+/-c(2 s) + (50 lb} !;-iri 30°(2 s) — , v,

32,2 fils' o 32,2 fth- "-

4.658 + 40 — 0.6Ne + 50 = 1.553v,,

50 lb The equation of equilibrium can bc applied in the y direction,Why?

( b ) F i g .

1 5 - 5

+T,..2„Fy = 0; żVe — $0 cos 30° lb =

Solving,

Nc = 43.30 lbZ2 = 44.2 flis Z A f2S.

NOTE; We can also solve this problem using the equation of motion. From

Fig. 15-5b,

+if£F, = ma,; 20t — 0.3(43.30) + 50 sin 50 30° = a322

a = 1 2 . 8 8 r + 7 . 7 3 4

Using kinematics

+V dv = a dt;v

dv = i (1.2.88t + 7.734)dt"3 ks

= 44.2 ft/s

By comparison. application of thc principle of impulse and momentum ehminates thc nced for using kinernatics (a = do/dr) and thercby yields an easier method for solution.

15.1 PRiNCIPLE OF 1.11,JAR IMpIJI.SW AND MoviEmum 227

EXAMPLE [15.3

A fis.(vB)2 = 35-8 m/s I Ta

= 19.2 N

Blocks A and B shown in Fig. 15-6a have a mass cif 3 kg and 5 kg, respectively, 1f the system is released from test, determine the velocity of block R in 6 s. Neglect the mass cif the pul leys and cord.

SOLUTION

Free-Body Diagram. See Fig. 15-6b. Since the weight of each błock is constant, the cord tensions will also be constant. Furthermore, since the mass of pul ley D is neglecteci, the card tensi(m TĄ = 27"B. Note that the blocks arc both assumcd to be moving downward in the positive coordinate directions, sA and ss.

Principle cif Impulse and Momentum, Block

•tzM(VA)] F,clr = m(.17,k)zJ

— 27'8(6 s) + 3(9.81) N(6 s) = (3 kg)(vA)2 (1)

rri(vB)E if F ydr = mr(vB)2

h

O + 5(9.$1) N(6 ś.) — T8(6 s) = (5 kg)(v8)2

Kinematics. Since the blocks are subjected to dependent motion, the vełocity of A can be related to that of B by using the kinematic analysis discussed in Sec. 12-9. A horizontal darum is established through the fixed point at C, Fig, 15-6a, and the position coordinates. sA and ss, are related to the constant total length d of the vertical segments °f the cord by the, equatiun

+ sB =

Taking the time derivative yields

= (3)

As indicated by the negative sign, when B moves downward A moves upward. Substituting this result into Eq, 1 and solving Eqs, 1 and 2 yields

NOTE: Realize that the positive (downward) direetion for vA and vs is consistenł in Figs. 15-6a and 15-6b and in Eqs, 1 to 3. This is important since we are seeking a simultaneous solution of equations.

(2)

Al, 3 kg

5 kg(a)

ł

s,,3(9.81) N

• a

5(9.81) N(b)

Fig. L—1

Datum

2 2 8 C H A R T E R 1 5 K I N E T [ C S O F A P A R T I C E E : I M P U L S E A N D M O M E N T U M

15.2 Principle of Linear Impulse and Momentum for a System of Particles

Y

Inertial courilinme system

Fig. 15-7

Thc principle of linear impulse and momentum for a system of particles moving rciative to aii inertial reference, Fig. 15-7, is obtained from the equation of motion applied to all the particles in the system, i.e.,

1FF = mi dr! (15-5)

The term on the left sicie represents oni}, thc sum of thc exłerrwl forces acting on thc particles. Rccall that the interna' forccs acting between particles do not appear with this summation, since by Newton's third law they occur in equal but opposite collinear pairs and therefore cancel out. Multiplying both sides of Eq. 15-5 by dt and integrating beiween the limits t = tt ,vi = (y1), and t = t7s vt = ("vi)2 yields

ini(v i) + F i ch = (15-6)

This equation states that the initial linear momenta cif the system plus the impulses of all the external forcs acting on the system from rl to t, is equal to the sys tem 's final linear momenta.from mrG = where in =rrrr is the total mass of alt the particles,Fig. 15-7, then taking the time derivative, we have

invG = Sirim

which states that the total Linear momentum of the system uf particles is equivalent to the linear momentum of a "fictitious" aggregate particie ofmass m -= moving with the veIoci ty of the mass center ot the system.Substituting intu Eq. 15-6 yields

m(vc,), 1.1 F idt = ni(v(;), (15-7)

fi

Here the inetiai lincar momcntum of the aggrcgatc particie plus the external impulses acting on the system of particles from rl to is equal to the aggrcgatc particle's fina] linear momentum. As a rcsult, the above equation justifies application of the principle of linear impulse and momentum te a system of particles that cornpose a rigid body.

15.2 PRINCiPLE OF LLNEAR 1MPULSE AND MOMENTUM FOR A 5V5'MM OF PARTICLES 229• FUNDAMENTAL PROBLEMS

F15-1. The 03-kg bali strikes the rough ground and rebounds with the velodties shown, Deterrnine the mag-

nitude of the impulse the ground exerts on the bali. Assume that the bali does not slip when it strikes the ground, and negicct the size of the hall and the impulse produced by the weight of the hall.

FLS-1

FIS-2. If the coefficient of kinetic friction between the150-lb crate and the ground = D.2, determine thespeed of the crate when r = 4 s. The trale starts Erom resi and is towed by the 100-lb force.

F 1 5 - 2

FIS-3. The motor exerts a fotce of F = (20r2) N on the cable, where r is in seconds. Determine the speed of the 25-kg crate when t = 4 s. The coefficients of stalic andkinetic friction between the crate and the piane are 0.3and p.k = 0.25, respectively.

FIS-4. The wheels of the 1.5-Mg car generale the traction force F described by the graf& If the car starta trom rest, determine its speed when r = 6 s.

FIS-5. The 2.5-Mg four-wheel-drive SUV tows the 1.5-Mg trailer. The traction fotce developed at the wheels is F0 = 9 kN. Detcrmine the speed of the truck in 20 s, starting from real. Also, determine the tension developed in the coupling between the SUV and the trailer. Neglect the mass of the wheels.

F n

F 1 5 - 5

F15—. The 10-Ib block A atiains er velocity of 1 ftis inseconds. starting erom test. Determine the tension in the

cord and the coefficient of kinetic friction between błock A and the horizomal piane. Neglect the weight of the puiley. Block B hasa weight (.3f 8 lb.

1001h

F ( k N )

a k N'

"›-

F 1 5 - 4

A

F15-3 F15-6

230 CHAPTER 15 KINET[CS OF A PARTICLE: IMPULSE AND MOMENTUM

PROBLEMS

15-1. A 2-lb bali is thrown in the direction shown wich aninitial speed = 18 flis. Determine the time needed forit to reach its highesł point B and Ule speed at which it is traveling at B. Use the principle of impulse and momentum for the solution.

Prob. 15-1

15-2. A 20-1h block slides down a 30' inclined piane with an initial velocity of 2 ft/s. Determine the velocity ot the Wek in 3 s if the coefficient of kinetic friction between the biock and the piane is juk = 0.25.

15-3. A 5-Ib błock is given an initial velocity of 10 ft f s up a 45 srnooth slope. Determine thc Linie it will take to łraveI up the slope before it stops.

`15-4. The I 80-1b iron worker is secured by a fali-arrest system consisting cif a harriess and lnnyard AB, which is fixed to thc beam.If the lanyard has a slack of 4 fł, determine the average impulsive force developed in the łanyard if be happens to fal] 4 feet.Neglect his size in the calculation and assume the impulse takes place in 0.6 seconds.

Prob. 15-4

15-5. A man bita the 50-,g golf bali such that it leaves the tee at an angle of 40° with the horizontal and strikes the ground at the same elevation a distanee of 20 m away. Determine the impuIse of the club C on the bali. Neglect the impuise caused by the ball's weight while the ciut is strśking thc bali.

Prob. 15-5

15-6. A tram consists of a 50-Mg engine and lbree cara. each having a mass of 30 Mg. 1f it takes 80 s for the tram to increase its speed uniformly to 40 knt/h. starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resuitant frictional traefive furce F wh ich gives the tram forward muflon, whereas the car wheels roll freely. Also, determine F acting on the engine wheels.

Prob. 15-6

15-7. Crates A and B weigh 100 Ib and 50 lb, respectively. 11 ihey start from rest, determine their speed when t = 5 s, Also, find the force exerted by crate A on crate 8 during the motion. The coefficient of kinetic friction between ihe crates and the ground is ,uk = 0.25.

A. r

P = 50 lb .....„.•"".-7/.... ..,....#,*

Proh. 15-7

B

15.2 PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES 231

*15-8. If the jeis exert a vertical thrust of T= (500t3/2) N. where t is in secondsdetermine the man's speed when t = 3 s. The total mass of the man and the jet suit is 100 kg. Neglect the Toss of mass due to the fuel consumed during the lift which begins from rent on the ground.

Prob. 15-8

15-9. Under a constant thrust of T = 40 kN, the 1.5-Mg dragster reaches its maximum speed of 125 m/s in 8 s storting from test. Deterinine the average drag resistance during ihis period of tine.

15-11. When the 5-kg block is 6 m from the wali. it is sliding at ul = 14 TnN If the coefficient of kinetic friction between the block and the horizon tal plane is jt = 0.3, determine the impulse of the wali on the błock necessary to stop the block. Neglect the friction impulse acting on the błock during tlte collision.

*15-11 For a short period of time, the frictional driving fotce acting on the wheels of the 2.5-Mg van is F n= (6001') N. where t is in seconds. If the van has a speed of 2b km/h when t = 0, determine its speed when r = 5 s

Prob. 15-11

uł - 14 m/s

T .= 4CI kN

Prob. 15-9

Prob. 15-12

15-10. The 50-kg crate is pulled by the constant furce P. If the crate starts from rest and achieves a speed of 10 m/s in 5 s, determine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is pk =

15-13. The /5-Mg van is traveling witli a speed of 100 km/11 when the brakes arc applied and alt four wheels lock. If the speed decreases to 40 km/h in 5 s, determine the coefficient of kinetic friction between the tires and the mad.

Prob. 15-13Prob. 15-10

232 CHARTER 15 KI NETICS OF A PARTICEE: iMP-LJESE AND MOMENTOM

15-14. The forte acting on a projectile having a mass m as it passes horizontally through the barrel of the cannon is F =, C sin Orifr'>. Determine the projectile's velocity when

z'. If the projectile reaches the end of the barrel at Ibis instant, determine the length s.

Prob. 15-14

15-15. During operation the breaker hammer develops on the concrete surface a forcc which is indicated in the graph. To achieve this the 2-lb spike S is fired from rest finto the surface at 200 flis. Determine the speed of the spike just after reboundin g.

Prob. 15-15

*15-16. The twitch in a nmscle of the arm develops a forte which can be measured as a funetion of time as shown in the graph. If the effective contraction of the muscle kasts for a fin-

le rU, determine the impulse developed by the mmdc,

15-17. A hammer head Jf having a weight of 0.25 Ib is moving verticaBy dovdnward at 40 ft/s when fil strikes the head of a nail of negligible mass and drives fil into a block of wood. Find thc impulse on the nail if it is assumed Chat the grip at A fis loose, the handle bas a negligible mass, and the hammer stays in contact with the nail wbite it comes to rest. Neglect the impulse caused by the weight of the hammer head during contaet with thc nail.

v .="- 40 ft/s

Prob. 15-17

15-18. The 40-kg slider block fis moving to the right with a speed of 1.5 m/s when it is acted upon by the forces F, wid F,. 1f [hese loadings vary in the manner shown on the graph. determine the speed of the block at t = 6 s. Neglect friction and the mass of the pul leys and cords.

F ( N )

3 0

1 0

F( 03) Ib 112.590

67-

5 45

22.5

O0.1 Q2 {13 U4

(rns)

26

Prob. 15-16 Prob. 15-18

15.2 PRINCIPLE OF LINEAR ł MPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES233

15-19. Determine the velociły of each block 2 s after the blocks are released from rosi. Neglect the mass of the pulleys and cord.

Prob. 15-19

*15-20, The particie P is acted upcm by its weight of 3 lb and forces F, and F2_ where r is in seconcis. 1f the particleorginaliy has a velocity of=+ 1 j -F 161k} ft/s,determine its speed aftcr 2 s.

z

F1 = {5iZtj+ rki lb

F2 = tr21} lb

Prob. 15-20

15-21. ff it takes 35 s for the 50-Mg tugboat to increase its speed unifornAy to 25 km /1-1, storting from reSl, determine the force of the rope on the tugboat.The propeller provides the propulsion furce F which gives the tughoat forward motion. whereas the barge moves freely. Also. determine Factingthe tuglx)at.The barge bas a mass of 75 Mg.

Prob. Li-21

15-22. Ii the furce Texerted on the cakle by the motor M is indicaled by the graph, determine the speed of the 500-lb crate when t = 4 s, si arting from rest. The coefficients of static and kinetic friction are tss = 0.3 and łuk = 0.25, respectively.

Prob. 15-22

15-23. The 5-kg block is moving downward at r, = 2.m/s when it is 8 rn from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents tato the sand and assurne the block does not rebound. Neglect the weight of the block during the Impact with the sand.

15-24. The 5-kg błock is falling downward at v, = 2 m/s when it is g m from the sandy surface. Determine the average impulsive force acting on the block by the sand if ihe motion of the block is stopped in 0.9 s once the block snikes the sand. Neglect the disłance the błock dents roto the sand and assume the block does not rebound, Neglect the weight of the block during the impact wich the sami.

1 vl = 2 mis

8 m

Probs. 15-23/24

30

T(1b)

51)

Y

t ( s )

2 3 4 C H A P T E R 1 5 K l N E T r C S O F A P A R T I C L E : I M P U L S E A N D M O M E N T U M

15-25. The 0.1-lb golf bali is struck by the club and then traveis along the trajectory shown. Deiermine the average impulsive force the club imparłs on the bali if the Club maintains contact with the bali for 0.5 ms.

5(10 ft

Prob. 15-25

15-26. As indicated by the derivation. the principle of impulse and momentum is valid for observers in any iner Lia l reference franri e. Show that this is so, by cansidering the 10-kg błock which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fLyed framc x, determine the finał speed of the block in 4 s if il bas an initial speed of 5 rnfs measured from ihe fixed frame. Compare the result with hiat obtained by an observer B, attached to the x" axis that moves at a constant vekłcity of 2 m/s relative to A.

A

B

2 mis

5 mis

6N

Prob. 15—.26

15-27. The winch delivers a horizontal towing force F to iis cable at A which varies as shown in the graph. Delermine the speed of the 70-kg bucket when / = g s. Originally the hucket is moving upward at vi = 3 m/s.

"15-28. The winch clelivers a horizontal towhtg force F to itr cable at A which varies as shown in ihe graph.Deiermine the speed of the 80-kg hucket when i = 24 s. Originally the bucket is released from rest.

12 24

Probs. 15-27f 28

15-29. The tram consists of a 30-Mg engine E, and cars A, B. and C. which have a mass of 15 Mg, 10 Mg, and 8 Mg, respeetively. If the tracka provide a traction force of F = 30 kN un the engine wheels, determine the speed of the train when i = 30 s, starting from resi. Also. find the horizontal coupling furce at D hetween the engine E and car A. Neglect rolling resistance.

Prob. 15-29

{s}

15.2 PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES235

15-30. The crate B and cylinder A have a mass of 200 kg and 75 kg. respectively. If ihe system is released from rest, determine the speed of the erałe and cylinder when t = 3 s. Negleet the mass of thc pulleys.

Prob. 15-30

15-33. Tli log has a mass of 500 kg and rests on thc ground for which ihe coefficients of stalic and kineticfriction are= 0.5 and p,k = 04. respecliveiy. The winchdelivers a horizontal towing forte T to its cabIe at A which varies as shown in the graph. Determine the speed of the log when t = 5 s. Originallv the tension in the eable is zero. Film: First determine the forte needed to begin moVillg the log.

r(N)

A T

T — 200 t2

3t (s)

15-31. Mack A weighs LO Ib and block B weighs 3 Ib, If B is moving downward with a velocity (m9), = 3 ft/s ai t = 0, determine the velocity of A when ł = I s. Assurne that the horizontal piane is smooth. Negleet the mass of the pulleys and cords.

*15-32. Block A weighs 10 lb and błock B weighs 3 lb.If B is moving downward with a velocity-= 3 fris at= 41, determine the velocity of A when r = 1 s. The

coefficient of kinetic friction between the horizontal piane and błock A is jr.i. A = 0.15.

Probs. 15-31/32

Prob. 15-33

15-34. The 50-kg błock is hoisted up the incline using ihe cable and motor arrangement shown. The coefficient of kinale friction hetwoen the block and thc surface is

= 0.4. If the block is initialty moving up the piane at

=

2 m/s. and ai ibis instant (t = 0) the motor deveiops a tension in the cord of T =- (300 + 120y) N. where t is in seconds, determine the velocity of the block when r = 2 s.

Prob. 15-34

236 CHARTER 15 KINET[CS OF A PARTICEE: IMPULSE AND MOMENTUM

The hammer in the top photo applies an impuisive lorce ki the stake. ()ming Ibis extremely short time of contact the weight of the stake can be considered rwnirnpuhsive, and providvd the slake is driver' into soft ground, the impulse of the grnund acting on the stake can also be considered nonimpulsive. By contrast, if the stake is used in a concrete chipper ło break concrete, ,hen iwo impulsive forces act on the stake one at its top due to the chipper and the olher on its butlom due to the rigidity of the concrete.

15.3 Conseryation of Linear Momentum for a System of Partides

When the sum of the external impulses acting on a system of particles is zeru, Eq, 15-6 reduccs to a simplified form, namely,

= £rni(vi)2(15-8)

This equation is referred to as the CO ri,servałion of iinear inornenfurn. It states that the total lincar momentum for a system of particles remainsconstant during the time period 1t to i2. Substituting myG rn,v, int°Eq, 15-8, we can also write

(va)i = (vG)3(15-9)

which indicates that ihe velociły v(; <,.)f the mass center for the system of particles does not change if no externa I impulses are applied to the system.

The conservation of lincar momentum is often applied when particles collide or interaet. For application, a careful study of the free-body diagram for the emire system of particles shouId be made in order to identify the forces which create either external or inłernal impulses and thereby determine in what direction(s) linear momentum is eonserved. As stated eariier, the inrernał Inipulses for the system will always cancel out, since they occur in equal hut opposite coilinear pairs. If the time period over which the motion is studied is very short. some of the external impulses may also be neglectecl or considered approximałely equal to zero. The forces causing these negligible impulses are called nonimpulsive forces. By comparison, forces which are very largo and act for a nery short period °f tirne produce a significant ehange in momentum and are called impulsive forces. They, of course, cannot be neglected in the irnpulse—momentum analysis.

Impuls-Ive forces normany occur due to an explosion or the striking of one body against another, whereas nonimpulsive forces may include the weight of a body, the tarce imparted by a slightly deform cd spring having a relatively smali stiffness, ar for that matter, any force that is very smali eompared to other larger (Im puIsive) forces.When making this distinction between irnpulsivc and nonimpulsive forces, it is important to realize that this only applies during the time ti to t2. To illustrate, consider the effect of striking a tennis bali with a racket as shown in the photo. During the cery shorł time of interaction, the torce af the racket on the bali is impulsive since it changes the ball's momentum drastieally, By comparison, the ball's weight will have a negligible effect on the change

15.3 CONSERVATkON OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES 237

in momentum, and thcreforc it is nonimpulsive. Consequently, it can be neglected from an impuise—momentum analysis during this firn°. If an impulse—momentum analysis is considered during the much longer time of flight after the racket—bal] interaction, then the impulse of the ba11's weight is important since it, ałong with air resistance, causes the change in the momentum of the bali.

Generally, the principle of linear impulse and momenturn or the,- conserxation of linear momentum is applied to a system of particks in order to detvrmihc thc finai velocities of the particfes just after the time period considered. By applying this principle to thc cntirc system, the internal impuises acting within the system, which may be unknown, are eliminałed from the analysis. For application it is suggested that the following procedure be used.

Free-Body Diagram.

• EstabIish the x. y, z incrtial framc of reference and draw the frec-body diagram for each particie M thc system in order to identify the internal and externai forces.

• The conservation of tinear momentum applies to the system in a direction which either has no external forces or the forces can be considered nonimpulsive,

• EstabIish thc direction and sensc M thc particIes' init iaI and final veiocities. If the sensc is unknown, assume it is along a positivc inertia] eoordinate axis.

• As an alternative procedura, draw the impulse and momentum diagrams for each particie M the system.

Momentum Equations.

• Apply the principle of linear impulse and momentum ur thc conservation of linear momentum in the appropriate dircetions.

• If it is necessary ło determine the internal impulse f F dr acting on only one particie of a system, then the particie must be holared (free-body diagram). and the principle of linear impulse and momentum musi be appfied to this particie.

• After the impulse is ca]culated, and provided the time At for which the impulse acts is known, then the average impuisive fora,' Fang can be determined from Fa = JF dt1 dł.

238 CHAPTER 1 ar KINET[CS OF A PARTICEE: IMPULSE AND MOMENTUM

The I5-Mg boxcar Ais coasting at 1.5 m/s on the horizontal track when it encounters a I2-Mg tank car B coasting at 0.75 m/s toward it as shown in Fig. 15-8a. If the cars collide and couple together. determine (a) the speed of both cars just after the coupling, and (b) the average force between them if the coupling takes place in 0,8 s.

I .5 rn /s...,. 0.75 misA

(a)

SOLUTION

Part (a) Free-Body Diagram.* Herc we have considered bod' carsas a single system, Fig. 15-8b. By inspection, momentum is conserved

xin the x direetion since the coupling force F is interna! to the systemand will therefore cancel out. It is assumed both cars, when coupled, move at v, in the posiłive x direction.Conservation of Linear Momentum.

—F(

b )

(± , ) m A ( v A ) i i n a ( v ł ( ) [ ( r n A ' ,Wy ,

(15 000 kg)(1.5 m/s) — 12 000 kg(0.75 m/s) -= (27 000 kg)v,

vZ = 0.5 m/s Ans.

Part (b). The avcragc (impuIsive) coupling forec. Faul, can be dctermincd by applying thc principle of lincar momentom to ehher one of the cars.

Free-Body Diagram. As shown in Fig. 15-8c, by isolating the hnxcar the coupling force is exrernal to the car.

Principle of impulse and Momentum. Since f F dt = Favg F,g(0.8 s), we have

(±) mA(vA), i F di = mAy2

(15 000 kg)(1.5 m/s) — s) = (15 000 kg)(0.5 m/s)

F „ g = 18.8 kN An

NOTE: S011160111 was possible bre since the boxcars finał velocity was obtained in Part (a), Try solving for F„s by applying the principic of impulsc and momentom to the tank car.

*Only horizontal forws arc shown on the free-body diagram.

V

I Lii -F(c)

Fig. 15-8

15.3 CONSERVATION OF LINEAR MO?vIENTUM FOR A SYSTEM OF PARTICLES 239

EXAMPLE r

(a)

150(9.81) N 150(9.81) N

(b)

Fig. 15-9

The bumper cars A and B in Fig. 15-9a each have a mass of 150 kg and are coasting with the velocities shown before they freely collide head on. If no energy is lost during the collision, determine their velocities after collision.

SOLUTIONFt ee- Body Diagram. The cars will be considered as a single system. The free-body diagram is shown in Fig. 15-9b.Conservation of Momentum.

rrig(yA)L + mA( va) = mA (7-',4)2 mn( v/3)2

(150 kg)(3 m/s) + (150 kg)( -2 rnfs) -= (150 kg)(vA)2 + (150 kg)(vB)2

(vA)2 = (VB)2 (1)

Conservation o# Energy. Since no energy is lost, the conservation of energy theorem gives

T1 + = T2 + V2

i 2 i ,2 n i 2 i—111A(VA)r + -1,18(vEth + u = -122A(v,4)2 + -,m,5(v8)i + O2 2 2 2

1 1-(150 kg)(3 rn/s)2 + -(150 kg)(2 m/s)'2 + O -= -(150 kg)(1).4).2 2 2

+ -1(150 kg)(vB); + O 2(vA):_f + (v8); = 13 (2)

Substituting Eq. (1) into (2) and simplifying, we get(vB)22 - (vs)2 - 6 = O

Solving for the two roots.

(vR)2 = 3 m/s and (vR)2 = -2 m/s

Since (v8)2 = -2 ni/s refers to the velocity of B just before collision, then the velocity of B just after the collision must be

(vB)2 = 3 mis -›

Substituting this result into Eq. (1), we obtain(vA)2 = 1 - 3 m/s = -2 m/s = 2 m/s .1/2


EXAMPLE

An 800-kg rigid pile shown in Fig.15-10a is driven into the ground using a 300-kg hammer. The hammer falls from rent at a height yo 0.5 ni and strikes the top of the pile. Determine the impulse which the pile exerts on the hammer if the pile is surrounded cntirely by loose sand so that after striking, the hammer does nor rebound off the pile.

SOLUTION

Conservation of Energy. The velocity at which the hammer strikes thc pile can be determined using the conservation of energy equation applied to the hammer. With the datum at the top of the pile, Fig. 15-10a. we have

To + •=. T1 + V1

2nr.H(vH)o WHYo 2 m r-r( t51-r) WITY1

O + 300(9.81) N(0.5 m) —1 (300 kg)(vH)2f + Oti = 0,5m 2batum (VH)1 = 3.132 niis

Sand Free-Body Diagram. From the physical aspects of the problem, thefree-budy diagram of the hammer and pile, Fig- 15-1 Ob, indicates that during the short dme from just before to just after the collision, the, wcights of thc hammer and pile and thc resistanec foree F, of thc sand arc all nonirnpuisive The impulsive foree R is internal to the system and therefore caneels. Consequently, momentum is conserved in the vertical direction during this short time.

COnSerVatOln of Momentum. Since the hammer does not rebound off the pile just after collision, then (vi)2 = (V p)? = V?.

(+ MI(r) H)] + irip<VAI InfiVz inpVz(30(1 kg)(3.132 mis) + O = (300 kg)vz + (800 kg)v?, .v

2 = 0,8542 mis

Principle of impulse and Momentum. The impulse which the pile imparts to the hammer can now be determined sincc v2 is known. From the free-body diagram for the hammer. Fig, 15-10c. we have

frr

(+ nuH(vtr), + f Fycfr = frzHvz

(300 kg)(3.132 mis) — J R dt = (300 kg)(0.8542 mis)

Y JR dr = 683 N • s 0%

(c) NOTE: The equal but opposite impulse acts on the pile.Try finding tltis

Fig. 15-10 impulse by applying the principle of impulse and mumenturn w the pile.

(a)

WH O

—

R

R

4 W F O Y

i Y

F y ~ O

( b )

W y s p

15.3 CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES241

EXAMPLE

(a)

F

( b )

Fig. 15-11

The 80-kg man can throw the 20-kg box horizontal3y at 4 m/s when standing on the ground. If instead he firmly stands in the 120-kg boat and throws the box, as shown in the photo, determine how far the bont will move in three seconds. Neglect water resistance.

SO LUTION

Free-Body Diagram. If the man, boat, and box are considered as a single system, the horizontal forces between the man and the boat and the man and the box become internat to the system, Fig. 15-11a, and so linear momentum will be conserved along the x axis.

Conservation of Momentum. When writing the conservation of momentum equation, it is important that the velocities be measured from the same inertial coordinate system, assumed herc to be fixed. From this coordinate system, we will assume that the boat and man go to the right while the box goes to the left, as shown in Fig. 15-1 lb.

Applying the conservation of linear momentum to the man, boat, box system,

O +O + O-= (nim+ mg} Vb - Mbox vbox

O = (80 kg + 120 kg) vb — (20 kg) vbox

vbox --- 10 vb (1)

Kinematics. Since the velocity of the box relative to the man (and baw), vbox/b, is known, then vb can also be related to vbox using the relative velocity equation.

()) Vixn = V b + Vbax/b

—vbox = Vb — 4 m/s (2)

Solving Eqs. (1) and (2),

vbox = 3.64 m/s 4-

vb = 0.3636 m/s --)

The displacement of the boat in threc seconds is therefore

ryt = (0.3636 m/s)(3 s) =1.09 m Ans.

242 CHAPTER 15 KlNETrCS OF A PARTICLE: IMPULSE AND MOMENTUM

The 1200-lb cannon show] in Fig. 15-12a fires an /3-.1b projectile with a muzzle velocity of 1500 ft/s measured relałive to the cannon. If firing takes place in 0.03 s. determine the recoil velocity of the cannon just after firing. The cannon support is fixed to the ground, and the horizontal rewii of the cannon is absorbed by twa springs.

SOLUT1ONPart (a) Free.Body Diagram.* As shown in Fig. 15-12b, we will consider the projectile and cannon as a single system, since, the impulsive forces, F and —F. between the cannon and projectile are internuj to the system and will therefore cancel from the analysis.Furthermore, duńng the time0.03 s, thc twa Duoli springs whichare attached to the support each exert a nonirnpuisjve forte F. on thecannon. This is because is very short, so that during this time thecannon only moves through a nery smalt distance Consequently,FS = ks 0, where k is the spring's stiffness, which is also consideredto be relatively smalL Hence it can be concluded that momentum for the system is conserved in the horizonial direalon.

Conservation of Linear Momentum.

rn,(/),)1 + mp(vp)I = —me(v,)2 mp(vp)2

( 1200 1h )(vJ2 + ( 1b )(vp)2ti 32.2 nis2 32.2

ftis-

(vp)2 = 150 (v, )2

O + O =

( 1 )

2F, —FF(b) Fig.

15-12

These unknown velocities are measured by a fixed observer. As in Example 15-7, they can also be related using the relative velocity equation.

(12p)2= (yr)2+(vp)2= —(vc)2+ 1540 ftis (2)

Solving Eqs. ) and (2) yiełds04)2= 9,93 ft/siq%0,,p)2 = 1490 ft/s

Apply the prineiple of impulse and momentum to thc projectile (or thc cannon) and show that thc average impulsive forte on the projectile is 123 kip.

NOTE; If the cannon is firmly fixed ta its support (no springs), the reactive forte of the support on the cannon musi be considered as an external impulse to the system, since the support would ali" no movement of the cannon. In this case momentum is not conserved.

herizontal forces are shown on the free-body diagram.

..- Rewii spring

15.3 CONSERVATCN OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES243• FUNDAMENTAL PROBLEMS

F15-7. The frcight cara A and B have a mass of 20 Mg and 15 Mg. respectively, Determine the velocity of A after collision if the cars collide and rebound. such that B moves to the right with a speed of 2 m/s. If A and 8 are in contaet for 0.5 s, find the average impulsive forte which acts bet ween [hem,

3 Inis 1.5 misA B

F15-10. The spring is fixed to błock A and block B is pressed against the spring. If the spring is compressed s --- 200 mm and then the blocks are released, determine their velocity at the instant No& 8 loses contaet with the spring, The masses of blocks A end 8 are 10 kg and 15 kg, respectively.

k = 5 ki \ /m

F15-7F15-10

F15-8. The cart and package have a mass of 20 kg and 5 kg, respectively. If the cert has a smooth surface and it is initially at rest, while the velocity of the package is as shown, determine the finał common velocity of the cart and paelcage after the Impact.

F15-11, Blocks A and B have a mass of 15 kg and 10 kg, respectively.If A issiationary and B has a velocity of 15 m/s jest before collision, and the błocka couple together after impast, determine the maximum compression of the spring,

k = i0 kNjm

15 m/s

F15-11.

F 1 5 - 8

F15-9. The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it coltides with the stationary block B cif mass g kg. If the twa blockscouple together after collision, determine their common velocity immediałely after collision.

v,, = 5 rni

F 1 5 - 9

F15-12. The cannon and support without a projectfie have a mass of 250 kg. If a 20-kg projedne is fired frorrk the cannon with a velocity of 400 m/s, measured reiadve to the eannon, determine the speed of the projectiIe as it leaves the bar-rel of the cannon. Negled rolling resistance.

FL5—I2

244CHAPTER 15 KINETrC5 OF A PARTICLE IMPULSE ANO MOMENTUM

PROBLEMS

15-35. The bus B bas a weight of 15 0001b and is traveling to the right at 5 ft/s. Meanwhile a 3000-lb car A is traveling at 4 ft/s to the łeft. If the vehicies crash head-on and becorne entangled. determine their common velocity OSI atter the collision. Assume that the vehicles are free ta roll during collision.

v, — 5 ftis

Prob. 15-35

*15-36. The 50-kg boy jurnps on the 5-kg skateboard with a horizontal velocity of 5 m/s. Detcrrnine the distance s the boy reaches up the rnclined piane before momentarily coming to rest.Neglect ihe skateboard'srolling resistance.

15-37. The 2.5-Mg pickup truck is towing the 1.5-Mg car using a cabłe as shown. If the car is initially at rest and the truck is coasting with a velocity of 30 kin/h when the Gable is slack. determine the common velocity of the truck and the car just after the cabIc bccomes taut. Also, find łhe Loss of energy.

15-38. A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m/s in the opposite direction. 1f the cars meet and couple together, determine the speed cif both cars just after the coupling. Find the difference between the tata] kinetic energy before and after coupling has occurred, and explain qualitafively whł.11 happened to this energy.

15-39. The car A bas a weight of 4500 Ib and is traveling to the right at 3 ft/s. Meanwhile a 3000-lb car B is traveling at t, ftis to the left. If the cars crash head-on and beconie entangled. determine Lilek cornmon velocity just after the collision. Assume that the brakes are not applied during collision.

v,i = 3 ft,is v, 6 ft7s

Prob. 15-39

15-40. The 200-g projectile is fired with a velocity of X300 m/s towards the center of the 15-kg wooden block. which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 300 m/s, determine the velocity of the block just after the projectile emerges. How long does the block shde on the rough surface, after the projectile emerges. hefore it coines to rest again? The coefficient of kinetic frietion between the surface and the block is µk =, 0.2.

900 tn / s

a rk

Be fere

Prob. 15-36

30 krnih

Prob. 15-37 A fter Prob. 15-40

15.3 CONSERVATON OF ilNEAR MOMENTLIM FOR A SYSTEM OF PARTICLES 245

15-41. The block has a mass of 50 kg and rests on thc surface of the carl having a rnass of 75 kg. If the spring which is attached to the cart and not the błock is compressed 0.2 m and thc system is reieased from rest, determinc the speed of the block relative to the ground after the spring becomes undeformed. Neglect the mass of the cart's wheels and the spring in the calculation. Also neglect frict łon. Take k = 300 N/rn.

15-42. The block bas a mass of 50 kg and rests an the surface of the carl having a mass of 75 kg. If the spring which is attached lo the cart and nol the block is compressed 0,2 m and the system is releascd from rest deterrnine the speed of the błock with respect to the can after the spring becomes undeformed. Neglect the mass of Ihe wheels and the spring in the calculation,Also neglect fricłion,Take k --- 300 Nim.

*15-44. Twa men A and B, each having a weight of 160 lb,stand on the słationarycarl. Each then runa with aspeed of 3 ftis mcasured relative to thc cart. Determine thc final speed of the cart if (a) A runs and jumps off. t hen B runa and jumps off the same end, and (b) bolh run al the same time and jurnp off at the same time. Neglect the masa of the wheels and assume the jumps are made horizontally.

Prob. 15-44

b 'WH1

Prohs. 13-11/4215-45. The block of masa tra travels at in the directionOj shown at thc top of the smooth slope. Detcrinine its speed and ds direchon O, when it reaches the bortom.

15-43. The three freight cars A. B. and C have masses of 10 Mg, 5 Mg, and 20 Mg, respectively. They are traveling along the track with the velochies showri. Car A collides with car B first, followed by car C. If the three cars couple together after coliision, determine the common velocity of the cars after the twa collisions have taken place.

Ii2

Prob. 15-43

Prob. 1545


15-46. The harge B weighs 30 000 lb and supports an automobile weighing 3000 Ib. If the barge is not tied tct the piet P and someone drives the automobile to the other side of the barge for unloading, determine how far the barge tnoves away from the pter. Neglect the resistance of the water.

15-49. The man M weighs 150 Ib and jumps onto the boku 8 which hasa weight of 200 lb. If he has a horizontal component of velocity rełative to the boru of 3 ft/s, just before he entcrs the boat, and the boat is traveling vg -= 2 ft/s away from the pier when be makes the jump. determine the resulting velocity of the man and boat.

15-50. The man M weighs 150 Ib and jumps onto the boat B which is originalIy at rest, If he bas a horizontal component of velocity of 3 ftjs jusu before he enters the how, determine the weight of the haat if it has a velocity of 2 ft/s once ihe man enters it.

200 ft

b . • _,Ił~••••••••~WE1 0 1 1

Prob. 15-46

15-47. The 30-Mg freight car A and 15-Mg freight car B are moving towards each other with the velocities shown. Determine the maximum compression of the spring mounted on car A. Neglect rolling resistance.

m

Probs. 15-4915020 km/h

A k 3 MN/m10 km/h

Prob. 15-47

*15-48. The harge weighs 45 040 lb and supports Iwo autcmnhiles A and B. which weigh 4000 Ib and 3000 lb, respectively. If the automobiles start from rest and drive towards cach other, accelcratIng at n4 = 4 ft/s2 and aN = 8 ft /s2 untfl they reach a constant speed of 6 ks eelalivgi to the barge, determine the speed of the barge junt before the automobiles collide. How mach time does this takt? Originally the barge is at rest. Negleet water resistance.

15-51. The 20-kg package has a speed of 1.5 m/s when it is deiIvered to the smooth ramp. After sliding down the ramp it landa onto a .10-kg cart as shown. Determine the speed of the cart and package after the package stopa sliding on the cart.

Prob. 15-51Prob. 15-48

15.3 CONSERVATCN OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES247

15-52. The free-rolling ramp bas a m ass of 40 kg. A W-kg crate is teleased from rest at A and slides down 3.5 1T1

to point B. If the surface of the ramp is smooth, determine the ramps speed when the crate reaches B. Also, what is the velocity of the crate?

Prob. 15-52

15-53. The /30-lb boy and 60-lb girl walk towards each other with a constant speed on the 300-Ib cart. If their velocities, measured relative to the cart, arc 3 ftis to the right and 2 ft/s to the left, respectively. determine the vełocities of ihe boy and girl during the ~non. Also, find the distance the cart bas traveled at the instant the boy and girl rieet.

15-54. The 80-1b boy and 60-Ib girl walk towards each other with constant speed on the 300-113 cart. If their velocities measured relative to the cart are 3 ftis to the right and 2 ft/s to the left, respectively. determine the velocdy of the carl while they are wałking.

15-55. A tugboat T having a masa of 19 ).4g is tied to a barge B having a mass of 75 Mg. If the rwe is "ciasne" such that it has a stiffness k = 600 kNitn. determine the maximum stretch in the rope during the initial towing. flUginally both the tugboat and barge are moving in the same direction with speeds (vr)i = 15 km/h and (yR) = 10 km/h, respectively. Neglect the resistance of the water.

(vB)1 1v7){

13T----

Prob. 15-55

15-56. Two boxes A and B. each having a weight of 160 lb, sil on the 500-ib conveyor which is free to roli on the ground. If the bell starts from rest and begins to run with a speed of 3 ft/s,. determine the finał speed of the conveyor if (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fali off together.

®I0101[01101[010~5.

Prob, 15-56

15-57, The 10-kg bloek is held at rest cm the srnooth inelined piane by the stop błock at A. If the 10-g bullet is traveling at 300 m/s when it becomes ernbedded in the 10-kg błock, determine the distance the block wili slidc up along the piane before momentarily stopping.

I - 2 0 1 1 - 1

Probs. 15-53/54Prob. 15-57


Piane of contact15.4 Impact

VA

15

V r t

Oblique impaet

( b )

Fig. 15-13

a aRequire

(L,A)1 (v8)1

Refore impact

( a )

Impact occurs when twa bodiesconlde with each other during a very skon vs

Line ofimpacl Period af time, causing relatively large (impulsive) forces to be exerted

between the bodies. The striking of a hammer on a nail, ar a golf club on▪bali, are common examples of impact loadings.In generał, there are two types of impact. Central impacr occurs when

the direction af triatlon of the mass centers af the twa colliding particles is along a line passing through the mass centers of the particles. This line is called the line of impact, which is perpendicular to the piane of contact, Fig. 15-13a. When the motion af one ar both of the particles make an angie with the line of impact, Fig. 15-131, the impact is said to

Line cif im ael be ublique impaer.

Central Impact. To illustrate the method for analyzing the mechanicsof impact, consider the case involving the central impact of the twa particles A and B shown in Fig, 15-14.

• The particles have the initial momenta shown in Fig. 15-14a.Provided (/‚,1)/ (ts)1, collision will eventually oecur.

• During the collision the particles must be thought of as deformable ar nonrigid. The particles will undergo a period of deformation such that they exert an equal hut opposite deformation impulse f P dt on each ather, Fig. 15-14b,

• Only at the instant of maximum deformation will both particles move with a common velocity v, since their relative motion is zero, Fig. 15-14c.

• Afterward a period of resiitation occurs, in which case the particles wili either return to their original shape ar remain permanentIy deformed. The equal but opposite resdiution impulse f R Lit pushes the particles apart tram one another, Fig. 15-14d. In reali ty, the physical propertics of any two bodies arc such that the deformation impulse will always be greater than that of restitution,c.1 P dr -› f R dł.

• Just after separation the particles will have the fina] momenta shown in Fig. I5-14e, where (t.3/3)2

rCentra impauti

( t)

Piane of contact

f P dr v IR di —JR dt m A(v .132 in B( v B)2— ł >

( A B

AEffect of A on B Effeetof B on A • OEffect uf A on B Effect of B on A

!Defom abon impulsel

A 13IVIdrriurn deformationj

Restitution impuise

A (1 '8)2> (v, i )z B

Atter impaet

(b) (c) (d) (e)

Fig. 15-14

15.4 ImpAcr 249

In most problenis the initial velocities of the particics will be knowa, and ii will be neeessary to detennine their finał velocities (vA)2 and (vy)2. In this regard, momentum for the sysrem of partieles is conserued sinee cluring collision the internat impulses of deformation and restitution cancel. Hence, referring to Fig. 15-14a and Fig. 15-14e we require

) A(I, A) t + n2B(v8)1 = nzA(v,4)2 + m8(2)8)2 (15-10)

In order to obtain a sccond cquation n ecessary ta solvc for (1),4)2

and (y8)2, we most apply the principie of impulse and momentum to each particie. For example, during the deformation phase for particie A. Figs. 15-14a, 15-14b, and 15-14c, we have

inA(VA)] — f P dr MAV

For the restitution phase, Figs. 15-14c, 15-14d. and 15-14e,

-) l mAv — (R dr = m4(vA)2

The ratio of the restitution impulse to the deformation impulse is called the coefficient of restitution, e. From the above equations. this value for particie A is

ł R ds v — (v.4)2

e i P d r ( v

In a similar rnanner, we can esłablish e by considering particie B, Fig. 15-14. This yields

R dse—

p d t

(V8)2 — V

If the unknown v is eliminated from the above twa equations, the coefficient of restitution can be cxpressed in terms of the partieles' initial and fina] velocities as

(v $)x — (v A)2e —

(2, A)] — (v 8)1

250 CHAPTER 15 KiNETlCs oF A PARTICLE iMPULSE ANO MOMENTUM

The guaiity af a manuraUured tennis bali is measured by the height of its holmu. which can be related to itscoefficient of restitution. Using the mechanice ol oblique impact, engineers can design a separation device to remove substandard tenni,s balis from a production linc.

The mechanice of pool depends uponapplication ot the conservation of

momentum and the coefficient of restituliM

Providcd a valuc for e is specifled, Eqs. 15-10 and 15-11 can be soived simultaneously to obtain (v,4)-2 and (DB)2. In doing sa, however, it is important to carefully establish a sign convention for defining the positive direction for both vA and v8 and then use it c.onsisteridy when writing both equations. As noted from the application shown, and indicated symbolically by the arrow in parentheses, we have defined the positive direction to the right when referring to the mołions of both A and B. Consequently, if a negative value results from the solution af eithcr (vA)2or (t,$)2, it indicatcs motion is to the left.

Coefficient of Restitution. From Figs. 15-14a and 15-14e, it is scen that Eq. 15-11 states that e is equal to the rafio of the relative velocity of the partieles' separation psl rrfrer impaer, (v8)2 — (vA)2, to the reIative velocity of the particles' approach :mm before impact, (vA)[ — (v8) y. By measuring these rełative velocities experimentally,it has heen found that e varies appreciably with impact veloeity as well as with the size and shape of the colliding bodies. For these reasons the coefficient af restitution is rebabie miły when used with data which closely approxirriate the conditions which were known to cxist when rricasurements of it were made. In generał e lias a value bctween zero and one, and one should be aware of the physical rneaning of these Iwo limits.

Elastic Impact (e = 1). If the collision between the two particlesisperfecity damie, the deformation impulse (.I P is equal and oppositeto the restitution impulse f R dr). Although in reality this can never be achieved, e = 1 for an clastic collision.

Plastic Impact (e = 0). The impact is sald to be inelasrie or plasric when e = O. In this case there is no restitution impulse lf R dr = O), so that after collision both partcles couple or stick rogether and move with a comm o n velocity.

From the above derivation it should be evident that the principle of work and energy cannot be used for the analysis of impact problems since it is not possible to know baw the internal forces of deformation and restitution nary er displace during the collision. By knowing the particle's velocities beforc and aftcr collision, however, the energy less during collision can be calculated on the basis of the difference in theparticle's kinetic energy. This energy less, :-.1/1 -2 = — ź T1, occursbecause some af the initial kinetic energy cif the particie is transformecl into thermal energy as well as creating sound and localized deformation of the materiał when the collision occurs. In particular, if the impact is perfeetly elasric, no energy is lost in the collision; whereas if the collision is plamic, the energy lost during collision is a maximum.

15.4 IlvipAcr 251

Procedure for Analysis (Central Impact)

In most cases the final velocities' nf twa s-mooth particles are to be determined just after they are subjected to direct central impact. Provded the, coefficient of restitution, the mass of each particie, and each particle's initial velocity just befr)re impact are, Icnown, the solution to this problem can be obtained using the foliowing two equations:

• The conservation of momentum applies to the system of particles,

• The coefficient of restitution, ,e = [(yE), — (24)1/1:(t5A)E — (vB)/I. relates the relative velocities of the particles along the line of impact, just before and just after cullision.

When applying these two equations, the sense of an unknown velocity can be assumed. If the solution yields a negative magnitude, the velocity acts in the opposite sense.

Oblique Impact. When oblique impact occurs between two smooth particies, the particles tnove away from each other with velocities having unknown dircctions as well as unknown magnitudes. Provided the initial velocities arc known, then four unknowns arc prescnt in the problem. As shown in Fig. 15-15a, these unknowns may be represented either as (v.4)2, (y8),, 02, and ci), or as the x and y components of the final velocities.

łmB{vity}.2

r i z ( + , , B , ) 1 _±on f r j o ( F n 1 3 . x ) 2+=

tIn Ii(V ny.

ntA(v.4-..)ilk

fFrit ln,t(A.)7= ••■X

(b )

Fig. 15-15

0 2 2 Line of impactx

(vAL (vs)i

Piane of contact

(a)

(v,t)? (v8)A

.ro edure for Analysis (Oblique

If the y axis is established within the piane of contact and the x axis along the line of impact, the irnpulsive forces cif defonnation and restitution act ordy in the x direction, Fig. 15-15b. By resolving the velocity or momentum vectors finto components along the x and y axel, Fig. 15-15b, -ił is then possihle to write four

independent scalar equations in order to dełermine(vBx)2, and (yR,,)2.

•Momentum of the system is conserved along the line of impaci, x axis, so that irpt(?),.)1 = łni(v,),,•The coefficient of restitution, e = 1.(1'8,)2 — Ax)2],[(v Ax)1 (vBx)1], relates the rełative-velocitycorcpernentsof the particles along the line of iinpaa (x axis).

• If these two equations are solved simultaneausly, we obtain (vAx)2 and (v8x)2.

• Momentum vf particie A is conserved along the y axis. perpendicular to the line of impact, since no impuiseacts on particie A in this direction, Asa result /it,k(vAv)t = mA (vA (VA,.)1 =(VA-s.)2

• Momentum of particie Bis conserved along the y axis, perpendicular to the line of impact, since no irnpulseacts on particie B in this direction. Consequently (vBy)1 (1)13y)2.

Application of these four equations is 'ilustrator] in Example 15.11.

252 CHAP -1 -ER 15 KINETICS QF A PARTICLE: IMPULSE AFJO MOMENTUM

EXAMPLE

1 5 . 9

b a t u m

/ 5 i b 3 f t

6 l b(b )

The bag A. having a weight of 6 ib, is released from rest at the position= 0°, as shown in Fig. 15-16a. After falling to O = 90°, it strikes an

18-lb box B. If the coefficient of restitution between the bag and box is e = 0.5, determine the velocities of the bag and box just after impact. What is the loss of energy during collision?

SOLUTIONThis problem involvcs central impact. Why? Before analyzing the mechanics of the impact. however. it is first necessary to obtain the velocity of the bag just before it strikes the box.Conservation of Energy. With the datum at O = O°, Fig, 15-1612, we have

To -ł- Vo = T1 + Vi(

O + 0 = 1 616— 6 lb(3 ft);(vA)1 = 13.90 ft/s2 32.2 ft/s2)(vA):1->

Conservation of Momentum. After impact we will assume A and B travel to the Ieft. Applying the conservation of momentum to the system, Fig. 15-16c, we have

( ) niR(yn)i + mA(vA)1 =MB(V B)2 A(V A)2

6 lb 18 1) 6 ibO+k32.2ft/s2)(13.90ft/s)=(32:2ft/s2) )8)2+\32.2ft/ s2) (V A)2

( v A ) 2 = 1 3 9 3 _ 3 ( : ) 2

(1)

Coefficient of Restitution. Realizing that for separation to occur after collision (v8)2 > (vA),, Fig, 15-16c, we have

( v 8 ) 2 ( v A ) 2 ( v 8 ) 2 ( V A ) 2 ( • 4 ) e — 0 . 5 —

(vA)i — (u8)1' 13.90 ft/s — O(vA)2 = (v8)2 — 6.950 (2)

Solving Eqs. 1 and 2 simultancausly yields

(vA)2 = —1.74 ft/s = 1.74 ftis —> and (v8)2 = 5.21 ft/s <— Ans.

Loss of Energy. Applying the principle of work and energy to the bag and box just before and just after collision, we have

£U1_2 ł Ul

_2 —

T2 — Ti;

I 18 lb ,+

2

6 lb,)(1.74 ft /s)21

2 32.2 ft/s-„ )(5.21 ft/s)-

( ( 32.2 ft/s

El 2 ( 6113 32.2 ft/s2)(13.9 f"2]

51/1_2 = —10.1 ft • lb Ans.

NOTE: The energy loss occurs due to inelastic deformation during the collision.

0 , $ ) , = 0

(uA), = 13.90 ft,is Just bcfore impact

B

(vH)2 0,5)2

Just after impact

(c)

Fig. 15-16

15.4 ImPAcr 253

EXAMPLE

15.10

Bali B shown in Fig. 15-17a has a mass of 1.5 kg and is suspencled from the ceiling by a 1-m-tang elastie card. If the cord is stretched downward 0.25 m and the hall is released from rest, determine how far the cord stretches after the hall rebounds from the ceiling.Thestiffness of the cord is k = 800 Nim, and the coefficient of restitutiion between the hali and ceiling is e = 0.8. The bali makes a central impast with the ceiling.

SOLUTIONFirst we must obtain the veioeity of the bali just before it strikes the ceiling using energy methods, then consider the impulse and momentum between the hall and ceiling, and finally again use energy methods to determine the streteh in the cord.

Conservation af Energy, With the datura Iocated as shown in Fig. I5-17a, realizing that initially y = ya = (I + 0.25) m = 1.25 m, we have

+ = + 7111

4m(2,8); WEiyo + = 2in(vi3)1 + OO – 1.5(9.81)N(1.25 m) + 4(800 Nim)(0.25 rn)2 =kg)(v8)i

(vB), -= 2.968 mis 1`

The interaction of the bali with the ceiling will now be considered using the principles of impact.*" Sinee an un known portion of the mass of the ceiling is involved in the impast, the conservation of momentum for the bali–ceiling system will not be written. The "velocity" of this po lon of ceiling is zero since it (or inc earrn) arc assumed to remain at rest bod/ before and after impacł.Coefficient of Restitution. Fig.15-17b,

evu)2 - O (v8)2 - (?).4)2 0.8 -(+ i) e -(yA)1 (1-,B)3 0 – 2.968 m/s

(v8)2 = –2.374 m/s = 2.374 m/s

Conservation af Energy. TIne maximum stretch s:, in the cord canbe determined by again applying the couservationof energy equationto the bali just after collision. Assuming that y = y3 = (1 + 53)m.

Fig. 15-17c. then+ V2 = T3 + V3

nz( /)8)3 ł O = .1"M(VB)Ii ł Wyy3 + ksi

.4g 1,5 kg)(2.37 mis)2 = O — 9.81(1,5) N(1 m + s3) + 4800 N/m)s-i

40(15-j – 4,715s3 — 18.94 = O

Solving this quadratic equation for the positive ront yields53 = 0.237 m = 237 mm

„The weight of the bali considered a nonimpulsive „orce.

D2t11M-

k = 8 f f l N i m y = ( 1 0 2 5 ) m

B _L (a)

(vs}zi (y8)1= 2.97 m/s

W a t o m

( c )

Fig. 15-17

k = 800 Nlm y ( 1 s 3 ) m

254 CHAPTER 15 KINETICS OF A PARTICLE: IMPULSE AND MOMENTUM

EXAMPLE 15.11

Two smooth disks A and B. having a mass of 1 kg and 2 kg, respectively, collide with the velocities shown in Fig. 15-18a. If the coefficient of restitution for the disks is e = 0.75, determine the x and y components of the finał velocity of each disk just after collision.

SOLUTIONThis problem involves oblique impact, Why? In order to solne it, wc have establishcd the x and y axel along the line of impact and the piane of contact, respectively, Fig. 15-18a.

Resolving each of the initial velocities into x and y components, we have

(vAx)i = 3 cos 30° = 2.598 m/s 3 sin 30° = 1.50 m/s

(vBx)1 = -1 cos 45° = -0.7071 mis (vEN)j = -1 sin 45° = -0.7071 mis

Thc four unknown velocity components aftcr collision are assioned to act in the positive direction. Fig. 15-18b. Since the impact occurs in the x direction (line of impact), the conservation of momentum for both disks can be applied in this direction. Why?

Conservation of "x" Momentum. In reference to the momentum diagrams, we have

( -±› ) tnA(v/Ix)1 + nt8(14,)i = in,4(v4,)2 + inB(vB,),

ms(v8,.) I kg(2.598 m/s) -l-. 2 kg( -0.707 m/s) = I kg(vA)2 + 2 kg(v8)2ine(vB.110JFdt ts„,„( ,..)2(v,.)2+2(vg,)2. 1.184 (1)

Coefficient of Restitution (x).( v B x ) 2 ( v u ) 2 ( v a x ) 2 ( v , 4 1 ) 2

±›)e = 0.75 -=(vA.,) - (yR,;) i 2.598 m/s - (-0.7071 m/s)

(%)2 - (vA,)2 = 2.482 (2)

Solving Eqs. 1 and 2 for (vA,)2 and yields

(vAi).2 = -1.26 m/s =- 1.26 m/s (va,)2 = 1.22 m/s —> Ans.

Conservation of "y" Momentum. The mornenturn of each disk is conserved in the y direction (piane of contact), sincc the disks are smooth and therefore no external impulse acts in this direction. From

= Fig. 15-18b,

02 = 30.1° ("F' ) n/A(VA3)1 = mA(vAy)2;

(+ In8(1,8,)t =JnB(Vsy)(c)

Fig. 15-18NOTE: Show that when the velocity components are summed vecłorially, one obtains the results shown in Fig. 15-18c.

( v 5 )3 =1 ni /s

A ,\= 450

1 1 . ■11, " x

= 30° I.ine of impactB

(v,i)( = 3 mis -Piane of contact

(a)

InA(v Ay) Ił

-IFdt in..1(vA.)2

nzA(vA3.4

rrta(v8r)2ł

(h)

(1J,L)2 = 1.96 m Js

(vB)7 = 1.41 m/s

(vAy)2 = 1.50 m/s t A K +

( % ) 2 = — 0 . 7 0 7 m / s = 0 . 7 0 7 m / s A r r . y .

15,4 IMPACT 255• FUNDAMENTAL PROBLEMS

F15-13. Determine the eoefficient of restitution e between hall A and bali B. The velocities of A and B beforc and after the collision are shown.

8 mis 2 mis

R 13

"Refom colllsion

l ints9 mis

7.AB

After cothsien

F1-5-13

F15-14. The 15-Mg tank car A and 25-Mg freight car B travel towards each other with the velocities shown. If the coefficient of restitution between thc bumpers is e -= 0,6, determine the velocity of each car jurt after Ule collision,

5 mis 7 m/sA — 1 '

IN • P!M

F15-14

F15-15. The 30-lb package A has a speed of 5 ftis when it enters the smooth ramp. As it slides down the ramp. it strikes the 81:/-3b package B which is initially at rest. 1f the coefficient of restitution between A and Bis e = 0.6, determine the velocity of B just after the impast.

F15-15

F15-16. The bali strikes the smooth wali wicha veiocity of (v,), = 20 m/s. 1f the coefficient of restitution hetween the bal] and the wali is e = 0.75, deterrnine the velocity of the bali j ust after the impast.

F15-1615F15-17. Disk A weighs 2 ib and slides on thc smooth hori7ontat piane with a velocity of 3 ft/s. Disk B weighs 1 I Ih and is initialty at rest. If :trier Impact. A lias a velocity of I flis, parafie] to the positive x axis, determine the speed of disk B after impast.

F15-17F15-18. No disks A and B each have a weight of 2 Ib and the initial velocities shown jest before they collide. 1f the coefficicnt of restitution is e = 0.5, dctermine thcir speeds jus! atter impast.

F15-18

256CHARTER 15 KINET[CS OF A PARTICEE: IMPULSE AND MOMENTUM• PROBLEMS

15-58. A bali h aving a mass of 200g is released from rest at a height of 400 mm above R nery large fixed metal surface. If t hc bali rebounds to a height of 325 mm above the surface, determine the coefficient of restitution between the balf and the surface.

15-59. The 5-Mg truck and 2-Mg car are traveling with the frce-rolling vclocities shown just heforc they collide. After the colIision. the car rnoves witki a velocity of 15 km/h to the right relntive to the truck. Delermine the coefficient of restitution between the truck and car and the lass of energy due to the collision,

30 km/h

10 km/h

Prob. 13-59

*15-60. Disk A bas a mass of 2 kg and is sliding forwardon the smooth surface with a velocity = 5 m/s when itstrikes the 4-kg disk B. which is sliding towards A at (z.38)1 = 2 rn/s. with direct central impast. If the coefficient of restitution between the disks is r = 0.4, compute the velocities of A and B just after collision.

(vA)i 5 171/s {vn)1 = 2 mis

aA

Prob. 15-60

15-61. Block A has a mass of 3 kg and is sliding on a mugh horizontał surface with a velocity (),4)1 = 2 m/s when it makes a direct collision with block 8, which bas a mass of 2 kg and is originally at rest. If the collision is perfectlyelastic= 1), determine the velocity of each bied( justafter collision and the distance between the blocks when they stop słiding. The coefficient of kinetic friction between the blocks and the piane is fck =- 0.3.

Prob. 15-61

15-62. If twa disks A and B have the same mass and are subjected to direct central impact such that the collision is perfectly elastic (e = l), prove that the kinetic energy before collision equals the kinetic energy after collision, The surface upon which they słide is smooth.

15-63. Each ball has a mass m and the coefficient of restitution between the balls is e. If tlicy ale moving towards one another with a velocity v, determine their speeds after collision. Also. deterrnine their cornmon velocity when they reach the state of maximum deformation. Neglect the size of each bali.

Prob. 15-63

15-64. Thc three balls each have a mass m. If A has a speed v just before a direct collision with B. determine the speed of C after collision. The coefficient of restitution between each pair of balia is e. Neglect the size of each bali.

C

Prob. 15-64

15-65. A 1-lig bali A is traveling horizontalły at 20 ftis when it strikes a er-fb blask B that is at rent. 'f the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic frictian between the piane and the biock is tik = 0.4, determine the time for the biock B to stop sliding.

15.4 IMPACT 257

15-66. ff the gni throws the bali with a horizontal velocity of v, = 8 ft/s, determine (he dislance d so (ha( (be bali bounces anse on the smooth surface and [hen lands in the cup at C. Takc e 0.8,

15-69. A 300-g bal] is kicked with a velocity of v4 = 25 m/s at point A as shown. If the coefficient of restitution between the bal] and Ihe field is e = 0.4, determine the magnitude and direction H of thc velocity of the rebounding bal] at H.

Prob. 15-69

A t a

3 f t

=-15 m/s

0°

A

Prob. 15-66

15-67. The three hans each weigh 0.5 Ib and have a coefficient of restitution of e = 0.85, ff hall A is released from rest and strikes hall B and then bal B strikes hall C, determine the velocity of each hall after the second collision bas occurred. The haIls sitek without friction,

15-70. Two smooth spheres A and B each have a mass m.If A is given a velocity of whilc sphere B is at rest,determine the velocity of Biust after it strikes the wall.The coefficient of restitution for any collision is e.

15

A B

Prob. 15-67

15-68. The gir] throws the bałt with a horizontal velocity of v, = ft/s. If the coefficient of restitution between the ball and the ground is e = 0.8, determine (a) the velocity of thc hall just after it rebounds from the ground and (h) thc maximum height to which the bali rises after the first bounce.

Prob. 15-70

15-71, Tt was ahserved that a tennis hall when served horizontaHy 7.5 ft above the ground strikes the srnooth ground al B 20 ft away. Determine the initial velocity v A of the bal] and the velocity vo (and O) of the bali just after it strikes the court at B, Take e = [1.7.

115-72.. The tennis bali is struck with a horizontal velocitystrikes the smooth ground at B. and bounces upward at O

--- 30°. Determine the initial velocity v,. the fina] velocity v,, and the coeffieient of restitution between the bali and the ground.

P roba. 15-71/72

J

Prob. 15-68

258CHAPTER 15 KINETCS OF A PARTICLE: IMPULSE AND MOMENTUM

15-.73. The I lb bali is dropped from rest and falls a distance cif 4 ft before striking the smooth piane at A. If e = 0.8, dctermine the distance d to wherc it again strikes the piane at R.

15-74. The 1 Ib bali is dropped from rest and falls distance of 4 ft before striking the smooth piane at A. If it rebounds and in ż = 05 s again strikes the piane at B, determine the coefficient of restitution e between the bali and the piane. Also, what is the distance d?

T Qi

15-75. The I -kg bali is dropped from rest at point A, 2 m above the smooth piane. If the coefficient of restitutionbetween the bali and the piane is e detennine thedistance d where the hall again strikes the ]zlane.

15-76. A bali of mass m is dropped vertieally from a height ho above the ground. If it rebounds to a height of hl. determine the coefficient of restilution between the bali and the ground. •

flll

Prob. 15-76

15-71. The cue bałt A is given an initiai velocity (v.4)1 = 5 m/s. If it makes a direct collision with bali B (e = 0.8). determinc the veiocity of B and the ansie H jest atter it rebounds from the eushion at C (e' = 0.6), > ach hall hasa mass of 0.4 kg. Neglect their size.

Prob. 15-77

15-78. USing a shngshol, the boy flres the 0.2-lb marbie at the concrete wali. striking it at B. If the coefficient of restitution between the marble and the wali is e = 0.5, determine Lite speed of the narhle after it rebounds from the wali.

Probs. 15-73f74

= 75 ttfs

45°511

A•

ITI

00 ttProb. 15-75 Prob. 15-78

O

s

Prob. 15-81

Prob. 15-79

15-82. The 20-lb box slides on the surface for which p. k = 0.3. The box has a velocity = 15 ft/s when it is 2 ft from the plate. If it strikes the smooth plate. which bas a weight of 10 Ib and is field in position by an unstretched spring of stiffness k = 4001h/ft. determine the maximum compression imparted to the spring, Take e = 0.8 between the box and the plate.A3511111C that the plate slides smooth ly.

u= 15 his

15,4 ImPAcT 259

15-79. The sphere of mass rn falls and strikes the triangular block with a vertical velocity z.If the biock rests on a smooth surface and has a ,lass 3 m, determine iłs velocity jurt after the colIision,The coefficient of restitution is e.

15-81. The gir] throws the 0.5-kg bali toward the wali with an initial velocity 1./.4 = 10 m/s. Delemiine (a) the velocity at which it strikes the wal] at B, (b) the velocity at which it rehounds from the wali if the coefficient of restitution e = 0.5, and (c) the distance from the wali to where it strikes the ground at C.

*15-80. Block A. having a mass m, is rekased from rest. falls a slistance h and strikes the plate B having a mass 2 m, If the coefficient of restitution between A and Bas e, determine the velocity of the pinie just after col lision. The spring has a stiffness k.

A

Prob. 15-80

2 ft

Prob. 15-8215-83. Before a cranberry can make ii to your dinner plate. it must pass a bouncing test which rates its quality. Ifcranberries having an 0.8 are to be accepted, determinethe dimensions d and h for the barrier so that when a cranberry falls from rent at A it strikes the incline at B and bounces over the barrier at C.

T

Prob. 15-83


*15-84. A hall is thrown onto a rough floor at an anglie H. If it rebounds al an angle 4 and the coefficient of kinetic friction is jc, determine the coefficient of restitution e. Neglect the size of the bail. flint: Show that during impact. the average impulses in the x and y directions are relatedby tr„ = Since the time arf impact is the same,F„ = ;F. At or Fx =

15-85. A bali is thrown onto a rough floor at an angleof H = 45°. If it rebounds at the same angle = 45°,determine the coefficient of kinetic friction between thefloor and the bali. The coefficient of restitution is e 0.6.

Show that during impact. the average impulses in thex and y directions are related by f, = Since the time ofimpact is the same, P, 6r = Ar or =

15-87. TWo smooth disks A and f? each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collicie, determine their finaI veiocities just aftercollision.The coefficient of restitution is e 0.75.

*15-88. Two smooth disks A and f3 each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide. determine the coefficient of restitution between the disks if after collision B travels along a lina. 30' counterclockwise from the y axis.

(744)z 6 m/s

(vi,)] = 4 mis

Probs. 15-87/88Prob. 15-84185

15-86. The "stone" A used in the sport of curling slides over łhe icc traek and strikes another "stane" B as shown. 1f each "stane" is smooth and lias a weight of 47 lb, and the coefficient of restitution between the -stones" is e = determine their specds just after collision. Iiiitially A lias a velocity of 8 ft/s and B is at rest. Neglect friction.

15-89. Two smooth disks A and B have the initial velocities shown just before they collide at O. If they havemasses = 8 kg and mn. = 6 kg. dctermine their specdsjust after impact.The coefficient of restitution is e = 0.5,

Prof". 15-86 Prob. 15-89

— :3 ft

15.4 IMPACT261

15-90. 1f disk A is sliding along the tangent to disk R and strikes B with a veioc-ity M. determine the velocity of B after the collision and compute the boss of kinetic energy during the coliision. Negłect friction, Disk Bis originally at rest. The coefficient of restitution is e. and each disk has the same sile and masa m.

15-91 TWo smoolh coins A and B. cad] having the same mass, siide on a srnooth surface with the rnotion shown. Determine the speed of ach coin after col1ision if they move off along the dashed paths. film: Since the line of impact has not been defined. apply the conservation of momentum along the x and y axel, respectively.

fi

Prob. 15-90

15-91. Two disks A and B weigh 21b and 51b, respectively. If they are sliding on the smooth horizon tal piane with the velocities shown, determine their velocities jurt after impact. The coefficient of restitution between the disks is e = 0,6,

Prob. 15-92

15-93. Disks A and R have a masa of 15 kg and 10 kg. respectively. H they arc sliding on a smooth horizonial piane with the velocities shown. determine their speeds jurt after impact. The coeffident of restitution bctween them is e -= 0,8,

Prob. 15-91Prob. 15-93

262 CHAPTFR 15 KINFT[CS O F A P A R T I C L E : I m F u L $ F A , N o M O M E N T U M

15.5 Angular Momentum

The angular momentum of a particie about point O is defined as the "moment" of the particle's linear moment urn about O. Sincc this concept is analogous to finding the moment of a force about a point, the angular momentum, Ho, is somełimes referred to as the moment of momenturn.

Scala r Formulation. If a particie moves along a curve lying in the x-v piane, Fig. 15-19, the angular momentum at any instant can be

7 ) 7 '

determined about point O (actually the z axis) by using a scalarmv formulation.The magnuude of Ho is

Fig. 15-19 (Ho), = (d)(my) (15-12)

o

z

Fig. 15-20

Flere d is the moment arm or perpendicular distance from O to the 'inc af action af mv. Common units for (110), are kg .m2is or sług ftĄ/s. The diredion of I1 is defined by file right-hand rule.As shown, łhe curl of the fingers of the right hand indicates the sense of rotation of mv about O, w that in this case the thurnb (or H0) is direcłed perpendicular to the x-y piane along the +z axis.

Vector Formulation. Tf the particie moves aiong a space curve, Fig. 15-20, the vector cross product can be used to determine the anffidar momentum about O. In this case

Ho r sny (15-13)

Here r denotes a position vector drawn from point O to the particie. As shown in the figure, Ho is perpendicular to the shaded piane containing r and mv.

In order to cvaluatc the cross product, r and mv should be cxpressed in terms of thcir Cartesian components. so that the angular momentum can be determined by evaluating the determinant:

i j k

Ho = r,-nmx

r,

mvY

r,rrlyz

(15-14)

15.6 RELATION BETWFEKI MOMENT OF A FORCE ANC) ANGULAR MOMENTUM 263

15.ó Relation Between Moment of a Force and Angular Momentum

The moments about point O of all the forces acting on the p-article in Fig. 15-2 I a can be rela led to the particle's angolar momentom by appIying the equation ot motion.11 the mass of the particie is constant, we may writc

= m v

The =lents of the forces about point O can be obtained by perfnrming a cross-prnduct multiplicałion of cech sicie of this equation by the position vector r, which is measured from the x, y, z inertial frarne of reference. We have

/Mo = r X !F = r X irriT

inc r tial cor>rdinatesyste in

(a)

Fig. 15-21

From A ppendix B, the derivative of r X mv can be written as

= d— (r X mv) = i. X mv + r X mvdr

The first term on the right side, r X mv = rnii X i.) = 0, since the cross product of a vcctor with itself is zero, Hance. the above equation becomes

2,1WIa=H0 (15-15)

whieh states that the resultant mytnem` about point O of aii the forces acting on the particie is equal to the tamte rate of change of the parricie's angular mornentum aboartpoofrt O. This result is similar to Eq. 15-1, i.e.,

SF =i. (15-16)

Here L = mv, sa that the resultant force acting on the particie is equal ro rhe drne rate of change of the particle's linear momenrum.

From the derivations, it is scen that Eqs. 15-15 and 15-16 are actually another way of stating Newton's second law of modom. In other sections of this book it will be shown that these equałions have many practical applications when extended and appbed to problems involving either system of part'icles or a rigid body.

264CHARTER 15 KINET[CS OF A PARTICEE: IMPULSE AND MOMENTUM

System of Particles. An equation having the, same form as E. 15-15 may be derived for the system of particles shown in Fig. 15-21b. The forecs acting on the arbitrary iti] particie of the system consist of a resultant external forte Fi and a rcsultant inłernal forte f,. Exprcssing the momcnts cif these forces about point O. using the form of Eq. 15-15, we have

Pi ;XF,)+(r ;Xf r)= (H i)0

Here (1:11)0 is the tirne race of change in the anguIar momentum of the ith particie about O. Similar equations can be written for each of the other particies of the system. When the results are summed vectorially, the result is

i(r. x Fi) + 5(r, X = £(14i)a

The second term is zero since the internal forces occur in equal but opposite colIinear pairs, and hence the moment of each pair about point O is zero. Dropping the index notation, the above equation can be written in a simpIified form as

łA1(..)1:10(15-17)

which states that the sum of the moments about point O of ali the externał forces acting on a system of partieles is equal ro the time race of change of the tołal angolar momentom of the sysrerrx cibora point O. A Ithough O bas been ehosen herc as the origin of coordinates, i t actually can represent any fixed point in the inertial framc of reference.

ertial coordinate system

( b )

Fig. 15-21 (cont.)

15.6 RELATION 8ETWEEN MOMENT OF A FORCE ANO AniGuLAR MOMENTUM 265

EXAMPLE

1113E.M.=~

The box shown in Fig. 15-22a has a mass m and travels down the smooth circuIar ramp such that when it is at the angle O it has a speed y. Determine its angular momentum about point O at this instant and the rate of increase in its speed. i.e.,

O

(a) (b)

Fig. 15-22

SOLUTION

Since v is tangent to the path, app]ying Eq. 15-12 the angułar momentum is

Ho = nnv Ans.

The rate of increase in its speed (dv i dł) can be found by applying Eq. 15-15. From the free-body diagram of the box, Fig. 15-22b, it can be seenthat only the weight W = contrihutes a moment about point O,We have

Ć+XMo = No;Jng(r sin O) = —dr(tlny)

Sinec r and tri aro congtan t,

dv,rigr sin O = rmdt

dv— g s i n Odr

A n s .

NOTE: Thi's same result can, of course, be ubtained from the equation of muflon applied in the tangential direction, Fig, 15-22b, i.e.,

mg sin B = m(dv)

dtdv = g sin dr

map;

2 6 6 CHARTER 15 KINET[CS OF A PARTICEE: IMPULSE AND MOMENTUM

15.7 Principle of Angular Impulse and Momentum

Principle of Angular Impulse and Momentum. IfEq.15-15 is rewritten in the form N40 d1 = d110 and integrated, assuming that attime t — r[, Ho — (H0)1 and at tiule t — Ho =- (Ho),, we have

2.i Mo = (1-10)2 — (Ho),

(H 0)1 + ł jr Mo dr (H0)2( 1 5 - 1 8 )

This equation is referred to as the principle of angular impulse and momentum. The initial and final angular momenta (Hn)] and (Ho), are defined as the moment of the Linear momentum of the particie (H0 = r x mv) at the instants i[ and t2, respectiveiy. The second term on the left side, lmodr, is called the artgalar impidse.It is determined by integrating, with respect to tirne. the moments of all the forces acting on the particie over the tinie period tl to t,. Since thc moment of a forte about point Cl is Mo = r X F, thc angular impulse may be expressed in vector form as

7 2 M o & = (

angtilar impulse = (r x F) dt,!, r

(15 -19 )

Here r is a position vector which extends from point O to any point on thc line of action of F.

In a similar manner. using Eq. 15-18. the principle of angular impulse and momentum for a system of particies may be written as

E

(11.0)1 + J Modł — 5(H0)2 (15-20)

153 PRINOPLE OF ANGULAR IMPUISE AND MOMENTUM 26 7

Here the first and third terrns represent the angular momenta of ali thc particles [S110 2(r, x inv,)] at the instants ti and t2. The sccond termis thc sum of the angular impulses given to all the parficies from tj to t2. Recail that these impulses are created onły by the moments of the external forces acting on the system where, for the ith particie, Mo = r, x F,.

Vector Formulation. Using impulse and momenturn principks, i t is therefore possible to write two equations which define the particie's motion, narady. Eqs. 15-3 and Eqs. 15-18, restated as

f(-r2

m y [ + ł F d i = n w ,,t,

tlio)1 + 2 f Mo& -"- (110)2

(15-21)

Scalar Formulation. In general, the above equations can beexpressed in x, y, z componcnt farni, yielciing a total of six scalar equations. If the particie is confinecl to move in the x-y piane, three scalar equations can be written to express the motion, narnely,

i,n i (v ,} 1 +i F xdł = rn(v ,} z

1,jrr,

in(vv)i + Fydi = m(vd2,

r,

(hro)[ + ł f Modt = (H 0 ) 2 r,

(15-22)

The first two of these equations represent the principle of linear impulse and momentum in the x and y direetions, which has Been discussed in Sec. 15-1, and the third cquation represents the principle of angular impulse and momentum about the z axis.

2 6 8 CHAPTFR 15 KINFT[CS OF A PARTICLE: ImFuL$F A,No MOMENTUM

x

Fig. 15-23

Provided air resistance is neglected, the passengers un this amusement-park ride are subjeeted to a conservation of angular momentum about the z axis of rotation. As shown on Lhe free-body diagram, the hiw or Kikut of the narmal lorce N d the seat on thc passenger passes lbrnugh :bis axis. and the passengers weight W is parallet Lo it.-rhus. no angular imptilse acts arointd the z axis.

Conservation of Angular Momentum. When the angular impulscs acting on a particie ate alt zero during the time 11 to t, Eq.15-18 redu,ces to the following simplified form:

(FIA = (H0)2 (15-23)

This equation is known as the conyerfstorn of anguiar momenturn. It statesthat from t1 the partiele's angular momentom rernains constant.Obviously, if na external impulse is applied to the particie, both linear and angolar momentom will bc conserved. In samC cascs. howevcr, the particie's angular momentom will be conservccl and lincar momentom may not. An example cif this occurs when the particie is subjected only to a central force (see Sec. 13-7). As shown in Fig. 15-23, the impulsive central force F is always directed toward point O as the particie moves along the path. Hence. the angular impulse (moment) ereated by F about the z axis is always zero, and therefore angular momentum of the particie isconseived about Ibis axis.

From Eq. 15-20, we can also write the conservation of angular momentum for a system of particles as

(1-16))1 = (I10)2(15-24)

In this case the summation most include the angular momenta of alt particies in tbe system.


When applying the principlcs of angular impulse and momentom, or the conservation of angular momentom, it is suggestcd that thc following proceclure be used.

Peee-Body Diagram.• Draw the particle's free-body diagram in order to determine any

axis about which angular momentom may be conserved. For this to oceur. the moments of alt the forces (or impulses) musi either be parallei or pass through the axis so as to create zero moment throughout the time period t, to t,.

• The direction and .rise of the partieles initial and final vełocities should also be estahlished.

•An alternative proceduro would be to draw the impulse and momentum diagrams for the particie.

Momentom Equations.

• Apply the principle of angular impulse and momentum, (Ho), + fir'Morit = (H0)2, or if appropriate, the conservation of angular momentum, (H0), = (I10)2.

15.7 PRiNCIPLE OF ANGuL.AR ImPIJI_S AND MOMENTLim 2b9

EXAMPLE

The 1.5-Mg car travels along the cireular ruad as shown in Fig. 15-24a. If the traetion force of the wheels on the mad is F = (150?) N, where is in seconds. determine the speed of the car when t = 5 s. The car initially travels with a speccl of 5 mis, Negeet the size of the car,

( a )

Free-Body Diagram. The free-body diagram of the car is shown in Fig.15-24ó.lf wc apply the principle of angular impulsc and momentum about the z axis, then the angular inapulse created by thc weight, norma] force, and radial frictional force will be eIiminated since they act parallel to the axis er pass through it.

Principte nr Angular Impulse and Momentum.

.r,(H , ) i j Af ,dr { . 1 - ) 2

rą+rF di = ruk.(v,.}2

(100 m)(1500 kg}(5 m/s.) + 000 in)[(1562) N] dłJ

= (100 m)11500 kg)(v,.),5$

750(i0) + 5000r3 = 1500113)1vc/2o

(v,)2 -= 9.17 m/s Ans (b)

Fig. 15-24

W 1500 (9.81)N

r= (150t2)N

2 7 0 CHAPTFR 15 KINEr4C5 OF A PARTICLE: ImFuL$F A,No MOMENTUM

EXAMPLE

15.14

The 0.8-1h hall B, shown in Fig. 15-25a, is attached to a cord which passes through a hole at A in a smooth takle. When the hall is r, = 1.75 ft from the hole, it is rotatrng around in a circie such that its speed is vł = 4 ftis. By applying the force F the card is pullcddownward through the hole with a constant speed= 6 Ft/s.Dctcrmine (a) the speed of the ban at thc instant it is r = 0,6 ft from the hole, and (b) thc amount of work done by F in shortening the radial distance from r1 ta r2. Negiect the size of the bali.

SOLUTION

Part (a) Free-Body Diagram. As the bali moves from r1 to r,, Fig. 15-25b, the cord force F on the hall always passes through the z axis, and the weight and NB are parallel to it. Hence the moments, er angular impulses created by łhese forces, are all zero about this axis. Therefore, angular momentum is conserved about the z axis.

Conservation of Angular Momentum. The ball's velocity v, is resolved inło twa components.The radial componerst, 6 ft/s, is known; however, it produces zero angular momentum about the z axis. Thus,

( a )

H i —

riniBv, = r2niRK

v',1.75 ft( 0.8 lb )4 fl/s = 0.6 ft(

0.8 lb 1\32.2 ftłs2132.2 ft/sa)

= 11,67 ft/s

r, — 1:75 ftThe speed of the bali is thus4ft/s

v2V(11.67 ftis)2 + (6 ft/s)2N!,

(l3)= 13.1 ftisFig. 15-25Part (b). The only force that does work on the bali is F. (The nourial

force and weight do not move vertically.) The initial and finał kinetic cnergies of thc bali can be determined so that from the principlc of work and cnergy we have

7', + =

" lb .,)(4 ft ł s)2 + F = 10.8 lb )03.1 ft/s)2

2 \ 32,2 flis-i 2 32.2 ft/s-,

UF = 1.94 ft • lbA N S

NOTE: The fotce F is not constanł because the -norma] component of acceleration, a, = v2/r, changes as r changes,

1 5 7 PRINCIPLE OF ANGuLAR IMPIJI_S AND MOmę.hkTum27 1

The 2-kg disk shown in Fig. 15-26a rests on a smooth horizontal surface and is attached to an elastic cord that has a stiffness

= 20 Nim and is initially unstretched. 1f the disk is given a velocity (v2,)1 = 1.5 m/s, perpendicular to the card, detcrmine the rate at which the cord is being stretchcd and the speed of thc disk at the instant the cord is stretched 0.2 ni.

SOLUT1ON

Free-Body Diagram. Atter thc disk has been launchcd, it slides along thc path shown in Fig, 15-266. By inspection,angular momentumabout point (Gr thc z axis) is conserved, since nono of thc forccsproduce an angular impulse about this axis. Also, when the distance is 0.7 m, only the transverse component (56), produces angular momentum of the disk about O.

Conservation al Angular Momentum. The camponcnt (v' f ),), can be obtained by applying thc conscrvation of angular momentum about O (the z axis).

(110111 = (H0)2

r D(v D) i = r-JrzE„(7.6)2

0.5 m (2 kg)(1..5 m/s) = 0.7 m(2 kg)(14,)2

= 1.071 m/s

Conservatinn cif Energy. The speed of the disk can be obtained by applying the conservation of energy equation at the point where the disk was launchcd and at the point whcrc thc cord is stretchcd 0.2 m.

9"1 ± V 1 +V2

+ = 7-1,Trip(vD)-2 + :Lzioc3

-J(2 kg)(1.5 m/s)2 + O = kg)(vD); + 4(20 N/m)(0.2 m2.)

(vD)2 = 1.360 m/s ,= 1.36 m/s Ans,

1-faving deterrnincd (vp), and its componcnt (iP;))9, thc rata of stretch of

the card. ur radial component, (1;).2 is deterrnined from the Pythagorean

theorem,

(vL)2 \7(vD):::: (vb):3.

= '1(1.360 n-2/5)2 — (1.071 mis)2

0.838 m/s rb

( 3 )

03)

Fig. 15-245

272CHAPTER 15 KINETrC5 OF A PARTICLE iMPULSE ANO MOMENTUM


F15-19. The 2-kg particie A has the velocity shown.Determine its angolar momentumobom point O.

10m,Is.

F15-19

F15-20. The 2-kg particie A bas the veiocity shown. Determine its angular momentum Hr about point P.

15 m s3 0

F15-22. The 5-kg block is rotating around the circular path centered at O cm the smooth horizontal piane when it is subjected to the furce F = (W]) N, where r is in seconds. If the błock starts from rest, determine its speed when i = 4 s. Neglect the size of the block. The (arce maintains the same constant angle tangent to the path.

F1S-22

F15-23. The 2-kg sphere is attached to the light rigid rod, which rotates in the horizomai płane centered at O. If the system is subjected to a couple moment M = (C1912) N • rn, where r is in seconds, determine the speed of the sphere at the Instant r = 5 s startmg from rest.

34‚142kg•

A

T

3m2inF15-20

M = (0.922)Nm, °°6 111"■,,yk i i l l i i i i i i t o % .

FIS-21. Initially the 5-kg block is rotating with a constant speed cif 2 m/s around the circular path centered at O on the smooth hori7ontal piane. lf a constant tangential fotce

= 5 N is applied to the błock, determine its speed when3:. Neglect the size of the block.

F15-21

F15-23

F15-24. Two identical 10-kg spheres are attached to the light rigid rod. which rotates in the horizontal piane centered at pin O. If the spheres are subjected to tangential forces of P = I ON, and the rod is subjected to a couple moment M = (Sr) N• m, where t is in seconds, determne the speed of the spheres at the instant i -= 4 s. The system starts iTUTIlrest,Neglect the size of the spheres.

P= IGN

P = 10 Al F15-24

273153 PRINOPLE OF ANGOLAR I MPULSZ AND MOMENTUM•PROBLEIVIS

O5 ft

15-94. Determine the angular momentum Ho of the particie about point O.

Prob. 15-94

15-95. Determine the angular momentum Ho of the particie about point a

Determine the angular momentum Hr, of the particie about point P

A@ 10 Ib

1 4 1 6 f t

3 ft2 ft

5 ft -4 ft

S ft9 ft

ProbL 15-95196

15-97. Determine the tata) angular momentum Ho for the system of three particies about point 0. A11 the particies arc moving in the x-y piane.

200 mm3 kg

6 m/s

C

OB f 2 mis,P001700 mm —r'

2.5 kg904)

Prob. 15-97

15-98. Determine the angular momentum HU of ezch ot the twa particies aboul point O. Use a scalar soIution,

15-99. Determine the angolar momentum Hp of each of the twa particies about point P. Use a scalar solułion.

P robs,. 15-98i99


*15-100. The smali cylinder C has a mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the france is subjected to a couple M = (8/2 + 5)N • m, where t is in secands, and the cylinder is subjecłed to a force of 60 N, which is atways directed as shffivn_ determine the speed of the cylinder when t = 2 s. The cylinder has a speed vo = 2 mis when t = 0.

Prob. 15-10.0

15-101. The 10-lb błock rests on a surface for which ‚J. = - 05. ft is acted upon by a radial fotce of 2 ib and a horizontal force of 7 lb. always directed at 30° from the tangent to the path as shown. If the block is initially movingin a eircułar path with a speed= 2 ftis at the instant theforces are applied, determine the time required before the tension in cord AB becomes 20 lb. Neglect the size of the block for the calculation.

15-102. The 10-1b błock is orighially at rest on the smoołh surface. It is acted upon by a radial fotce of 2 lb and a horizontal force of 7 lb, alwavg directed at 30' from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T = 30 lb. What is the speed of the błock when this occurs? Neglect the size of the block for the calculation.

P robs. 15-101/102

15-103. A 4-lb bali B is traveling around in a circle of radius r] = 3 ft with a speed fuR), = 6 fr/s. If the attached cord ispulled down through the hole wich a consłant speed v, = 2ft/s, determine the ball's speed at the instant

= 2 ft. How much work bas to be dane to pull down the cord? Neglect friction and the size of the bali.

*15-104. A 4-1b hall B is traveling around in a circle of radiu ri = 3 R with a speed (vB)i = 6 ft/s. If the attached cord is pulled down through the hole with a constant speed v, = 2 ftis, determine how much time is required for the bal/ to reach a speed of 12 ft/s, How far r, is the hall from the hole when this occurs' Neglect friction and the size of the bali.

Probs. 15-1031104

15-105. The [nur 5-lb spheres are rigidly attached to the crossbar frame having a negłigible weight. If a couplemoment M = (<1.51 0.8) lh • ft, where t is in secands. isapplied as shown. determine the speed of cuch of the spheres in 4 secands starting from rest. Neglect the size of the spheres.

Prob. 15-105

15.7 PRINGPLE OF ANGULAR rMPULSE ANO MOMENTUM275

15-106. A smali particie having a mass m is placed inside the semicircular Lube. The particie is placed at the position shown and released. Apply the principle ol angular momentum ahout point .0 (M() l j). and show that the motion of the particie is governed by thedifferential equation(g/Ri sio B = O.

15-11.16

95-108. A &liki having a mass of 50 kg holds her legs upas shown as she swings downward from rent at= 311r.Her center of mass is iocated al point Gi. When she is at thebonom position H = she suddenly lets her legs eomcdown. shifting bet center of mass to posit ton Gc. Deterrnine her speed in the upswing due to ibis sudden movement and the angle fh to which she swings before morrientarily = ling to rest, Trem the child's body as a particie.

Prob. 15-108

15-107. The bali B bas a weight of 5 lb and is originally rotating in a cirelc, As shown, the cord AR has a length of 3 ft and passes through the hole A. which is 2 ft above the piane of morion. If 1.5 ft of cord is puIled through the hole, determine the speed of the bali when it moves in a circular path at C.

15-109. The 150-lb car of an amusement park Tido is connected to a rotating telescopic boom. When r = 15 R, the car is moving on a horizontal circular path with a speed of 3€1 ft/s, If the boom is shortened at a rate of 3ft/s, determine the speed of the car when r = 10 R. Also, Pind the work dane by the axial forte F along the boom. Neglect the size of the car and the mass of the boom.

2 ft

Prob. 15-107 Prob. 15-109

276CHAPTER 15 K[NETkCS OF A PARTICI.: IMPUL$E AND MOMENTUM

15-110. An antusernent park ridc conslsts of a car which is attached to the cabie 0A. The car rotates in a horizon tal circutar path and is brought to a speed vj = 4 ft/s when r =- 12 ft. The cable is then pulled in at the constant rate of 0.5 ft/s. Determine the speed of the car in 3 s.

Prob. 15-110

115-112. A smal błock h aving a mass of 0.1 kg is gtven a horizontal velocity of y, -= 0.4m/s when r1 -= 500 mm. ft slides idong the smooth conical surface. Delermine the distance h i# must descend for it ta reach a speed of

=

2m/s, Also, what is the angle of descent B. that is, the angle measured from the horizontal to the tangent of the path?

Prob. 15-112

15411. The $0[3-łb roller-coaster car starts from rest an the track having the shapc of a eyIindrical hetix. if the liclix clescends 8 ft for every one rev~ion, determine the speed of the car in t = 4 s. Also, how far bas the car descended in ibis dme? Neglect friction and the size of the car.

8Ft

Prob. 15-1.11

15-113. An earth satellite of mass 700 kg is launched finto a free-flighl irajeclory about the earth with an initial speed of vA = 10 kinis when the distance from the center of theearth is rq15 Mm. If the launch angle at this positkin isef3A -=determine the speed v8 of the satellite and itsclosest distance r8 from the center of the earth. The earth lias a mass M, = 5.97N1024) kg. Hiw: Under these conditions, the satellite is subjected oni)/ to the earth'sgravitational furce, F = Eq.. 13-1. For part of thesol uti.con. use the conservation of energy.

Prob. 15-113

15,8 STEADY FLOW OF A FLUID STREAM ,277

15.$ Steady Flow of a Fluid StreamUp to this point we have restricted our study of impulse and momentum principles to a system of particles contained within a ciosed volutne. In this section, howcvcr, we will apply the principle af impulsu and momentum tu the stcady mass flow of fluid partcics entcring into and then out of a control volume. This volume is defined as a region in space where fluid particles can flow into ar out of a region. The size and shape af the control volume is frequently made to coincide with the solid boundaries and openings of a pipe, turbine, or pump. Provided the flow of the, fluid into the control volume is equal to the flow out, then the flow can be classified as steady flow.

Principle of irnpuise and Momentum. Consider the steady flow of a fluid stream in Fig. 15-27a that passes through a pipe. The region within the pipe and its openings will be taken as the control volume. As shown, the fluid flows into and aut of the control volume with velocities vA and v

i?, respectively. The changc in the dircction of thc fluid flow within the control volumc is caused by an impulsc produccd by the resultant external fotce exerted on the control surface by the wali of the pipe. This resultant force can be determined by applying tle principle of impulse and momentum to the control volume.

( a )

Fig. 15-27

The eonveyor hett musi_ supply frictional forces to the gravel [hal falis upon it in order w change the momentom of the gravel stream, so that it begim to travel along the beli.

The air on one side of Ibis fan is essenlially M resł. and as it passes through the Hades its momentum is inereased.To changc the momentom of the air flow in this manner, the blades musi cml a horizontal chrust on the air stream. As the hlades turn raster, the equal bul opposite Chrust ni the air on the htatJes coold overcome the rolling resistance of Ihe wheel S on the ground and begin to move the frame of the fan.

278 CHARTER 15 KINEr[cs OF A PARTICEE: IMPULSE AND MOMENTUM

r ,

O Tirrie O Time dr O Timc i + dr

( b )

As indicated in Fig. 15-27b, a smali amount of fluid having a mass dm is about to enter the control volume through opening A with a velocityuf vA at time t. Sincc thc flow is considered stcady, al tinic dt, thesame amount of fluid will leave the control volurrie through opening B with a velocity vp.. The momenta of the fluid entering and leaving the control volurne are therefore dni vĄ and dm y8, respectiveiy. Also. dli ring the time dr, the momentum of the fluid mass within the control volurne remains constant and is denoted as mv. As shown on the center diagram, the resultant external force exerted on the control volume produces the impulse iF dt. If we apply the principle of linear impulse and momentum. we have

dm y + mv + ,Flit — dm ys + mv

If r, rA, riy are position vectors measured from point O to the geometrie centers of the control volume and the openings at A and B. Fig. I5-27b, then the principle of angular impulse and momentum about O becomes

rA x dni vA + r x mv + r ' X F dt = r x mv + X dm y8

Dividing both sides of thc above two equations by dt and simplifying, we get

dm= ( v — v A ) (15-25)

t . c )

Fig. 15-27 (cont.)

dni= dr —(r8 3C vB — rA 3C PA) (15-26)

15,8 STEADY FLOW OF A FLUID STREAM 27 9

The term dn r idt is called the mass flow. ft indicates the constant amount of fluid which flows either into or aut a the control volume per unit of time. If the cross-sectional areas and densities of the fluid at the entrance A arc AA, pA and at exit B, A8, p8, Fig. 15-27c. then for an incompressible fluid, the comMuiły of mass requires dni = pdV W px(dsAAA) = )138(dsBAB). Henee, during the time dr, sił= vA = dsAidi and yfi = dsiddr, we have dm/dr = pAvAA

A = pBV IB or in general,

dm—d r = pvA = Q

(15 -27 )

The term Q = vA measures the volurne of fluid flow per unit of time and is refcrred to as the discharge ar the vohimełric Pow.


Problems involving steady flow can be solved using the following procedure.

Kinematic Diagram.

• Identify the control volurne. If it is ~Mg, a kinemade diagram may be helpful for determining the entrance and exit velocities of the fluid flowing inni and aut of its openings since a reladve-morion analysis of velocity will be involvecl,

• The measurement of velocities vA and y$ must be made by an observer fixed in an inertial frame of reference.

• Once the vełocity of the fluid flowing into the control volurne is determined, the mass flow is calculated using Eq. 15-27.

Free-Body Diagram.

•Draw the free-body diagram of the control volurne in order to estahlish the forces SF that act on ił. These forces will include the support readions, the weight of al I solid parts and the fluid contained within the control volume, and the stanc gauge pressure forces of the fluid on the entrance and exit sections. The gauge pressurc is the pressure mcasured above atmosphefic pressure. and so if an opening is exposed to the atmosphere. the gauge pressure there will be zero.

Equations of Steady Flow.

•Apply the equations of steady flow, Eq. 15-25 and 15-26, using the appropriate components of velocity and force shown on the kinematic and free-body diagrams.

In the SI system, pressure is measurecl using the pascal (Pa). where 1Pa = I N/m2.


0.1m

-r

(a )

F,

20(9.81) N

0.125 m0.3 m

(b)

Fig. 15-2.8

Deterrnine the compunents of reactłon which the fixed pipe joint at A exerts on the elbow in Fig. 15-28a. if water flowing through the pipe is subjected to a static gauge pressure of 100 kPa at A. The discharge atB is Qa = 0.2 m3/s. Water has a density -= 1000 kg/m3, and thewater-filled elbow has a mass of 20 kg and center of mass at G.

B S O L U T I O N

v B We will eimsider the control vohime to be the (Juter surface of theclbow, Using a fixed inertia1courdinate system. thc velocity of now al A and B and thc mass now rate can bc obtained from Eq. 15-27. Sincethe density of water is constant, tQ, = = Q. Hence,

r f r i rdr = p„Q 0000 kg/m3)(0.2 ni3/s) --- 200 kgis

Q 0,2 tr3/s

▪—,Afi 7r(0.05 m)' 25.46 m/s 4.

Q 0,2 m3/s = 6.37 mis ->

▪Ay.7:-(0.1

F ree-Body Diagram. As shown on the free-body diagram of the control vołume (elbow) Fig. 15-28b, the fixedeonnection at A exerts a resultant couple moment Mo and force cornponents F, and F. on the elbow. Due to the static pressure of water in the pipe, the pressure kirce acting un the open control surface at A is FA = 12,1

A„. Since 1 kPa = 1000 Nim2,FA = rp,4,4m = [100(103) N/m21{7r(0.1 m)21 = 3141.6N There is

no static pressure acting at B, since the water is discharged at atmospheric pressure: i.e., the pressure measured by a gauge at B is equal to zero, pE = 0.

Equations of Steady Flow.

SF, =ch/z %); 3141.6 N = 200 kg/s(0 - 6.37 m/s)dt

= 4.41 kN /7.Y.

+ r F, —dt(va- vA3);-F- 20(9.81)N = 200 kg/s(- 25,46 mis -

F. = 4.90 kN Ans.

1f moments arc summed about point O, Fig. 15-28b, then F„, F,, and the static pressure FA arc eliminated, as wen as the moment of momentom of the water entering at A, Fig. 15-28a. Hcnce,

C -1-0 = 'ILdr"(dosvs J daAvA)

Mo -i- 20(9.81)N (0,125 m) = 200 kg/s[(0.3 m)(25.46 m/s) - 0]

Mo = 1.50 kN • m Ans.

15.8 STEADY FLOW ()F A FLUID 5TREAM2$1

EXAMPLE

A 2-in.-diameter water jet having a velocity of 25 tt f s impinges upon a single moving blade, Fig. 15-29a. If the blade movcs with a constant velocity of 5 ft/s away from the jet, determine the horizontal and vertical components of furce which the blade is exerting on the water. What power does the water generate on the blade? Water has a specific weight of y„ = 62.41b/ft3.

SOLUTION

Kinematit Diagram. -Herc the control volume wili be the, stream of water on the blade. From a fixed inertial coordinate system,Fig.15-29b, the rato at which water enters the control volume at A is

v A - {251} ftis

The reltiliv.e.-flow velociły within the control volume is v„,„, = vN. - vc.„ --T-- 25i - 51 = {20i} ft/s. Since the control volume is rnoving with a velocity of v,.V = {5i } ft/s, the velocity of flow al B measured from the fixed x, y axes is the vector sum, shown in Fig.15-29b. Here.

= {51 + 20j} ftis

Thus, the mass flow of watcr °nt° the control volume Chat undcrgoes a mornenium change is

alit 2.4= = 632_2) (20)Hr(12)21= 0.8456 slug/s

Free-Body Diagram. The free-body diagram of the control volumeis shown in Fig. 15-29c.The weight of the water will be neglected inthe calculation, sinet this force will be smalt compared to the reactivecomponents F, and F.

.Equations of Steady Flow.

ciutEF = dr (v,8 - vA)

+ F,.f = 0.8456(51 + 20j - 251)

Equating the respective i and j components gives

F, = 0.8456(20) = 16.9 lb Ans.= 0.8456(20) = 16.9 lb t Ans.

The watcr exerts equal but opposite forces on łhe blade.Simo the water (orce which causes the blade to move forward

horizontalły with a velocity of S ft/s is Fz = 16.9 lb, then from Eq. 14-10 the power is

= 25 ftis

yx

B

L,-F,i

F j r (

c )

Fig. 15-29

P F 16.91b(5 ft/s)

P - - 0.154 lip550 hp/(ft • lb/s)


*1 5.9 Propulsion with Variable Mass

A Control Volu me That Loses Mass. Consider a device such as a rocket which at an instant of time lias a mass m and is moving forward with a velocity v, Fig. 15-30a. At this same instant the amount of mass rrap is expelled from the devicc with a mass [law velocity v,. For the analysis, the control volume will include botki the mass m of the device and the expelled mass m,. Thc impulse and momentum diagrams for thc control volume arc shown in Fig. 15-30b. During the time dr. its velocity is increased from y to v + dv since an amnunł of mass dni, lias been ejected and thereby gained in the exhaust.This increase in forward velocity, huwever, does not changc the velocity v, of the expelled mass, as scen by a fixed observer, since this mass ITIQVCS with a constant velocity once it has been ejected. The impulses are created by iF,„, which represents the resultant of all the externai forees, such as drag or weight, that act on the control volume in thc direction of motinn. This force resultant does not inchtde the force which causes the control volume ta move forward, since this force (called a chrust) is interna! to the control uolume; that is, the thrust acts with equal magnitude but opposite direction on the mass m af the device and the expelled exhaustmass .* Applying the principle af impulse and momentum ta thecontrol volurne, Fig. 15-30b, we have

) mi) — dr = — dtii,)(v + dv) — (cne

VCont ro lV,Iume

(a) nr

dt = —v dm, + m dv — dm, dv — v, dm,

(rn — (v + dv)m Y

‚11,V,

n t ,T i m e t

F. dr

Time dr

( m d m > ,

M — dm,.

T ime r+dr

( b )

Fig. 15-30

*F represents the external resultant tarce «ring on rhe connoif vohnne, which is different Erom F. the Feudum rurce aciing only on the device.

15.9 PROPULSION WITH VARIASŁE MASS283

Without loss of accuracy, the third term on the right side may be neglected since it is a "second-order" differential. Dividing by dż gves

dv dm,= ł - ( v v , ) d t

The velocity of the device as scen by an observer maving with the particles of the ejected mass is vafC (v + v,), and so the final result can bewritten as

Coutrol Wurne1

dv dniam—dłdt

(15-28)

Here the term dm,./dt represents the race at which mass is being ejected.To illustrate an application of Ey. 15-28, consider the rocket shown

in Fig. 15-31, which lias a weight W and is moving upward against an ałmospherie drag forte FD. The control volume to be considered consists d the mass of the rocket and the mass of ejected gas Appiying Eq. 15-28 givcs

(+T)-Fn - W = W dv dme

The last term of this equation represents the thrusł T which the cngine cxhaust exerts on the rocket. Fig. 15-31. Rccognizing that dvi& = a, we can therefore write

(+T) T - F» - W = —wag

If a free-body diagram of the rocket is drawh, it becomes obvious that this equation represents an application of F = ma for the. rocket,

A Control Volurne That Gains Mass. A device such as a sconp or a shovel may gain mass as it moves forward. For example, the device shown in Fig. 15-32a lias a mass m and n-toves forward with a velocity v. At this instant, the device is collecting a particie stream of mass nap. The flow velocity vi of this injectecl rnass is canstant and independent of the velocityv such that v .24. The control vołume to be considcred herc includesboth the mass of the device and the mass of the injcctcd partieles

Gont r ol Volum eFig. 15-31y ż

(a)

Fig. 15-32


m V I

m,v,

mi F„. cit

Time

Time d i (b )

Time 1 i dr

Fig. 15-32 (coal.)

15

(c )

The seraper hux behintl this tractor represents a device that gains masa. If the tractor maintains a constant velacily 7v, then ein/dt = O and, becnuse the spił is originalłyat resł, =v. AppIying Eq. 15-29. thehorizontal lowing kirce on the seraper kot is then T = O + o(dmieli). where dm Ici$ is the race of &ojt aceumulated in the box.

The impulse and momentum diagrams are shown in Fig.15-32b. A king with an increase in mass clinf gained by the device, there is an assumed increase in velocity dv during the time interval dt.This increase is causedby the impulse created by the resultant of aIl the cxternal forcesacting on the conrroi voturne in the direction of motion, The (orce summation dGes not include the retarding furce of the injected mass acting on the device. Why? Applying the principle of impulse and momentum to the control volume, we have

) mv + In;vi SF„ di = (m 4. drzr;)(v + dv) (mi dmi)vi

Using the same proceduro as in the previous case, wc may write this equation as

d v dmir n i + ( v - v i ) —

Since the vełocity of the device as scen by an observer movMg with thepartides of the injected mass is vDp = - .v), the finał result can bewritten as

C15-29)

where dmij dr is the rate of mass injected into the device. The last term in this equation represents the magnitude of forte R, which the injected mass exerts on the device. Fig. 15-32c. Since dv fdt = a, Eq, 15-29 becomes

- R = rrza

This is the application of LF = ma.As in the case of steady floty, problems which are solved using Eqs.

15-28 and 15-29 should be accompanied by an identified control volume and the necessary free-body diagram. With this diagram one can then determine SF,„ and isolate the forte exerted on the device by the particie słrearn.

F, /7- -J Rm

m—dv drni+ Oni

dł dł

15.9 NOPULSIOM WITH VARIASŁE MASS285

EXAMPLE MEI gr

The initiaI combined mass of a rocket and its fuel is mo. A total mass nz, of fuel is consumed at a constant rate of dm,./dt = c and expelled at a constant speed cif 71 relative to the rocket. Deterrnine the maximum velocity of the rocket, i.e., at the instant the fuel runs out. Neglect the change in the rocket's weight with altitude and the drag resistance of the air.Thc rocket is fired vertically front rent.

SOLUTIONSince the rocket lases mass as it =yes 12pward, Eq. 15-28 can be used for the solution. The ony extertud forte acting on the watrof rołutne consisting of the rocket and a portion of the expelked mass is theweight W.Fig, 15-33, Hence,dv dv• F,,, nr— — —W = in— — uC (1)

dr f dr dr

The rocket's velociry is obtained by integrating this equałion.At any giver/ instant t during the flight, the mass of the rocket can be

expressed as m = irap — (dm,/art), = ira0 — cł. Since W = mg, Eq. I becomes

dvct)g = (m71 —— uedt

Separating the variables and integrating, realizing that v = O at t = 0, we have

( )= rr ln gtino — ct

( 2 )

Note that liftoff requircs the first term on the right to be greater than the second during the initial phasc of motion. The linie t' needecl to consume all the fuel is

( din, )

inf — =Hence.

d v = l o ( I n o — c r

M C

g) di

v = —u ln(tno — co — gl

Substituting ino] Eq. 2 yieids

u In—)ine grot


A chain of length 1, Fig. 15-34a, has a mass m. Deterrnine the magnitude of force F required to (a) raise the chain with a constant speed v„, starting from rest when y = 0; and (b) ]ower the chain with a constantspecd starting from rest when y = L

SOLUTION

Part (a). As thc chain is raised, all the suspended links arc given a sudden downward impulse by each addcd link which is lifted off the ground. Thus, the siispended portion of the chain niay be considered as a device which is gaining mass. The control volUMe to be considered is the Iength of chain y which is suspended by F at any instant, including the next link which is about to be added hut is stili. at rest, Fig. 15-34h. The forces acting on the control vołume exchlde the internal forces P and -P, which act between the added link and the suspended portion of the chain. Hence, "XF,„ = F - nłg(y i I).

To apply Eq, 15-29. it is algo nccessary to firui the rato at which mass is bcing addcd to thc systenri.Thc vclocity v,. of the chain is equivalcii t to Why? Since yt. is constant, dv, j‚dt = O and dyidł = v.. Integrating,

using the initial condition that y =- O when t = O. gives y Thus,the mass of the control volume at any instant is m,.„ = m(y//) = m(v„,11), and therefore the race at which mass is added to the suspended choin is

d n 4 v „dł )= on(—1

( a )

—P Applying Eq. 15-29 using this data, we have (1.02

(h) + = rn—cir f VD/i—cit

F - m g ( - y T l

F ) = O + v ,m(—)

}lence,

F (mi 1)(gy ‹'.) A‚iN.

my(2!.-)i

C.

1;

(c)

Fig. 15-34

Part (b). When the chain is being lowered, the links which arc expelled (given zero velocity) da nor kwart an impulse to the rem4iningsuspended links. Why? Thus. the control volunnc in Part (a) will not be considered. 1nstead, the cquation of motion win be used to obtant thc solution. At time t the portion of chain stili off the floor is y. The free-body diagram for a suspended portion of the chain is shown in Fig. 15-34e.Thus,

+ T F =ma; F - mg(-) =

F = Ing(-11)

15.9 PROPULSICN WITH VARIASŁE MASS287• PROBLEMS

15-114. The fire boat discbarges iwo strearns of seawater, each at a flow of 0.25 m3/s and with a nozzle velocity of 50 m/s. Determine the tenskm developed in thc anchor chain, needed to seeure the boat. The density of seawater is p, = 1020 kg /m3.

15-116. The 200-kg bc_lat is powered by the fan which develops a slipstrearn having a diameter of 0.75 ni. If the fan ejects air with a speed of 14 m/s, measured relative to the bont, deterniine the initial acceleration of the boal if it is initially at rent. Assume that air has a constant density cif p,, = I 22 kg/m3 and that the entering air is essentially at test. Neglect the drag resistance of the water.

Prob. 15-116

15-117. The chute is used to divert the flow ol water. Q = 0.6 m3/s. If the water has a cross-sectionai area of 0.1)5 tn1. determine the force components at the pin D and ratler C necessary for equilihrium. Negleet the weight of the chale and weight cif the +ower on the diuk, Ply i I

Mg/m3.

Prob. 15-114

Prob. 15-117

15-115. A jet of water having a cross-sectional arca of 4 in'- strikes the fixed blade will] a speed of 25 ft/s. Determine the horizontal and vertical components of force which thc blade exerts on the water, -y, = 62.4 lb/ft 3.

25 ttis

Prob. 15-115

130'

288 CHAPTER 15 KiNuncs OF A PARTICLE IMPULSE ANO MOMENTUM

15-118. The buekets nn the Pelton whed ara subjected to a 2-in-diameter jat of waier, which hasa velocity of 150 ft/s. If each bucket is traveling at 95 ftp when the water strikes it, determine the powcr developed by thc bucket. y,. -= 62,4 lb/ft3,

Prob. 15-118

15-119. The blade divides the jet of water having a diameter of 3 in. If one-fourth of the water flows downward whi1e the other łhree-fourths flows upwards, and the tota1 flow is Q = 0.5 ft3/s, determine the horizontal and vertical components of furce exerted on the blade by the jat, Y, = 62.4 lb/ft3.

15-120. The fan draws air through a vent with a speedi nf 12 ft/s, If the cross-sectional arra of the vent is 2 ft2. determine the horizontal thrust on the Nade. The specific weight of the air is y„ = 0.076 lb/&.

Prob. 15-120

15-121. The gauge pressure of water at C is 40 Ib/in2. If water flows nut of the pipo at A and B with vełneities ttg = 12 flis and vi? = 25 ft/s, determine the horizontal and verłical components of fotce exerted on the elbow necessary to hold the pipc assembly in equilibrium. Negleet thc weight of water within the pipa and the weight of the pipe.The pipa has a diameter of 0.75 in. at C. and ai A and B the diameler is 0.5 in. y„. = 62.4 113/ft3.

Prob. 15-119

3 in.

= 12 flis

4;..5

3

Iv$ = 25 f i is

v r

Prob. 15-121

15.9 PROPULSION WITH VARIASŁE MASS289

15-122. A speedboat is powered by the jet drive shown. Seawater is drawn roto the pump housing at the rate of 20 ft-3

/s through a 6-in.-diameter iniake A. An irnpeller accelerates the water flow and forces it aut horizontally through a 4-in.-diameter nozzle B. Determine the horizontal and vertical components of thrust exerted 011 the speedb-oat,The specific weight of seawater is = 64.3 11«.

15-125. Walor is discharged from a nozzle with a vełocity of 12 m/s and strikes the blade mounted on the 20-kg cart. Deternnite the tension developed in the card. needed to hold the cart stationary, and the norma reaction of the wheels on the cart. The nozzle has a diameter of 50 mm andthe density of water is1000 kg/mg3.

Prob. 15-125

15-123. A plow located on the front of a locomotive scoops up snow at the rate of 10 ft3/s and stores il in the tram. If the locomotive is travehng at a constant speed of 12 ft/s, determine the resistanee to motion eaused by the shoveling.The specifie weight of snow is y, = 6 Ihift3.

*15-124. The boat hasa mass of 180 kg and is traveling forward on a. river with a constant velocity of 70 km/h. measured relative to the river, The river is flowing in the opposite direction at 5 km/h.lf a tufie is placed in the water. as shown, and il collects 40 kg of walter in the boat in 80 s, determine the borizonłal thrust T on the Lube that is required ło overcome the resistance dugi to the water collection and yet maintain the constant speed of the boat, p„. = 1 Mg/m3.

Prob. 15-124

15-126. Water is flowing from the 150-mm-diameter firehydrant with a velocity = 15 m/s. Determine thehorizontal and vertical eomponenłs of farne and the moment developed al the bace joint A. il the stalic (gauge) pressure at A is 50 kPa. The diameter of the fire hydrant at A is 200 mm, p . W I Mg/m'.

Prob. 15-126

Prob. 15-122

290CHAPTER 15 KlNETrCS OF A PARTICLE: IMPULSE AND MOMENTUM

15-127. A cdi of heavy open chain is used to rednec the stopping clisiance of a sJed that has a mass M and travels at a speed of yfj. Determine the required mass per unit length of the chairs needed to slow down the sled to (1/2)% within a distance x = s if the sled k hooked to the chain at x = 0. Neglect friction between the chaM and the ground.

'15-128. The car is used to scoop up walec that is Iying in a trough at the tracks. Determine the force needed to puli the car forward at constant velocity v for each of the dirce cases. The scoop has a cross-sectional area A and the density of water is p

w..

15-130. The mini hovercraft is designed so that air is druwn in at a constant tale of 20 m3/s by the fan blade and channeled to provide a vertical thrust F, jurt sufficieni to lift the hovercraft off the water, while the remaining atr is channelecl to produce a horizontal thrust T on the hovercraft, 1f the air is discharged horizonłally at 200 m/s and vertically ai 800 mis, deterrnine the ihrust T produced. The hovercraft and its passenger have a total masa of 1.5 Mg. and the dcnsity of the air is pu= 1,20 kg/m3.

111 , •V

(a) (b)V

( c )

Prob. 15-128

15-129. The walet flow enters helow the hydrant at C at the rate of 0.75 m3/s.II. is then divided equally between the twa outlełs at A and B. If the gauge pressure at Cis 300 kPa, dctermine the horizontal and vertical furce rcactions and the moment reaction on the fi xe d suppart at C. Tli e di am e ter of ihe iwo outlełs at A and B is 75 mm. and the diarneter of the inlet pipe at C is 150 mm. The density of water is p„, 1000 kg/m3, Neglect the masa cif the eontained water and the hydrant.

250 mm

C

Prob. 15-129

F

Prob. 15-130

15-131. Sand is discharged from the siło at A at a rate of 50 kg/s with a vertical velacity at 10 m/s anto the conveyor beli, which is moving with a constant velocity of 1.5 m/s. If the conveyor system and the sand on it have a łotai masa of 750 kg and center of masa at point G, determine the horizontal and vertical carnponents of reaction at the pin support B and roHer support A. Neglect the ihickness of the conveyor.

Prob. 15-131

15.9 PROPULSION WITH VARIABLE MASS291

15-132. The fan blows air at 6000 frimin. If the fan has a weight of 30 lb and a center of gravity at G. determine the smallest diameter d of its base so that 11 will not tip over. The specific weight of air is y = 0.076 lb/W.

Prob. 15-132

15-133. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 1 ft and it carries a discharge of 50 ft3is,determine the horizontal and vertical components of forte reaction and the moment reaction exerted at the fixed base D of the support. The Lotni weight of the bend and the water within it is 500 lb. with a masą center at point G. The gauge pressure of the water at the flanges at A and f3 are 15 psi and 12 psi, respectively. Assurne that no forte i5 transferred to the flanges at A and B. The specific weight of water is y„ = 62,4 111./&,

15-134. Fach of the two stages A and Raf the rocket has a mass of 2 Mg when their fuel lanka aTe empty. They each carry 500 kg of fuel and are capable of consuming it at a rate of 50 kg/s and eject it with a constant velocity of 2500 m/s, measured with respect to the rocket,The rocket is launched vertically from rem by first igniting stage B. Then stage A is ignited inimediately after all the fuel in B is consumed aud A has separated from B. Deterniine the maximum velocity of stage A. Neglect drag resistance and the variation of lhe rocket's weight with altitude.

Prob. 15-134

15-135. A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of 6 m/s ihrough au intake unii A. which hasa cross-sectional arca of AA T 0.25 rn2, and then discharging it at the ground, B, where the cross-sectional arca is Ak = 0.35 m2. lf air at A is subjectecl mity to atmospheric pressure, determine the alt pressure which the lawn mower exerts on the ground when the weight of the mowci is freely supported and no bad is placed nn flit handle. The mower has a masa of 15 kg wich center of mass at G. Assume that air has a constans densiły of p = 1.22 kgim3.

Prob. 15-135

G1.5 it

411

D

Prob. 15-133

45'

292 C H A R T E R 1 5 K I N E T [ C S O F A P A R T I C E E : I M P U L S E A N D M O M E N T U M

*15-136. The 12-ft-long opery-link chain bas 2 ft of its end suspended from the bole as shown. If a (orce of P = 1 U lb is applied to its end and the chain is released from rent, determine the vclocity of the chain at thc instant the entirc chain has heen extended.The chain basa weight of 2 Ibift.

F= 10 Ib

Prob. 15-136

15-138. The second stagc of thc two-stage rocket weighs 2000 lb (empty) and is launched from the first stage with a velocity of 3000 mi/h. The fuel in the second stage weighs 1000 Ib. H it is consumed at thc rate of 50 lb /s. and ejected with a relative velocity of 8000 ftis, determine the acceleration of the second stage just after the engine is fired. What is the rockers acceleration just before Ml the fucl is consumed? Neglect the effect of gravitadon.

15-139. The missile weighs 40 000 lb. The constant thrust provided by the turbojel engine is T = 15 000 Ib. Additional thrust is provided by hvo rocket boosters B. The propellant in each booster is hurnec1 at a constant rate of 150 lb/s, with a relative exhaust velocity of 3000 flis, 1f the mass of the propeilant tost by the turbojet engine can be neglected, deterrnine the velociły of the missile after the 4-s buro time of the boosters. The inidal velocity of the missile is 300 mifh.

Prob. 15-139

2 f i l

15-140. The 10-Mg helicoptcr carrk...s a bucket containing 500 kg cif water, which 1.5 used to fight tires. If hovers over the land in a fixed position and then releases 50 kg/s of water at 10 m/s, measured relative to the helicopter„deterrnine the initial upward acceleration thc liclicopter experiences as the water is being released.

15-137, A chain of mass anty per unit lengih is loosely coiled on the floor. If one end is subjected to a constant ford P when y = 0, determine the velocity of the chain as a tuncdon of y.

Prob, 15-137 Prob. 15-140

15.9 PROPULSION WITH VARIABLE MASS293

15-141. The earthmover initially carries 10 m3 of sand hav'mg a der sity of I521=I kg/m3, The sand is unloaded horizontally through a 2.5 m2 dumping port P al a raje of 900 kgis measured reiative to the port. Determine the resultant tractivc force F at i łs front whcels if the accelerafion of the earthmover is 0.1 m /s2 when half the sand is dumped. When empty, the earthmover has a mass of 30 Mg. Neglect any resistance tv forward motion and the mass of the whcels. The rear wheels are free to roll.

15-1.43. The jet is traveling at a speed of 500 mi/h. 30° with the horizontal. If the f uel is being sperli at 3 lb/s. and the engine takes in air at 400 113/s„ whereas the exhaust gas (air and fuel) has a reiative speed of 32 800 ft/s. determine the acceieration of the piane at this instant. The drag resistance of the air is Fr, = (0.7v2) Ib, where the speed is measured in ft/s. The jet bas a weight of 15 000 lb. flint: See Prob. 15-142..

Proh. 15-141

Prob. 15-143

15-142. The I2-Mg jet airplane has a constant speed of 950 km/h when it is flying along a horizontal siraight iine, Air enters the intake scoops S at the rate of 50 m3/s, If the engine burns fuef at the rate of 0.4 kgis and the gas (air and fuel) is exhausted relative to the piane with a speed of 450 mis, determine the resultant drag force exerted on the piane by air resistance. Assumc that air has a constant density of 1.22 kg/m3. Hitu: Since mass both enters and exits the piane, Eqs. 15-28 and 15-29 musi be combined to yield

dm dmi£F , = m— -

dr dt

*15-144. A four-engine commercial jumbo jet is cruising at a constant speed of 800 km/h in Level Mglił when alf four engines are in operation. Each of the engines is capable of discharging combustion gases with a vełocity of 775 m/s reiative to the piane. If during a test Iwo of the engines, one on each side of the piane, are shut off, determine the new cruising speed of the jet.Assurne that air resistance (drag) is propordonal to the square of the speed, that is, Fa = CI. where c is a constant to be determined. Neglect the losy of mass due to fuel consumption,

Proh. 15-142 Prob. 15-144

294 C H A R T E R 1 5 K I N E T [ C S O F A P A R T I C E E : I M P U L S E A N D M O M E N T U M

15-145. The car has a mass Fno and is osad to tow the smooth chain havtng a tota) length ł and a mass per unit oflength If the ehain is originafiy piled up, determine thetractive lorce F that musi be supplied by the team wheels of the car, necessary tn maintain a constant spced while the chain is being drawn aut.

V

F

Prob. 15-145

15-146. A rocket bas an emptyweight of 500 Ib and carries 300 ib of fuel. If the fuel is burned at the rate of 1.5 th/s and ejected with a velocity of 4400 Et./s relative to the rocket, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation vn the rocket.

15-147. Deterrnine the magnitude of fotce F as a funcdon af dme. which musi be app]ied to the end of the cord at A to raise the hauk H wśth a constant speed v = 0.4 m/s. Initialły the chain is at rest on thc ground. Neglect the mass of the card and the hook.The chain lias a mass of 2 kg/m,

'15-148. The truck has a mass af 50 Mg when empty, When n is unloading 5 rn3 of sand at a constant rato of 0.8 rn3/s, the sand flows aut the back at a speed of 7 m/s, rneasuredrelative to the truck, in the direcdon shown. If the truck is free to roli. determine its initial acceleration just as the load begins to empty. Neglect the mass of the wheels and any frictional resistance to motion. The density of sand is pt = 1520 kg/ml.

Prob. 15--148

15-149. The chain has a latał length L < d and a mass perunit length of If a portion h of the chain is suspendedover the table and released, determine the veioeity of its end A as a function of its position y. Neglect ffictian.

Prob. 15-147 Prob. 15-149

15.9 PROPULSIONJ WITH VARIASŁE MASS295


P15-1. The bali travels to the left when it is struck by the bat. If the bali then rnoves horizontally to the right, deterrnine which measurements you could make in order to determine the net impulse given to the bali. Use numerical values to give an example of how this can be done.

P15-1

P15-2. The steel wrecking -har is suspendcd front the boom using an olei rubber tire A, The crane operator lifts the bali then allows it Lo drop freely to break up the concrete. Explain. using appropriate numerical data, why it is a good idea to use the rubber tire for this work.

P 1 5 - 2

P15-3. The train engine on the left, A. is at rest, and the one on the right. B. is coasting to the left. If the engines are identical. use numerical values to show how to determine the maximum eompression in each of the spring bumpers chat are mounted in the front of the engines. Each engine is free to roli,

P 1 5 - 3

Threc train cars each have the sanie mass and are rolling freely when they strike the flxed bumper. Legs AA and BC on the bumper are pin connected at [heir ends and the angle BAC is M' and BCA is 60°. Compare the average impulse in each leg needeci to stop the motion if the cars have no bumper and if the cars have a spring bumper, Use appropriate numerical values to explain your answer.

P 1 5 - 4

296CHAPTER 15 KiNuncs OF A PARTICLE IMPtJLSE ANO MOMENTUM

CHAPTER REVIEW

Impulse

An impulse is defined as the product of force and time. Graphically it represcnłs the arca under the F—t diagram. If the force is constant, [hen the impulse becomesł = F,(r2 ti).

Principle of Impulse and Momentum

When the equation of mottom F = ma, and the kinennałic equation, a = dv f d t, are combincd, we obtain the principlc of impulse and momentum. This is a vector equation that can be resolved finto rectangular components and utsed to solne prohlems that involve furce, velocity, and time. For appiication, the free-body diagram should be drawn in order tn account for ali the irripulses that act on the particie.

F

r= 1:" F(i)dt

r2

Przy, + f r F dr mv-2r ,

Conservation of Linear Momentum

If the principle af impulse and momentuni fis applied to a system of pardeles. then the collisions between the particles produce interna im pulses that are equal, opposi te, and coiline ar, and łhe-refore cancel tram the equation. Furtherrnore, if an exte rnai impulse is small,that is, the force is smali and the Lima is short. the n the impulse car be classified as nonim pul sive and can be neglecied. Consequently, momentum for the system of particles is conscrved.

The conservation-of-momentuni equation is useful for finding the final velocity of a particie when interna] irnpulses are exerted between twa particies and the initial veIocities of the particles is known. If the internal impulse is to be determined, then one of the particles is isołated and the principi e af impulse and momentum is applied to this particie.

2,tni(v,)1Smi(vi),

Impact

When twa particles A and B have a direct impact, the interna) impulse between them is equai, opposite, and col linear, Consequently the conservati on of momentum for this system applies along the line of irnpact.

mAtvti)1 + ni84,811 = ntA(-0A)2 + nta(t>8)z

CHA PTER R EVI EVV297

"If the final veiocities are unknown. a second equation is needed for solu hon. We musi use the coefficient of restitution, e. This experimentally determined coefficient depends upon thc physieal properties of the colliding particles. It can be expressed as the ratio of their relative velocity after collision to [heir relative velocity bef ore collision. If the callision is elastic. no energy is lost and e = 1. For a plastic collision e = O.

H the impact is obrique, then the conservation of resfitution equation apply along the line of impact. Also, conservation of momentum for each particie applies perpendicular ta [hit line (piane ol contact) because no impulse acts on the pardcles in this direction.

(ts)2 (uhlae

(y4), —

Central impact

Piane of cuntadB mLe of impact

I.111C ()t impacti V t

VA YR

Piane uf contact

Obliqueimpact

(H0), = («MV)12

(}10)1 +51 Mo - (110)2 f i

Y■ H o

Principle of Angular Impulse and Momentum

The moment of thc linear momentum abcmt an axis (z) is called the angular momentum.

The principle of angolar impulse and mornenturn is often used to eliminate unknown impulses by summing the momenłs about an axis through which the lines of action of these impulses produte no moment. For (hit reason, a free-body diagram should accornpany the solution.

Steady Fluid Streams

Impulse-and-momentum methods are often used to deterrnine the forces that a device exerts on the mass flow of a fluid —liquid nr gas. To do so, a free-body diagram of the fluid mass in contact with the device is drawn in order to identify these forces. Also, the velocity of the fluid as it fiows into and out of a control volume for the device is calculated. The equations of steady flow involve summing the forces and the moments to deterniine these reactions.

Propulsion with Variable Mass

Some devices, such as a rocket, lose mass as they are propelled fonvard, Others gain mass, such as a shovel. We can account for łhis mass lass nr gain by applying the principle of impulse and momentum to a control volume for ihe device. From this equation. the forte cxerted on the device by the mass flow can then be determined.

dmY — [vy - vA)

_Srvlo d, (rs x vs - ra X V,x)

dv dm,= m—dr -

Loses Masa

dydm,--dr

Gains Mass

Review

Kinematics andKinetics of a Particie

The topics and problems prese-ntecl in Chapters 12 though 15 have all been categorizM in order to provide a elear fueus for learning the variom problem-solving principles involved. In engineering practice, however, i t is most important to be able w identify an appropriate method for the soltnion of a particular problem. In this rcgard. one must fully understand the linnitations and we cif the equations of dynamics, and be able to recognize which equations and principles to use for the problem's soluiion.For these reasons, we will now summarize the equations and principles of particie dynamics and provide the opportunity for applying them to a variety of problems.

Kinematics. Problerns in kinematics require a study of the geometry of motion, and do not account for the forces causing the motion. When the equations of kinematics are applied, one should dearly establish a fixed origin and select an appropriate coordinate system used to define the position d the particie. Once the positive direction of each coordinate axis is established, thcn the directions of the components of position, velocity, and acceleration can be determined from the ilgebraic sign of their numerical quantities.

Rectilinear 'Action. Varhible Acceleration. If a mathematical {ar graphical) relation #s established hetween any rwo of the foiLr variables s, v, a, and r, thcn a (hird variable can be determined by using one of the following equations which relates all three variables.

ds dvv = — a a ds =vdv

drdr

Consram Accelerathm. Be absolutely certain that the acceleration is constant when using the following equations:

s = so + vol + -4a4t2 v = vo

242,(s —

Curvilinear MOtiOn. x. y, z Coordinates. These coordinates arc often used when the motion can be resolved into rectangular components. They arc also uscful for studying projectilc motion sincc the accełeration of the projectile is always downward.

REVIEW 1 KINEMATICS AND KINETICS CIF A PARTICLE 299

= i a = x v= j ,

aY = vZ1._= Ża_=

t. b Coordinates These coordinates are particularly advantageous for studying the particie's acceleration along a known path. This is because the I and n cornponents of a represent the separate changes in the magnitude and direction of the velocity, respectively, and these components can be readily formulated.

v =

d v= v —

d s1)2

a, = —P

whcre

[1 + (dyiri.r)2 P/2

P — d2v ł clx'when the path y =,,f(x)is given.r ,

O. z Coordinałes, These coordinates arc used when data regarding the anguiar motion of the radial coordinate r is given to &serii-Je the particle's motion. Also, some paths of motion can conveniently be described using these coordinates.

v, = r a, = T — rÓ2

= rÓ a f, = rB + 2i-9

v, = ża, =

Relative Motion. If the origin of a translating coordinate system is cstablished at particie A, then for particie B.

rB = ± r/3/

A

— v.4 ±

HB = alVA

Here the relative motion is mcasured by an obscrver fixed in the translating coordinate system,

Kinetics. Problems in kinetics invoive the analysis of forces which cause the motion. When applying the equations of kinetics, it is absolutely necessary that measurennents of the motion be made from an inerttal coordinate system, i.e„ one that does not rotate and is either fixed or transiates with constant velocity, if a problem requires sirnuitaneous solution d the equations of kinetics and kincmatics, then it is important that the coordinate systems selected for writing each of the equations define the positive directions of the axes in the same manner.

300 REVIEW 1 KINEMATICS AND KINETICS OF A PARTICLE

Equations of Motion. These equations arc used to solve for the particie's acceleration ur the forces causing the motion, If they arc used to determine a particies position, velociły, or time of motion, then kinematics will also have to be considered to complete the solution. Before appiying the equations of motion, always draw a free-body diagram ta identify ali the forces acting on the particie. Also, establish the direction of the particies acceleration or its components. (A kinetic diagram may accompany the solution in order to graphically account for the ma vector,)

= ma, ma„ £F,.= mir„ F, = ma, Fy = md„

F, = „nu_ 1F = O 1F, = mu-

Werk and Energy. The cquation of work and energy represcnts an ntegratcd form of the tangentiał equaton of motion, F, = mat, combined with kinematics (a, ds = t dv), It is used w salve probłems invoiving forte, velociły, and displaremenł. Before appłying this equation, always draw a free-body diagram in order to identify the forces which do wark on the particie.

IU1_1 = T2

whereT = m-L, 2

Ur = 1 S F cos O ds s,= F,cos f3(s, — BI)

liw = —W dy= —6ks; - .4ks?)

(kinetic energy)

(wark of a variable furce)

(work of a constant furce) (

work of a weight)

(work of an ełastie spring)

If thc forces acting on the particie are conserl?ative forces, i.e., thosc that do not cause a dissipation of energy, such as friction, then apply the conservation of energy equation. This equation is easier to use than the equation of work and energy since it applies at only owo points on the path and dres not require calculałion of the work done by a furce as the particie moves along the path,

Tł ± VL = T2 + V2

where V = V2 -ł- V, and

V6. Wy (gravitational potential energy)

4 ks2 (e lastic putential energy)

If the power developed by a force is to be detcrrinned, usedU

P = — = F • vdt

where v is the velociły of the particie acted upon by the force F.

EViEW 1KINEMATICS AND KINETICS OF A PARTICLE301

Impulse and Momentum. The equation of fineur irtIpuLeE and znomeitłum is an integrated form of the equation of motion, F =combined with kincmatics (a = dvidt). lt ts itsed to solve problems Mvolving force, velociły, and fiole Before applying this equation, one should always draw the free-body diagram, in order to identify all the forces that cause impulses on the particie. From thc diagram thc impuisive and n °nim pulsi ve forces should be identified, Recall that the non impulsi ve forces can be neglected in the analysis during the time of impacł. Also, estab]ish the direction of the particies velocity just before and just after the impulses are applied. As an alternative procedure. impulse and momen Wrn diagrams may accompany the solution in order to graphically ac.count for the terms in the equation.

rp

tnvl + 2.ir Frit = mv ,• rL

several parlides are invoived in the problem, consider applying the conservation ofmomenlum to the system in order to elirninate the interna! impulses from the analysis. This can be done in a specified direction, provided no extcrnai impulses act on the particles in that direetion.

mu l =

If the problem involves impact and the coefficient of restitution e is given. then apply the following equation.

(vn), — (,4)2e = (along line of impact)

(vA)1 (vo..11

Remember that during impact the principle of work and energy cannot be used, since the partiŁ:les deform and therefore the work due to the interna] forces will be unknown.The principle of work and energy can be used, however. to determine the energy loss during the col lision once the particleS initial and fina] velocities are determined,

The prindpie of angular impulse and momentum and the conservation of angidar niomentnnt can be applicd about an axis in order to elintinate same of the mikito» impulses acting on the particie during the period when ils motion is stuclied. Investigation of the particie's free-body diagram (ar the impulse diagram) will aid in choosing the axis for application.

(Ho), + J F M,dl = (110},-,

(H0)1 = (Ho)2

The following probiems provide an opportunity for applying the above coneepts, They are presented in random order so that practice may be gained in idenlit`ying the various types of problcros and developing the skills necessary for their soiution.

302REVIEW 1 KINEMATtCS AND KINETiCS OF A PARTICLE• REVIEW PROBLEMS-WF

R1-1, An automobilc is traveling with a constara sNed along a horizonta] dreular curve that has a radius p = 750 ft. If the magnitude of acceleration is a --- 8 ft/s2., determine the speed at which the automobile is traveling.

R1-2. Block B rests on a smooth surface, 1f the coefficients of frietkm between A and B arc p, = 04 and lik = 03, dc termine the acceleradon of each Hack i f (a) F -= 6 lb, and (b) F =- 50 lb.

2U113

)1.- F

If5o lh

Prob. R1-2

R1-3. T'he smali collar starting from rest at A slidesdown along lhe smooth rod. DurMg i he motion. the collar is acted upon by a forte F = (10i -i- 95yj + 2;.k 411x, where x, y, z are in feet, Determine the collars speed when it strikes the wali at B.

Prob. R1-3

*R1-1. The automobile travels from a parking deck down along a cylindrical spiral ramp at a constant speed of v = 1.5 rn/s If Ule ramp descends a distanee of 12 m for every fuli revoIution, B = 27r rad, determine the magnitude of the ears aceeleration as it moves along the ramp, r = 10 m. !hut For part of the solution, note that the tangent to the ramp at any point is at an angle of

-= tan 1 (12/127r(10)1) = 10,81' from the horizontal. Usethis to determine the veloeity componcnts i and.which in turn are used to determine B and ż.

Prob. R1-4

R1-5. A rifle has a masa of 2.5 kg. If it is loosely gripped and a 1.5-g bul]et is fired from il wilh a horizontal muzzle veloeity of 1400 m/s, determine the reeoil veloeity of the rifle just after firing.

R1-6. If a 150-lb crate is released from rest at A, determine its speed after it slides 30 ft down the plane.The coefficient of kinetic frietion between the erate and piane is j. = 0.3.

Prob. R1-6

R EVIEW 1 KINEMAT1CS AND KINETICS OF A PARTICLE 303

R1-7. The van is traveling at 20km/h when a coupling of the trailer at A fails.If the trailer hasa mass of 250 kg and caasts 45 m before coming to rest, determine the constans horizontal force 1= crcałed by rolling friction which causes the trailer to stop.

R1-10. Packages having a mass of 6 kg slide down a srnooth chute and land horizontally with a speed of 3 m/s R1 on the surface of a conveyor bclt. If the coefficient of kineticfriction between the beli and a package is = 0.2,determine the time needed to bring the package to rest on the bell if the beli is moving in the same direction as the package with a speed a = 1 iu/s.

Prok R I -M

20 km/h

Prob. R1-7

*R1-8, The position cif a particie along a straight line isgiven by s = (t3 - 9r2 15t) ft, where r is in secund&Determine its maximum accełerafion and maximumvelocity during the time intervalt s 10 s.

R1-9. The spool. which has a masa of 4 kg, slides along the rotating rod. At the Instant shown, the angolar rate of rotation cif the rod is 6 = 6 rad/s and Ibis rotation is increasing at ft = 2 rad/s2. At this same instant. the spool bas a veloeity of 3 rn/s and an acceleration of I m /s2, bulb measured relative to the rod and directed away from the center O when r = 0.5 m. Dctermine the radial frietional force and the norrnal force, both exerted by the rod on the spooI at Ibis instant.

R1-41. A 20-kg block is originaliy at rest on a horizontal surface for which the coefficient of stalic friction is ja, = 0.6 and The eoefficient of kinetic friction ispk = 0.5. If a horizontal force is applied such that lit varies with timc as shown, determine the speed of the block in 10 s. flint: First determine the time nceded tn overcome friction and start the btock moving.

r• ▪ = 0_5 m

F

Fl = 2 raci/s2"--

F {N)

va,= 3 mis fi.< = i mt's

200

r (s)

5 I0

Prob. R1-9 Prob. R1-11

3134 REVIEW 1 KINEMATICS AND KINETICS Of A PARTICLE

*R1-12, The 6-lb hall is f, red from a tufie by a spring having a stiffncss k = 20 lb/in. Dcterrnine how far the spring musz be compressed to tirc thc hall from thc compre,ssed position to a hcight of 8 ft, at which point it has a velocity of 6 ft/s.

v = f t / s 1 O

k= 20 lb/in.

Prob. R1-12

R1-13. A train car, having a mass of 25 Mg. travels up incline with a constant speed of $0 km/h. Determine the power required to overcomc the force of granity.

R1-14. The rocket sled has a mass of 4 Mg and travels from rosi along the sirnooth horizontal track such that it maintains a constant power output of 450 kW, Neglect the loss of fuel mass and air resistance, and dctermine how far itmusi travel to reach a speed = 60 m/s.

uR1-15. A projcetile, initially at the origin, mew-es vertically downward along a straight-Iine path through a fluid mediumsuch that its velocity is defined as v .= teł, m/s,where t is in seconds. Plot the position s of the projedne during the first 2 &Uw the Runge-Kutta rnethod to evaluates with incremental values of h = 0.25 s.

*111-16. The chain has a mass of 3 kg/m_ If file coefficient of kinetic friction between the chain and the piane is FLk = t1.2, detcmine the velocity at which the end A will pass point B when the choin is rełeased from rent.

Prob. R1-16

R1-17. The motor M pulis in lis attached rope with an acceleralion ur, = 6 roisz. Deterrnine the towing form exerted by M on the rope in order to move the 50-kg crate np the inclined plano.. The cocfficient of kinetic friction between the. crate and the piane is ;.tk = 0.3. Neglect the mass of the pulleys and ropo.

T

11 II II II U II

Prob. 111-14 Prob. R 1-17

EviE w 1 KINEMATICS AND KlMETIC$ OF A PARTICLE305

R1-18. The drinking fountain is designed such that the nnzzle is locate.d from the edge of the basin as shown. Determine thc maximum and minimum speed at which water can be ejected from the nozzzie so that it does not spIa.sh over the sicles of the basin at B and C.

50 mm

Prob. R1-18

R1-19. The 100-kg crate is subjected to thc action cif two forces, Ft = 800 N and F2 =, 1.5 kN, as shown. If it is originally at rest. determine the distance it slides in order ki attain a speed of 6 m/s. The coefficient of kinetic friction between ihe crate and the surface is 1ea. = 0.2.

Prob. R1-19

*R1-.20. b` a particie has an initial velocity y0 = 12 ft/s to the right, and a constant acceleration of 2 ft/s2 to the

lett. determine the partiele's displacement in 10 s. Originally = 0.

R1-21. The ping-pong bali has a mass of 2 g. if it is stnick with the velocity shown, determine how high h it rises above the end of the smooth labie after the rebound.Take e = OS.

Prob. R1-21

R1-22. A sports car can acceierate at 6 misi and deeelerate. at 8 rri/s. If the maximum speed it can attain is 60 m/s, determine the shortest time it takes to travel 900 m storting from resl and then stopping when s = 900 m.

R1-23. A 2-kg particie rests on a smooth horizontal piane and is acted opon by forces F,. = O and f; = 3 N. If x = 0, y = 0, t‚, = 6 rn,/s, and y, = 2 m/s when r = 0, determine the equation y = f(x)which describes the path,

*R1-24. A skier starts from rest at A (30 ft, O) and descends the smooth sIope.which may be approximałed by a parabola. If she has a weighit of 120 lb, determine the norma forte she exerts on the ground at the instant she arrives at point B.

Pruli. R1-24

306REVIEW 1 KINEMATtCS AND KINETiCS OF A PARTICLE

R1-25. The 20-1b block B rests on the surface of a labiefor which the coefficient of kinelic friction is 0.1.Determine the speed of the 10-lb block A a fter it has rnoved downward 2 ft from rent. Negleel the mass of the pulleys and cot ds.

201h

Prob. R1-25

R1-26. At a given instant the 10-lb block A is moving downward with a speed of 6 flis. Determine its speed 2 s later, Bloek B has a weight of 4 Ib. and the coefficient of kinełic friction between it and the horizonłal piane is ;.tk = 0.2. Neglect the mass of the pulleys and cord,

Two smooth billiard balfs A and R have ar ecitial mass of rn = 200 g. if A strikes B with a velocity of (vA)i = 2 m/s as shown, determine thcir finał velocities jurt after colłision, Bali B is originaily at rent and the coefficient of restitution is e = 0.75.

Prob. R1-27

l`R1-28. A crałe has a weight of 1500 lb. TF ii is pulIed along the ground at a constant speed for a chstance of 20 ft, and the towing cable makes an angie of 15' with the horizontal. determine the tension in the takle and the work done by the towing fasce. The coefficient of kinetic frictionbetween the crate and the ground is = 0.55.

R1-29, Thc collar of negligible size has a masa of 0.25 kg and is attached to a spring having an unstretched length of 1(X) mm. 1f łhe cetnar is released from rest at A and traveis along the smooth guide, determine iłs speed just bef= it strikes B.

Pro r. R i -2{; Prob. R1-29

REVIEW 7 KINEMATICS AND KlNETIC$ OF A PARTICLE 3 0 7

R1-30. Determine the tension developed in the two cords and thc acceleration of each hloek. Neglect the mass of the pulleys and cords. ihnt Sinoe the system consists of tow cords, relatc łhc motion of block A to C, and cif błock B to C. Then, by elimination, relate the motion of A to B.

4 kg

Prob. R1-30

R1-31. The baggage truck A hasa mass of 800 kg and is used to pull each of the 300-kg cars. Determine the lension in lhe couplings al B and C if lhe tractive force F on the truck is F = 480 N. What is the speed of the truck when

= 2 s, tarting from the rest? The car wheels are Free to roll. Neglect the mass of the wheels.

„1(1-32. The baggage truck A has a mass of 800 kg and is used to puli each of the 300-kg cars. If the tracłive force F on the truck is F = 480 N. deterrnine the initial acceleration of the truck. What is the acceleration of the truck if the coupłing al C suddeniy fails'? The car wheels are Free to roli. Neglect the mass of the wheels.

Probs. R1-31132

ER1-33. Packages having a mass of 2.5 kg ride on the surfacc of thc conveyor bełt. If the bełt starta from rest and R1 with constant acceleration increases to a speed of 0.75 m/s in 2 s, determine the maximum angle of tilt, B, so that none of the packages slip on the inchned surface AB of the bełt. The coefficient of stalic friction between the belt and each package is kr, = 0.3. Al what angle rb do the packages first begin to slip off lhe surface of the belt if the, bell is moving at a constant speed of 0,75 m/s?

Prob. R1-33

R1-34. A particie travels in a straight line such that for a short lime 2 s s r -s- 6 s its motion is described by = (

4/a> fł/s where is in ft/s2 . If = 6 ft/s when = 2 s. determine the partiele's acceleration when / = 3 s,

R1-35. The b lo cks A and B weigh 10 and 30 I b, respectively. They are connecled together by a light cord and ride in the frrclionless grooves. Determine the speed of each block after bIock A moves 6 fl up along the plane.The blocks are released Erom resl.

Prob. R1-35

3 m

308REVIEW 1 KINEMATICS AND KINETICS OF A tARTIClE

R1-36. A motorcycle starts from rest at I = O and travels along a straight mad with a constant acceleration of 6 11/s2 unii it reaches a speed of 50 ft/s. Aftenvards it maintains ]his speed. Also, when r = 0, a car locateci 6000 ft down the mad is travefing loward the rnotocycle at a constans speed of 30 ft/s. Determine the time and the distance traveled by the motorcycle when they pass each other.

R1-37. The 5-1b hall, attached to the ctird, is sftuck by the boy. Determine the smallest speed he must impart to the bali so that it will swing around in a vertical circle, without causing the cord lo become slack.

Prob. R1-37

R1-38. A projectile, initially at the origin, moves atong a straight-lirce path through a fluid medium such that itsvelocity is1800( i - e-°31) mm/s where f is in seconds.Determine the displacemenl of the projectile during the first 3 s.

R1-39. A particie travels along a straight line with a veloeity tr = (12 - 3t2) m/s„ 9,vhere t is in seconds_ When r = 1 a, the particie is located 10 ni to the Ieft of the origin. Determine the acceleration when t = 4 s, the displacementfrom r = 0 to r = s, and the distance the particie travelsduring ]his time period.

R1I-40. A particie is rnoving along a circular path ()f 2-m radius such that its position as a function of firne is given by t# = (5/2) rad, where r is in seconds. Determine the magnitude of the particles acceleration when H = 30. The particie starts from rest when & = T.

R1-41. If the end of the uhle at A is pulled down with a speed of 2 m/s. dctcrminc the speed at which hlock 13 rises.

Prob. R1-41

R1-42. The bottle rests at a distance of 3 fl from the center of the horizontal platform. If the coefficient of stalic frictionbetween the bottle and the platform is g, = determinethe maximum speed that the bottle can attain tocfore slipping. Assume the angolar motion of the platform is slowly increasing.

R1-43. Work Prob. R1-42 assuming that the platform staris rotating from rest so that the speed of the botile is increased at 2 ft/s2.

3 I'l

Probs. R1-42/43

REVIEW 1 KtNEMAT1CS AND KINETICS OF A PARTICLE309

*R1-44. A 3-lb Wek, initially at rest at point A, slides along the smooth parabulie surface. Determine the normal force acting on the błock when it reaches B. NeOect the sine of the block,

Prob. R1-44

R1-45. A car starts from rest and moves ałong a straight line with an acceleration of a = (3a•-1r3) m/s2. where s is m meters. Determine the cat's acceleration when r = 4 s.

R1-46. A particie travels along a caryc defined by the equation s = Q3 - 3ż2 + 21) m, wherc is in seecinds. Drawthe ,r-t, v-r, and u-a graphs for the particie for D r s 3 s.

R1-47. The crate, having a weight of 50 lb, is hoisted by the pulley system and motor M. Tf the erate starts from rest and, by constan t acceleration. attains a speed of i 2 ftis afte r rising 10 ft, determine the power that must be supph.ed to the motor at the instant s = 10 ft. The motor has an efficiency E = 0.74.

Prob. R1-47

*R1-48. The block bas a mass of 0.5 kg and moves within the smooth vertical slot. If the block starts from rest when the anached spring is in the unstretched position at A. determine the consrarir vertical force F which must be applied to the cord so that the błock attains a speed vp = 2.5 m/s when it reaches B: s8 = 015 m. Neglect the mass

of the sard and pilicy.

0.3 m

Prob. R1-48

R1-49. A bali having a mass of 200 g is rdeased from mst at a height of 400 mm above a very large fixed metal surface. 1f thc bali rebouncls to a height of 325 mm alit-we the surface, determine the coefficient of restitution betwcen the hall and the surface.

R1-50. Determine the speed of block B if the end of the cable at Cis pulled downward with a speed elf 10 ft/s. What is the relative velocity of the block with respect to C.

Prob. R1-50

Chapter

Kinematics is Tmpnrtant for the design of the mechansrn used on this durno truck_

Planar Kinematicsof a Rigid Body

CHAPTER OBJECTIVES

•To classify the yarious types of rigid-body planar motion.

•To investigate rigid-bady translation and angular motion abouta fixed axis.

▪To study planar motion using an absolute motion analysis.

•To provide a relative motion analysis of velocity and acceleration using a translating frame of reference.

•To show how to find the instantaneous center of zero velocity and determine the velocity of a point on a body using this method_

•To provide a relative-motion analysis of velocity and acceleration using a rotating frame of reference.

1 6.1 Planar Rigid-Body Motion

In this chapter, the planar kinematics of a rigid body will be discussed. This study is important for the design of gears, cams, and mechanisms used for many mechanical operations. Once the kinematics is thoroughly undcrstood, thcn wc can apply thc cquations of motion, which rclate the force5 on the body to the body's motion,

The planar motion of a body occurs when all the partieles of a rigid body move along paths which arc cquidistant from a fixed piane. There are three types of rigid body planu motion. In order of increasing complexity, they are

312 CHAPTER 16 PLANAR KINEMATICS O f A R IWD. BODY

Path of rectilinear translation Path of curvihncar translation

(a) L(1))

t

Rotation about a fixed axis General plant motku'(c) {d}

Fig. 16-1

• Transiation. This type of motion occurs when a line in the body rernains parallel to its original orientation throughout the motion. When the paths of motion for any two points on the body are parafie] lines, the motion is called rectilinear translation, Fig. 16-1a. U the paths of motion arc aIong curved lincs. the motion is calledcurvilinear translation, 16—lb.

• Rotation about a fued axis: When a rigid body rotates about a fixed axis, ali the particies of the body, except those which lie on the axis of rotation, move along circular paths, Fig. 16-1e.

• General piane motion. When a body is subjected to generał piane motion, it undergoes a combination of translation mul rotation, Fig. 16-1d. The translation occurs withiri a reference piane, and the rotation occurs about an axis perpendicular ta the reference piane.

In the foIlowing sections we will consider each of these motions in detail. Examples of bodics undergoing these mations arc shown in Fig. 16-2.

Curvilinear translationGeneral plant motion

Rectilinear translationR.cuallion about a fixed axis

Fig. 16-2

16.2 TRAMSLATION313

1 6.2 Translation

Consider a rigid body which is subjected to either rectilinear or curvilinear transiation in the x-y piane, Fig. 16-3.

Fiw:decordinate system

x'Translatingcoordina te system

Fig. 16-3

Position. The locations of points A and B on the body arc defined with respect to fixed x, y reference frame using poyinon imcwrs rA and rB The translating x`, y' coordinate system is jxed itr the body and has its origin at A, hercafter reform d to as thc bale point. The position of B with respect to A is denoted by the reladve-position vecior rBi„k ("r of B with respect to A"). By vector addition,

ry = r.4

Velocity. A relation between the instantaneous veIocities of A and B is obtained by taking the timc dcrivative of this cquation, which yicidsvB

T vA drBiA ftir. Here v.4 and vB denote abxolute velficiżies sincethese vectors are measured with respect to the x, y axes. The term drsiA idt = 0, since the rnagsingde cif r81,4 is constanł by dcfinItion of a rigid body, and because the body is translating the direction of i-

87A is also consiunt. Therefore,

Acceleration. Taking the time derivative of the velocity equation yields a similar relationship betwecn thc instantaneaus accelcrations of. A and B:

a8 a,4

The above two equations in dicate that ali points in a rigid body subjected to eidier reciilinear or curvilinear trarrstutłon inove with the sanie veloeity and aceeleration. As a result, the kinematics of particie motion.cliscussed in Chapter 12,can also be used to specify the kinematics of points located in a translating rigid body.

Riders this ainusernnt ride aresubjected to curvilinear translation, since the veh.ide moves in a circułar path yet fft always remains in an upri.g1bl position,

3i4CHAPTER 16 PLANAR KINEMATIC5 OF A RIGID BODY

1 6.3 Rotation about a Fixed Axis

When a body rotates about a fixed axis, any point P located in the bady travels along a ećrezdar path. To study this motion it is first ncccssary ta discuss the angular motion of the body about the axis.

Angular Motion. Since a point is without dimension, it cannot have angular motion. Only litwy or bodies undergo angular nunion. For example, consider the body shown in Fig. 16-4a and the angular motion of a radial line r located within the shaded piane.

Angular Position. At thc instant shown, the anguiar posidon of r is defined by the angle O, measured from a fixed reference line lo r.

Angular Displacement. The change in the angular position, which can be measured as a d1fferentlal tiO, is called the ungidar dispiaceiraenur This vector czas a rnagnitude of £10, measured in degrees, raclians, or revolutions, whcrc 1 rev = 2/r rad. Since motion is about a fixed axis, thc direction of dB is always along this axis. Specifically, the direction is determined by the right-band rulet that is, the fingers of the right band are curled with thc sense of rotation, so that in this case the thumb. or dO, ponts upward,Fig. 16-4a, In 'Iwo dimensions, as shown by the top view Uf

the shaded piane, Fig. 16-4b, bołh O and dO are counterclockwise, and so the thumb points outward from the page.

Angular Velocity. The time rate of change in the angular position is ealled che angrdar velocity w (omega). Sincc de occurs during an instant of time dt, then,

( C + )dO

(16-1)

This vector tras a rnagninule which is often measured in radis. It is expressed herc in scalar form sincc its direction is also along the axis of rotation, Fig. 16-4a, When indicating the angular motion in the shaded piane, Fig. 16-4b, we can refer to the sense of rotation as clockwise or counterclockwise. Here we have a rbitrardy chosen counterelockwise rotations as positive and indicated this by thc curl shown in parentheses next to Eq. 16-1. Realize, however, that the di rectional sense of r is actually outward from the page,

( b )

de

4,1)

*It is shown in Scc. 20. I that finitc rotations or finitc angular displacerncnts arc nvi vccłorFig. 16-4quantittes, al tho ugh differential rotations d9 are vcctors.

1L3 ROTAT4QN ABOUT A Fb<ED AXIS315

The gears used in the operafion of a craneall notat fixed axel. Engineersmust be able to rełate their angular mot_ions in order to properly desgn this gear system.

Angular Acceleration. The angular acuderarion cv (alpha) measures the łime rate of change of the angular velocity. The magnirude of this vector is

( C + ) d w

= d t

Using Eq. 16-1, it is also possible to express or

d 2 0 =

d r

The line of action of or is the same as that for w., Fig. 16-4a; however, its sense of direcrtoit depends on whether w is increasing or decreasing. If w is decreasing, thcn a is called ara angtdar deceleration and thereforc has a sense of direction which is opposite to w.

By eliminating dt from Eqs. 16-1 and 16-2, we obtain a differential relation between the angular acceleration, angular velocity, and angular displaccmcnt, namely.

( +) I a dO - w dw j (16-4)

(16-2)

(16-3)

The similarity between the diffcrcntial relations for angular motionand thow developed for rectilinear rnotion of a particle dsidt.a dvicil, and a ds = v dv) shoulcl be apparent.

Constant Angular Acceleration. If the angular acceleration ofthe body is constanł. a = then Eqs. 16-1, 16-2. and 16-4, whenintegrated, yield a set of formulas which relate thc body's angular velocity, angular position, and thne.These equations arc similar to Eqs, 12-4 to 12-6 used for rectilinear motion.The results are

ra t r w o +

O = Oo wcyrw2 = wó + 2a,.(0 - Oo)

Constant Angular Acceleration

( 1 6 - 5 )

( 1 6 - 6 )

( 1 6 - 7 )

Here Oo and WC are the initial values of the body's angular position and angular velocity. respectively.

3 1 6 CHAPTER 16 PLANAR KINEMATIC5 OF A RIGID BODY

( d i

Motion of Point P. As the rigid body in Fig. 16-4c rotates, point P travels along a drcular path of radius r with center at point O. This path is coniained within the shaded piane shown in top view. Fig. 16-1d.

Positio n and D ispla cement. The position of Pis clefmed by the position vector r, which extends from O to P. If the body rotates,c/H then P will displace ds = tyk).

Vei0City. The velocity of P lias a magnitude which can be found by dividing ds = rt-10 by di' so that

( c )

Fig. 16-4 (cmvi.)

= cdr (16-8)

As shown in Figs. 16-4c and 6-4d, the diredion of v is łangent to the circular path.

Both the magnitude and direction of v can also be accounted for by using the cross product of w and rp (sec Appendix B ). Here, is directed from any poitzł on the axis of rota Ilon to point. P. Fig, 16-1c, We have

= co X r, (16-9)

The order of the vectors in this formulation is important, since the cross product is not commutative. i.e,. w x rp = rp X w. Notice in Fig. 16-4c how the correct direction of v is established by the right-hand rule,The fingers of the right hand are curled from w toward rp (w "cross- rp). The thumb indicates the correct direction of y, which is tangent to the path in the direction of motion. From Eq. B-8, the magnitudc of v in eq. 16-9 is v = (urpsin 0, and since r = rp sin ,cP, Fig. 16-4c, then v = off, which agrees with Eq. 16-8. As a special case, the position vector r can be chosen for rp. Here r iies in the piane of motion and again the velocity of point P is

v=wXr (16-10)

16.3 ROTATM ABOUT A FłXED AMS31 7

A ccel eration. The acceleration of P can be expressed in terms of its norma' and tangential components. Since rar = dv i dł and a„ = v21 p, where p = r, y = wr, and a = dwIdi, we have

a, =cer

a„ (16-12)

The tangentrat component of acreleratran, Figs, 16-4e and 16-4f, represents the time rate of change in the velocity's magnitude. I f the speed of P is increasing, then ar acts in the same direction as v; if the speed is decreasing, a, acts in the opposite direction of v; and finaliy, if the speed is constant, a, is zero.

The norrnal cornpanena acceleration represcnts thc time rate ofchange in the velocity's direction.The direction of a„ is always łoward O, the center of the circular path, Figs, 16-4e and 16-4f.

Like the velocity, the acccleration of point P can be expressed in terms of the vecłor cross product,Tak ing ihe time derivative of Eq. 16-9 we have

( 0 )

a a,dv dco drpdr dra = = — X n, + dr X — ,P

RecalIing that a = diroidt, and nsing Eq. 16-9 (drp/dr = v = w x rp), yi el ds

a = a x rp + a) x (w x rp) (16-13)

From the definition of the cross product, the first term on ihe right has magnitude cr, = arpsin d = ar, and by the right-hand rule, a X rp is in the direction of ar, Fig. 16-4e. Likewise, the second term has a magnitude

= w2rpsin40 = w'-r, and applying the right-band role Lwice, first to determine the resułt vp = co X rp then t_d X vp, it can be scen that this rewii is in the same direction as a„, shown in Fig. 16-4e. Noting that this is also the sanie direction as -n which lies in the piane of morion. we can express a„ in a much simpler form as a, = -w2r. Hence, Eq. 16-13 can be identifi e d by its two components as

(f)

Fig. 16-4 (eoni.)

a ar + a„. - ( r X r - w 2 r

(16-14)

Sinet a, and a„ are pc rpendicular to one another, if needed thc magnitude of accekration can be determined from the Pythagorean theorem; nameiy,

= Fig.16-4f

318 CHAPTER 16 PLANAR KINIEMATICS OF A RIID• BODY

If two rotating bodies contacł one another, then the poinrs in CO,/ taci rnove along differeta circular paths. and the velocity and łattgential components of acceleration of the points will be the same: however, the normal componetas of acceleration will not be the same. For example, consider the two meshed gears in Fig. 16-5a. Point. A is located on gcar B and a coincident point A' is located on gear C. Due to ehe rotational motion,!JA = V k, Fig. 16-513, and as a result, coBrB = or .(118 = toc(rci rb.). Also,from Fig. 16-5c, (aA), (ak), so that oB ac(rcirwi however, sinee bothpoints follow different circular paths, (aA),, (ajt.")„ and therefore. asshown,

Important Points

• A body can undergo two types of translation. During rectilinear translation all points follow paralici straight-lino paths, and during curvilinear translation the points follow curved paths that are the same shape.

• Al] thc points on a translating, body move with the same velocity and acceleration.

• Points located on a body that rotates about a fixed axis follow circular paths,

• The relation re c/(31 = mico is derived from a = &ai& and w = (16 jdt by eliminating dt.

• Once angular mot.kms w and a are known, ehe velocity and acccleration of any point on the body can bc determined.

• The velocity always acts tangent to the path of motion.

• The acceleration has two components.The tangential acceleration measurcs the rate of changc in thc magnitude of thc vcIocity andcan be deterrnined from fxr. 'The norrnal accelerationrneasures the rate of change in the direction of the velociły and can be determined from = ai=p.

a-4 (an.)r

(c) Fig, 16-5

16.3 ROTAT4QN ABOUT A Fb<ED AXIS319


The veloci ty and acceleration of a point located on a rigid body that is rotating about a fixed axis can be detcrmined using the lollowing procedura.

Angular Motion.

• Establish the positive sense of rotation about the axis of rotation and show it alongside each kinematic equation as it is applied.

• 11 a relation isknown between any two of the kiur variables a, ca, O, and i, then a third variable can be obtained by using one of the following kinematic equations which relates alt three variables

dOdww = — drdt a cli? = w dw

• If the body's angular acceleration is COnSłani, then the following equations can be used:

= aeł

=<Jod + •1-3

Gó 2 = wó + 2«,.(0 Bo}

• Once the solution is obłained, the sense of 0, w, and a is detcrminecl from the algcbraic signs of their numerical quantitics.

Motion of Point R

• In most cases the velocity of P and its two eornponcnts of acceleration can be determined from the scala r equations

= wrc r r = a r

a , w 2 r

• If the geometry of the problem is clifficult to vistialize the following vector equations should be used:

v = c o X r p = w . o X r

a , = a X r p = a X r

ar, = to X (co X rp) = –cd2r

• Here rp is directed from any point on the axis of rotation to point P. whercas r lias in the piane of motion cif P. Either of (hese vectors, aiong with w and a. should be expressed in terms of its i, j, k cnmponents, and, if necessary, the cross products determined using a determinant expansion (see Eq. B-12).

320 CHAPTER 16 PLANAR KINEMATICS Of A RIWD. BODY

EXAMPLE

16.1

Fig. 16-6

A cord is wrapped around a wheel in Fig. 16-6, which is initially at resł when O = 0. If a force is appiied to the cord and givcs it an aceelcration

= (4i) m/52, where t is in 5CCOndS, detCrMilrle, as a function of time, (a) the angular velocity of the wheel, and (b) the angular position of /inc OP in radians.

SOLUTION

Part (a). The wheel is subjected tu rotation aboul a fixed axis passing thrt.

iugh point 0.Thus,poinł P an the wheel has motion ahout a circular path, and the acceIeration of this point has both tangcntial and normal components. The langentiał componerił is (ap), = (4t) mis2, since the card is wrapped around the wheel and moves tangent to it. Hence the angular acceleration of the wheel is

(C+)(ap), --- ar

(4t) m/s2 a(0.2

= (20t) radis2.)

Using this result, the wheel's angular velocity w can now be determincd from a = (kołdr, since this equation rclates a, t, and w. Integrating, with the initiaI condi lim that w = O viThen t = O, yieids(C+) a = — = (2Ot) radis2

dw dt

f, dw = 20t cft

w = 10,2radis

Part (b). Using this result, the angular position O of OP can be found from w = dH f dt, 3ince this equation relałes 0, w, and r. Integrating, withthe in1tiaI condition O = O when t 0, we have

dodt = w = (1012) rad/s

dO = 1012 dt0

= 3.3313 rad Ans.

NOTĘ: We cannot use the equation of constant angular acceleration. since a

is a function of time.

16.3 ROTATlON ABOUT A F1XED Axis 321

EXAMPLE 1 6 . 2

The motor shown in the photo is used to turn a wheel and attached blower contained within the housing. 'The details arc shown in Fig. 16-7a. 1f the pulley A connected to the motor begins to rotate from rest with a constant angular acceleration of otA = 2 rad/s2, determine the magnitudes of the velocity and acceleration of point P on the wlieel, after the pulley has turned iwo revolutions. Assume the transmission beli does not slip on the pulley and wheel.

SOLUTION

Angular Motion. First we will convert the two revolutions to radians. Since there are 2ir rad in one revolution, then

21r radda = 2 rev =- 12.57 rad

I rev

Since aq is constanł, the angular velociły of pulley A is therefore

(C +) W2 = (08 + 2a,(8 — do)

«),2t = O + 2(2 rad/s3)(12.57 rad — 0)

= 7.090 radis

The pelt has the same speed and tangential component of acceleration as it passes over the pulley and wheel. Thus,

= «,ArA = idBra; 7.090 rad/s (0.15 m) = wB(0.4 m)

coR = 2.659 rad/s

a, = aArA = aBrB; 2 rad/s2 (0.15 m) = aB(0.4 m)

aB = 0.750 rad/s2

Motion of P. As shown on the kinematic diagram in Fig. 16-7b, we have

015 mA = 2 rad/s2

P

( a )

Thus

v p = w a r t / = 2 . 6 5 9 r a d / s ( 0 . 4 m ) = 1 . 0 6 m / s ( a p ) ,

= c y p r B = 0 . 7 5 0 r a d / s 2 (0 . 4 m ) = 0 . 3 m / 5 2 ( a ) , ( a p ) , w

= Br8 = (2.659 rad/s)2(0.4 m) = 2.827 m/s 2 Vp

( b )

ap = 1(0.3 m/s2)2 + (2.827 ni/s2)2 = 2.84 in/s2 Ans. Fig, 16-7

322CHAPTER 16 PLANAR KINEMATICS Of A RIWD. BODY• FUNDAMENTAL PROBLEMS

F16—L When the gear rotates 20 revolutions. it achieves an angular velocity of w -= 30 rad/ s, storting from rest. Determine its constant angular acceleration and the time required.

F16-4. The bucket is hoisted by the ropo that wraps around a dram wheel. If the angular displacement of the wheel is

=

4- 150 rad, where r is in seconds, deterrnine thevelocity and acceleration of the buckel when r = 3 s.

F16-1

F16-2. The fly-wheel rołates with an angular velocity nf w = (0.00502) rad/s. where H is in radians. Dctcrminc the angular aceeleration when it has rotatecl 20 revolutions.

F16-3. The tlywheel rutates with an angular velocity of w = (4 j']/2) rad/s. whete H is in radians,Determine the time rt takes to aehieve an angular velocity of (.1 = 150 rad/ s. When r = 0, A = 1 rad.

F16-3

F 1 6 - 4

F16-5. A wheel has an angular aceeleration of o = (0,5 (3)radis2, virhere A is in radians. Determine the magnitude ur the velocity and aceeleration of a point P located on its firn after the wheel has rotated 2 revolutions. The wheei has a radius of 0.2 m and starts at (dl) = 2 rad is.

F16-6. For a short period of time. the motor turm gear A with a constant angular acceleration of rra = 4.5 rad/s2. storting from rest. Determine the velocity of the cylinder and the distance it travels in three seeonds, The eord is wrapped around puliey D which is rigictly attached to gear B.

225 mm75 mm 125 mm

1 1 6 - 6

aA -= 4.5 rad/s

F 1 6 - 2

16.3 ROTAT4QN ABOUT A Fb<ED AXIS323• PROBLEMS

16-1. The angular veloeity of the diak is defined by w = (5t2 + 2) rad/s, where t is in seconds_ Deterrnine

the magnitudes of the velocity and aceeleration of point A on the disk when t = 0.5 s.

Prob. 16-1

16-2. A fIywheel has its angular speed increased uniform iy from 15 radjs to 60 rad/s in 80 s, If the diameter of the wheel is 2 ft. determine the magnitudes of the norma and tangential components of aceeleration of a point on the dm of file wheel when t = H s, and the tata] distanee thc point travels during the time period,

16-3. The disk is odginany rotating at wo = 8 rod/s. If subjeeted to a constant angular aeceleration of tY = 6 rad/s2, determine the magnitudes of the veloeity and the n and t coraponents of aeoeleration of point A at the instant r - 0.5 s.

16-4. The disk is odginany rotating al wo = 8 rad/s. If it is subjected to a constant angular aceeleration of cr =

6 rad/s2, determine ihe magnitudes of ihe velocity and the n and t components of aceeleration of point 8 just after the wheel undergoes 2 revolutions.

Prob 16-314

16-5. Initially the motor on the eircular saw turns its drive shaft at w = (20/2") rad/s, where t is in seconds. If the radii of gears A and B aro 0.25 in. and 1 in.. respectively. determine the magnitudes of the velocilv and aoceleration of a tooth C on the saw blade after the drive shaft rotates

= 5 rad starting from rent.

Prob. 16-5 16-6. A wheel has an initial elockwise angular velecity of

10 rad/s and a conslani angular acceleration of 3 rad/s2, Determine the number of revolutions it must undcrgo to acquire a elockwise angular veloeity of 15 radfs. What time is required?16-7. 1f gear A rotates with a constant angular aceeleration of rt, = 90 radis2, starting from rest. determine the linie required for gear D to attain an angular velocity rit 600 rpm. Also, find the number vf revoluiions of gear D to attain this angular veloeity Gears A. B. C, and D have radu of 15 mm, 50 mm. 25 mm, and 75 mm. respectively.

*16-8. If gear A rotates with an angular veloeity oftuA = (6.4+1) rad/s, whereis the angular displacement ofgear A , m easured fin radians, determine ihe angular aceeleration of gear D when BA = 3 rad, starting from rest. Gears A, B, C, and D have radli of 15 mm, 50 mm. 25 mm, and 75 mm, respectively.

324CHAPTER 16 PLANAR KINEMATICS Of A RIWD. BODY

16-9. The verłical-axis wind t-ul-bine consists of twa blades that have a parabolic shape_ If the blades arc odginany at rest and begin to turn with a constant angular acceleration of Er, = 03 rad/s?, deterrnine the magnitude of the velocity and acceleration af points A and B on the blade after the blade has rotated through two revolutions.

16-W. The vertical-axis windturbine consists of two blades that have a parabolic shape. If the bladcs arc odginany at rest and begin to turu with a constant angularacceleration of0.5determine the magnitude ofthe velocity and acceleration of points A and B on the blade when i = 4 s.

Prob& 16-911016-11. Tf the angular velocity of the drurn is increased uniformły from 6 rad/s when r = U to 12 radis when r = 5 s. determine the magnitudes <O the velocity and acceleration of points A and B on the beli when i = ł s. At Ibis instant the points arc located as shown.

Prob. 16-II

*16-12. A motor gives gear A an angular acceleration ofay = (0.250303) radisz,wherc O is in radiany. If this gearis initially turning at (£0,4)u = 20 rad /s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

16-13. A motor gives gear A an angular acceleration of ElEA = (4/3) rad/S2, where / is in seconds.If this gear is initialty

turning at (wri.)0 = 20 rad/s. dełermine the angular velocity of gear B when r = 2 s.

Prohs. 16-12/13

16-14. The diak starts from rest and is given an angular acceleration a -= (2t) radfs, where t is in seconds. Determine the angular vefocity of the disk and its angular displacement when t = 4 s.

16-15. The diak starts from rest and is given an angular acceleration a = (5/9 rad/s2, where r is in seconds. Determine the magnitudes of the normal and tangcntia] componcnts ot acceleration of a point P on the rim of the diak when ł = 2 s.

*16-16. The disk starta at w,y = 1 radfs when O = 0. and is given an angular acceleration a = (0.30) rad/s2, where O is in radiany. Determine the magnitudes of the nurmal and tangential components of acceleration af a point P on the rim of the disk when fl = l WW.

Pratis. 16-14115/16

16.3 ROTA-T-4QN ABOIJT A F4XED AXIS 325

16-17. Surfing at (w4)0 = 3 radis, when O = 0, s -=- policy A is given an angular acceierafion = (0.60) rad /s2, where O is in radians. Determine the speed of błock B when it has risen s = 13.5 m.'The pulley has an inner hub D which is fixed to C and turns with

16-18. Starting from rest when s- = 0. pulley A is given constans angular aeceleration cre = 6 rad/s2. Determine The speed of błock B when it has risen s = 6 m. The pulley has an inner hub D which is fixed to C and turris with it.

Probs. 16-17/18

16-19. The vaeuum cleaner's armature shaft S rotates with an angular acceieration of er = 4(0314 radis2, where w is in rad/s. Desermine the hrosh's angular velflcity when r = 4 s, starling from wo = I rad /s, at O = 0. The raduj of the shaft and the kwh are 0.25 in. and 1 in., respectively, Neglect the ibickness of the drive bel!.

16-20. The operałion (.4 "reverse- for a three-speed automotive transmission is illustrated schematicaily in the figuro_ If the crank shaft G is turning with an angular speed of 60 rad js, detc=inc the angular speed of the drive. shaft Each of the gears rotates about a fixed axis. Note that gears A and B, C and D, E and F are in mesh. The raclii of each of these gears are reported in the figure.

16-21 . A motor g ives d i sk A a c lockwisc angu la racceleration of = (0,6/2 -i- 0.75) radisz, where r is in

seconds. If the initial angular vełocity of the disk is = 6 radfs, defermine the magnitudes ot the velocity and acceleration of

błock B when / = 2 s.

Wc; -- 60 rad/sw r r

G

r,, - 90 mmrc = 30 mm

rb = 50 rarn= 70 mm

rF = 60 mm

Prob. 16-20

- 6 Tdidis

Prob. 16-19 Prob. 16-21

326CHAPTER 16 PLANAR KIKEMATICS Of A RIGID BODY

16-22. Fora short time the motor turns gear A with anangular acceleratian of erA(30$141rad/s2., where t is insecond.s. Deterrnine the angular veloeity of gear D when = 5 s. starting from rest. Gear A is initially at rest. The radli of gears A. B, C, and D are rA = 25 mm, rr,IM mm,rc = 40 mm, and rp -= 100 mm, respeetively.

16-23. The motor Tunis gear A so that its angular veloeity inereases uniforrnly from zero to 3000 rev/ min after the shaft tunis 200 rev.Determine the angular veloeity of gear D when

= 3 s. The radli of gears A, B. C, and D are rr = 25 mm.100 mm, ft- = 40 mm, and f, T= 100 mm, respeetively

Probs. 16-22/23

16-24. The gear A on the drive shaft of the autboard motor lias a radius rA = 0.5 in. and lhe mesheci pinion gear B on the propeller shaft has a radius rs. = 1.2 in. Determine the angular veloeity of the propeller in t = 1.5 s. if the drive shaft rotates with an angular aceeleration n = (401.1/3) rad/ where r is in seconds. The propeller is originally at rest and the motor frarne does not move.

16-25. For the autboard motor in Prob. 16-24, dełermine the magnilude of the vełocily and acceleration of point P located on the lip of lhe propeller at the instant r = 0.75 s.

Probs. 16-24/25

16-26. The pinion gear A on the motor shaft is given a constant angular acceleration re = 3 rad/s2, 1f the gears A and B have the dimensions shown, determine the angular veloeity and angular displaeement of the output shaft C, when i = 2 s starting from rest. The shaft is fixed te B and ttirns with it.

Prob. 16-26

16-27, For a sheirt time, gear A of the automobile starter rotates with an angular aceeleration of

=

(450rg + 60) rad/s2, where r fis in seconds. Determine the angular veloeity and angular displacement of gear B when t = 2 s, storting from rest, The radli of gears .A and B are 10 mm and 25 mm. respectively.

16-28. For a short time. gear A of the automobile starter rotates wilh an angular aceeleration of viA = (50w L/2) rad/s2, where w is in radfs.Detetmine the angular velocity of gear Bwhen t = I s. Orginally= I rad is when t = O. Theradli of gears A and B are 10 mm and 25 mm, respeetively.

J:J.1,.,1uI

.1:111 ui

Probs. 16-27128

16.3 ROTAT10N ABOLIT A F4XED AXI5327

16-29. A rrill in a textile plant uses the beli-and-pulley arrangement shown to transmit power. When t = O an eleetrie motor is turning pulley A with an angular veloeify of

= 5 rad/s. If this pulley is subjeeted to a constant counterelockwise angular aceeleration of tr1 = 2 rad/s2, determine the angular velocity of pulley B atter B tunis 6 revolutions. The hub at Dis rigidły coren.ecr& to pulley C and turns with it.

Prob. 16-29

16-30. A tape having a thiekness s wraps around the wheei wh ich is turning at a consta nt rate w. Assuming the unwrapped portion of tape rernai ns horizontaideterrnine the aectIeration of point P of the unwrapped tape when the radius of the wrapped tape is r. flint: Since yp = wr, take the finie derivative aud note that dr/dr = Ws12171,

P

w

Prob. 16-30

16-31. Due to ihe screw at E. the aclualor provides linear ~tlen to the arm al F when the motor turns the gear at A, 1f the gears have the radli listed in the figure, and the screw at F hasa pilch p = 2 mm, determine the speed at F when the motor torus A at 0.1.4 .= 20 rad/s. Hint: The screw piteh indicates the amount ni advance of the screw for each ful' revolution.

Prob. 16-31

16-32. The driving bełt is twisted so that pulley i3 rotates in the npposite direetion to that of drive wheel A_If A has a constant angular aceeleration of nA = , 30 rad/s. determine the tangential and norma components of acceleration of a point located at the lim of B when r = 3 s, starting from rest.

16-33. The driving bełt is twisted so that pulley B rotates in the oppeske direction to Chat of drive wheel A. 1f the angular displaeernent of A is BĄ = (5/3 + 10/2) rad, where is in seoonds, determine the angular velocity and angular acceieration of B when r = 3 s.

Prtibs. 16-32/33

328CHAPTER 16 PLANAR KINEMATICS of A RIWD. BODY

16-34. The rope uf diarneter d is wrapped arouncl the tapered drutu whieh has the dśmensions shown, It the dram is rotating at a consfant rato: of w, determine the upward aczcterafion of the bibek. Negleet che smali horizontal dispłacernent of the błock.

Prob. 16-34

16-37. The rod assembly is supparted by bali-and-so,eket joints at A and B. At the instant shown it śs rotating about the y axis with an angular velocity w = 5 rad/s and has an angular awvIeration n .= S rad jsz. Deterrnine the magriitudes of the velocity and acceleration of point C at this instant. Solne the problem using Cartesian vectors and Eqs. 16-9 and 16-13.

Prob, 16-37

16-35. If the shaft and plate rotates with a censtant angular velocity of w = 14 rad/ s, determine the velocity and acceleration of point C located on the corner of the plate at the Instant shown. Express the resull in Cartesian vector form.

*16-36. At the instant shown, the haf and plate rotates with an angular velocity nf w = 14 radis and angular acceleration of rx = 7 rad/s2. Deterrnine the velocity and acceleration of point D located on the canner of the plate at this instant. Express the result in Cartesian vector form.

Probs. 116-35/36

16-38. Rotation of the robotic arm occurs due to linear rnovement of the hydraulic cylinders A and B.If this ma tten causes the gear at D to rotate clockwise at 5 ra.d/s,determine the magnilude of velocity and acceleration of the part C held by the grips of the arm.

Prob. 16-38

16.4 ABsaLuTE MOTION ANALYIS329

1 6.4 Absolute Motion Analysis

A body subjected to generał piane Morion undergoes a simultaneous transłation and rotation. If the body is represcnted by a Chin slab, thc slab translates in the piane of the slab and rotates about an axis perpendicular to this plane,The motion can be completely specified by knowing burki the angular rotation of a line fixed in the body and the motion of a point on thc body, One way to relate these niotions is to use a rectilinear position coordinate s to locate the point along its patki and an angular position coordinate Q to specify the orientation of the line.The two coordinates are then reIatecl using the geometry of the problem. By direct application ofthe lime-differential equations w = dsfdt, a = dv /art, = cfojdt, and

= dw/dr, the morion of the point and the angular motion of the line can then be related.This procedure is similar to that used to solve dependent motion probierris involving pulleys, Sec. 12.9. In some cases, this same procedure may be used to relate the motion of one body, undergoing either rotation about a fixed axis or translation, to that of a conneeted body undergoing general piane motion.


The velocity and acceleration of a point P undergoing rectilinear motion can be relatecl to the angular velocity and angular acceleration of a line contained within a body using the following procedure.

Position Coordinate Equation,

• Locate point P on the body using a position coordinate s, which is measured From a fixed origin and is directed along the słraighł-line pal* of rnotion of point P.

• Measurc from a fixed reference line the angular position H of a line lying in the body.

• From the dimensions of the body, relate s to O, 3 = f", using geometry andfor trigonometry.

Tirne Derivatives.

• Take the firsł derivative of s -= flO) with respect to time to get a relation between r and OJ.

• Take the second time derivative to get a relation between a and

• In each case the chain rule of calculus musi be used when taking the time derivałives of the position coordinate equation. See Appendx C.

The dumping bin on thc truck rotates about a fixed axis pasing through the pin at A. IL is uperated by the extewiun uf the hydraulic cylinder B C. The angular position of the bin can be specified using the angular position coordinate O, and the position of poinł C on the bin is specified using the rectilinear position coordinate s,Sinee a and b are fixed lengths, then the two coordinates can be rclatcd by thecosineraw, s = a2 + b2 — 2ab cos H.The time derivative uf this cquation relates the speed at which the hydrauiic cylinder extends to the angular velocity of the bin.


EXAMPLE

16.3

The end of rod R shown in Fig. 16-8 maintains contact with the cam by means of a spring. If the cam rotates about an axis passing through point O willi an angular acccicration u and angular velocity determine the velocity and accełeration of the rod when the G41TI the arbitrary position O,

W

Fig. 16-8

SOLUTION

Position Coordinate Equation, Coordinates O and x are chosen in order to relałe the rotrifloraal ozatiorr of the line segment OA on łhe cam to the rectiliitearrrao.slatiari of the rod.Thesc coordinates arc measured from the fixed point O and can be rrlaLed to each other using trigonometry. Since OC = CB = r cos 0, Fig. 16-8, then

= 2r cos OTime Derivatives. Using the chain tule ol' calculus, we have

dx dOdt - —2r(sin

• = —2rrr sin f# Ans.v - = —Zr dłdł

{—(1w) sin O — 2rw(cos 0}—dr

•=sin + W2 cos 0) Ans.NOTE: The negative signs indicate that y and a are opposite to the direction of positive x. This seems reasonable when you visualize the motion.

16.4 AB$OLLJTE MOTION ANALYSIS 331

EXAMPLE

At a given instant, the cylinder of radius r, shown in Fig. 16-9, has an angular veloeity w and angular acceleration oz. Determine the velocity and acceleration of its center G if the cylinder rolls without slipping.

SOLUTION

Position Cuordinate Equation. The cylinder undergoes general piane motion since it simultaneously transIates and rotates. By inspection, point G moves in a straight /iw to the left, from G to G', as the cylinder rolls, Fig, 16-9. Consequently its now position G' will be specilied by the hofizontal position coordinate sG, which is measured from G to G'. Also, as the cylinder rolls (without slipping), the arc length A'B on the rim which was in contact with the ground from A to B, is equivalent to sG. Consequently, the motion requires the radial line GA to rotatc O to the position GrA Since the arc AS = rO, thcn G travels a distance

S G = r 0

Time Deriaatlues. Taking successive time derivatives of this equation, realizing that r is constant, w = deidt, and a = d.o/dt, gives the neeessary relationships:

Ans .

Ans :

NOTE; Remember that these relałionships aru valid only if the cylinder wheeI, bal], etc.) rolls without slipping.

332CHAPTER 16 PLANAR KIKEMATICS OF A RiGi l ) BODY

EXAMPLE

2m

nlg. 16-10

The large window in Fig.16-10 is opened using a hydraulic cylinder AB. If the cylinder extends at a constant rate of 03 m/s, determine the angular velocity and angular aceeleration of the window at the

instant =

SOLUT1ON

Position Coordinate Equation. The angular motion of the window can be obtainedusing the coordinate d, whereas the extension ar motionałonghydraulic cylinder is defined using a coordinate s, whichmeasures its Iength from the fixed point A to the moving point B. These coordinates can be related using the law of cosines, namely,

s2 = (2 rn)2 + (1 m)2 – 2(2 m),(1 in) cos

s2 = 5 – 4 cos (1)

When O = 30°,

s = 1.239 m

Time Derivatives. Taking the time derivatives of Eq. I. we have

ds dH2s— – 4(–sin 0)—

s(v5) = 2(sin 0)w (2)

Since 41, = 0.5 m/s, then at O = 30%

(L239 m )(0.5 m/s) = 2 sin 30'ww = 0.6197 radis = 0.620 rad/s A.

Taking the time derivative of Eq. 2 yields

aLs dr, d

v, + s—dt = 2(cos 0) dt Ow 4- 2(sin 0)—dw

dt v2; + sa, = 2(cos (3'){ó2 2(sin 0)a

Since us = dv,' dt = 0. then

(0.5 m/s)' + O -= 2 cos 30°(0.6197 rad/s) + 2 sin 30"a.

a = –0.415 radis2

Because the resuit is negative, it indicates the window has an angular cleceIcration.

16.4 AB$atu-rE MOTION ANALYIS333

■ PROBLEMS

16-39. The bar DC rotales uniformly about the shaft at D with a constant angular velocity w. Uetermine the velocity and acceleration of the bar AB. which is confined by the guides tri rnove vertically.

16-40. The mechanism is used lo convert the constant ciKular motion w of rod A B lnto translating motion of rod CD and the attached verticaI slot. Determine the velocity and acceleratism of CD for any angle- B ofAB.

16-41. At the instant B = 50', the slotted guide is moving upward with an acceleration of 3 rrt/s2 and a velocity of 2 mis, Determine the anguiar acceleration and 4rgular velocity af link A B at this instant.Nore:The upward motion of the guide k in the negative y direetion.

Prob. 16-41

16-42. The mechanism shown is known as a Nurembcrg seissors_ 1f the hook at C rnoves with a eonst ant wiochy of v. determine the velocity and acceleration af collar A as a function of 0. The collar slides freely along the vertical guide.

Prob. 16-39

Prob. 16-42

334 CHAPTER 16 PLANAR KIKEMATICS Of A RIGID BODY

16-43. The crankshaft AB is rotaling at a constant angolar velocity of w = 150 rad/s. Determine the veloeity of the piston P at the instant H = 30°.

Pruli. 16-43

*16-44. Determine the velocity and acceleration of the follower rod CD as a function of H when the contact between the cam and fo1lower is a1ong the straight region AB nn the face of thc cam. The cam rotates with a constant counterclockwise angular velocity w.

16-45. Deterrrine the velocity of rod R for any angle O cif the cam C if the cam rotates with a constant angolar velocity or. The pin connection at O does not causc an interference with the motion of A on C.

Prali. 16-45

16-46. The bridge girder G of a basenie bridge is raised and lowered using the drive mechanism shown. If the hydraulic cylinder AB shortens at a constant rate of 0.15 m/s, determine thc angolar velocity of the bridge girder at the Instant O = 60°.

Prob. 16-46

16-47. The circular cam of radius r is rotating dockwise with a constant angolar velocity w about the pin at Q, which is at an eecenłrie distance e front the center of the earn. Determine the velocity and acceleration of the folIower rod A as a funcdon of B.

Prob. 16-47

*16-48. Peg B mounted ran hydraulic cylinder BD slides freely along the slot in link AC. If the hydraulic cylinder extends at a constant rate of 0.5 m/s, determine lhe angular veIocity and angular acceleration of the link at the instant

= 45°.

Prob. 16-44

D

Prob. 16-48

16.4 ABSOWTE MOTION ANALYSIS335

16-49. Bar AR rotates uniforrnly about the fixed pin A with a euristant angular velocity w. Dełermine the velocity and acceleration of błock C. at the instarit H = 60°.

16-51. The bar is confined tv rrove alung the yertica] and inchned plaues.lf thc velocity of thc miler at A is u), =i, 6 ft/s when 9 = 45°, determine the bars angular velocity and the velocity of miler Bat this instant.

Prob. 16-49

Prob. 16-51

*16-52. Arm AB has an angular velocity of w and an angular acceleration of rx. If no slipping occurs between the disk and the fixed curved surface, determine the angular velocity and angular acceleration of the disk.

16-54. The błock moves to the left with a constans velocity vc. Determine the angular velocity and angular acceleration of the bar as a funetion of B.

yJ

Prob. 16-50 Prob. 16-52

336 CHAPTER 16 PLANAR KIKEMATICS Of A RIGID BODY

16-53_ Tf the wedge moves to the lett with a constant velocity v. &tern-line the angular velocity of the rod as a function of N.

16-54. The slotted yoke is pinned at A while end B is used to rnove the ram R horizontally. H the disk rotates with a constant angular velocity w, determine the velocity and acceleration of the ram.Thc crank pin Cis fixed to the disk and tunis with it.

16-55. The Genewa wheel A prQvides i nitrrnittent stary motion wA for continuous motki!, w = 2 radis of disk D. 13y chonsing d = 1[H5V mm, the wheei has zero angular velocity at the instant pin B .enters or leaves one of the four slots. Determine the magnitude of the angular velecity wA of the Genewa wheeł at any angle d for which pin B is in cantacl with the slot.

d— 100'4'2 mm

Prob. 16-55

16-56. M the instant shown, the disk is rotating with an angular velocity trf w and has an angular acceleration of a. Detcrminc the velocity and acceleration of cylinder Bat this insłant. Neglect the size of the pulley at C.

Prob. 16-53

Prob. 16-54 Prob. 16-56

3 3 716.5 RELATIVE-MOTiON ANALYSIS: VELGCITY

l 6.5 Relatiye-Motion Analysis: Velocity

Translating reference

Fixed reference

(a}

Fig. 16-11

The general piane motion of a rigid body can be described as a eombinalion of transiation and rotation. To view these "component" motions separately we wili use a reladve-motionanalysis invoIving two sets of coorctinate axes. The x, y coordinate system is fixed and measures the absołu.re position of twa points A and B on the body, herc represented as a bar, Fig. 16--11a. The origin of tiw x', y coordinałc system wili be attached to the selected "base point" A. Which generally has a knowa motion. The axes of this coorclinate system 1nm:5-Jare with respect to the fixed frame but do not rotate with the bar.

Position. The position vector rA in Fig. 16-11a specifies the location of the "base point" A. and the relative-position vector rsiA iocates point B with respect to point A, By vector addition, the position of B is then

= rA rBm

Displacement. During an instant of film dt. points A and B undergo displacernents drA and drB as shown in Fig. 16-11b. 1f we consider the general piane motion by its component parts then the entire bar first fransiałes by an amount drA so that A. the basc point, moves to its finał posinan and point B moves to B', Fig, 16-11c, The bar is then rower] about A by an amount dH so that h" undergoes a rekaive displacement dram and thus moves to its fina] position B. Due to the rotation about A, dram = r"4 dO, and the displacement of Bis

dra = drA + dram

due to rotation about A due Co transiation of A

due to translation and rotation

drR fS

Translation Rotation

tcJ

Time r Time Ę + dr

Generał piane motinn

(.13)

1 6


As slider błock A =yes hori2ontally to the left with a velocity it causes crank CB Lo rotate counterclockwise, such that v is directed tangent to its circular path, i.e., upward to the left. The connecting rod AB is subjected ta generał piane =Ilon, and at the instant shown it has an angular velocity w.

Vei0City. To deterrnine the relałion between the velocities of points A and B. it is nccessary takc thc timc dcrivativc of thc position equation, or simply divide the displacement equation by dr. This yields

dr 8 d r A d rB iA

dr dr dr

The terms drsjdt yR and drAirit-= vA are measured with respect to the fixcd x, y axes and represent the absolute velireiries of points A and B. respectively. Since the relative efisplacement is caused by a rotation, the magnitude of the third term is drRłA /dr = rRIA (N i& = rBiAÓ = r81Aw, whcre w is the angular velocity of thc body at the instant considered. We will denote Ibis term as the reiative velocity va/A, since it represents the velocity of B with respect to A as measured by ari observer fixed to the translating x', y' axes. In other words, rhe bar appears to rnove as if it were rotafing with ara angular vetority w abour the axis passing through A. Consequently, vBiA has a magnitude of vB/A = wr4.1 and a direction which is perpendicular to rB,A. We therefore have

Vn = VA VB iA

(16-15)

where

= velocity of point BvA = vclocity of the hase point A vBiA

= velocity of B with respect to A

116.5 RELATIVE-MOTtON ANALYS1S: VELOCITY 339

P a t h o f p o i n t A

Translation

( c ) ( g )

VA

Path of point B

Generał piane motion

(d)

Rołation abont the base point A

(t)

Fig. 16-11 (ront.)

What this equation states is that the velocity of B, Fig. 16--11d, is determined by considering the entire bar to translate with a velocity of v/t , Fig. 16-11e, and rotate about A with an angular velocity w, Fig. 16-11f Vector addition oF thcse two Mccl& apphed to B, yields vB, as shown Fig. 16-11g.

since the relative velocity vB,,A represenłs the effect of circular motion, about A, this term can be expressed by the cross product vftłA = w X r131,1, Eq. 16-9. Hence, for application using Cartesian vector analysis, we can also write Eq. 16-15 as

v11= + c a rum (16-16)

where

vB velocity of B't."A = velocity of the base point A w = angular velocity of the body

rom = position vectur directed from A to B

The velocity equation 16-15 or 16-16 may be used in a practical manricr to study the gencral piane motion of a rigid body which is c,i ther pin connected to or in contact with other ~ing hodies. When applying this equation, points A and B should generally be selected as points an the bady which arc pin-connected to othcr bodies,or as points in contact with adjaccn t bodics which have a known motion. For example, point A on link AB in Fig. 16-12a must rnove along a horizontal path, whereas point B moves on a circular path. The direetions of vA and vB can thercfore be cstablished since they are always tangent to their paths of motion. Fig. 16-12b. In the case of the wheel in Fig. 16-13, which rolIs wahoui slipping, point A on the wheel can be selected at the ground. Here A (momentarily) has zero velocity since the ground does not rnove, Furthermorc. the center ol the wheel, f3, vrioves along a horizon tal path so that vB is horizon

V fi

V A

(b)

Fig. 16-12

Fig. 16-13


Procedure far Analysis ■

The relative velocity equation can be applied either by using Cartesian weto' analysis,or by writing the x and y scalar component equations directly. For application, it is suggested that the following procedure be used.Vec-tor Analysis

Kinematic Diagram.

• Establish the directions of the fixed x, y coordinates and draw a kinematic diagram of the body. Indicate on it the velocities vA

ol points A and B. the angolar velocity w, and the relative-posłtion vector rB/A.

•If the magnitudes of vA, vu, or w are unknown, the sense of dircetion of these vectors can be assumed.

Velocity Equation.•To app]y VB = FA (.1.1 X r,R/A., express the vectors in Cartesian

vector form and substitute them into the equation. Evaluate the cross product and then equate the respective i and j components to obtain two scalar equations.

• If the sol ution yields a negative answer for an ufflknown magnitude. it indicates the sense of direction of the vector is opposite to that shown on the kinematic diagram.

ScaJar Analysis

Kl nernatic Diagram.

• lf the velocity equation is to be applied in scalar form, then the magnitude and direction of the relative velocity vii/A must be cstablishcd. Draw a kincmatic diagram such as shown in Fig.16-11g, which shows the relative motion. Since the body is considered to be -pinned" rnomentarily at the base point A, the magnitude of v,9/A is un/A = torem. The sense of dircction of VajA is always perpendicular to 1-B/A in accordance with the rotational motion w of the body.*

Velocity Equation.

• Write Eq, 16-15 in syrnbohc form, vB = vA vB/A, andunderneath each of the rerms represent the vectors graphically by showing their magnitudes and directions.The scalar equations arc detcrrnined from the x and y components of thcsc vectors.

'1"be notation W8 = VA + V urny be helpful in reealling that A is " pinned:"

16.5 RELATIVE-MpT[ON ANALYSiS: VELOCiTY 341

The link shown in Fig. 16-14a is guided by two blocks at A and B. which move in the fixed siots. If the velocity of A is 2 m/s downward, dctcrminc the velocity of B at the instant O = 45°.

SOLUTION I (VECTOR ANALYSIS)

Kinematic Diagram. Since points A and I3 are restricted to move along the fixed slots and vA is directed downward, then velocity vB must be directed horizontaliy to the right. Fig.16-14b,This motion causcs the 1Mk io rotate counterclockwise: that is. by the right-band tule the angular velocity w is directed outward. perpendicular to the piane of motion.

velocity Equation. Expressing each of the vectors in Fig. 16-14b in terms of their i, j. k components and applying Eq. 16-16 to A, the base point, and B, we have

vB -= 4- w X rgiA

Vgi = -2j + [wk X (0.2 sin 45°i - 0.2 cos 45j)] vBi = -

2j + 0.2w si n 45°j -F 0.2w cos 451

Equating the i and j components gives

vB = 0.2w cos 45° O = -2 + 0.2w sin 45°

Thus.= 14.1 rad/s) vs = 2 mis —> Ars_

SOLUTION II (SCALAR ANALYSIS)

The kinematic diagram of the relative "circular motion" which produces vB7

, is shown in Fig. 16-14c. Hece vBir, = w(0.2 m).Thus,

V8 = VA VBJA

v8 = O + w(0.2) cos 45° O

= -2 + w(0.2) sin 45°

The solution produces the above results.It should be emphasized that these results are valid only at the

instant H = 45°. A reca1culation for O = 44° yie1ds v8 = 2.07 m/s and w = 14.4 rad/s; whereas when H = 46°, v8 = 1-93 m/s and w =13.9 rad/s, etc.

NOTE: Since vA and w are known. the velocity of any other point on the Iink can be determined.A.s an exercise, see if you can apply Eq,16-

16 to points A and C or to points B and C and show that when O -=45°, = 3.16 m/s, directed at an anglc of 18.4' up from the horizontal.

e„ = 2 misA

4 °

( b )

Re lativc ~Liant e )

Fig. 16-14

342 CHAPTER 16 PLANAR KIKEMATICS O f A R IGID BODY

EXAMPLE _ 16.7

(aJ

The cylinder shown in Fig. 16–I 5a roik without sIipping on the surface of a conveyor beli which is nioving at 2 ft/s. Determine the velocity of point A.The cylinder has a cIockwisc angular velocity w = 15 rad/s at the instant shown.


Kinematic Diagram. Since no slipping occurs, point B on the cylinder has the same velocity as the conveyor, Fig. 16-15b. Also, the augulm-

velocity of the cylinder is known,so we can apply the velocity equałion to B. the base point, and A to determine vA.

Velocity Equation

so that

VA/B = wrAff3= (15 rad/s) 0.5 ft ) = 10.6 ft/s

45°Thus,

Rcła nye motion(c)

Fig. 16-15

VA = vB vA/B

f(Vxl ł ENA),.1ft/S1 + [10.6 flis]

45°

vA v8 + W X rAi8

(vA),i+ (vA)yi = 2i + (-15k) X (-0.5i + 0.5j)

(vA).,) (vA),j = 21 + 7.50j + 7.50i

= 2 + 7.50 = 9.50 ft/s (N)

, = 7.50 ft/s

vA =V(9.54)2 + (7.50)2 = 12.1 ftis

7.50= tanH —9.50 = 38.3°

Equaling the x and y components gives the same results as before, namely,

-±0 (vA), = 2 + 10.6 cos 45° = 9.50 ft/s

(+t) (vA)y. = 0 + 10.6 sin 45' = 7.50 ft/s

= 2 flis

SOLUTION II (SCALAR ANALYS1S)As ali allernative proccdure. the scalar componcnis of vA + vAiscan be obtained directly. From the kinernatic diagram showing the relative "circular" motion which produces vA18, Fig. I6-15c, we have

(b)

Tbus,

16.5 RELATIVE-MOTtON ANALYSlS: VELOCITY343

t—+-›)

(-1)'BR = O + rik8( 0.2 .\.7 cos 45°)

O = —2 + aics(0.2-`vi sin 45')

ve = 2 m/s

WCEI

= 2 m /s

The Miar C in Fig. 16-16a is =ing downward wich a velocity of 2 m/s. Determine the angular velocity of CB at this instant.


Kinematic Diagram, The downward motion of C causes B to move to the right along a c-urvcd path. Also, CB and AB rotate cnun terclock wise.

Ve/ocity Equation, U n k C8 (general piane motion): See Fig.16-16b.

vB = vc toci3 X ra/c

tai = —2j + wcfik X (0.2i — 0.2j)

ust = —2j + 0.2cocEj + 0.2wcyj

V,r; = 0.2COcii

O = —2 + 02wcs

wc] , = 1.0 rad/s )

v s = 2 mis

SOLUTION II (SCALAR ANALYSIS)The scalar component cci uations of vs = vc vs,..c. can be obtained directly. The kinematic diagram in Fig. 16-16e shows the relative "circułar" motion which produces VB/c. We have

v B = v c v B ( c

vs [2 m/s i Etors(.2 \72 m)]

Resolving these vectors in the x and y directions yields

which is the same as Eqs.1 and 2.

NOTE: Since link AB rotates about a fixed axis and vs is known,Fig. 1.6-16d, its angular velocity is found from = w/kBr As. car2 mis = .wAs (0.2 m), wAs = 10 rad/s.

(1)

(2)

Arar Ve = 2m s Z rkIC

m o i ] - 7 m -

(a)

v+t

(b)

Bkelative motion

( c )

344 CHAPTER 16 PLANAR KINEMATICS OF A RIGID BODY

EXAMPLE

The bar AB of the linkage shown in Fig. 16-17a has a clockwise angular velocity of 30 rad/s when O = 60°. Determine the angular velocitics of member BC and the wheel at this instant.

u.lSOLUTION (VECTOR ANALYSIS)

Kinematic Diagram. By inspection, the velocities of points B and C are defined by the rotation of link AB and the wheel about their fixed axes.The position vectors and the angular vclocity ol each member arc shown on the kinematic diagram in Fig.16-17b.To solve, we will write the appropriate kinematic equation for each member.

(a) Velocity Equation. Link AS (rotation about a fixed axis):

= to" XrB

= (-30k) x (0.2 cos 601 + 0.2 sin 60°j) =

{5.20i — 3.0j} m/s

vsr0.2m 0.2 m

(o.. H = 30 rad/s

60'A

(b}

Fig. M-17

vc Link BC (general piane motion):

Vc = VB WBC X rciB

vci = 5.20i — 3.0j + (conck) X (0.2i)

vci = 5.201 + (0,2%r — 3.0)j

vc = 5.20 m/s

O = 0.2w8c — 3.0

(oBt. = 15 rad/s'5 .4/1s.

Ans.

Wheel (rotation about a fixed axis):

Ve = 6JD X rc

5.20i = (wDk) X (-0.1j)5.20 = 0.1% ra a

= 52.0 rarl is

34516.5 R ELATIVE-MOT[ON ANALYSiS: VELOCiTY

— 3 mis

FUNDAMENTAL FiROBLEMS

F16-7. If miler A rnoves to the right with a constant velocity of trA = 3 m/s, determine the angular velocity of the link and the velocity ot miler 8 at the instant O =.

0,1 3 m fs

F16-7

F16-8. The wheel roik without slipping with an angular velocity of w = lU rad/s. Determine the magnitude d the velocity of point B at the instant shown.

F16-11

F16-9. Determine the angular velocity of the spool. The capie wraps around the inner core, and the spool does not slip on the platform P.

F16-12

F16-10. If crank OA rotates with an angular velocity of w = 12 rad/s, determine the velocity of piston B and the angular veiocity of rod AB at the instant shown.

F16- LOF16-11. If rod AB slides along the horizon tal slot with a velocity of 60 ft/s, determine the angular velocity of link BC at the instant shown.

F16-11

F16-12. End A of the link has a velocity of u4 = 3 m/s. Determine the veiocity of the peg al B al 1.1-lis Instant:rue peg is constrained to move along the słot.

346CHAPTER 16 PLANAR KINEMATIC5 OF A RIGID BODY

PROBLEMS

16-57. If h and O are k nowi. and the speed of A and B is vp = i3. determine the angular veioeity ed of thc body and the direetion tfi of v8.

16-59. The velocity of the slider Wek C is 4 ftis up the inelined grotwe.Determine the angular vełocity of Iinks AB and BC and the veloeiły of point B at łhe instant shown.

Pp

Prob. 16-57

If thc błock at C is rnoving downward at 4 ft/s, deterrnine thc angular velocity d har AB ał the instant shawa.,

I ft

= 4 Ft/s

Prob. 16-59

16-60. Thc epicyciie gear train eonsists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the conne,eting Iink DE pinned to B and Cisrotating at toDE1/3 rad/s about the pin at E, determinethe angular veleeities of the planet and sun gears.

Lic = 4 ft/s

Prob. 16-58 Proh. 16-60

16.5 R ELATIVE- MerT[ON ANALYSiS: VELOCiTY347

16-61, Determine the velocity of the biock at the instant 60°, if link AR is rotating at 4 radis.

Prob. 16-61

16-63, 1f the angolar velocity of link AR is wA = 3 radis, dcterminc the veloeity of the hlock at C and the angular velocity of the connecting link CR at thc Instant A = 45° and ep -

Prob. 16-63

16-62. Ef the flywheel is rotating with an angular velocity of ("JĄ = 6 rad/s, determine the angular velocity of rod 13C at the instant shown.

Prob. 16-62

*16-64. The winion gear A rolls on the fixed gear rack B with an angular velocity w = 4 radis. Determine thc velocity of the gear rack C.

16-65. The 'Anion gear roils on the gear racks. If B is moving to the right al 8 ft/s and C is moving in the iefł at4 ft/s. determine the angular velocity of thegear andthe velocity of its center A.

1 3

ProlN. 16-64/65


16-66. Determine the angular veloeity of the gear and the velocity of its center O at the instant shown.

16-67. Determine the velocily of point A on th.e rim cif the gear at the Instant shiawn.

16-69. If the gear rotates with an angular velocity of = li1 n1d/a and the gear rock moves at ve --- 5 m/s,

determine the velocity of the &lider błock A at the instant .shown.

Probs. 16-66/67

*16-68. Part of an automatic transmission consists of a fixed ring gear R, three equal planet gears P. the sun gear S, and the planet earrier C. which is shaded. If the sun gear is rotating at w, = 6 rad/s, determine the angular veloeity cac of the planet' earrier. Note that C is pin-connected to the center of each of the planet gears.

Prob. 1.6-69

16-70. If the slider block C is rnoving at vc = 3 m/s. determine the angular velocity of BC and the crank AB at the instant shywn.

Pruli. 16-68 Pryh. 16-70

16.5 RELATIVE-MQT4ON ANALYS1S: VELOCiTY349

16-71. The two-cylinder engine is designed sa that the pistons are conneeted to thc erankshaft BE using a master rod ABC and articulated rod AD. If the crankshaft is rołating at w = 3n radiN, cleterrnine the velocities of the pistons C and D at che Instant shown.

16-74. If crank A B rotates with a constant angular velocity of rt,4 -= 6 rad/s, deterrnine the angular veIodty of rod Be and che; velocity of thc slidcr błock at the instant shown. The rod is in a horizontal position.

Prob. 16-71

16-72. Determine thc velocity of the center O of the spool when the cable is pulled to the right with a velocity of v.The spod rolls without slipping.

16-73. Determine the veiocity of point A on the outer rim of the spool at the Instant shown when che rabie is pulled to the right with a veloeity of v.The spool roik without slipping.

16-75. If the slider block A is moving downvvard at vA = 4 m/s, deterrnine the velocity cif block B at the

instant shown.

16-76. If the slider błock A is moving downward at vA = 4 m/s, determine the velocity of block C at the instant shown.

Prob. 16-74

Probs. 16-72/73 Probs. 16-75/76

350 CHAPTER 16 PLANAR KIKEMATICS OF A RiGil) BODY

16-77. The pianetary gear system is used in an autornatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, wR = O. and the sun gear S is relating al

= 5 radfs. Determine the angular velocity ol each of the planet geai s P and shaft A.

16-79. The ditferenłial drum uperates in such a manner that the rope is unwound from the smali drutu B and wonnd up on the large Brum A. If the radli of the largo and smalt drums ara R and r, respe.ctively. and for the pulIey it is (R +r)/2, determine the speed at which the bucket C rises if the man rotates the handle with a constans ang-ular velocity of w. Neglect the thickness of the rope.

40 min

Prob. 16-77

15-78. the ring gear D rotates counterclockwise '91,rith anangular velocity of wp = 5 radis wbite link Al3 rotates eIockwise with an angular veloeity of wAs = 1 O rad/s, determine the angular velueity °f gear C.

Prob. 16-78

Prob. [6-79

16-80. Mechanical loy animals °hen i2e a walking mechanism as shuwn idealized in the figure, 1f the driving crank AB is propelled by a spring motor such that w„,,R = 5 radis, determine the velodty of the rear foot E at the instant shown. Although not part of this problem, the upper end of the foreleg has a skitted guide which is constrained by the fixed pin at G.

Prob. 16-4tft

16.6 INSTANTANECIUS CENTER OF ZERQ VELOCiTY 351

6.6 Instantaneous Center ()f Zero Wiochy

The velocity of any point B located on a rigid body can be obtained in a very direct way by choosing the base point A to be a point that has zero velocity at the, instant considered. In this case, vA = O. and therefore the velocity equalion. y8 = vA + w X r8/A, becomes v8 = r..‹, X r$/,4, For a bady having generał piane motion, point A so chosen is cal]ed the instanłaneous center oj zero velocity //C), and ił lies on the insłanłaneous axis of zero velocity. This axis is always perpendicular to the piane of motion, and the interseetion of the axis with this piane defines the ]ocation of the [C. Since point A coincidcs with the /C, then v8 = w x rfiiic and so point B movcs momentarily about the ir in a circular park in othe r words, the body appears to rotate ahout the instantaneous axis. The magnitude of v8 is simply v8 -= wr8fic, where w is the angular velocity of the body. Due to the circular motion, the direction of y8 must ałways be perpendicular to r8fic.

For example, the /C for the bicycle wheel in Fig. 16-18 is at the contact point wich the ground.There the spokes are somewhat visible,whereas at the top of the wheel thcy becomc blurred. If one imagincs that the whecl is mornentarlIy pinned al this point, the velocities of various points can be found using y wr. Here the radial distances shown in the photo. Fig. 16-48, must be determined from the geometry of the wheel.

Fig, 16-18

352 CHAPTER 16 PLANAR KINEMATIC5 OF A RIGID BODY

Cent rade

IC = O

Loemion of TC knowing vA and w

(a)

16Locatien of IC

knowing the direetioffl of and yx

A

( b )

Fig, 16-19

Location of the IC. To locate the IC we can use the fact that the velocity of a point on the body is always perpendicular to the relative-position vector directed from the IC to the point. Several possibiliłies exist:

■ I C

• The vełoeity v A of a pobił A on the body and the angular velocity au of the body are known, Fig. 16-19a. In ibis case, the /C is located along the line drawn perpendicular to vA at A, such that the distance from A to the IC is rArEC = vAltki. Notę that the /C iics up and to the right ot A sirwe vA must eause a clocicwise angular velocily rl about the IC.

• The litwy ofaction of rwo nonparallet velocities vA and vB are known,rAgc 16-19b. Construct at points A and B line segtnents that are

perpendicular to vA and vB, Extending these perpendiculars to

rFffic their point of interseergan as shown locates the IC at the instant

\ • fconsidercd.

• The magnitude and direction of rwo para/lei velocin-es v A and vR are known. Here the loeation uf the [C is cletermined by proportionał triangles.Examplcs are shown in Fig. 16-19e and d. In botki casesrAhre = vAicó and = vB/cd. If dis a known distance betweenpoints A and B, then in Fig. 16-19e, t-A/ic ł rBirc = d and in Fig. 16-19d. Nic, -- rAire -= d.

Location uf /C knowing vA and v',

( d )

16.6 INSTANTANEOLJ5 CENTER OF ZERQ VELOCITY353

--9~1M111 As the board slides down ward to the left it is subjected to general piane motion. Since the directions of the velocities of its ends A and B arc known. the 1C is located as shown. Ai this instant the board will momentarily ruta te about this point. Draw the board in several olher pusitions and etablish the IC fur each case.

Realize that the point chosen as the instantaneous center of zero velocity for the body can ony be used at the instant considered sincc the body changes its position from one instant to the next.The lotus of points which define the location of the IC during the body's motion is called a ceturode,Fig.16-19a, and so each point on the centrode acts as the W for the body only for an Instant.

Although the, IC may be conveniently used to determine the velocity of any point in a body, it generally does noł have zero acceleratton and therefore it shoutd not be used for finding the accelerations of points in a body.


The velocity of a point on a body which is subjected to general piane motion can be determined with reference to its instantaneous center of zero vcłocity provided the Iocation of thc /C is first established usin g one of the lhree methods described above.•As Shiffirri on the kinematic diagram in Fig. 16-20, the body is imag-

ined as "extended and pinned" at the IC so that, at the instant considered, it rotates about this pin with its angular velocity w.

• The magniłude of velocity for each of the arbitrary points A, B, and C on the body can be determined by using the equation wr, where r is the radia! distance from the IC to each point.

■The line of action of each velocity vcctor v is perpendicrdar to its associated radial line r, and the velocity hasa SCFge of direction which tends to move the point in a manner consistent with the angular rotation w of the radial line, Fig. 16-20. Fig. 16-20


EXAMPLE

16.10

Show how to determine the location of the instantaneous center of zero vclocity for (a) member BC shown in Fig. 16-21a; and (b) the ]ink CB shown in Fig.16-21 c,

( b )

SOLUT' ONPart (a). M shown in Fig. 16-21a. point 8 moves in a circular path such chat v2 is perpendicular to AB. Therefore, it acts at an angle from the horizontal m shown in Fig. 16-21 b. The motion of point B causcs the piston to move forward horizautahry with a vclocity ve. When lines are drawn perpendicular lo va and vc, Fig. 16-21 b, they intersect at the /C.

Part (b). Poims B and C follow circular paths of motion since Iinks AB and DC are each subjected to rotation about a fixed axis. Fig. 1.6-21c. Sincc the velocity is always tangent to thc path, at thc instant consiclered, cm rod OC and v,s, on rod AB arc both directed vertically downwarcl, along the axis of łink CB, Fig. 16-.21d, Radial lines drawn perpendicular to these two velocilies form parallel lineswhich intcrsect at "infinityi" ?kik—) and r•Bifc—>x. Thus,mes = (re/rc.pe)---> O. As a result, link CB mornentarily transiales. An instant iater, however,CB will rnove to a tilted position, causing the /C to move to same finite location.( c l

t d i

Fig, 16-21

16.6 INSTANTANE01JS CENTER OF ZER() VELOCiTY355

An.%.v p 3 m i s

= — 5.30 rad/sfiWc 0.5657 m

= 2 .R

i t r u

= = 5.30 rad/s (0.4 m) = 2.12 mis '45°

From Fig. 16-22c., the angular veloci ty of AB is

1)11 2.12 m/s(L)A8 = 5.30 rad/s?

rspi 0.4 m

(c)

Fig. 16-22,

NOTE: Try and solvc this problem by applying vi) = vD/R tomember nn,

E X A M P L E

Błock D shown in Fig. 16-22a moves with a speed of 3 mis. Determine the angular velocities of links BD and AR, at the instant shown.

SOL UTIONAs D moves to the right, it causes AB to rotate clockwise about point A. Hence, is dirccted perpendicul ar to A B. The instantancous center of zero veIocity for BD is located at the intersection of the line segmenis drawn perpendicular to vii and vi), Fig. 16-22b. From the geometry.

rEiic = 0.4 tan 45° m = 0.4 m

0.4 mrD/re cos 45° = 0.5657 m

Since the magnitude of vp is known. the angular velocity of link BD is

The velocity of Bis therefore

(b )


EXAMPLE 16.12

The cylinder shown in Fig. 16-23a rolls without slipping between the two moving plates E and D. Determine the angolar velocity of the cylinder and the vcIoeity of its center C.

1= 0.25 m/s

A = 0.25 ra/s

0.25 m

(b)

Fig. 16-23

SOLUTI ONSince no slipping occurs, the contact points A and B on the cylinder have the same velocities as thc plates E and D. respectively. Furthermore, the velocities vA and vB arc paraltel, S13 that by thc proportionality of right lriangIes the /Cis located at a point on line AB, Fig. 16-23b.Assuming this point to he a distanee x frorn B, we have

V8 = 0.4 m/s = .cox

V A = w(0.25 rn — x); 0.25 m/s = «0.25 In —

Dividing one equation juto thc other eliminates w and yields

0.4{0.25 — = 0.25x

0.1x = — 0.1538 m 0.65

Hance, the angułar veloeiły of the cylinder is

vB 0.4 m/s(1.> = x = 0.1538 m — 2.60 rad/s

The veloci ty of point Cis therefore

ve = wrryir. = 2.60 radis (4.1538 m — 0.125 m) =

0.0750 mis

16.6 INSFANTANEOUS CENTER OF ZERO VELOCiTY357

EXAMPLE

1 6 . 1 3

The crankshaft AB turns with a ciockwise angular velocity of 10 rad /s, Fig. 16-24a. Determine the velocity of the piston at the instant shown.

tal

SOLUT1ONThe crankshaft rotates about a fixed axis, and so the velocity of point B is

vB = 10 rad/s (0.25 ft) = 2.50 ft/s .45°

Since the directions of the velocities of B and C arc known, thcn the location of the /C for the connecting rod BC is at the intersection of the lines extended from these points, perpendicular to vB and vc, Fig.16-24b.The magnitudes of rplic and rcjic can be obtained from the geometry of the triangle and the law of sincs, i.e..

0.75 ft NIC

sin 45° - sin 76.4°r-Birc = 1.031 ft

0.75 ft rc_rcsi n 45° - sin 58.6°

rciic = 09056 ft

The rotational sense of a■Bc must be the same as the rotation causedby v8 about the K, which is counterclockwise. Therefore,

vB 2.5 ft/s4 ) 8 c = — = — - 2 . 4 2 5 r a d / s

rBitc 1.031 ftI r5rog this result, the velocity of the piston is

vc = cóBerrirc = (2.425 rad/s)(0.9056 ft) = 2.20 ft/s l 11%.

(b)

Fig. 16-24

vc.

C

0,75 ft

2.50 irls

tC45.0'

58_6'

358CHAPTER 16 PLANA12 KINEMATICS Of A RIG16 BODY


F16-11 Deterrnine the angular veloeity of the rod and the veloclty of point C at the instant shown.

F16-13

F16-16. If cable AR is unwound with a_ speed of 3 mis, and the gear rack C bas d speed of 1.5 m/s. dełerrnine the angular velocity of the gear and the velocity of its center O.

F16-17. Deterrnine the angular velocity ❑f link RC and the veloeity tlf the piston C at the instant shown.

F16-16

F16-14. Deterrnine the angular velocity of link RC and velocity «the piston C at the instant shown.

— 0 . 6 m - 1.2 mF16-17

F16-14

F16-15. If the center O of the wh lis moving with aspeed of = 6 m/s. determine the velocity of point A onthe wheel.The gear rack B is fixed.

0.6 m Le rEk

A — ' 1 . 4 5 6 m i s

-knrerk.nr"

F16-15

F16-18. Determine the angular velocity of links BC and CD at the instant shown.

F16-18

16.6 INS.TANTANE01,115 CENTER OF ZER() VELOCiTY 359

■ PROBLEMS

16-81. In eaeh ease show graphically how to loeate the16-83. At the instant shown. the disk is Totating Atinstantaneous center of zero vclocity of Iink AR, Assumera = 4 rad/s. Deterrnine the veloeities of points A, B, and C.the geometry is

Prob. 16-81

16-82. Determine the angular velocity of link AB at the instant shown if biock C is moving upward at 12 in/s.

= 4 rad/s

0\_1-;5 r n 7' B

F.

Prob. 16-83

*16-84. If link Cl) bas an angular velocity of tira) = 6 rad/s. determine the velocity of point B on Iink BC and the angular vełocity of Iink AB at the instant shown.

16-85. If link CD bas an angular velocity of web = 6 rad/s. determine the velocity of point E on link BC and the angular velocity of Iink AR at the instant

Prob, 16-82 Probs. 16-84/85

360CHAPTER 16 PLANAR KINEMATICS OF A RIGID BODY

16-86. At the Instant shown, the truck travels to the right*16-88. If link AR is rotating at wAa = 6 rad/s. dc-tem-lineat 3 m/s, while the pipc rolls counterclockwise atthe angular velocities of links BC and CD at the instantw --- 6 rad/s without slipping at B. Deterrnine the velocity6of the pipe's center G.

25D mmwig = 6 rad is

Prob. 16-86

C31}0 mm

3 m400 mm

Prob. 16-88

16-87. If crank AB is rotating with an angular velocity of wA❑ = 6 rad/s, dełerrnine the velociły of the center O of the gear at the Instant shown.

16-89. The nil pumping unit consists of a walking beam AB. connecting rod BC, and crank CD. If the crank rotales at a constant rate of 6 radis.deterrnine the speed of the rod hanger H at the instant shown. Unit. Point R kill"s circular path about point E and therefore the velocity of is not vertical.

Prob. 16-87 Prob. 16-89

16.6 INSTANTANEWS CENTER OF ZER() VELOCiTY361

16-90. Due to slipping, points A and H on the rim of the disk have the velocities shown. Determine the velocities of the center point C and point D at this instant.

16-91. Due to slipping, points A and B on the rirn of the disk have the velocities shown. Determine the velocities of the center point C and point E at this instant.

z+B y 10 l'I/N B

U.8 h>3°'

F°A = 5 ft/ Prolp,. 16-90/91

16-92. Knowing thal the angular velceity of ]ink AB is c.d.AB = 4 radjs, determine the velocity of ihe oollar at C and the angular velocity of Iink CB al the instant shown. Link CH is horizontai at fbi5 instant.

16-93. Ef the coilar at Cis moving downward to the lcft at = 8 in /s, determine the angular velocity of Iink AB at the instant siwym.

A

Probs. 16-92/93

16-94. Tf the roller is given a velocity of UA = 6 flis to the right, detcrmine the angular velocity of the rod and the velocity of C at the instant shown.

Prob. 16-94

16-95. As the car travels forwat d at 80 ftis on a wet ruad. due to slipping, the rear wheels have an angular velocity =

100 raelis. Determine the speeda of points A, !Land C eaused by the rnotion.

Prob. 16-95

16-96. Determine the angular velocity of the douhle-tooth gear and the velocity of point C on the gear.

AU , L - 4 m i s

0_3muf̀ 0 I:7 m

= F M

Prob. 16-96


16-97. The wheel is rigidly attached to gear A. which is in mesh with gear racks D and E. If D hasa velocity of

= ó ft/s to the right and the wheel rolls on track C without slipping, determine the velocity of gear rack E.

16-98. The wheel is rigidly attached to gear A, which is in rnesh with gear racks D and E. If the racks have a velocity

of = 6 flis and (4. = 11) ft/s, show that it i s necessary forthe wheel to slip on the fixed traek C. Also find the angolar velocity of the gear and the velocity of its center O.

Probs. 16-97/98

16-99. The epicyclic gear train is driven by the rotatinglink DE, which has an angular velocity = 5 rad/s. 11 thering gear is fixed, determine the angular vełocitics ofgears A. B, and C.

Prob. 16-99

*16-100. The simiiar links AR and CD rotate about the fixed tri n5 at A and C. If AR has an angular velocity osĄs =

S rad/s, determine the angolar velocity of BDP and the velocity of point P.

Prob. 16-11Hi

16-101. If rod AB is relating with an angular velocity co,tp -= 3 rad/s, deterrnine the angular włochy of rod BC at the instant shown.

16-101 If rod A B is relating with an angular włochy wca = 3 rad/s, deterrnme the angolar veloci ly of rod CD at

the instant

Probs. 16-101/102

16.7 RELATiVE-MOTION ANALYS1$: ACCELERATIgN 363

16.7 Reiative-Motion Analysis: Acceleration

An equation that relates the acceierations of two points on a bar (rigid body) subjected to generał planc motion may be determined by differentiating vjg = vA + vBlA with respect to tirne.This yields

dvi? dvA (IV BMd r -

- dr dr

The terms dynjdt = an and dvAidt = are measured with respect to a set of fixed x, y axel and represent the absoime aceelerations of points B and A , The last term represen bi the acceleration of R with respect to A as measured by an observer fixed to łranslating x', y' axel which have their origi n at the base point A. In Sec. 16.5 it was shown that to this observer point B appears to move along a cirmiar arc that hal a radius of curvature rB/A. Consequently, aBiA can be expressed in terms of itstangenłial and normal components; i.e., aB/A = (aRiA), (a B/A)„, where(aB/A), = arBiA and (aBi,l)„ = w2rsiA Hence, the relative-acceleration equation can be wrilten in the form

ag aA (aRiA)! (aB/A)n (16-17)

whe re

acceleration of point B

acceleration of point A

tangential acceleration component of B with respect to A, The rnagnitude is (crBfA}, = arBIA, and thedir e e t i vn i s perpenc l i cu lar Esm .

norrnal acceleration component of B with respect to A , The Magnitude is (a814), = w2r13.1,1, and the direciian is always prom B towards A.

The terms in Eq. 16-17 are represented graphically -in Fig.16-25. Here it is scen that at a given instant the acceleration of B, Fig. 16-25a, is determined by consiciering the bar to translate with an acceleration aA, Fig. 16-25b, and simultaneously rotate about the base point A with an instantaneous angular vclocity w and angular acceleration a, Fig. 16-25c, Vector addition of these two effects, applied to B. yields aB, as shown in Fig. 16-25d. should be noted from Fig. 16-25a that since points A and B move along curved paths, the acceIerations of thcsc points will have bod; tarrgentFaI and norma! eQmponents. (Recall that the acceleration of a point is =geni to dir pad] ordy when the path is recioliriear ar when it is an inticction point on a eurve.)

P a t h o f p o i n t A \

a ż i

Ratation about thc base point A

(c)

(aram)..

a A

(d)

Fig. 16-25

n a =21A (

aB/A)/

(as/A),, =

Path ofGenterai piane morion point 13

( a )

a,

364 CHARTER 16 PLANAR KINEMATICS OF A RIWD. BODY

(b)Fig. 16-26

(ą)

A(b)

Fig. 16-27

Since the relative-acceleration components represent the eftect of circular nrotion observed from translating axes having their origin at the base point A, these terms can be expressed as (sum), = a X rs,A and (ap/A),, = -ar=ram, Eq. 16-14. Hcnce, Eq. 16-17 becomcs

ab, = aA + a X 1.13m - riiirs/A 1(16-18)

w h c r c

aR = acecleration of point B

aA = acceleration of the base point A

a = angular acceleration of the body

w = angolar veiocity cif the body

rB/A = position vectur directed from A to B

f f Eq. 16-V7 or 16-18 is applied in a practical manner lo study the accelerated motion of a rigid body which is pin conneded to twa other bodies, it should be realized that points which are conicidenr ar the phi movc with the sanie acceleration, sinee the palli of motion over which they travel is the same. For example, point B lying on either rod BA or BC of the crank mechanism shown in Fig. 16-26a bas the same acceleration, since the rods are pin connected at B. Herc the motion of B is along a circular path, so that aB can be expressed in terms of its tangential and norma! components. At the other end of rod BC point C moves along a sirafght-lined path, which is defined by the piston. Flence, ac is horizontaI, Fig. 16-26b.

Finally.consider a disk that roik without slipping as shown in Fig.16-27a. As a result, vA = O and so from the kinematic diagram in Fig. 16--27b, the velocity of the mass center C is

vG = v A + W 5 ( rGiA = O + (-wk) x (rj)

So that

vG = cor (16-19)

This same result can also be determined using the IC method where point A is the ./C.

Since G moves along a straighr lirze. its acceleration in this case can be determined from the time derivative of its velocity.

dvG Lir")- r-

ar; = ar (16-20)

These twa important results were also obtained in Example 16-4.Th ey apply as well to any circular object, such as a ball, gear, wheel, etc, thatrolis slipping.

d r

16.7 RELATIVE-MOTION ANALYsis: ACCELERATION365

Procedure for Analysi_

The relativc acceleration equation can bc applied between any two points A and E on a body either by UN' ng a Cartesian vector analysis. or by writing the x and y scalar component equations directly,

Velocity Analysis.• Determine the angular velocity w of the body by using a velocity

analysis as discussed in Sec. 16.5 or 16.6. Also, determine the vclocities yA and vg of points A and B if these points ~e along curved paths.

Weto r Analysis Kinematic Diagram.▪Establish the directions of the fixed x, y coordinates and draw the

kinematic diagram of the body.Indicatc on it aA , ag, w, Lx. and rg/A.

• If points A and B movc along CZ4 rve d paths, then their acceIcrations should be indicated in terms of [heir tangential and normal components, i.e., aA = (aA), + (a„,),, and ag = (a8), +

Acceleration Equation.

• To apply aB = aA + a X rym — w2rBIA, express the vectors in Cartesian vector form and substitute thern into the equation. Evaluate the cross product and then equate the respective i and j components to obtain twa scalar equations.

• If the soIntion yields a negative answer for an unknown magnitucle, it indicatcs that thc sense of dircction of the vector is opposite to that shown on the kinematic diagram.

Scala r Analysis Kinematic Diagram.

• 1f the acceleration equation is applied in scalar form, then the rnagnitudes and directions of the relative-acceleration components (aBiA), and (agiA)„ must be established. To do this draw a kinematic diagram such as shown in Fig. 16-25c, Since the body is considered to be momentatily -pinned" at the base point A, the maguitudes of these components ara (aRłA), = cugla and (aBb4)„ = w2ropt. Thcir sense of direction is cstablishcd from the diagram such that (aB/A), mas perpendicular to ram, in aceordanee with the rotational motion a of the body, and (agiA), is directed from B towards A,*

Accele ration Equation.

• L'Qepresent the vectors in ag = a4 + (aam), + (ag/A)„ graphically by showing łheir rnagnitudes and directions underneath each term. The scalar equations ara determined from the x and y components of these vectors.

*The notation afi = ay -+- (a",,„,,,), -Fmay be hełpful in reealling that Ais assurned to be pinned.

The mechanisrn for a window is shown. Here CA

rotates about a fixed axis through C, and A8 undergoes general piane motion. Since point A mavs dong a cuNed path it bas two components of acceł eration, where as point B moves a[ong a straight track and the dIrealon of its accelerafion is specified.

3 6 6 CHARTER 16 PLANAR KINEMATICS OF A RIWD. BODY

EXAMPLE 16.14

10 ni

( a )

Y

r y , A . . 4 5 °

45° w = 0.283 ritc3/

= 3 Tn/$2

( h )

The rod AB shown in Fig. 16-28a is confined to move along the inclined p]anes at A and B. If point A has an aceeleration M 3 mis2

and a velocity of 2 m/s. both directcd down the piane at the instant the rod is horizon tai. determine the angolar accekration uf the rod at this instant.

SOLUTION I (VECTOR ANALYSIS)We will apply the acceIeration equałion to points A and B on the rod. To do so it is first necessary to determine the angolar velocity M the rod. Shaw that it is w = 0.283 rad/s j using cither the velocity equalian or the method of instantaneous eenters.

Kinematic Diagram. Sincc points A and B both morfo aIong straight-line paths, they have no components of acceleration norma] to the paths, There arc two unknowns in Fig. 16-28b, namely. (43 and cx.


= aA + er X rB/A — Wiry/A

as cos 45°i + rr8 sin 45°j = 3 cos 459 — 3 sin 451 + (ak) x (10i) — (0.283)2(10i)

Carrying out the cross product and equating the i and j components yields

cep cos 45° = 3 cos 45° — (0.283)2(10) (1)

45° —3 sin 45' + a(10) (2)

Solving, we have

as = 1.87 ra/s2.45°

a = 0.344 radis2 Ans.

SOLUTION II (SCALAR ANALYSIS)From the kinematic diagram, showing the relative-acceleration

(a8/4)( = r8+.71 components (3.8./A), and (aB,A),, Fig. 1&-28e, we have

as = aA OSIA + (21B/A).re

°R =[3n1/1 + „, [Ce(10 + {(0.283 rad/s).:(10

. 2 4 5

(c) Equat ng the x and y components yields Eqs. 1 and 2, and the solutionFig. 16-28 procceds as before.

ma

rirlAlt B

--- 0.283 rad/s

'16.7 RELATIVE-MOTION ANALYSIS: ACCELERATION 367

EXAMPLE

16.15

The disk rolls without slipping and has the angular motion shown in — 6 rad/s= 4 rad/s'-

441111111111G 05

ft

Fig. 16-29a. Determine the accelcration of point A at this instant.


Kinematic Diagram. Since no slipping oceurs, app]ying Eq. 16-20,

w 6 rad/s a =4 rad/s;(74

ftls' c;

( b )

= 6 ractisił, a = 4 rad/si2

= = (4 rad/s2)(0.5 ft) = 2 ftis2


Wc will apply the acceleration equation to points G and A, Fig. 16-291).

aA = aG a x rAiG — w2rA/G

aA = —2i -F (4k) x (-0.5j) — (6)2(-0.5j)

= i 18.i) ft/s2

SOLUTION II (SCALAR ANALYSIS)

Using the result for a G = 2 ft/s2 determined above, and from the kinematic diagram, showing the relative motion aA/c, Fig. 16-29c, we have

aA = ac + (flAfG)Y

<—

(a A),—2 + Z—O

[(a Al + [(aA),, _ [2 ft! s21 "[ (4 rad s)(0.5 ft)] ± [(6

rad/s)2(0.5 ft)]A(a.1.....))..-

(d) Fig. 16-29 +

1 (C1A)y = 18 ft/s2

Therefore,

aA = NI(0)2 + (18 ft/s2)2 = 18 ftis2 Am.

NOTE: The fact that a A = 18 ft/s2 indicates that thc instantaneous center of zero velocity, point A, is not a point of zero acceleration.

368 C H A P T E R 1 6 P L A N A R K I N E M A T I C S O f A R I W D . B O D Y

EXAMPLE

16.16

( a )

,T;

(n wed lw2rsx

ar .= 2 ft,I2

vra3 rad iis = 4 rarlis

( b )

= tt.7i lit

i -= 3 rad/g -= 4 radfs2

Fig. 16-30

(c)

The spool shown in Fig. 16-30a unravels from the cord, such that at the instant shown it kas an angolar velocity of 3 radis and an angular acceleration of 4 rad/s2. Detcrmine the acceleration of point B.

SOLUTION I (VECTOR ANALYSIS)The spool "appears" to be rolling downward without slip ping at point A. Therefore, we can use the results of Eq. 16-20 to determine the acceleration of point G, i.e.,

aG = ar = (4 rad/s)(0.5 ft) = 2 ft/s2

We will apply the acceleration equation to points G and B.

Kinematic Diagram. Point B moves along a eurved puch having an unknown radius of curvaturc,* Its acceleration will be represenlcd by its unknown x and y components as shown in Fig. I16-30b.

Acceleration Equati

an = aG + tx x rjg/G - i.o2r8JG

(<2B),i (08),j = + (-4k) X (0,75j) - (3)2(0.75j)

Equating the i and j tcrms. the componcnt cquations arc

(aB)„ = 4(0.75) = 3 ft/s2 (1)

(aB)„. = -2 - 6.75 = -8.75 ft/s" = 8.75 ft is" (2)

The magnitude and direction of aa are therefore

ua = V(3)2 + (8.75)2 = 9.25 ft/s2

8.75= t a t r 3 = 7 1 l ° )

SOLUTION II (SCALAR ANALYSIS)This problem may be solved by writing the scalar component equations direcłły. The kinematic diagram in Fig. 16-30c shows the relative-acceleration components (aR/G), and (aNG)„.Thus,

= aG + {asie), (asfa).

[2 ft/s21 E4 rad/s2 (0.75 ft)] [(3 rad/s)2(0.75 ft)]

The x and y components yield Eqs. 1 and 2 above.

,'Realize that the path-S radius of CUTtiatura p is not equal to the radius of the spool sutce the spoci] is not rotating about potni G. Furtherrnore, p i5 not defined ;:is the tlIshinoe from A (1C) to B, siwe the location of the /C deponds anty on the veioeity of a point and not the geometry of its path.

16.7 RELATIVE-MOTION ANALYSIS: ACCELERATION 369

EXAMPLE

The collar C M Fig. 16-31a moves downward wich an acceleration of I rafs2. At the instant shown, it hasa speed of 2 nt/s which gives links CB and AB an angular vclocity wzy = ffir t i = 10 radis. (See Example l6.8.) Determine the angular accelerations of CB and AB at this instant.

uC=lmfs2 = 2 i n Is

( a )

SOLLITION (VECTOR ANALYSIS)

Kinematic Diagram. The kinematic diagrams of both 'Mks AB and CB are shown in Fig. 16-31b, To solne, we will apply the appropriate kinematic equation to each Iink.

Accel arat ion Eq uationac

Link AB (rotation about a fixed axis):

ag = aAB X rB - 01,3r / j

aB (aABk) x (-0.2j) - ( I 0)2(-0.2j)

aB = 020ŁABi + 20j

Not e hiat aB ha5 n and t companentssince it moves along a circular path.

Link BC (gencral plant motion); Using the resuit for aB and applying Eq. 16-18, we have

a ik = ac + acB X ry je - #.6rBic

0.2am i + 20j = - l j + (cr c Bk) X (0.21 - 0.2j) - (10)2(0.2i - 0.

2j) 0.2am i + 20j = l j 0.2a c B j 0.2a c B i - 20i ł 20j

Th1.1.9,0.2aAB = 0.2acB - 20

20 = - I + 0.2<xcli + 20

Solving,5 radis2

orAi, = -95 racifs= = 95 rad/s2

wc!? "— 10 rad/s

(h)

Fig. 16-31

wAE 10 radfsaAR

fj 0.2 m

f s

370 CHAPTER 16 PLANAR KINEMATICS OF A RIGID BODY

EXAMPLE

16.18

0.75 ft

rufte = /43 rad/s

(Dm,= 10 rad/s O t AB = 20 rad/s2 A

(a)

The crankshaft AB turns with a clockwise angolar acceleration of 20 rad/s2, Fig. 16-32a. Determine the acceleration of the piston at the instant AB is in the position shown. At this instant wAB = 10 rad/s and (i)BC = 2.43 rad/s. (Sce Example 16.13.)

SQLUT1ON (VECTOR ANALYS1S)

Kinematic Diagram. The kinematic diagrams for both AB and BC arc shown in Fig. 16-32b. Herc ac is vertica1 since C moves along a straight-line path.

Acceleration Equation. Expressing each of the position vectors in Cartesian vector form

1.8 = {-0.25 sin 451 + 0.25 cos 45°j} ft = { -0.177i + 0.177j } ft

rus = {0.75 sin 13.61 + 0.75 cos } ft = {0.1771 + 0.729j } ft

Crankshaft AB (rotation about a fixed axis):

aB = a,(B X rB - <0;7(Bra= (-20k) X (-0.1771 4- 0.177j) - (10)2(-0.177i + 0.177j) = {

21.211 - 14.14j} ft/s2

Connecting Rod BC (general piane makiem): Using the result for ap and noting that ac is in the vertical direction, we have

ac = aB aBc X re ls (4cre lBacj = 21.21i - 14.I4j + (a8 k)x (0.1771 + 0.729j) - (2.43)2(0.1771 + 0.729j)

acj = 21.21i - 14.14j + 0.177aBcj - 0.729aBci - 1.041 - 4,30j O =

20.17 0.729a8c

ac = O. I 77aBc - 18.45

Solving yiel dsaBE = 27.7 rad/s2

ac = -13.5 ft/s2 Ans.

NOTE: Since the piston is moving upward, the negałive sign for acindicatcs that the piston is decclerałing, i.e.. ac = { -13.5j} ft/s2. This causes the speed of the piston to decrease until AR becornes verfical, at which time the piston is momentarily at rest.

r( k

05 cos 45 'N a m t 20 rad/s '

wAk = 10 rad/s2 lr.

(13)

Fig. 16-32

0.75 cos 13.6° ft

37116.7 RELATIVE-MOTION ANALYSIS: ACCELERATION

4 in

FUNDAMENTAL PROBLEMSF16-19. At the instant shown, end A of the rod lias the velocity and acceleration shown. Determine the angular acceleration of the rod and acceleration of end B of the rod.

F[6-19

F16-20. The gear rotls on the fixed rack with an angular velocity of a, = 12 rad/s and angular acceleration of a = 6 rad/S2. Determine the acceleration of point A.

6 radis2

w 12 rad/s

F1.6-20

F16-21. The gem- rolls on the fixed rack B. At the instant shown. the center O of the gear moves with a velocity of vo = 6 m/s and acceleration of ao = 3 m/s2. Determine the angular accelerałion of the gear and acceleration of point A at this instant,

1116-22. At the instant shown, cabie AR has a velocity of 3 m/s and acceleration of E.5 m/s.., while the gear rack lias a velocity of I.5 m/s and ac,ceieration of 0.75 ni/s. Determine the angular acceieratkin of the gear at this instant.

F16-22

F16-23. At the instant shown, the wheel rotates wth an angular velocity of w = 12 rad/ s and an angular acceleration of a = 6 radis2. Determine the angular acceleration of link BC and the acceleration of piston C at this instant,

1.2 ni

6 rad/s"!

= 12 rad/s

F16-23

F16-24. At the instant shown. wheel A rotates with an angular velocity of w = 6 radfs and an angular accełeration of cx = 3 rad/52, Determine the angular acceleration of link BC and the acceIcration of piston C.

F16-24

ac=0.75mfs2 vc — 1.5 mis

372CHAPTER 16 PLANAR KIKEMATICS OF A RiGil) BODY

PROBLEMS

16-103. At a given instant the top end A of the bar has the veloeity and acceIeration shown, Detemtine the acet Ierałion of the bottom B and the bars angular acceleration at this instant,

16-106. Crank Af3 is rotating with an angular veIocity of c,J,5 = 5 radis. and an angular acceleration of npR = 6 radf:sz, Determne the angular acceleration of Be and the

acceleration of the slider blask C at the Instant s hown.

= 5 kis aA = 7 ft/:.;

Prob. 16-103Prob. 16-106

16-104. At a given Instant the bonom A of the ladder has an acceleration aA = 4 ft/s2 and velocity t?, = 6 ft/s, both acting to ihe lefl. DeterM1Fle the acceleration of ihe top of Ehe ladder, B, and the ladders angular aceeIeration at this same Instant.

16-105. At a given instant the top B of the ladder bas an acceleration ub --- 2 ft/s2 and a veloeity of yg = - 4 ft/s, both aeting down ward. De te FIlli ne the acceleration of the botiorn A of the ladder, and the ladders angular acceleration al Ibis instant.

16-107. At a given instant, the slider bloek A has the velocity and deceleration shown.Determine the acceleration of block B and the angular acceleration of the link at this instant.

.4— A = 1.5 mis aA = 16 m/s2

Probs. 1 —104/105

Prob. 16-107

16.7 RELATIVE-MOTION ANALYSIS: ACCELERATION 373

*16-1011. As the cord unravels from the cylinder, the 16-110. At a given instant the wheel is rotating with thecylinder has an angular acceleration of a =--- 4 rad/s2 and an angular motions shown. Determine the acceleration of theangular velocity of w = 2 rad/s at Ehe instant shown. collar at A at this instant.Determine the aceelerations of points A and Bat this instant.

Prob. 16-108

60̀

m=8 rad/s ce = 16 rittils2„0-

:N00 nim

rr

Prob. 16-110

16-109. The hydraulic cylinder is extending with a velocity of oc = 3 ft/s and an acceleration of ac = 1_5 ft/s2. Determine the angular acceleration of links RC and AR at the instant shown.

16-111. Crank AB rotates with the angular velocity and angular acceleration shown. Determine the acceleration of the slider block C at the instant shown.

Prob. 16-109 Prob. 16-111

374CHAPTER 16 PLANAR KINEMATICS Os A RIWD. BODY

*16-112. The wheel is moving to the nght such that it has an angular velocity w = 2 rad/s and angular

accelerałion = 4 rad/s2 at the instant shown. If it does not slip at A, determine the acceleration of point B.

16-115. A cord is wrapped around the inner spool of the gear.If it is pulled with a constant velt-scity determine the velocities and accelerałions of points A and B. Thc gear rołls on the fixed gear rack.

16-113. The disk is ~ing to the left such that it bas an angular acederation u = 8 rad/s2 and angutar veloc

sty = 3 rad/s at the instant shown, If it does not slip at A. determine the aeccieration of point B.

16-114. The disk is moving to the left such that it has an angular acceleration u = 8 rad/s` and angular veloci ty OJ = 3 rad/s at the instant shown. H it does not slip at A. deterrnine the accełeration of point D.

Probs. 16-113/114

Prob. 16-115

6-116.. At a Oven instant, the gear racks have the velocities and accelerations shown. Dełermine the aceeleration of point A.

16-117, At a Owen insłan t, the gear raeks have the veloo kies and accelerations shown. Determine the aceeleration of point B.

Probs. 16-116/117

Prob. 16-112

= 4 racl/s2 w 2nd/s

6.7 RELATIVE-MOTION ANALYSIS: ACCELERATIQN375

16-118. At a given instant gears A and B have the angular motion shown. Determine the angular acceleration of gear C and the acceleration of its center point D at this instant. Notę that the inner hub of gear C is in rnesh wich gear A and its muter rim is in mesh with gear B.

Prob. 16-11.8

16-119. The whed rolls without slipping cuch that at the instant shown it has an angular velocity fal and angular acceleration ti. Determine the velocity and aceeleradon of point B on the rod at this instant.

*16-120. The center O of the gear and the gear rack P move with the veloeities and accelerations shown.Determine the angular acceleration of the gear and the acceleration of point B located at the rim of the gear at the instant shown.

16-121. The tied crank and gear niechanism gives rocking motion to crank AC, necessary for the operation of a prinfing press. If link DE has the angular motion shown, determine the respective angular velocities of gear and crank AC at this instant, and the angular acceleration of crank AC.

Prob. 16-121

16-122. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the angular acceleration of puiley B at Ihe instant shown.

16-123. Pulley A rotates with the angular velocity and angular acceleration shown. Detmninc thc acceleration of hloek E at the instant shown.

Probs. 16-122/123

Prob. 16-119

= 2 r51/5 ap = 3 m is2

Prob. 16-120


*16-124. At a given instant, the gear has the angular motion shown. Determine the acceleratiens cif points A and B on the link and the link's angular acceleration at this Instant.

16-126, At a given instanł, the cables supporting the pipo have the mołions shown. Determine the angular velocity and angular aceeIeration of the pipc and the velocity and acceleration of point B lecated on the pipe.

Prob. 16-124

= 6 ft/s

— 2 ft/s2

A

v=5 ftls

= 1.5 fi/s2

Prob. 16-126

16-125. The ends of the bar A8 arC egnfind ta move along the paths shown. At a given instant, A has a velocityof 4 ftis and an acceleration of L I A = 7 ft/s2.Determine the angular velocity and angular acceleration of A B at this Instant.

16-127. The s/kier błock moves with a velocity of vs = 5 ftis and an acceleration of aB = 3 ftis2. Determine the angular acceleration of rod A.B at the instant shown.

*16-128. The slider błock moves with a velocity of oh = 5 ftis and an acceleration cf. ak = 3 ft/s7. Determine the acceleration of A at the instant shown.

Prob. 16-125

t)s = 5 ft/su8— 3 ft/s'

Probs. 16-127/128

16,8 RELATiVE-MOTION ANALYSIS US1NG ROTATNG AXES 377

l 6.8 Relative-Motion Analysis using Rotating Axes

In the previous sections the reladve-motion analysis for veIocity and acceleration was described using a translating coordinate system. This type of analysis is useful for dctcrmining thc motion of points on the same rigid body, or the motion of points iocated on several pin-corinceted bodies. In some problerns, however. rigid bodies (mechanisms) are construeted such that sliding will oceur at their connections. The kinernatic analysis for such cases is best perforrned il' the rnotion is analyzed using a coordinate system which both Iranslaies and nnates. Furthermore, this frame of reference is useful for analyzing the motions of two points on a meehanism which arc not locatcd in the same body and for specilying the kinematies of particie motion when the particie moves along a rotating path.

In the following analysis two equations will be developecl which relate thc velocity and acceieradon of two points, one of which is the origin of a moving frame of reference subjeeted to both a translation and a rotation in the piane.*

Position. Consider the two points A and B shown in Fig. 16-33a, Their location is specified by the position vectors rA and rB, which are measured with respect to the fixed X, Y, Z coordinate system. As shown in thc figurc, the "base point" A represents thc origin of the x, y, z coordinate system, which is assumed to be both transiating and totating with respect to the X, Y, Z system.The position of B with respect to A is specified by the relative-position vector rB,,,A. The components of this vector may be expressed eithcr in terms of unit vectors along the X, Y axes, i.e.. I and J, or by unit vectors along the x, y axes, i.e., i and j. For the developrnent which follows, rB/A will be measured with respect tothe moving x, y frame of reference. Thus. if B bas coordinates yB),Fig, 16-33a, then

r a f y = x g i Y a i

Using vector addition, the thrcc position vectors in Fig. 16--33a arc related by the equation

(a)

Fig. 16-33

r8 = rA rB/A (16-21)

At the instant considered, point A has a velocity vA and an acceleration a,k, while the angular velocity and angular acceleration of the x, y axes are fl (omega) and fl = dS1 Jdt, respeetively.

*The nicre gcneraL, Jhrec-dimcnsionał motion or the pointa is dc,.e[eped in See. 20.4.

3 7 8 C I - E A P T E R 9 6 P L A N A R K I N I E M A T I C S O F A R I G H : J . B O D Y

Velocity. The veloeity of point B is determined by taking the time clerivative of Eq. 16-21, which yields

V = A +drkm dt

(16-22)

The last term in this equation is evałuated as follows:

d r 3 / A

d ld

(-1,11 Yai)

di .i + + + yB7

)( 43d) a . di + _di

= —d t i + —dt 'I + I D 7 1 Y 8 ch

(16-23)

Y

y ■

J

Fin, 16-33 <amij

The two terms in the first set of parentheses represent the components of velocity of point B as measured by an observer attached to the moving x, y, z coordinate System.These terrns wili be denoted by vector (v8łA)_„,, In the second set of parentheses the instantaneous time rate of change of the unit vectors i and j is measured by an observer located in the fixed X, Y, Z coordinałe system. These changes, di and dj, are duc ordy to the roładorr d6 of the x, y, z axel, causing i to become i' = i + di and j to become j' = j + dj, Fig. 16-33b. As shown, the rrragrritudes of both di and dj cqual 1 dO, since i = i' = j = j' = 1. The directiorz of di is defined by +j, since de is tangent to the path described by the arrowhead of i in the limit as i1t dr, Likewise, dj acts in the -i direcłion, Fig. 16-33b. Hence,

d i d O d j _d dz = = 111 = 71(—i) =

t

Vicwing the axes in three dirnensions. Fig. 16-33c. and noting that = ak, we can express the above dcrivatives in tcrms of the cross product as

= II x j (16-24)

Substituting thcsc resuits into Eq. 16-23 and using the distributive properly of the vector cross product, we obtain

— (vBiA),y, Si x (xBi .vBj) (vBiA),),śl x ryp% (16-25)dr

di = Iż x itijdi

drBłA

16,8 RELATNE-MOTIQN ANALYSIS US1NG ROTATNG AXES 379

Hence, Eq. 16-22 becomes

Vg - NA + 11 X r,v, + (vB/A),(3, (16-26)

where

v,s = velocity of B, measured from the X, Y, Z reference

v, = velocity of the origin A of the x, y, z reference, measured from the X, Y, Z reference

velocity of "B with respect to A,' as measured hy anobserver attached to the rotating x, y, z reference

angular velocity of the x, y, z reference, rneasured from the X, Y, Z reference

Nt,= position of B with respect to A

Comparing Eq. 16-26 with Eq. 16-16 (v8 = vA + 11 X NA), which is valid for a translating frame of reference, it can be scen that the only di iference betweerl these Iwoequations is represeri led by the term (

BtAWhen applying Eq. 16-26 it is often useful to understand what each of the

terms represents. In order of appearance, tł:wy arc m follows:

{obsoInte velocity luf B

(equals )

} mołion of B observedfrom the Y, Y, Z framc

NA { absolute velocity of the origin of x, y, z

frame

II X rBtA

(VB(A)xyz

(p lus )

{ angular velocity effect caused by

retation of x, y, z (rome (

płus)

(velocity of B witki respect to A

mot lon of x, y, z

franie observed from

the X, V. Z frame

}rnotion of B observedfrom the x, y, z frame

380 CHAPTER 16 PLANA12 KINEMATICS Of A RIG16 BODY

Acceleration. The aceeleration of B, observed from the X, Y, Z coordinate system, may be expressed irl terms of its motion measured with respect to the rotating system of coordinates by taking the time derivative of Eq. 16-26.

dv3 dvA „Ił drB/A d(vivA)x.,7_fiżA= x r ± X

dłdrdrdtdtdrBIAd{v 8/A),„

aB = aA 1.1 X rs/A + — + dt (16-27)

Here = i5 the angular accelerałion of the x, y, z coordinatesystem. Since n is always perpendicular to the piane of rnotion, then measures only the change in niagnande ni R. The derivative drB/A f dt is defined by Eq. 16-25. so thatn x

drB/dtA - n x (yR„) + n x (n x r,,,A) (16-28)

Finding the time derivative of (vB/A),, (vB/A),1 (vB/A),j,

diNg/A 11.x3 z _ r Ci(VR/A )7f i + at(15RIA )3' di dj]ji -i- f(vBiA) -- -V (vivA),—

dr dr dt '' dr drThe twa terms in the first set of brackets represent the components of acceleration of point B as measured by an observer attached to the rotating coordinate system.These terms win be denoted by (aBIA),0.,. The termy in the seeond set of brackets can be simplified using Eqs. 16-24.

— (aB/4),„y, + n x (vRiA),„dt

Substituting this and Eq, 16-28 into Eq. 16-27 and rearranging terms,

aB aA + 1"). r" + x (n x 4- 20 x + (aBiA),,

(16-29)

w h e r e

ap = acceleration of B, measured from the X, Y, Z reference

aA = acceleration of the origin A of the x, y, z reference, measured from the X, Y, Z reference

(14B/A )_ry-.: (V1VA)A3.z accelerałion and velocity cif B with respect to A, asmeasured by an observer attached to the rotating x, y, z reference

- angolar acee!eration and angular velocity of the x, y, z reference, measured from the X, Y. Z reference

rB,TA = position of B with respeet to A

d(vB/A),,,„

16.8 RELATIVE-MOTIQN ANALY51S US1NG ROTATNG AXES 3B 1

If Eq. 16-29 is coinpared with Eq. 16-18, written in the form aB = aA -4- 1"'"/ X rByA f .1"/ X (n X rsiA), which is valid for a translating frame of reference, it can be scen that the difference between these two cquations is represented by the terms 2.11 x (vB/A),„ and (alim),,. In particular, 251 x (vivA).„, is called the Cariedis acceleration.numud atter the Freneh engineer G. C, Coriolis, who was the first to determine This term represents the difference in the acceleration of B as rneasured from nonrotating and rołał1ng x. y, z axes. As indicatcd by the vcctor cross produci, the Coriolls acceleration will afways be perpeildicular tu buth and (vB/A)x,. Ił is an important component of the acceleration whieh must be considcred whenever rotating reference frames arc used. This often occurs, for example, when studying the accelerations and forces which act on rockets, Iong-range projectiles, or other bodies having motions whose measurements are significantly affected by the rotation of the earth.

The following interpretation of the terms in Eq. 16-29 may be utseful when applying this equation to the solution of problems.

{ absolute acceleration of B

(equals)

{ absolute acceleration of the

origin of x, y, Z frame (

plus)

{ angular acceleration

effeet eaused by

rotation of x, y, z frame (plus)

morion ()f B observed}

from the X, Y, Z frame

motion of

x, y, z franieobserved from the X, Y, Z frame

} t -tleraeting mOtiOn

{ angliku' velocity effect caused by

rotation of x, y, z frtime (

plus)

{ combined effect af-B maving

rclatiwc to x, y, z

coordiHates and rotation of

x, y, z frame (plus)

{acceleration of B with morion of B observed

respect to A from the x, y, z frame

aR

aA

h x rBtA

3 8 2 C H A P T E R 16 P L A N A R Kl K E M A T I C S O f A R I G I D B O D Y


Equations 16-26 and 16-29 can be applicd to the solution of problems involving the planar motiort of partieles or rigid bodies using the folIowing procedure,

Coardinate Axes.

• Choose an appropriate location for the origin and proper orientation of the axes for both fixed X, Y, Z and moving x, y, z reference frames.

• Most often solutions are easily obtained if at the instant considered:

1, the origins arc coincident2. the corresponding axes are coi1inear3. the corresponding axes are parallel

• The moving fratrie should be selccted fixed to thc body or device aiong witch the relative motion occurs.

Kinennatic Equations.

• Aftcr defining the origin A of the moving reference and spccifying the moving point B, Eqs. 16-26 and 16-29 should be written in symbolic form

Vg = vA + S>r x rBłAae = aA + i). x rB/A + D, x (5) x rBiA) 211 X (vB/A),,,, (as/A),,

• The Cartesian components of alt these vectors may be expressed along either the X, Y, Z axes or the x, y, z axes. The choice is arbitrary provided a consistcm set of unit vectors is uscd,

• Motion of the moving reference is expressed by vA, aA , H. and £1; and rnotion of B with respect to the moving reference is expressed by rgiA (Vg/A)xy_-_-, and (as/Aj,,,.

The rotation of tle dumping bin of the truck about point Cis operated by the extension of the hydr aulic cylinder A8. To determine the rotation of che bili due to this extension, we can use che equations of relative motion and fix thc x, v axes to the cylinder so that the relativ e morion of the cylinder's exiension occnrs alnng the y axis.

16.8 RELATNE-MOTION ANALYSIS USING ROTATNG AXES 383

At the instant B = 60°, the rod M Fig. 16-34 has an angular velocity of 3 rad/s and an angular acceleration of 2 rad/s2. At this same instant, collar C travels outward along the rod such that when x -= 0.2 m the velocity is 2 m/s and the acceleration is 3 m/s2, both measured relative to the rod. Determine the Coriolis acceleration and the velocity and acceleration of the collar at this instant.

SOLUTIO N

Coordinate Axes. The origin of both coordinate systems is locatcd at point O, Fig. 16-34. Since motion of the collar is reported relative to the rod, the moving x, Y, Z frame of reference is anached to the rod.

Kinematic Equations,

vc = 111 X rcio (Vc/0)s, (1)

ac = 110 + 11. X rcyo + fi X (Li X rcio) 2f/ X (vc/o),„+(2)

It will be simpler to express the data in terms of i, j, k component vectors rather than I. J. K components. Hence,

Fig. 16-34

Motion af Motion of C with respectrnoving reference to moving reference

vo = O rcio = {0.2i} m

ao = O (vc/0),,„ = {21} m/s

= {-3k} rad/s (acio)),yz = {31} m/s2

{ -2k } rad/s2

The Coriolis acceleration is defined as

acan = 211 X (vcio),„ = 2(-3k) x (2i) = {-12j} m/s"

This vector is shown dashed in Fig. 16-34. If desired, it may be resolved into I, J components acting along the X and Y axes, respectively.

The velocity and acceleration of the collar are determined by substituting the data into Eqs. 1 and 2 and evaluatinR the cross products, which yieldsvc = V 0 ± 11 X reja + (V cio) .,y,

= O + (-3k) x (0.2i) + 21= {21 - 0.6j} m/s Ans.

ac = ao + IŻ X rejo + tł X (fl X rejo) + 211 X (vc/0),,,, + (acio),o,

= O + (-2k) x (0.2i) + (-3k) x [(-3k) x (0.21)1 + 2(-3k) x (2i) +

31 = O - 0.4j - 1.80i - 12j + 3i

{ 1.201 - t 9J di } m/s' Ans.

384 C H A P T E R 1 6 P L A N A R K I N E M A T I C S O F A R I G I D B O D Y

EXAMPLE

16.20

Rod _4B, shown in Fig. 16-35, rotates clockwise such that it bas an angular velocity wAĘ = 3 rad/s and angular acceleration aAB = 4 rad/s2

when O = 45°. Determine the angular motion of rod DE at this instant.B

0.4 m The collar at Cis pin connected to A B and slides over rod Dv

• A' ' SOLUTION

?ODE, Coordinate Axes. The origin of both the fixed and moving frameswłis = 3 radis of reference is located at D, Fig.16-35.Furrhermore, the x, y, z referenceami 4 racl/s2 is attached to and rotates with rod DE so that the relativc motion of

O = 45° the collar is easy to follow.

Kinematic Equations.

vc = VD n rCYD (VCID)xyz (1)Fig. 16-35 ac = ap + x rcip + n X (11 x !kin) + 2.1-a X (v,D)„ + (acip)xyz

(2)Al] vectors will be expressed in terms of i, j, k components.

Motion of Motion of C with respectmoving reference to moving reference

VD = O r c i p = ( 0 , 4 1 )

SD = O (vcip),,yz == —wDEk (ac,D)xy, = taco,„,zi

SŻ=

Motion of C: Since the collar moves along a circular pach of radius AC, its velocity and acceleration can be determined using Eqs. 16-9 and 16-14.

vc = cei,u3 X rem = (0(-2 3k) X (0.41 + 0.4j) (1.21 — I .2j } m/s

ac — "A13 X relli AOCIA

= ( -4k) X (0.41 + 0.4j) — (3)2(0.41 + 0.4j) = —21 — 5.2j} m/s'

Substituting the data into Eqs. 1 and 2, we have vc =

vi) + 11 X rcip (vCID)uz

1.21 — 1.2j = 0 + (—coDEk) x (0.4i) +1.21 — 1.2j = 4 — 0.4wDEj +(vc/D),,,2 = 1.2 m/s

w 3r a / s )DE = Ans.ac = a

_ D + ił X [kip +x (1-1 X rc,D) + 211 X (vc/D),„, + (ac/p)xyz—2i — 5.2j = O + (—erDEk) x (0.41) + (-3k) x [(-3k) x (0.41)] + 2(-3k) X (1.21) + (aciD)„,i

—21 — 5.2j -= —0.4a,DE j — 3.61 — 7.2j + (ac/D),„) (ac/D)xvz = 16 mis2

aDE = —5 rad/s2 = 5 rad/s'") Ans.

0 . 4 m

16.8 RELATIVE-MOTIQN ANALYS1S US1NG ROTATING AXES 385

EXAMPLE

Planes A and B fly at the same elevation and have the motions shown in Fig. 16-36. Deterrnine the velocity and aceeleration of A as measured by the pilot of B.

SOLUTION

Courdinate Axes. Sb-we the relatlye motion of A with respeet to the pilot in B i5 being sought, the x, y, z axel are attached to piane B, Fig. 16-36. At the :trwam considered, the origin B coincides with the origin of the fixed X, Y, Z frame.

Kinematic Equations. v

A – v s + n X rA/B

a4 – as + ii x rA,B śl k (fi X rAm)211. X (vAis).„ (11A}B),.Z( 2 )

P 400 krn – 0.25 rad/h2

fl = {0.25k} rad/h2 Fig. 16-36Motion of A with Respect to Moving Reference: r

k s = –4i } km (v A i s )„ = ? (a A i s )„ , = '?

Substitułing the data into Eqs.1 and 2, realizing that vA = { 700j } km/h and a, = {50j} km/h', we have

v , — v , n , X r A / 8700j =- 600j + (-1.5k) X (-4i) + (vAis

},y, = {94j} km/h

an = a B + rAfB + n (£1rnJ8) -F 211 X (vA„s)xyz (ak/)xyz

50j = (9001 – 10(d) + (0.25k) x (-41)(-13k) X [(-1.5k) x (-4f)] + 2(-1.5k) x (94j)

= i-119 Li + 15 Ij krn/h2

NOTE The solution of this problem should be compared with that of Example 12.26, where it is scen that (?)51A),y, g (74/8),„ and (aB/A).,:v2

(1 )y , Y

Motion of Moving Reference: vs

= {600j} krn/h

v2 8(600)2

( a B ) „ — p = 4 0 0 = 900 km/h 2 a s = ( a

s ) „ + ( a s ) , = { 9 0 0 1 – 100 j} km/h i

z5B 600 km/hfl – 1.5 rad/h )

p 400 km= (as), 100 km/h2

386 CHAPTER 16 PLANAR KIKEMATICS O f A R IGID BODY

PROBLEMS

16-129. Hall C rnoves along the stert from A to 13 with a speed of 3 ftis.which is increasing at 1.5 ft/s2,hoth measured relative to the circułar płatu. At this same instant the platc rotates with the angular velocity and angolar deceleration shown. Dełermine che velocity and acceleration of the hall at this instant.

Prob. 16-12916-130. [he cranes ielescopic boom ro tales with the angolar velocity and angular acceleration shown. At ihe same instant. the boom is etending with a constant speed of 0.5 ft /s, measured relative to che boom. Determine the magnitudes of the velocity and acceleration of point B at this instant.

16-13L While the swing bridge is ctosing with a constant rotation of 0.5 radjs, a man runs aiong the roadway at a constant speed of 5 ft/s relative to the roadway. Deterrnine his velocity and acceleration at the instant d = 15 ft.

*16-132. Wbite the swing bridge is closing with a constant rotation of 0.5 rad/s. a man runs along the roadway such that when d = 10 ft he is running outward from the center at 5 ftis with an acceleration of 2 his2, hoth measured relativt to the roadway. Determine his veiocity and acceleration at this instant.

Probs. 16-131(132

16-133. Collar C moves ałong rod BA with a velocity of 3 mis and an acceleration of 0.5 m/s2,both directed from B towards A and measured relative to the rod. At the same instant, rod AU rotałes with the angułar velocity and anguIar acceleration shown. Deterrnine the collars velocity and acceleration at ibis instant.

60 ft

Prob. 16-130

Ć.,›A0.02 rad/s= 0.01 rad As

0.5 m

Prob. 16-133

x w = 6 rad/sa = 1.5 radfs

3 m/s0.5 m/s=

16.8 RELATNE-MOTIQN ANALYSIS US1NG ROTATNG AXES387

16-134. Błock A, which is attached to a corel, rnoves along the slot of a horizontal forked rod.At the instant shown, the cord is pulled down through thc hole at O with an acceleration of 4 ro/s2 and its velocity is 2 m/s. Detcrmine the acceleration of the hlock at this instant.The rod rotates about O with a constant angolar velociływ = 4 rad/s.

16-137. At the iostant sbown, rod AR lias an angular velocity WAR -= 3 rad/s and an angular acceleration apg =

5 rad/s2. Determine the angolar velocity and angolar acceleration of rod CD at this instant.The collar at Cis pinconnecłed to CD and slides over AR.

Proh. 16-134

16-135. A girl stand al A on a platform which in rotating with a constant angular vełocilv w = 0.5 rad/s. If she walka at a constant speed of u = 0.75 m/s measured rełative to the platform. deterrnine her acceleration (a) when abc reaches point D in going along the palb A OC, d =

ł m; and (b) when she reaches point B if she foliowa the path ABC, r = 3m.

16-136. A girl stands at A on a platform which is rotating with an angular acceleration rx = 0.2 rad/s2 and at theinstant shown has an angular velocity ry0.5 rad/s. H abcwalka at a constant speed v = 0.75 m/s measured relatśve to the platform, de:termine her acceleration (a) when sile reaches point D in going along thc path A DC, d =

l re, and (b) when she reaches point B if she foliowa the path ABC, r= 3m.

....-.-Dii'...i......--

B

x

Pr€ Ak'4. 16-135/136

Prob. 16-137

16-138. Wiar R moves to the left with a speed of 5 nr/s, which is inereasing at a constant rato of I.5 m/s2.relative to the hoop, whiie the hoop rotates with the angular velocity and angular acceleration shown. Determśne the magnitudes of the velocity and acceleration of the collar at this instant.

w - 6 railis ar - 3 rad/s2

450 mm

Prob. 16-138

388 CHAPTER h b PLANAR KINEMATICS OF A RIGID BODY

16-139. Błock B of the mechanism is confined to move within the slot member CO. 1f AB is rotating at a constant race of Eva = 3 rad/s, determine the angular velocity and angular acceleration of member CD at the Instant shown.

16-141. The "quick-return" mechanism consists of a crank AD. slider block B, and slotted link CD. If the crank has thc angular motion shown, determinc the angular motion of the siotted link at this instant.

Prob. 16-139

*16-140. At the instant shown rod AD has an angular velocity wAti = 4 rad/s and an angular aceeleration crAs = 2 rad/s2. Determine the angular velocity and angular aceeleration of rod CD at this instant.The collar at Cis pin connected to CD and slides freely along AB.

D

Prob. 16-141

16-142. At the Instant shown, the robotic arm AD is rotating counterclockwise at w = 5 rad/s and has an angular acceleration cr = 2 rad/sz. Simultaneously, thc grip BC is rotating counterclockwise at w' =- 6 rad/s and u' = 2 rad/s2, both rneasured relatiue to a fixed reference. Determine the velocity and acceleration of the object held at the grip C.

wAB = 4 radfs = 2 radfs2

6tr

0/5 m

Prob. 16-140 Prob. 16-142

16.8 RELATIVE-MOTIQN ANALY51S US1NG ROTKPNG AXES389

16-143. Peg R on the gear slides freely along the slot in link B. Tf the gears center O moves with the velocity and aoceleration shown. dełermine the angular velocity and angular acceleration of the link at this Instant.

16-144. The cars on the arnnsement-park ridc; rotate around the axle at A with a constant angular velocity &JAJ =

2radfs, measured relative to the fratrie AB. At the same dme the frame rotates around the main axle support at B with a constant angular velocityc~1= I radjs.Determine the velocity and aceeleration of the passenger at C at the instant shown.

16-145. The cars on the amusement-park ridc rotate around the axle at A with a constant angular velocity ar4 = 2 radis, measured relałive to the frame AB. At the same time the frame rotates around the main axle support at B with a constant angular velocity wf = l radis.Deterrnine the velociły and acceleration of the passenger at D at the Instant shown,

16-146. If the slotłed arm AB rotates about the pin A with a constant angular velocity of WĄ5 = 10 rad/s. de fen/line thc angular velocity of link CD at the instant shown.

B

Prob. 16-146

16-147. At the instant shown, boat A travels with a speed of 15 mis. which is decreasing at 3 rn /s2, wbije boat B travels with a speed of 10m/s, which is increasing at 2 rn/s2.Determine the velocity and accelerationbeat A withrespcct to boat B at this instant.

416-148. At the instant s hown, boat A traw ls with a speed of 15 mis. which is decreasing at 3 rn f sz, whi le boat B łravels with a speed of lUm js, which is increasing at 2m is2. Determine the velocity and acceleration of boat B with respect to boat A at (bis instant.

Prob. 16-143

/7

—30 ni—

15 m/s

rn

,in /s3 ni /s-

2 mis2

Probs. 16-147/148

f - 5 0 m

x

Probs. 16-144/145


16-149, If the piston is rnoving with a vel celty of Up = 3 misand acceleration of = 1_5 misa, &tern-line the angularvełocity and angular acceleration of the slotted link at the Instant shown. Link AB slides freely along its slot on the fixed peg C.

16-150. The twolink mechanism serves to amplify angular inotion. Link AR hasa pin at B which is confined to move Within the slot of link CD.lf at the Instant shown.AB (input) has an angular velocity of = 2.5 rad/s.determine the angular veloeity of CD (autput) at ihis instant,

16-151, The gar has the angular moti o n s.h own, D ełermine the angular velocity and angular aceele.ration of the slotterl link BC at this instant.The peg at A is fixed to the gear.

= 2 radjs= 4 rad/s 2

Prob. 16-151

*16-152. The Geneva mcchanism is used in a packaging system to convert constant angular motion into intermittent angular motion, The star wheel A makes one sixth of revolution for each fuil revolution at the driving wheel B and the attached guide C. To do this, pin P, whieh is attached to 8, slides into one of the radia' slots of A. therehy turning wheel A, and then exits the slot. 1f B lias a constant angular veIocity or wa = 4 rad/s, determine o..¥5 and aA of wheel A at the instant shown.

vA = 3 ntis a., = 1.5 misi

• 30'

5 m

Prob. 16-1.49

mH = 4 rad/s

Prob, 16-152

,F)

45̂

— 2-5 rad /s i

ł

Prob. 16-150

16_8 RELATNE-MOTIQN ANALYSIS US1NG ROTATiNG AXES391

CONCEPTUAL PROBLEMS

P16—L An eIecti-ic motor turns the tire at A at a consłant angular velocity. and friction then causes the tire to roli without slipping on the inside rim of the Ferris wheel. Using appropriate numerical values, determine the magnitude ot the velocity and acceleration of passengers in one of the baskets. Do passengers in the other baskets exper•ienee this same motion? Explain.

P1.6-1P16-2. The crank AB tunis counierclockwise at a eonstant rate w eaushig the eannecffilg arm CD and roeking beam DE to move. Draw a sketeh showing the Ineation of the FC for the connecting arm when

= 0°, 90°, M', and 270°. Also, how was the eurvature of the head at E determined. and wliy is ił curved in this way?

P16-3. The bi-foid hangar door is opene.d by cablcs that move upward at a constant speed of 0.5 m/s. Determine the angular velocity of BC and the anguIar velocity of AR when 9 = 45°. Panel BC is pinned al C and has a height which is the same as the height of BA. Use appropriate numerical values to explain your result.

P 1 6 - 3

P16-4. If the TireS do not slip on the pavement. determ ine the points on the tire that have a maximum .ind minimum speed and the points that have a maximum and minimum aceeleration. lise appropriate numerical values for the car's speed and tire size to expIai n your result.

PH. 2 P16-4


CHAPTER REV1EW

Rigid-Body Pisuar Mafiom

A rigid body uldergoes three types of pkurar motion: translation, rotation abouł a fixed axis, and general piane motion.

Translation

When a body lias rectilinear translation_ ai the particie of the body travel along parallei s traight-iine paths. If the paths have the Same radius af curvature, then curv-ilinear trans lation aceurs..Frovided we know the motion of one of the particies, then the motion of ali of the others is algo known.

Rotation about a Fixed Axis

For this type ot motion, alt of the particles move along circuiar paths. Here, alt line segments in the body undergo the same angular dispiacement, angular wiodły, and angular acceleration.

fince the angular motion of the body is known, then the vełocity of any particie a distance r from the axis can be obtained,

The acceleration of any particie has two oornponents. The tangential componcnt aecounts for the change in the magnitude of the velocity, and the nermal component accounts for the change in the velociły's direction.

General Piane Motion

When a body undergoes generał piane motion, it simultaneously transiates and rotates. There are several methods for arialy7Ing this riwtion,

Absofure Modo?! AnalysisIf the motion of a point on a body or the angular motion of a line is known, lhen it may be possible to relate lhis motion to that of another point or line using an absolute motion analysis. To do sa, linear position coordinates s or angular position coordinates H are established (measured from a fixed point or line). These position coordinates ara then related using the geometry of the body_ The time derivative of this eguation gives the relationship hetwecn the velocitics andior the angular velocitics. A second time derivative relates the acc.elerattons andior the angular accelerations.

zf= wr

Rotation about a fixed axis= ćffijdr=aci

cx =orO = Ot, ł wp! +4,a

a d6 = w do)(92 = Lo2, 2a,(0 — Oo)

Constant o.

= ar. aF=nir

Path af recullnear translatton

Path of eury'linear translation ł

General piane niobem

CHARTER REVIEW393

Refative-Motion usingTranslaring AxesGeneral piane motion can also be analyzed using a reta_tive-motion analysis between two points A and 8 located on the body.This method considers the motion in parta: first a translation -of the selected base point A. ihen a relative "rotation" (if the body about point A, which is measure-d from a translating, axk Since the relative motion is viewed as circutar motion about the basu point, point B will have a velocity vo" that is tangent to the circle. It also has Iwo cornponents of acceleratio n „ (aR,,,,), and (up/A),,. It is also important to realize that a4 and ak will have tangential and norma) componeMs if these points move along curved paths.

= „JĄ + 03 X r8,..,4

= aitOI X rk,A L72 r gm

Instantaneaus Center of Zero VełociłyIf the base point A is selected as having zero velocity, then the relative ve/ocity equation becornes VB = w x rB,A . In this case, rnotion appears as if the body rotates abouł an instantaneous axis passing through A.

The instantanecus center of rołation (TC) can be establishcd provideel the directiona of the velocities of any two points on the body arc known, or łhe velocity of a point and the angular vełocity aru known. Since a radia line r will always be perpendicuIar to each velocity, then the [C is at the point of inłersection of these two radiał lines. Es measured location is determined from the geometry cif the body. Onuc it is estabłished, )hen the velocity of any point P on the body can be deterinined from v = wr, where r extends from the TC to point P.

Refarive Morion using Rottaing AxesProblem s that involve connected members that slide relative to one another or points not łocaled on the same body can be analyzed using a relative-motion analysis referenced from a rotating frame, This gives rise to the term 2fi x (vBiA) that is called the Corialis aceeleration,

VB = VAli X TEM(V5,44 )53,r.

+ dŻ x r8,5% ± n x (n 2£1. (ałiłA).,y,

Chapter 7

Tractors and other heavy equipment can be subjected to severe Ioadings due to dynamie ioadings as they accelerate. In this chapter we will shaw how to

determine these loadings for planar moon.

Planar Kinetics of aRigid Body: Forceand Acceleration

CHAPTER OBJECTIVES

■To introduce the methods used to determine the mass moment of inertia of

a body.

■ TO deve lap the Planar ki netic equations of motion for a symrnetric rigid body.

■To discuss applications of these equations to bodies undergoing translation, rotation about a fixed axis, and generał piane m otion.

17.1 Mass Moment of Inert ia

Sincc a body lias a dcfinitc size and shapc, an applicd nonconcurrcnt tarce system can cause the body to both translate and rotatc-The translational aspects of the motion were studied in Chapter 13 and are governed by the equation F = ma. It will be shown in the next section that the rotational aspects, causcd by a moment M, are govcrned by an cquation of the form M = fa. The symbol I in this equation is termed the mass moment of inertia. By comparison, the moment of inertia is a measure of the resistance of a body to angular acederation (M = Jot) in thc same way that mass is a measure of the body's resistance to aceekration (F = ma).

3907 CHAPTER 17 PLANAR KINETICS OF A RIOID BODY: FORCE AND ACCELERATION

The flywheel on the engine of this tractor hasa largo moment of inertia about its axis of rotation. Once it is set into rnotion, it will be difficult to stop, and this in turn wili prevenł the engine from staIling and instead will allow it to maintain a constant powe r.

Fig. 17-1

We define the moment of inertia as the integral of the "second moment" about an axis of alt the elements of mass dm which compose the body.* For example. the body's moment of inertia about thc z axis in Fig. 17-1 is

I= f r2 drn (17-1)

Here the "moment arm" r is the perpendicular distanee from the z axis to the arbitrary element dm. Since the formulation involves r, the vaiue of f is differenl for each axis about which it is computer', In the 5tudy of planar kinetics, the axis chosen for analysis generally passes łhrough the bodys mass center G and is always perpendicular to the piane of rnotion. Themoment of inertia about this axis will be denoted as Since r is squaredin Ey. 17-1, the mass moment of inertia is always a posifive quarkłity. Common units used for its measurement arc kg - m2 or sIug -112.

If the body consists of materiał having a variable clensity, p = p (x,y, z), thc elernental mass dm of the body can be expressed in terms of its density and volume as dm = p dV. Substituting dm into Eq. 17-1, the body's moment of inertia is then computed using voiume elements for integration; i.e.,

= fr2p dV

(17-2)

v *Another property of the body, which rneasures the

symmetry of the body's mass willi

respeet to a coordinate system, is thc product of inertia.This property applies to the threedimensional morion of a body and will be discussed in Chapter 21.

17.1 MASS MOMENT OF INERTIA 397

In the special case of p being a constant,this term may be factored out of the integral, and the integration is 'hen purely a function of geometry.

I = p f r"2 dV (17-3)

. v

When the volume element chosen for integration has infinitesimal dimensions in ali three directions, Fig, 17-2a, thc moment of inertia ofthe body most be determined using "triple integration."" The integration procesy can, however, be simplified to a .single integration provided thechosen volume element has a differential size or thickness in only one direedon. Shell or disk elements ara &ten used for this purpose.

m = p dV

( a )

( b )

(c)

Fig. 17-2


To obtain the moment of inertia by integration, we will consider only symmetric bodies having volumes which arc generated by revolving a curve about an axis. An example of such a body is shown in Fig. 172a.Two types of differential elements can be chosen.

Shell Element.

• If a siwił element having a height z, radius r =, y, and thickness dy is chosen for integration, Fig. t7-2b, then the volume is dV = (2Try)(z)dy.

• Th.i8 Clerrient may be used in I q,17-2 or 17-3 tor dełermining the moment of inertia of the body about the z axis, since the emire ciernem, due to its "thinness," lies at the same perpendicular distance t- = y from thc z axis (see Example 17.1).

Disk Element.

• If a disk element having a radius y and a thickness dz is chosen for integration. Fig. 17-2c, then the volume is dV = (iry2)dz.

• This element is finiłe in the radia] direction, and consequently its parts do not all lie at the same radia] distance r from the z axis. As a resul t, Eq. 17-2 or 17-3 cannot be used to determine directly. Instead, to pertom the integration it is first necessary to determine the moment of inertia of flte. eiernenł about the z axis and then integrate this result (sce Examplc 17.2).

Y

3 9 8 C H A P T E R 1 7 P L A N A R K I N E T I C S O F A R I O I D B O D Y : F O R C E A N D A C C E L E R A T I O N

EXAMPLE 117,1

Determine the moment of inertia of the cylinder shown in Fig. 17-3a about the z axs.The density of the materiał, p, is constant.

b2

( a ) {b}

2

Fig. 17-3

SOi_UTIOIV

Shell Element. This problem can be silved using the shell element in Fig. 1.7-3h and a single inłegration. The vołume of the element is

=

(2irr)(h) dr, so that its mass is dni = pdV = p(2irhr dr). Since the claire element lies al the same distance r from the z axis, the moment of inertia of the ciernem is

diz = r2drn = p27rhr3 dr

Integrating over the entire region of łtie cylinder yields

R= f r2 dm = p27rh J r3 dr = —P7 R4h

O 2

The mass of the cylinder is

Rrn = ł dm = p2-rrh f rdr = hR2

LrE O

so that

= 1 — naR.-

I

17.1 MASS MOMENT QF iNERTIA399

If the density of the materia] is 5 sług/ft-3. determine the moment of incrtia of the solid in Fig. I7-4a ahout the y axis.

1 ft

(L)

Fig. 17-4

1 ft x

(b )

SOLLiTl0 N

Disk Element. The moment d inertia will be found using a disk element, as shown in Fig. 17-4h. Herc the element intersects the curve at the arbitrary point (x,y) and has a mass

tint = p dV = p(ira') dy

Although alt portions of the element arc not located at thc same distance from the y axis, it is stili possihle to determine the moment of inertia dJ, of the element about the y axis. In the preceding example it was shown that thc moment of inertia of a cylinder about itslongitudinal axis is where m and R arc the mass and radiusof the cylinder. Since the height is not involved in this forrnula, the disk itself can be thought of as a cylinder. Thus, for the disk element in Fig. 17-4b, we have

dl, = 4(dni)x.2 = 41-p(irx2)(1Ax2

Substituting x = y2, p = 5 slugift3, and integrating with respect ło y, from y = O to y = 1 ft, yields the moment of inertia for the entire solid.

17(5 slug/ftł)i li 7r(5)dy = — y dy = 0.873 slug • ft2 ft riN_

2

400 CHAPTER 77 PLANAR KIKETICS OF A R I ID BODY: FORCE AND ACCELERATION

Fig. i 7-5

Parallel-Axis Thearem. If the moment of inertia of the body abouł an axis passing through the bodys mass center is known, hen the moment of inertia about any other parallel axis can be determinecl by using the partzilet-axis theorern :This the orem can be der kred by considering the body shown in Fig. 17-5. Herc the z' axis passes through the mass center G, whereas the corresponding parallei z axis lies at a constant disłance d away. Sclecting the differential element of mass dssi. which is focated at point (x', y'). and 41.5ing the Py th go reali theorem, s-2 = (d + x')2 y'2, we can express the moment of inertia of the bady about the z axis as

= f r2drss = f [(d + xr)2 y ' '] dm

= f (x"2 y'2)dm + 2d x" drn + d2f drn m

Since = x'2 y'2, the first integral represents /G. The secundintegral equals zero, since the z" axis passes through the body's masscenter, i.e., [x'din = .Tr'n? = 0 since = O. Finally, the third integraf


represents the tota] mass m of the body. Hence, the moment of inertia about the z axis can bo written as

(17-4)

where

= moment of inertia about the z' axis passing through the mass center G

m = mass of the bodyd = perpendicular clistancc bctwcen thc paraitel z and z' axes

R ad iu s of Gyratio n. Occasionaliy, the moment of inertia of a body about a specified axis is reported in handbooks using the radna uf gyration, k. This is a geometricał property which has units of length. When it and the, bodys mass m are known, the bodys moment of inertia is determined from the equation

= mk 2 o r k =

(17-5)

Note the similariły be,twecn the definition of k in this form ula and r in thc equation dl = r2dm, which dcfincs the moment of inertia of an clemental mass drri of thc body about an axis.

Composite Bodies. If a body consisłs of a number of simple shapes such as disks, spheres, and rods. the moment of inertia of the body about any axis can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the axis. A Igebraic addition is riccessary sincc a composite part must be considercd as a ncgative quantity if it has already becn counted as a piece of anothcr part—for example, a "hole" subtracted from a solid plate. The parallelaxis theorem is needed for the calculations if the center of mass of eaeh composite part does not lie on the axis. For the ea]culation, then, I = £(TG -t- rrid'). Here /G for each of the composite parts is determined by integration, or for simple shapes, such as rods and disks. it can be found from a table, such as the one given un the insi& back cover of this buok.

402 CHAPTER 17 PLANAR KfNETrOS OF A RIGID BODY: FORCE AND ACCELERATJON

1 7 . 3

If che plate shown in Fig. 17-6a has a density of 8000 kg/m3 and a thickness of 10 mm, determine its moment of inertia about an axis directed perpendicular to the page and passing through point ❑.

125 mm125 mm

O Thickness 10 mm

(b)Fig, 17-6

50LUTIONThe plate consists of two composite parts, the 250-mm-radius disk minus a 125-mm-radius disk,Fig.17-6b.The moment of inertia about O can be detennined by computing the moment of inertia of each of Ulew parts about O and [hen adding the results algebraically. The calculations are performed by using the parallcl-axis theorem in conjunction with the data listed in the table on the inside back cover.

Disk. The moment of inertia of a disk about the centroida) axis perpendicular to the piane of the disk is IG = 4 Mr2 . The mass center of the disk is located at a distance of 0.25 ni front point O. Thus,

rnd = Mi = 8000 kgłrn3 [7r(0.25 m)2(0.01 rri)] = 15.71 kg

a)o = "ilndr3 Iradd2

= 1(15.71 kg)(0.25 m)2 + (15.71 kg)(0.25 in)2 =

1.473 kg • rn2

Hole. For the 125-mm-radius disk (hole), we have

znp, = [JIM, = 80130 kg/rn3 [17(0.125 m)2(0.01 in)] = 3.927 kg

(4)0 = mhd2

= 2 (3,927 kg)(0.125 m)2 + (3.927 kg)(0.25 m)2 =

0.276 kg- rn2

The moment caf inertia of the place about point O is therefore

/0 = (/4)0 (b.)0= 1.473 kg • in2 — 0.276 kg • nn2

1.20 kg•m2

17.1 MA55 MOMENT OF INERTiAd03

EXAMPLE

The pendulum in Fig. 17-7 is suspended from the pin at O and consists of two thin (-ods. Rod OA weighs 10 lb, and DC weighs 8 lb. Determine the moment of inertia of the pendulum about an axis passing through (a) point O. and (b) the mass center G of the pendulum.

S O UTION

Part <a). Using the takle on the inside back cover, the moment °f inertia of rod 0A. about an axis perpendicular to the page and passing through point O of the, rod is fo = 3 mi-2. Hence,

1 1 ( lb (ioA)o '5111- =

3 3210 .2 ft/s2)(2 0)2 = 0.414 siug • ft2

This same value can be obtained using „G = mi` and the parallel-axis theorem.

10 lb 2)(2 foz 10 Ib )(1 fo,(foA)o = md2 — 112 (32.2 ft/s12 32.2 ft/s2)

I 4 slug ft2

For rod BC we have

( ł B c ) c ) m d „ 1 ( 8 f l 3 ) 0 , 5 f t ) ? 8 l b 2,)( ft)2

12 12 \ 32.2 ft/s2) ik,32.2 ft/s-

= 1.040 sług • ft2The moment of inertia of the pendulum about O is therefore

ło = 0.414 + 1.040 -= 1.454 -= 1.45 slug • ft2 Ans:.

Part (b). The mass center G will be Ioeated reIative to point O. Assurning

this distance to be "5-,, Fig. 17-7, and using the furii-uda for

deterrnining the mass center, we have

27m 1(10/322) + 2(8/32.2)— 44

Sm (10/32.2) + (8/322) — 1.4 ft

The moment cd inertia Il may be found in the same manner as 10, which requires successive applications of the parailel-axis theorern to transfer the rnoments of inertia of rods OA and BC to G. A more direct solution, however, involves using the result for to, i.e..

lo = fG + nac12; 1.454 slug • ft2 =32.2 ft/s2

181b )(1.444 f02

„G. 0.288 slug • ft2 Ans,

0.75 fl •0.75 ft

Fig. I7-7

404 CHAPTER 17 PLANAR KINETICS OF A NIGIO BODY: FORCE ANO ACCELERATICN

■ PROBLEMS

17-1. Delermine the moment of inertia i, for the slender rod. The rod's density p and cross-sectional area A are canstant. Express the result in terrns of the rod's totaI mass m.

17-3. Determine the moment of Me/1'1a of the thin ring about the z axis.The ring has a mass m.

Prob. 17-3

Proh. 17-1

17-2. The solid cylinder has an ouier radius R, height h, and is made from a material having a density that varies from its center asp = k + ar 2'. where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis.

*17-4. Determine the moment of inertia of the semiellipsoid with resped to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density p.

Prob. 17-2

Prob. 17-4

17.1 MASS MOMENT O F i NERTIA 405

17-5. The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia f, and express the resułt M terms of lhe totai mass m of the sphere. The material bas a constant density p.

17-7. The solid is formed by revolving the shaded area around the y axis. Determine the radius of gyration The specific weight of the material is y = 380 lb/ fti.

Prob. 17-53 i i n ,

3

Prob. 17-7

17-6. Delermine the mass moment of inertia f, of the cone formed by revolving the shaded area around the z ams.The total density of the materiał is p. Express the result in terms of the mass m of the cone,

*17-8. The concrete shape is formed by rotating the sliaded area about the y axis. Determine the moment of inertia 1,, The specific weight °f concrete is -y =-- 150 thift3,

Prob. 17-6 Prob. 17-S

4 0 6 C H A P T E R 1 7 P L A N A R < / N E T I C S O F A R I G I D B O D Y : F O R C E A N D A C C E L E R A T i O N

17-9. Determine the moment of inertia f. of the torus. The mass of the torus is m and the dersity p is eonsłant. Suggestion: Use a shell element.

Prob. 17-9

17-111. Determine the mass moment of inertia of the pendulum about ar axis perpendicular to the page and passing through point O. The slender rod has a mass of 10 kg and the sphere has a mass of 15 kg.

17-11. The slender rods have a weight of 3 Ibift.Determine the moment of hienia of the assembly about an axis perpendieular to the page and passing through the pin at A.

Prob. 17-11

*17-12. Determine the moment of inertia of the solid steel assembly about the x axis. Sieci has a speeific weight of 7.„ '=

490 Ibift3.

O 25 ft _L0.5 ft

x

2f-t 3 it

Prob. 17-1217-13. 'The wheel consists of a thin ring having a mass of 10 kg and four spokes made from słencler rods and each having a mass of 2 kg. Determine the wheers moment of inertia about an 4?{"ś5 perpenctteular to the page and passing through point A.

Prob. 17-10 Prob. 17-13


17-14. 1f the large ring. smali ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb. respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.

*17-16. Determine the mass moment of inertia of the Chin plate about an axis perpendicular to the page and passing through point O. The material lias a mass per unit area of 20 kg/m2.

Proh. 17-14

Prob. 17-16

17-15. Determine the moment of inertia about an axis perpendicular to the page and passing through the lin at O. The thin plate has a hole in i ts center, fts łhickness is 50 mm, and the material hasa density p = 50 kg/m3.

Prob. 17-L5

17-17. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg/m. Determine the length L of DC so that the center of mass is at the hearing O. What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through

17-18. The assembly consists of a disk having a mass of 6 kg and słender rods AB and DC which have a mass of 2 kg/m_ lf 1. = 0,75 m, determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through O.

Probs. 17-17/18

408 CHARTER 17 PLANAR KrNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON

17-19. The pendulum eansists of two slender rody AR and OC which have a mass of 3 kgirn.The lhin circular plale has a mass of 12 kg/m2. Determine the location v of the center of masa G of the pendulum, then calcułate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

*17-20. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin circular plate has a mass of 12 kg/m2. Determine the moment of Media of the pendulum about an axis perpendicular to the page and passing through the pin at O.

Probs. 17-19120

17-21. The pendulum consists of the 3-kg slender rod and the 5-kg thin place. Deterrnine the location p of the center of mass G of the pendulum: łhen calcuIate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

•

0 5 m

Prob. 17-21

17-22. Determine the moment of inertia of the overhung crank aboul ihe x axis.The materia] is sieci having a &silny of p = 7.85 Mg/m3.

90 mm

m m

20 mm

50 mm30 mrn

Prob. 17-22

17-21 De trrnine the moment of inertia of the overhung crank about the x axis.The materiał is steei having a destiny of p = 7.85 Mg/m3.

20 mm50 mm i30 mm1-1

Prob,. 17-23

20 mm

20 mmJ

50 mm

20 mm

90 mm

50 mm

mm

20 mm

17.2 PLANAR KNETIC EQUATIONS OF MOTION 409

1 7 2 Planar Kinetic Equations of Motion

In the following analysis we will limit our study of planar kinetics to figiel hodies which. aiong with their loadings, ara considered to be symmetrWat with respeet to a fixed reference piane,* Since the motion of the body can be viewed within the reference plane.alI the forces (and couple moments) acting on the body can then be projected onto the piane. An example of an arbitrary body of this type is shown in Fig. 17-8a. Herc the inertial frame of reference x, y, .z has its origin coincident with the arbitrary point P in the body. By definition, these axes do not rotate and are eiłher fixed or rranslate with constant vełocity

W

P

7 F ]

(a)

Fig. 17-8

Equation of Translational Motion. The external forces acting on the body in Fig. 17-8a represent the effect of gravitational, electrical, magnetie, or contact forces between adjacent bodies. Since this (orce system bas been considered previousy in Sec. 13,3 for the analysis of a system of particles, the resulting Eq. 13-6 can be used herc, in which case

= inau

This equation is referred to as the tratislariormi equation of molton for the mass center of a rigid body. It stater that the sum of ail the external forces acting on dw bady is equal Fa the body 's mass times the acceleradon of its mass center G.

Fot rrsotion of the body in the x—y piane, the translational equation of niotion may be written in the forni of twa independent scalar equations. namcly,

Fs = m(aG), FV

= in(aG),

*By doing this, the rotational ermation of motion rednces to a rather simphfied form. The moce generał CEISC of body shape and loading is considered in Chapter 2

41 0 C H A P T E R 17 P L A N A R < / N E T I C S O F A R I G I D B O D Y : F O R C E A N D A C C E L E R A T i O N

Particie free-body diagram

( b )

1 1

Particie kinetic diagram

( c )

r r

( d )

Fig. 17-8 (gont)

Eqution of Rotational Motion. We will now determine the effects caused by the moments of the external furce system computecl about an axis perpendicular to thc piane of motion (the z axis) and passing through point P. As shown on thc free-body diagram of thc ith particie, Fig. 17-8b, F, represents the restihant external forte acting on the particie, and f; is the resultant of the inłernal forces caused by interactions with adjacent particles. If the particie has a mass mi and its acceieration is łhen its kinetic diagram is shown in Fig. 17-8c. Summing moments about point P, we require

r x F ; + r X = r x m i k

o r

(Mp)r = r X mik

The moments about P can also be expressed in terms of the acceieration of point P, Fig. I7-8d. If the body has an angular aceeleration or and angular veloci ty w, then using Eq. 16-18 we have

(MA = mir x (ap + cr X r — 0)2r)

= mjr x ap + r x (ar x — w`fr x r)1

The Iast term is zero, since r X r = O. Expressing the vectors with Cartesian components and carrying aut the cross-product operations yields

(MAik = sni((xi yj) X [(ap),i + (ctp)yj]

+ (xi + yj) X [ak X (xi + yj)1)

(Alp)ik = inii—y(aP)s -r(ap)y + fYC? + ay2jk

C. (M p)i = y(ap), + x(ap), + ar2]

Letting m —) dni aud integrating with respeot tv the entire mass m of the body, we ob ł ain the resultan ł moment equation

C Skip—(if y clin)(ap), + (if x dra)(ap)y + (j r2c7r1)cx

Here £34 represents ordy the moment of the emernal forces acting on the body about point P. The resultant moment of the internal forces is zero, since for the, entire body these forces oecur in equal and opposite collinear pairs and thus the moment of cach pair of forces about P eaneels, The integrals in the first and second terms on the right aro used to locałe the bodys center of mass G with respect to P, since urn = f y dm and xm = J x dm, Fig. 17-8d. Also, the Last integral represents the body's moment of inertia about the z axis. i.e., fp = fr2dm. Titus,

1A/p = + :rni(ap), + Ipoc (17-6)

17.2 PLANAR KNETIC EQUATIONS OF MCMON 41 1

It is possible to reduce this equałion to a simpler form if point P concideswith the mass center G for the body. If this is the case, then x = y = O, F,and therefore*

łcid (17-7)

This rołałional equałion of mołion states that the sum of the moments of all the external forces about the bodyś mass center G is equal to the product of the moment ofinertia of the body elbom` an axis passing łhrough G and the body's angular acceleration.

Equation 17-6 can also be rewritten in terms of the x and y components of ac and the body's moment of inertia „G. If point G is located at (.7r, 5"), Fig. 17-8d, then by the parauel-axis łheorem, Ip = IG + in(51-2 + y?). Substituting into Eq. 17-6 and rearranging terms,, we geł

Sitip++ Yanr<ap)y + :va] loa (17-8)

From the kinernatic diagram of Fig. 17-8d, ap can be expressed in terms of a

c; as

x

G

\ < 2

Free-body diagram

(e.)

m(ae)5

Kirietic diagram

( 0

a c = a p - F r . r x f — t o 4

= (ap)xi (ap)„.j ÷ cyk X (7vi 3.7i) — w2(x; + :Fj)

Carrying out the cross product and equating the respective i and j components yields the two scalar equations

(ac)x (ap)x — yee -1N2

(aG)y (ap),, + xer

From these equations. [—(ap)„ + ycr] = 1—(aG) and[(ap)x, + -;cor] = Rac), + Y(02]. Substituting these results into Eq. 17-8 and simplifying gives

Zmp —Y3n(aG), 7-m(0G), + 'ca (17-9) Fig, 17-8 (cunt.)

This importom wsi& indicates that when mornents of the external ,forces shown on the free-body diagram are summed about point P Fig. 17-8e, they are equivalent to the sum of the "kinetic moments" of the components of mac about P plus the "kinetic moment" of Ica, Fig. 17-8f. In other words, when the "kinetic mornents," S(.+RA),., are computcd, Fig. 17-8f, the vectors rn(aG).,, and in(ac)., are treated as sliding vectors; that is, they can act at ony point along their hne of action. In a simiiar manner, /ca can be treated as a free vector and can therefore act at any point. It is important to keep in mind, however, that mar and /Gar are not the same as a force or a couple moment. Instead, they are caused by the external effects of forces and couple moments acting on the, body. With this in mind we ran therefore write Eq. 17-9 in a more generał form as

Zmp = ł (*Op

'Ft also reduees to this same simple form 2.Mp = Iped if point P is a ,fixed point (see 17-16) or the acceleration ()f point P is direeted along the firn PCi_

41 2 CHAPTER 17 PLANAR </NETICS OF A RIGID BODY: FORCE AND ACCELERATiON

G e n e r a l A p p l i c a t i o n o f t h e E q u a t i o n s o f M o t i o n . T oFI summarize this analysis, łhree independent scalar equations can be written

w dcscribc the general plano motion of a sy-mmetrical rigid body.

Fx = m(ac)_,

Iw Fy = m(a,c)y

Free-body diagram F2 Gr ‚WF = (.14k)E. (17-11)

(e)

m(oc)y:

1•

Kinetic diagram

Fig. 17-8 (cmii.)

(a)

When applying these equations, one should always draw a free-body diagram, Fig. 17-8e, in order ta account for the terms involved itr XFx,

ur SM P. In some problems it may also be helpful to draw the kinetic diagram for the body, Fig, 17-8f.This diagram graphically accounts for the terms m(ac),, rii-(ac),,, and /cot. It is especially convenient when uscd to detcrmine the componcnts of InaG and the moment of these components in 1(.41,k)p.*

17.3 Equations of Motion: Translation

When the rigid body in Fig.17-9a undergoes a rranshnion, all the particles of the body have the sanie aceelerarion Furthermore, a = O. in which case the rotational equation of motion applied at point G reduces to a simplifiecl form, namely, A,fc = O. Application of this and the force equations of motian will now be discussed for each of the two types of transIation,

Rectilinear Translation. When a body is subjeeted to recrilinear translation, all the partieles of the body (slab) travel along paranel straight-Iine pałhs. The free-body and kinetic diagrams are shown in Fig. 17-9b.Sinee-= 0, only mac is shown on the kinetic diagram. Hence, theequations of motion which apply in this case becorne

Fk = on(a5),

F y = tn(a(;)y (17-12)2.MG = 0

*For this reason, the kinetic diagram will be used śn the solutlon of an example problemFig. 17-9whenever £Mt. =is applied.

173 EQUATIONS OF MOTION: TRA,NSLATION.41 3

Y17111G

(h )

It is also possible to sum moments about other points on or off the body, in which cale the moment °f mac must be taken hitu aceount, For example, if point A is ehosen,which lies at a perpendicular distance d from the line of action of ina,-;, the foilowing moment equation appb'es;

+5MA = 5{,44)A; 5.kirA = trmiG)d

Here the sum of moments of the external forces and couple moments about A (5MA, free-body diagram) equals the moment of mac about A(ł, kinefic diagram).

CurviIinear Translation. When a rigid body is subjected to curvilinear transiation, all the particles of the body have the same aceelerations as they travel along ctirved paths as noted in Sec.16-1. For analysis, it is oftcn convcnient to use ort inertial coordinatc system having att origin which coincides with the body's mass center at the instant considered. and axes which are oriented in the norma' and tangential directions to the path of motIon, Fig. 17-9c.The three scalar equations of motion are then

F. m(ac)„5F, = m(aG), (17-13)

5,214L-; = O

If moments are summed about the arbitrary point B, Fig. 17-9c, then it is necessary to aceount for the moments, (,14k)g, of the two components ni(aG)„ and m(aG), about this point. From the kinetic diagram, h and e represent the perpendicular distances (or "moment arms") from 1-3 to the blies of action of the components. The required moment equation therefore becomes

c +2,fida = :(-44)B; ?d = e[m(aG),] — h[m(aG),,]

If

(c)

Fig, 17-9

414 CHAPTER 17 PIANAR KiNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON


Kinełic problerns involving rigid-body transiaturn can be solved using the following procedure.

Free-Body Diagram.

• Establish thc x, y or n, r inertial coordinatc system and draw the frec-body diagram in order to account for alt the extcrnal forces and couple moments that act on the body.

• The direction and sense ()f the. acceleration ()f the body's mass cent« a

c; should be cstablished.

• Identity the unknowns in the problem.

• If it is decided that the rotational equation of motion Mp = (Atk)p is to be used in the soluticin, then consider drawing the kinetic diagram, since it graphically accounts for thecomponents nr(ac), or m(ac)„ in(aG)„ and is thereforeconvenient for "visualizing" the terma needed in the momentsum .


• Apply the thrce equations of motion in accordanee with the established sign convention.

• To simplify the, analysis, the moment equation ,MG = O can be replaced by the more genem! equation SMp = 1Gfri,k)p, whcre point P is usually located at thc intersection cif the lines of action of as many unknown forces as possible.

• If the body is in contact with a rough surface and slipping occurs, use the friction equation F = pk1V. Remember, F always acts on the body so as to oppose the motion of the body relałive to the surface it contacts.

Kinematics.•Use kinematics to determine the veIocity and posirion of the body.•For roctilincar translation with variable acceteration aG =

duo! dł aads0 = vGdv0

• For rectilinear translation with constant acceleration

vG. (7)c)o+ aGt V J (vG).(} + 2aG[sG — (sG)01

sc = (s{;){ (v,-;)01 acs2

• For curvilinear translation

(cto),. v:ći P

(ac) , = dvddt (aadsG = vGrive

The free-body and kinetic diagrams for this boat and irailer are drawn first in order to apply the equalions of motion_ Here the forees {In the free-body diagram eause the e ffeet shown on the kinetic diagram. [f mornents are surnmed about the mass center, G. then MG = However,if momenty are surnmed aboul point B lhen C +2A4 = rnadd).

T

17.3 EOIJATIONS OF MOTION: TRANSIATION. 41 5

The car shown in Fig. 17-10o has a mass of 2 Mg and a center of mass at G. Determine the acceleration if the rear -`driving" wheels are always slipping. whereas the front wheeis are free to rotate. Neglect the mass of the wheels.The coefficient of kinetic friction between the wheels and the road is jet = 0.25.

SOLUTION I

Free-Body Diagram. As shown in Fig. 17-10b, the rear-wheel frictional force FN pushes the car forward, and since slipping oecurs, FR

= 0.25N8. The frictional forces acting on the front wheels arc zero, since thesc wheels have negligible mass.* There arc threc unknowns in the problem, NA, Na, and u G. Here we will sum mornents about the mass center. The car (point G) accelerates to the left, i.e., in the negative x direction, Fig. 17-10b.

Equations ci Motion.

Fz = in(aG),; -0.25% = -(2000 kg)a(; (1)

± = tn(ac,),; NA + Na - 2000(9.81) N = O (2)

+.dri = -N,,k(1.25 m) - 0.251,(0.3 m) N/3(0.75 m) = O (3)

Solving,

aG = 1.59 mis2

NA = 6.88 kN

Na =- 12.7 kN

SOLUTION II

Free-Body and Kinetic Diagrams. If the "moment" equation is applied about point A, then the unknown NA will be eliminated fromthe equation, To "visualize" the moment of about A, wc will indu&the kinctie diagram as part of the analysis. Fig. 17-10c.

Equation af Motion.

+Sm, = (J14.0,1; NB(2 m) - [2000(9.8 I) N1(1.25 m) =

(2000 kg)aG(0.3 m)

Solving this and Eq. 1 for oc; leads to a simpIer solution than that obtained from Eqs. 1 to 3.

,` With negligibk wheel masa, Ia = 0 and the- frietionał furce at A required tu turn lhe wheeł is zera. tf łhe wheels mass were ineluded, then the solution would be MOIC involved, siwce a genetal-płant-motion analysis of the wheels wota( have to be considered (see Ser. 173).

1.25 m.0 .75 m

(b)

-.1d1 Jr •

2000 (9.81) N

,Arb,

_ = 0-25 Nh

2000 (9.81) N

A 0 . 3 i n

(c)

Fig. 1.7-10

416CHAPTER 17 PLANAR KfNETrCS OF A RIGID BODY: FORCE AND ACCELERATJON

E X A M P L E

The matorcycle shown in Fig. 17-11a has a mass of 125 kg and a center of mass at G , whi le the -lider has a mass of 75 kg and a center of mass at G2. Determine the minimum coefficient of static friction between the wheels and the pavement in order for the rider to do a "wheely," i.e., Iift the front whecl off the ground as shown in the photo. What acceleration is necessary to do this? Negleet the mass of the wheels and assume that the front wheel is free to roll.

735.75 N1226.25 N

0.4 m 0.4 rit

11 SOLUTION

Free-Body and Kinetic Diagrams. In this problem we will consider both the motorcycle and the rider as a single sysrent.It is possible first tor cletenriine the location of the center of mass for this "system" by

using

5 kg,

125 kg aiG 0 m the equations _Tr = S.-x-/n/Sni and y = n Sar. Here, however, we willconsider the weight and mass of the motorcycle and rider seperate as

0.6mshown on the free-body and kinetic diagrams. Fig. 17-11b. Botki of these,parts rnove with the same acceleration. We have assumed that the frontwitce' is aboin to leave the ground, so that the norrnal reaction NA G.The thrcc unknowns in the problem arc NR, FR, and aG.Equations of Motion.

S F, = in(a),; FB = (75 kg + 125 kg)aG(1)

+ Fy = tri(aG),; - 735.75 N - 1226.25 N = O

C = S.G14)8; -(735.75 N)(0.4 m) - (1226.25 N)(0.8 m) =

-(75 kg aG)(0.9 m) - (125 kg ac)(0.6 ni) (2)Solving,

eiG = 8.95 mis?- Ans.

NB = 1962 N

FB = 1790 N

Thus the minimum coefficient of static friction is

FB 1790 N(g,)mt. = N& 1962 N - 4.912

(h )

F. 17-11

173 EQUATIONS OF MOTIoN: TRANSLATioN41 7

EXAMPLE

A uniform 50-kg crate rests on a horizontal surface for which the coefficient of kinetic friction is fck =" 0.2. Determine the acceIeration if a force of P = 600 N is applied to the crate as shown in Fig. 17-12a.

H

(a)

SOLUTION

Free-Body Diagram. The force P can cause the crate either to slide or to lip over. As shown in Fig. 17-121). it is assumed that the crate slides, so that F = 1.6kNc = 0.2Nc. Also, the resultant norma] fotce Na acłs at 0, a distance x (where O < x 03 m) from the crate's center line.* The three unknowns are Wc, x, and ac.

Erivations of Motion.

L. F in(z4G),; 600 N - 0.2Nc = (50 kg)aG

(1

)

Nc - 490.5 N = O

(2

) +ZA/G =—600 N(U.3 m) + Arc(x) — 0.2Ne({1.5 ni) = O

(3

)

Solving,

Arc = 490.5 N

x = 0.467 ni

ao = 10.0 mis2 -› rr:s.

Since x = 0.467 m c 0.5 m, indeed the crate slides as originally assumed.

NOTE: H the solution had given a value of x .> 0.5 m, the problem ~Id have to be rewurked since tipping occurs. Tf this were the case, Ncwould act at the corner poinł A and F c 0.2Nc.

line of aenon oF Nc does not neeessarily pass through the mass center G (), = 0). srace Nc musi e<yunieracI the tendeney for tipping caused by P. Sce Sec. 8.3 of Engineering Mechanics: Stati.c.s.

0.5 m

0.5(b)

Fig. 17-12

41 8 CHAPTER 17 PLANAR 10ENETECS OF A RICID BADY: FORCE AND ACCELERAT1ON

— 0.4 0.4 m—

(a)

The 100-kg beam BD shown in Fig. 17-13a is supported by two rods having negligible mass. Deterrnine the force developed in each rod ifat the instant fl = = 6 rad/s.

SOLUTION

Free-Body Diagram. The beam maties with etrrvrliHertr łranstation sinee all points on thc beani movc along eirettlar paths, each path having the same radius cif 0.5 ni, but different eenters of eurvature. Using norma' and tangential coorclinates, the free-body diagram for the beam is shown in Fig. 17-13b. Because of the łranslation, G has the same motion as the pin at B, which is connected to both the rod and the beam. Note that the tangential compranerat of acceleration acts downward to the left due to the cIockwise direction of a. Fig. 17-13c. Furthermore, the normal connponent ()f acceleration is always directed toward the center of curvature (toward point A for rod A B), Since thc angular velocity of A& is 6 radi's when 9 = 30°, then

(aG), = (o7r = {6 radis)2(0.5 m) = 18 m/s2

The tliree unknowns are TE, T0, and (c I G), The directions of (aG)„ and (ac), have been estabIished, and are indicated on the coordinate axel.

T81 1TD

G

4-4 m 0.4 m-1

V981N

(b) (c)

Fig. 17-13


4".2.F„ = rra(aG;),,; TE ± T - 981 cos 30° N = 100 kg(18 infs2)(

1 )

= rn(ac),; 981 sin 30° = 100 kgac),(

2 )

C +EMU = 0; -(7'a cos 30')(0.4 m) + (To ras 30°)(0.4 m) = O(

3 )

Simultaneous solution of rkese three equations giyes

Tg = TD = 1.32 kN ,1?r,

(aG), = 4.905 m/s2

NOTE: It is also possible to apply the equations cif motion along horizontal and vertical x. y axes. but the sołution becomes more involved.

173 EQUATIONS OF MOTION: TRANSLkTION419• FUNDAMENTAL PROBLEMS

F17—L The cart and itr load have a total mass of 100 kg. Determine the areeleration of the eart and the normal reactions on the pair of wheels at A and B. Neglect the mass of the wheels.

[00N

F17-1

F17-2. If the 80-kg cabinet is allowed to roll down the iriclined piane, deterniine the acceleration of the cabinet and the norrnal rcactious on the pair of rollers at A and B that have negłigible mass.

1.5m

0.5mmF17-2

F17-3. The 20-]b link AB is pinned to a moving frame at A and hołd in a vertical position by means of a string BC which can support a maximum tension of 10 lb. Determine the maximum acceleration of the france without breaking the string. What are the corresponding components of reaction at the pin A?

F17-4. Determine the maximum acceleration of the truck without causing the assemWy to move relative to the truck. Also what is the corresponding norma' reaction nn legs A and 8? The 100-kg labie has a mass center at G and the coeffieicnt of static friction hetween the legs of the takle and the bed of the truck is p,. = 0,2.

F17-4

FI7-5. At the instant shown both rods of negligible mass swing with a countercloekwise angular velocity of (.0 = 5 rad/s, while the 50-kg bar is subjected to the 100-N horIzontal force. Determine the tension devetoped in the rods and the angular acceleration of the rods at this instant.

F17-5 F17-6. At thc Instant shown, link CD rotates with an angular velocty of = ó rad/s, If it is subjected to a couple -

moment M = 450 N- nit deterrnine the forte developed in ]tuk AR. the horizontal and verticai component of reaction on pin D, and the angular acceleration of link CD at this instant,The block bas a masa of 50 kg and center of mass at G. Neglecl the mass of links AB and CD.

F17-6

17

420 CHAPTER 1 7 P L A N A R 1 0 E N E T E C S O F A R I C I D B O D Y : F O R C E A N D A C C E L E R A T 1 O N• PROBLEMS

'17-24. The door bas a weight of 200 lb and a center of granity at G. Determine how far the door rnoves 1n 2 s, starting from rest, if a man pushes on it at C with a horizontaI force F = 30 lb. Aiso. find the vertical reactions a-1 the pollers A and B.

17-25. The door hasa weight of 200 lb and a center of granity at G. determine the constant furce F that musz bo applied to the door to push it open 12 ft to the right in 5 s, starting from rest. Also. find the vertical reactions at the rollers A and B.

Probs. 17-24125

17-26. The uniform pipe bas a weight of 500 lb/ft and diarneter of 2 ft. Ifil is hoisted as shown with an acceleration of 0.5 ft/s2. determine the internat moment at the center A of the pipe due to the lift.

17-27. The Brum truck suppnrts the 600-1b drutu that has a center cif granity m G. If the operator pushes it forward with a horizontal force of 20 Ib. determine the acceleration of the truck and the nomial reactions at each cif the four wheels. Neglect the mass of the wheels.

*11-28, If the cart is given a constant aceeleration °f a = ft/ s2" up the incrined piane, determine the force devełoped in rod AC and the horizontal and vertical components of force at pin B.The crate hasa weight of 150 Ib with center of granity at G. and it is secured on the platform. so that it does not slide. Neglect the platform weight.

17-29. If the strat AC can withstand a maximum compression force of 150 lb before it fails. determine the cart's ma.xtrzium permissible acceleradon. The crate lias a weight of 1501b with center of gravity at G, and it is sccured on the platform, so that it does not shde. Neglect the pl a norm 's weight.

1 ft

2 ft

P roba. 17-28129

2016

4 ft

? ft1

A

0.5 ft 1 ft

Prob. 17-27

10 5 ftls2

1

,

5 ftft

5 ft-l- 5 ft 5 n _4_ 5 ft —P- ft

Prob, 17-26

17.3 EQUATIONS OF MOTION: TRAINSLATION421

17-30. The drop gale at the end of the trailer bas a mass of 1.25 Mg and mass center at G. 1f it is supported by the rabie AB and hinge at C. determine the tension in the rabie when the truck begins to accelerate at 5 m/s2. Also. what are the hortzontal and vertical components of rcaction at the hinge C?

Prob. 17-30

17-31. The pipe bas a length of 3 m and a mass of 500 kg. It is atłached to the bark of the truck using a 0.6-m-

long chain AB. If the coefficient of kinetic Friction at Cis = 0.4. determine the acceleration of the truck if the

angie d = with the road as shown.

Prob. 17-31

*17-32. The mountain bike has a mass of 40 kg with center of masa al point G1, while the rider has a mass of 60 kg with center of mass al point G2. Determine the maximum dcccleration when the brake is applied to the front whecl. without eausing the rear wheel i4 to leave the ruad. Assume that the front wheel does not slip. Neglect the mass of all the wheels.

17-33. The mountain bike has a mass of 40 kg with center of mass at point Gi, while the rider bas a mass of 60 kg with center of mass at point G2. When the brake is applied to the front whed, it eauses the bike to decelerate at a constant rate ❑f 3 mis2. Determine the normal reaction the road exerts on the front and rear wheels. Assume that the rear wheel is free to roli. Neglect the mass of all the wheels.

Probs. 17-32/33

17-34. The trailer with its load has a mass of 150 kg and a center of mass at G, If it is subjected to a horizontal force of

= 600 N. determine the trailer's aceeleration and the norrnal furce on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

0.75m. 125 ni-

Prob. 17-34

422 CHAPTER 17 P L A N A R 1 0 E N E T E C S O F A R I C I D B O D Y : F O R C E A N D A C C E L E R A T 1 O N

17-35. At the start of a race. the rear drive wheels 8 of the 1550-lb car slip on the track. Deterrnine the car's acceleration and the normal reaction the track exerts on the front pair of wheels A and rear pair of wheels B, Thecoefficient of kinetic friction is = 0.7. and the masscenter of the car is at G. The front wheels are free ti3 roli. Neglect the mass of ali the wheels.

*17-36. Determine the maximum aeceleration that can be achieved by the car without having the front wheels A leave the track or the rear drive wheels B slip on the track. Thecoefficient of static friction is = 0.9. The ear's masscenter is at G, and the front wheels are free to roli. Neglect thc mass of alt thc wheels,

A 6 ft ---^-475

Probs. 17-35/36

17-37. If the 4500-119 van Tras front-wheel drive, and the coefficient of static friction betwcen the front wheels A andthe mad is = 0.8, determine the norma' reactions on thepairs cif front and rear wheels when the van has maximum acceleration. Also. find this maximum accelerati on. The re ar whecls arc free to roli. Neglect the mass of the wheels.

17-38. Tf the 4500-lb van bas rear-wheel drive, and the coefficłent of stalic frictiori between the front wheels B andthe ruad is 0.8, determine the norma' reactions on thepairs of front and rear wheels when the van has maximum acceleration. The front wheels are free to roli. Neglect the mass of the wheels.

A6 ft

33 h

Probs. 17-37/38

17-39. The uniform bar of mass m is pin connected to the Miar. which slides along the smooth horizon lal rod. If ihe collar is given a constans acceleration of a, determine the bar's incIin ation angle H. Neglect the collar's mass.

Prob. 17-39

17-40. The lift truck has a mass of 70 kg and mass center al G. If ii lifis the 120-kg spooI with an acceieraiion of 3 mis2, determine the reactions of each of the four wheels on the ground. The loading is syrnmetric. Neglect thc mass ot the movable anri CD.

17-41. The lift truck has a mass of 70 kg and mass center al G. determine the largest upward acceleration of the 120-kg spool sra that no reaction of the wheels on the ground egtceeds i00

Probs. 17-40/41

17.3 EQUATIONS OF MOTION: TRANSIATION. 423

17-42. The uniform crate has a mass of 50 kg and rests on the carl having an inclined surface.Determine the smalIesi acceleration that will cause the crate either to tip or slip relative to the eart. What is the rnagnitude of this acceleration? The coefficient ot statie friction between thetrale and the carl is = 0.5.

6 I T I

Prob. 17-42

17-43. Determine the acceleration of the 150-Ib cabinet and the normal rcaction under the legs A and B if P 35 lb. The coefficients of static and kinetic friction between the cabinet and the piane are n = 0.2 and p.* = 0.15.respectively.The cabinet's center of graeity is located at G.

ft—t— I ftl

Prob. 17-43

17-44. The assembly has a mass of 8 Mg and is hoisted -

ming the boom and puliey system. If the winch at B draws in the cable with an acceleration of 2 m/s2, determine the compressive furce in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G.

Prob. 17-44

17-45. The 2-Mg truck achieves a speed of 15 m/s wlth constans acceleration after it has traveled a distance of 100 m. starting from rest. Detcrmine the normal furce cxerted on each pair of front wheels R and rear driving wheels A. Also, find the traction (orce on the pair of wheels at A.The front wheels are free to roli. Neglect the mass of the wheels.

17-46. Determi ue the shortesł time pnssihlti for the rear-wheel drive.2-Mg łruck to achieve a speeci of 16 mis with a constant acceleration starting from rent. The coefficient of stalic friction between the wheels and the ruad surface is p, = 0.8.The front wheels arc free to roll. Neglecł the mass of the wheels,

Pro h s. 17-45146

424 CHAPTER 17 PLANAR K[NETfCS OF A RICID BODY: FORCE AND ACCELERAT1ON

17-47. The snowrnobile has a weight of 250 lb, centered at G1, while the fider has a weight of 150 lb. centered at G2. If the accełeration is a = 20 ft/s2, determine the maximum height h of G, of the rider so that the snowmobiie's frant skid does not lift off the ground. A150, what are the traction (horizontal) {orce and normai reaction ander the rear tracks at A?

*17-48. The snowmobile lias a weight of 250 lb. centered at Gl, winie the rider lias a weight of 150 lb. =tel-cel at G,. If h = 3 ft. determine the snowmobile`s maximum permissible acceleration a 50 that its front skid does not lift off the ground. Also, find the traction (korkowa') force and the norma] reaction undcr the rear tracks at A.

0.5 ft

Prohs. 17-47148

17-49. If the cart's mass is 30 kg and it is subjected to a horizontal force of P = 90 N. determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC.

17-511 If the carf's mass is 30 kg, determine the horizontal lorce P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC lias a mass of 15 kg.

Probs. 17-49150

1.7-51. The pipe has a mass of 800 kg and is being towed behind the truck. If the acceIeration of the truck is a, = 0.5 m/sł, determine the tingle O and the tension in the cable. The coefficient of kinetic friction hełween the pipe and the ground is 1uk = 0.1.

'17-52. The pipe hasa mass of 800 kg and is being towedbehind a truck. If the angle O determine theacceleration of thc truck and the tension in the cable. The coefficient of kinetic triction between the pipe and the ground is Juk = 0.1.

a ,

Probs. 17-51152

1753. The arched pipe hasa mass of 80 kg and rests on the surface of the platform.As it is hoisted from one level tothe nex-t. a 3).25 radb?" and w = 0.5 radfs at the instantO = 30r. 1f it does not slip, determine the normal reactions of the arch (m the platform at ibis instant.

17-54. The arched pipe has a mass of 80 kg and rests on the surface, of thc platform for which the coefficient of sta.ticfriction ;s p, = Determine the greatcst angularacceleration rr of the platform. starting from rest when O = 45°, whhout causing the pipe to slip on the platform.

Prabs. 17-53154

17.4 EQUATIONS QF MOTION: ROTAllON ABOUT A FIXED Axis 42S

17-55. At the instant shown. link CD rołates wich an angular vełoeity of wc]) = 8 radi:, If link CD is subjected W a cnuple moment of W = 650 lb. ft, deterrnine the forte developed in link AB and the angular acceleration of the links at this instant. Neglect the weight of the links and the platform, The erate weighs 150 Ib and is fully seeureei on the platform.

Dełerm/ne the forte developed in the links and the acceleration cif the bars mass center immediateły after the cord fails. Neglect the mass of IM ks AB and CD. The uniform bar bas a mass of 20 kg.

— 3

0.6 m

Pro b. 17-55Proli. 17-56

17.4 Equations of Motion: Rotation about a Fixed Axis

Consider the rigid body (or slab) shown in Fig. 17-14a, which is constrained to rotate in the vertical plane about a fixed axis perpcndicular to the page and passing łhrough the pin at O. The angular vełocity and angular acceleration are caused by the external force and couple moment system acting on the body. Because the bodys center of mass G moves around a circular path, the acceleration cif this point is best represented by its tangential and norrnal cornponents, The tangential carnponent of acceleration has a magniffide of (ac), = arc and must act in a direction which is wrisisteni with the body's angular acceleration The magnitude of the norma! eamponent xfAttticration is [at;)„ =This component is always directed from point G to O. regardless cif the rotational sense of

(a)

Fig. 17-14

426 CHAPTER 17 P'LANAR </NETICS OF A IR 'GID BODY: FORCE AND ACCELERATiON

( a )

The free-body and kinetic diagrams for the body are shown in Fig. 17-14b. The two components m(a6), and m(aG)„, shown on the kinetic diagram, are associated with the tangential and normal components of acceleratian of the body's mass center, The i`c, a vector acts in the same cip'ectkm as a and hal a magnirude (.)f „Ga, where irG is the body's moment of inertia calculated about an axis which is perpendicular to the page and passes through G. From the derivation given in Sec. 17,2, the cquations of mot ion which apply to the body can be written in the form

1-F„ = m(ao)„ = mw2rG1F, m(aG), = marc, (17-14)

"X141G -= Ga

1 F

r 4 1 . —

The moment equation can be replaced by a moment summation about any arbitrary point P on ar off the body provided one accounts for themomcnts produced by /c, a, m(aG)„ and Jrf(aG),, about the point,

Moment Equation About Point O. Often it is convenient to sum momenty about the pin at O in order to elirninate the unknown forte Fo. From the kinetic diagram, Fig. 17-14b, this requires

C +1Mo = 2(.04)0; SM0 = rGm(au), ± to« (17-15)

Nate that the moment of ni(ac,)„ is not included here since the line of action of this vector passes through O. Substituting (aG), = raa. we mayrewrite the above equation as C (irc + mi-2G)a. From the paraIlel-axis theorem, fG = IG + rild2, and therefore the term in parentheses represents the moment of inertia of the body abotił the fixed fixis of rotation passing through O." Consequendy, we can write the three equations of motion for the body as

.P.'„ = m(aG)„ = nuo2rG

.5F, = Jiz(aG), = marG (17-16)

Wo --- locić

IG.r When using these equations, remember that "fija" accounts for thex

m(26)+ "moment" of botki in(ac), and IG a about point O. Fig. 17-14b. In otherwords, 2-Ado = 1.(.44t)0 = foa, as indicated by Eqs, 17-15 and 17-16,

n z

( b )*The. rvsult 1..M0 = ioco can also bc oblained direcily horn Ey. 17-6 by selteting

Fig. 17-14 (cont.) point P to coincide with O. realizing lhal Cap), = (ar),. = 0.

17.4 EQUATIONS OF MOTION: ROTAllON ABOUT A FIXED Axis 427

• Procedure for Analysis

Kineric problems which involve the rotation of a body about a fixed axis can be solved using the following procedure.

Free-Body Diagram.

• Estabiish the inertial n, r coordinate system and specify the direction and sense of the acccIerations (aG),, and (ac,), and the angular acccIeration a of the body. Recall that (aG), m ust act in a direction which is in accordance with the rotational sense of a, whereas (aG), always acts toward the axis of rotation, point O.

• Draw the free-body diagram to account for all the external forces and couple mornents that act on the body.

• Detcrrninc the moment of inertia 1G or J.

• Idenfify the unknowns in the problem.

• I f i t i s d e c i d e d t h a t t h e r o t a t i o n a l e q t t a t i o n o f m u f l o nLW p 5.(44)F, is to be uscd, i.e., P is a point other thau G or 0,then consider drawing the kinetic diagram in order to help -visualize" the "trioments" developed by the components n (aG),. m(aG),, and fc,« when writing the terms for the moment sum

(Atóp

Equations of Mot ion.

• Apply the three equations of motion in aceordance with the established sign convention.

• If mon-lents are surnmed about the body's mass center, G, thcn SMG = l

Ga, since (mac), and (rade;)„ create no moment about G.

• If moments are surnmed about the pin support O on the axis of rotation, then (niaG)„ creates no moment about 0, and it can be shown that ,kfty = 1oa.

Kinematics.

• Ust kincmatics if a complete solution cannot be obtained strict ly from the cquations of motion.

• If the angular acceleration is variable, use

dwa = — a £ 1 6 = w d w w =drdt

• If the angular acceleration is constant, use ar

= u,ty + 11,1

O = Oo wot -a tt2 = 0.1

2) + 2a,(0 — Oo)

I I

the crank an the oil-pumping rig undergoes rotation about a fixed axis which is eaused bya driving torque M of the loadingsshown on the free-body diagram cause the effects shown on Ihe kinetic diagram, lf rnaments are =med about the mass center, G, Ulen SMG = fory_ However, if rnornems are summed about point O. noting that

(ac,), = dcl, then = /6-a++11(AG)rd m(aG)„,(0) = + trid)a = 100,

428 CHARTER 17 PLANAR KrNETICS OF A RICID BODY: FORCE AND ACCELERAT1ON

EXAMPLE

The unbalaneed 50-lb flywhee] shown in Fig. 17-15a has a radius of gyration of kG =- 0.6 ft about an axis passing through its mass center G. 1f it is released from rest, detetmine the horizontal and vertical components of rcaction at the pili O.

SOLUTION

Free-Body and Kinetic Diagrarns. Sinct:,. G ~ves in a circular path. it will have both norrnal and tangential components of acecieration. Also, since a, which is causcd by the tlywhecrs weight, acts clockwise, the tangential component of accelcration must act downward. Why? Since w = 0, only,i2(a(;), = marc; and ket are shown on the kinematie diagram in Fig. 17-15b. Here, the moment of inertia about G is

„G = mk2G = (50lb/32.2 ft/s2)(0.6 ft) ' - = 0.559

slug • ft2 The three unknowns are O. 0,, and ot. Equations

ni tActlon.

= orco2rr; 0,, = 050 lb Atu,

+ marli; + 501b = ,)(a)(0.5 ft) (1)\, 32.2 ft/s-

C +./1,/G = fax; 0,(0.5 ft) = (0.5590 sług • ft2)aO 5 ft ")\% t

- 4 - 101- G Solving,

0„ cr =. 26.4 rad/s2 0, = 29.5 lb/

Moments can also be summed about point O in order to el im inateAO'',.()' i0 lb

and O, and thereby obtain a dired solution for a, Fig. 17-15b. This can be done in one of rwo ways.

- ica

O G )

50 Ib(0.5 fi) = 0.9472a (2)1f 5,140 = Iwa is applied. then by the parallel-axis theorem themarc V moment of inertia of the flywheel about O is

= IG + =. 0.559 + 5° )(0.5)2 =x.9472 sług • ft232.2

(b)ficnco,

Fig. 17-15 +2.Mo = /0«: (50 lb)(0.5 ft) = (0.9472 sług • ft'--)re

which is the same as Eq. 2. Solving for a and substituting into Eq. 1 yields the answer for O, obtained previously.

j r

(_,14),(3;

(501b)(0.5 ft) = (0.5590 slug • ft2)a + [( 50 ib 03.5 ft) (0.5 ft)

32.2 ft./52

17.4 EQUATIONS OF MOTION: ROTATION ABOUT A FIXED Axis429

EXAMPLE 17.10

At the instant shown in Fig. 17-16a, the 20-kg slender rod has an angular velocity of = 5 rad/s. Determine the angular acceleration and the horizontal and vertical components of reaction of the pin on the rod at this instant.to

60 N - mf = 5 rad/s

3 _

(a)

SOLUT1ON

Free-Body and Kinetic Diagrarns. Fig. 17-16b. As shown on the kinetic diagram, point G moves art-y(1nd a circular path and so it has two components of acceleration. it is important that the tangential component a, = ar( act downward since it must be in accordance with the rotational sense of a.The three unknowns are 0„, O„ and a.

Equation of Motion.

SF„ = mwrG; 0„ = (20 kg)(5 rad/s)2(1.5 m)

60 N • m

— 1 . 5 m -

i i

G

20(9S1) N

muirG

= mara;C +1' MG = IGa;

Solving

+ 20(9.81)N = (20 kg)(a)(1.5 m) 0,(1.5

m) + 60 N • m = [71-2,(20 kg)(3 m)2 i a

O, = 750 N 0, = 19.05 N a = 5.90 rad/s2

A more direct solution to this problem would be to sum moments about point O to eliminate 0„ and 0, and obtain a direct solution for a. Here,

C +SM° = (.«k)0; 60 N • m + 20(9.81) N(1.5 m) =

[1(20 kg)(3 m)21 a + [20 kg(a)(1.5 in)[(1.5 ni)

= 5.90 rad/s2

Also, since Io =3m12 for a slender rod, we can apply

C +5M0 = ł oa; 60 N • m + 20(9.81) N(1.5 m) = B(20 kg)(3 m)21 a

a = 5.90 rad/s' Ans.

NOTE: By comparison, the last equation provides the simplest solution for a and does not require use of the kinetic diagram.

(b)

Fig. 17-16

430CHAPTER 17 PLANAR i<NETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON

The Brum shown in Fig. 17-17a has a mass of 60 kg and a radius of gyration ko = 0.25 m. A cord of negligible mass is wrapped around the periphery of the drum and attached to a block having a mass of 20 kg. 1f the błock is released, determine the drum's angular acceleration.

SOLUTION IFree-Body Diagram. Here we will consider the drutu and block separateiy, Fig. I 7-17b..Assuming the, block accelerates downward at a. it creates a counterelockwise angular acceleration a of the drum. The moment of inertia d the drum is

1.0 = onk20 (60 kg)(0.25 m)2 = 3.75 kg • m2

There are five unknowns, namely 0,, O y, T, a, and a.

Equations of Motion. App/ying the t ranslat ional equat ions ofmotion QFa = m(aa), and = m(aG)). to the drum is of noconsequence to the solution, since these equations involve the unknowns O, and O. Thus, for the drum and błock, respectively,

Cr -1-111f0 = ioa;T(0.4= (3.75 kg • m2)a(1)

+I 1.F„ = m(aG},; —20(9.81)N + T = —(20 kg)a (2)

Kinematics. Since the point of contact A between the cord and drum has a tangential component of acceleration a, Fig. 17-17a, then

+a = ar; a = a(0.4 rri) (3)

Solving the above equations.

T = 106 N a = 4.52 m/s2

11,3 radis2" Ans.SOLUTION [1

Free-Body and Kinetic Diagrams. The cable tension T can be eliminated from the analysis by considering the drum and block as a single system, Fig. 17-17c. file kinctic diagram is shown sincc moments will be summed about point O.

Equations of Motion. Using Eq. 3 and applying the moment equation about O ta eliminate the unknowns 0, and 0,, we have

C +.1v/0 = 120(9,81)Ni (0.4 m) =(3.75 kg • m2)a [20 kg(a 0.4 ni)](0.4 m)

= I 1.3 rad/52

NOTE: If the block were rernoved and a force of 20(9.81) N were applied lo the card, show that a = 20.9 rad/s2. This value is larger since the błock has an inertia, or resistance to acceleration.

( a }

60 (9.81) N

20{9.8t} N

( h )

20(9_81) N (20 kg)a

(c)

Fig. 17-17

17.4 EQUATl4N5 GF MOTFON: ROTATION ABOUT A FIXED Axis 431

EXAMPLE

The slender rod shown in Fig. 17-18a has a mass m and length i and isreleased from rest when O = Deterrnine the horizontal and verticalcomponents of force which the pin at A exerts on the rod at the instant O = 9(1°.

SOLUTION

Free-Body Diagram. The free-body diagram for the rod in the general position B is shown in Fig. 17-18b. For convenience, the force components at A a re shown acting in the n and t directions, Note that a acts cłockwise and so laa)r acts in the +ł direction,

The moment of inertia of the rod about point A is ł Ą

Equations of Motion. Moments wi1[ be summed about A in order to eliminate A„ and A,.

+1". SF„ = inco2rG; A„ — mg sin B = inw-1(172), (1)

mara; A, + mg cos O = ma(1/2) (2)

= iAa; mg cos 1/2) = 0.m12-)a (3)

Kinematics. For a given anglc O there are four unknowns in thc above three equations: A A„ w, and a. As shown by Eq. 3, a is nar con,słanr; rather, it depends on the position O of thc rod.The necessary fourth equation is obtained using kinematics. where a and w can be rełated to O by the equation

(C -F) w dw = a di9 (4)

Note that the postive, clockwise direction for this equation agrees with that ofEq,3.This is important since we arc seeking a simulłaneous saIution.

In order to solvc for w at 8 = 90°, climinate a from Eqs. 3 and 4, which yields

w dw = (1.5g i l) cos O dO

Since w = O at O = we have9fr

E r o = ( I . 5 g 1 1 ) i c o s O d 8

o°= 3g/i

wz Substituting this value into Eq. 1 with O = 90° and solving Eqs. 1 to 3 yields

a = O

A, = O A, = 2.5,ng Ans.

NOTE: If 2MI = £(...14)A is used, one musi a.ecounł for the moments of IGOL and m(aG), about A.

( a )

432 CHARTER 17 PLANAR KfNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON• FUNDANIENTAL PROBLEMS

F17-7, The 100-kg wheel has a radius of gyration about its center O of kr, = 500 mm. 1f the wheel starts from rent, determine its angular velocity in i = 3 s.

F 1 7 - 7

F17-8. The 50-kg disk is subjected to the couple moment of M = (9.0 Ni • m, where I is in seconds. Dełermme the angular velocity of the disk when i = 4 s starting front rest,

(9t)N-m

F 1 7 - 8

F17-9. At the instant shown, the uniform 30-kg slender rod has a totntercłockwise angular velocity of w = 6 rad/s. Determine the tangential and normai components of reaction of pni O on the rod and the angular acceleration of thc rod at this instant,

-- 0.3 m 0.6m

Arch_

° 7tą = 60 N•in

F 1 7 - 9

F17-10. At the instant shown, the 30-kg disk hasCOUT1terclockwise angolar velocity of lOradfs.Determine Lite tangential and norma' components of reaction of the pin O on the disk and the angular atce1eratian of the disk at this instant.

F17-10F17-11. The uniform slender rod has a mass of 15 kg. Determive the borizontal and vertical components of reaction at the pin O. and the angular acceleration of the rod just after the cord is cut.

0.6 m 0.3 inF17-11

F17-12. The uniform 30-kg slender rod is bcing pulled by the cord Chat passes over the smalł smooth peg at A. ii the rod has a counterclockwise angular vełocily ol w = 6 radis at the instant shown, cletermine the tangential and norma components vf reaction at the pin O and thc angular acceleration of the rod.

0.6 nr—F17-12

03 m-

17.4 EQUATIONIS OF MOTION: ROTATION Al;s0UT A FIXFP AXIS433• PROBLEMS

17-57. The 10-kg wheel lias a radius ()f gyration kA = 200 mm, If thgi wheel is subjected to a moment M -= (5t) ls m. where r is in seconds, determine łrs angular velocity when r = 3 s starting from rest. Also, col npute the reactions which the fixed pin A exerts on thc sylwo' during the rnotion.

17-58. The 80-kg disk is supported by a pin at A. If ii is released from rest from the position shown, determine the initial horizontal and vertieal components of reaction at the pin.

Prob. 17-5N

li-59, The uniform slender rod has a mass ni, 1f it isrełeased from rest when O = deterrnine the magnitudeof the reactive force exerted on it by pin B when O = 90°.

Prob, 17-59

*17-60. The drutu has a weight of 80 lb and a radius of gyration kc) = 0.4 ft, If the eable, which is wrapped around the drutu. is subjected to a vertical force P = 15 lb, deterrnine the trute needed to increase the drum's angular velocity from 'roi = 5 rad/a to w, = 25 rad/a. Ne.glect the mass of the cable.

Prob. 17-60

17-61. Cabfe is unwound from a spool supported on smalt rollers at A and B by exerting a lorce of 7' = 300 N on the cable in the direction shown. Compute the tlnie needed to unravel 5 m of cable from thc spool if the spooI and eable have a total mass of 600 kg and a oentroidal radius of gyration of k, = 1.2 in. For the calculation. neglect the mass of the eable being unwound and the mass of the rollers at A and R. The rollers turn with no frietion,

Prob. 17-57

T = 300 N

30̀ :

Prob. 17-61

434 CHARTER 17 KANAR Kr NETrCS OF A RICID BODY: FORCE AND ACCELERATION

17-61 The 10-Ih bar is piimed at its center O and connected to a torsional spring. The spring has a stiffness k = 5 Ib • ft/rad, su that the torque developed is M = (50'ib • ft. where O is in radians. If the bar is released from rest when śt is vertical at O = - 90% determine its angular ve'ocity ał the #nstant 9 =

17-63. The 10-lb bar is pinned al its center O and connected to a torsional spring. The spring has a stiffness k z-- 5 lb• ftfrad, so that the torque developed isM = (50) łbwhere H is in radians. If the bar is releasedfrom rest when it is yertiea at O = 90°, determine its angular wiochy ał the instant O = 45°.

17-65. Determine the verticai and horizontaI components of reaclion al the pin support A and the angular acceleration of the 12-kg rod at the instant shown, when the rod has an angular vcioeity of rs = 5 rad/s,

0.6 m

Bw --- 5 rad/s-

Probs, 17-62163

''17-44. If shaft BCis subjected to a [orli ue of M = (0.45ti /2) N - m. where t is in seconds, determine the angolar velocity of the 3-kg rod AB when r = 4 s. starłing from rest. Neglect the mass of shaft

Prob. 17-65

17-66. The kinetic diagram representing the genera] rotational ruotion of a rigid body about a fixed axis passing through O is shown śn the figure, Show that /G& may be eliminated by movirig the veetors rn(a0}, and rn(aG)„ to point P,located a distance rep = r„G from the center of mass Gof the body. Here kG represents the radius of gyration of the body about an axis passing through G. The point P is called the center of percussion of the body

"Prob. 17-64 Prob. 17-66

17.4 EQUATION5 QF MOTION: ROTAllON ABOUT A FIXEd Axis 435

17-67. Determine the position rp of the center of percussion P of the I0-lb slencler bar, (See Prob, 17-66.) Whal is the horizontal component of force that the pin at A exerts on che bar when h is struck at P whki a force of F .= 201h?

Prob. 17-67

*17-68. The disk bas a mass M and a radius R. If a block of mass m is attached to the cord, determine the angular acceleration of the disk when the block is released from rest. Also, what is the vełocity of che block after it falls u distance 2R starting from rest?

Prob, 17-6/1

17-6. The door will close automatically using torsional springs mounted on the hinges. Each spring hasa stiffness k = 50 N • miracl so that the torque on cach hinge is M =, (500) N m, where U is measured in radians. If the dooris released from rest when it is opon at O =, determineits angular velocity at the instant łt = 0°. For the calculation, treat ihe door as a thin piate having a mass of 70 kg.

17-711. The door will close aut omaticaily using torsional springs mounted on the hinges, If the, torque on each hinge is M = kO, where O is measured in radians, deterrnine the required torsional stiffness k so that the door will close <0 = 0°) with an angular velociły w 2 radfs when it is released from rest at O = 90°. For the calculation. treat the door as a thin plate having a mass of 70 kg.

Probs. 17-69170

17-71. The pendulum consists of a 10-kg uniform slcnder rod and a 15-kg sphere. If the pendulum is subjected to a torque of M=50 N • m, and has an angular velocity of 3 radis when O = 45°. detennine the magnitude of the reacłive force pip O exerts on the pendulum at this instant.

Prob. 17-71

436 C H A P T E R 17 K A N A R K r N E T r C S O F A R I C I D B O D Y : F O R C E A N D A C C E L E R A T 1 O N

*17-72. The disk lias a mass of 20 kg and is originally spinning at the end of the strot with an angolar veloci ty of w = 60 rad/s. If it is then placed against the wali. for whiehthe coefficient of kinetic friction is = 0.3 determine thefime required for the morion to stop. What is the (orce tn sirut BC duriTig this linie?

1.7-14. The 5-kg cylinder is inftiaIly at rest when it is placed in contact with the wali B and the rotor at A. II the rotor always maintains a constant clockwise angolar velocity w = 6 rad/s. determine the initial angular acceieration of the cylinder_ The eoefficient of ltffnctie friction at the contacfing surfaces f1 and C śs µk = 0.2.

Prob. 17-74

Prob. 17-72

17-73. The slender rod of length L and mass m is released from rest when U = T. Determine as a function of O the normai and the frictional forces which are exerted by the ledge on ihe rod at A as it falis downward. At what angle O does the rod begin to slip if the coefficient of słatie ffletion at A isp?

Prob. 17-73

17-75. The wheeI has a mass of 25 kg and a radius of gyration kK = 0, I5 m. It is originally spinning at w l -= 40 radfs. If it is placed on the gTound, for which thecoefficient af kinetie frietion is = 0.5. determine thedme required for the motion to stop. What are the horizonłal and vcrticaI componcnts of reaction which the pin at A exerts on AR during this time? Neglect the mass of Aft,

0 . 4 i n - 1

}.2 in

Prob. 17-75

17.4 EctuATioNs OF MOTICN: ROTAllON ABOUT A FiXED Ams 437

.117-76. A 40-kg boy sits on top of the large wheel which hasa mass cif 400 kg and a radius of gyration kc = 5.5 m. Ii the boy essentially stans from rest at H = 0°, and the wheel begins to rotałe freety,determinc the anglc at which thc boy hegins to slip. The coefficient af static friction between the wheel and the boy is µ,„ = 0.5. Neglect the size of the boy in thc ealeulation.

17-78. Błock A bas a mass m and rests on a surface having a coefficierd of kirietic friction pk, The cord attached to A passes over a pulley at C and is attached lo a block B having a mass 2m.If B is released, determine the acceleration of A. Assume that the card does not slip over the pulley. The pulley can be approximated as a thin disk of radius r and mass 4m. Neglect the mass of the cord.

Prob. 17-78

Prob. 17-76

17-77. Gears A and .8 have a mass of 50 kg and 15 kg, respectivety. Their radli of gyration about their respective centers of mass aro ke = 250 mm and icD = 150 mm. 1f a torque of M = 200(1 — N • m, where is in seconds, isapplied to gear A, determine the angular velocity of both pars when t=3 s, starłing from rest.

17-79. The two blocks A and B have a mass of 5 kg and 17 10 kg. respectiveIy. 1f the pulley can be treated as a disk of masa 3 kg and radius 0.15 m, deterrnine the acceleration of block A. Neglect the masa of the cord and any slipping on the pulley.

*17-80. The two Nocka A and B have a mass mA and ma, respectiveły, where mB mA. 1f the pulley can be treated as a disk of masa M. determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley,

Probs. 17-79/80

M=200(1 — e'.2ż)N•rn

B

Prob. 17-77

438CHAPTER 17 PLANAR KrNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON

17-8L Determinc the angular acceleration of the 25-kg diving board and the horizonlal and vertical components of reaction at the pin A the instant the uran junips off. Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum arnount of 200 mm, w = 0, and the board is horizontal. Take k = 71cNjm.

17-83. The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass and length I.

Prob. 17-83

Prob. 17-81

17-82, The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontaI it lias ara angular velocity of 15 rad/s and a cłockwise angular accekralion of 8 rad/s-. Determine the interna norma forte, shear furce, and moment at a seetion through A. Assumc the rotor is a 50-m-Iong Slender rod. having a mass of 3 kg/m.

417-84. The armature (sIender rod) AB hasa mass of 0.2 kg and can pivot Elbom the pin at A. Movement is controlled by the cIcctromagnet E, which exerts a horizonłal attractive tarce on the armature at B of Fs = (0,2( ł 0-3>F2) N, where l in meters is the gap between the armaiure and the magnet at any instant. If the armaturo lies in the horizontal piane, and is origMaily at rest, detcrmine thc speed of the contaei al B the instant = 0.01 ni. Originally 1 = 0.02 m.

Prob, 17-82 Prob. 17-84

4 3 917.4 EQUATIONS OF MOTICN: ROTATION ABOUT A FIXED Axis

17-85. The bar has a weight per length of w. If it is rotating in the vertical piane at a constant rate w about point O, determine the internat normaf furce, shear furce, and moment as a function of x and O,

Prob. 17-85

17-86. A lorce = 2 Ib is appTied perpendicular to the axis of the 5-lb rod and moves from O to A at a constantrata of 4 ft/s. If the rod is at rest when O = and Fis at Owhen r = 0, determine the rod's angular velocity at the Instant the (arce is at A. Through what angle has the rod rotated when this occurs? The rod rotates in the horizontal piane.

Prob. 17-86

17-87 The 3-kg block A and 20-kg cylinder R aTC connectecl by a light cord that passes over a 5-kg pulley (disk). If the system is released from rest, determine the eylinder's velocity after its has traveled downwards 2 m. Neglect friction hetween the piane and the block, and assume the cord does nut slip over the pulley.

*17-88. The 15-kg block A and 20-kg cylinder E are connected by a light cord that passes over a 5-kg pulley (disk). 1f the system is released from rest, determine the cylinder's velocity atter its bas traveleci downwards 2 m. The coefficient of kinetic friction between the block and the hori7ontal piane is juk = 03. Assume the cord does not slip over the pulley.

A

Probs. 17-87188

17-89. The "Catherine w heel" is a firework that consists of a coiled lube of powder which is pinned at its center. If the powder burns at a constant rate of 20 gis such as that the exhaust gases always exert a fotce having a constant magni lude of 0.3 N. directed 113.11gęlit to the witce!, deterrnine the angular velocity of the wheel when 75% of the mass is burncd off. Initialiy, the whccl is at rest and has a mass of ł 00 g and a radius of r = 75 mm. For the calculation.consider the wheei to aiways be a thin disk.

Prob. 17-89

440 CHAPTER 17 PLANAR </NETICS OF A RIGID BODY: FORCE AND ACCELERATiON

( b )

Fig. 17-19

Fig. 17-20

17.5 Equations of Motion: General Piane Motion

The rigid body (nr slab) shown in Fig.17-19a is suhjected to general piane rriotion caused by the externally appiied furce and couple-moment system. The free-body and kinetic diagrams for the body aro shown in Fig.17-19b. If an x and y inertial coordinate system is established as shown, the three equations of motion ara

£F,- in(aG),= irk(aG)y (17-17)

SNIG. = Ga

In some problems it may be conveniont to sum moments about a point P otlier than G in order to eliminatc as marsy unknown forces as possible from the moment sumniation. Whcn used in this more generał case. the three equations of motion arc

in(aG),1F, = 3n(aG)7 (17-18)

łA1p = 2(Jtit1p

Here £(..41.k)p represents the moment sum of arca and maG (or its componcnts) about P as determined by the data on the kinetic diagram.

Moment Equation About the IC. There is a particuIar type of prohieni that involves a uniform disk, nr body of circular shape, that mus on a rough surface withaur slipping, Fig 17-20. If we sum the moments about the instantaneous center of zero velocity, then (.44).[c. becomes fica, so that

Mrc = Irca(17-19)

This result compares with iMo = ł oa, which is used for a body pinned at point O, Eq. 17-16. See Prob. 17-90.

17.5 EQUATIONS OF MOTtON: GEKJERAL PIANE MCinON441

H

As the snil =pelot-. or "sheep's funt miler" moves forward. the ratler bas generał piane motion. The forces shown bn its free-body diagram cause the effects.shown on the kinezie diagram. II mornents are sumrned about ihe mass center, G. then 111.G = 1GO. However. if moments are summed about point A {ihe /C)

then C 1-'5A/4 = fccitmar.)d = (4«.


Kinezie problems invołying generał piane motion of a rigid body can be solyed using the following procedure.

Free-Body Diagram.•Establish the x, y inertiaI coordinate system and draw the freebody

diagram for the body.•Specify the direction and sense cif the acceleration of the mass

center, aG , and the angolar acceleration a of the body.•Detennine the moment of inertia P.

• ldentify the unknowns in the problem.

• If it is decided that the rotational equation of motion Nip = 2:(44..op is to be used, then consider drawing the kinetie diagram in

order to help"yisualize" the "moments" deyeloped by the components m(ac),, in(aG),, and irca when writing the terma in

the moment sum i(J/Lk)p,


• Appły the three equations of motion in aceordance with the established sign conyention.

•When friction is present, there is the possibility for motion with no slipping ar tipping. Each possibility for motion should be considered.

Kinernatics.•Use kinematics if a complete solution cannot be obtained strictly from the

equations of motion.•If the bodys motion is constrained duc to its supports. additional

cquations may be obtained by using aB = aA a$/A, which relates the acceierations of any twa paints A and B on the body.

• When a whcci, disk. cylinder, or bali rolls withouł slipping, then aG =

1 7

442 CHAPTER 17 PLANAR KfNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON

I M N iv r'+) ac = ar; = cr (0.5 rri) (3)

-ł a= 10.3 radis2

= 5.16 mis2 T

= 19.8 N

EXAMPLE

Determine the angolar acceleration of the spool in Fig. 17-21a. The spool has a mass of 8 kg and a radius of gyration of kG =- 0.35 m. The cords of negligible mass arc wrapped around its inner hub and ower rim.

SOLUTION i

Free-Body Diagram. Fig, 17-21 b. The 100-N (orce causes ac to act upward. Also, a acts clockwise, since the spool w-inds around the cord at A.

There arc threc unknowns T, au, and a. The moment of inertia of the spool about its mass center is

/c = ink?; = 8 kg(0.35 m)2 = 0.980 kg • m2


+ fiF,, = In(aG)y; T + OO N - 78.48 N = (8 kg)ac (1)

C +1,MG IGa; 100 N(0.2 m) - T(0.5 m) = (0.980 kg • m2)a (2)

Kinematics. A complete solution is obtained if kinernatics is used to x rclatc ac, to a. In this case thc spool "rons without slipping" on the cord at A. Hence, we can usc the results of Example 16.4 or 16.15 so that,

Solving Eqs. 1 to 3, we have

SOLUTION 11

Equations of Motion. We can eliminate the unknown T by summing moments about point A. From the free-body and kinetic diagrams Figs, 17-21 b and 17-21 c, we have

100 N(0.7 m) - 78.48 N(0,5 ni)

= (0,980 kg • m2)a [(8 kg)ac1(0.5

Using Eq. (3).

= 103 rad/s2 An.+.

SOLUTION III

Equations of Motion. The simplesł way to solve this problem is to realize that point A is the IC for the spool. Then Eq. 17-19 applies.

= (100 N)(0.7 m) - (7848 N)(0.5 m)

= [0.980 kg • ni2 + (8 kg)(0.5 m}'-]a

a =- 10.3 rad/s2

C +xmA 2.-(Atk)A

100 N

(a)

(b)

(8 kg) a r ,

(c)

Fig. 17-21

17.5 EQUATIONS OF MOTION: GENERAL PLANE MOTION443

EXAMPLE 17.14The 50-lb wheel shown in Fig. 17-22 has a radius of gyration kG = 0.70 ft. If a 35-1b • ft couple moment is applied to the wheel, determine the acceleration of its mass center G. The coefficients of static and kinetic friction between the wheel and the piane at A are

= 0.3 and lik = 0.25, respectively.

35 lb. ft1 . 2 5 f t

50 lb

„G = = 50 lb , (0.70 f02 = 0.7609 slug • ft2

32.2 ft/s-

The unknowns are NA, FA, 0G, and a. G1.25 ft


= in(0G)r;

= m(aG)y; =

IGa;

( 5 0 1 b )FA i

32.2 ft/5- aG

NA - 50 lb = O

35 lb • ft - L25 ft(FA) = (0.7609 slug ft2)a

(1

)

(2

)

(3

)

N , ,

( b )

Fig. 17-22

A fourth equation is needed for a complete solution.

Kinematics (No Slipping). If this assumption is made, then+) aG = (1.25 ft)a (4)

Solving Eqs. 1 to 4,

NA = 50.0 Ib FA = 21.3 lb

a = 11.0 rad/s2 aG = 13.7 ft/s2

This solution requires that no slipping occurs, i.e., FA p.,NA

However, since 21.3 lb > 0.3(50 lb) = 15 lb, the wheel slips as it rolls.

(Slipping). Equation 4 is not valid, and so FA = ,ukNA, orFA = 0.25NA (5)

Solving Eqs. 1 to 3 and 5 yields

NA = 50.0 lb FA = 123 lba = 25.5 rad/s-2

aG = 8.05 ft/s2

SO LUTION

Free-Body Diagram. By inspection of Fig. 17-22b, it is scen that the couple moment causes the wheel to have a clockwise angular acceleration of a. As a result, the acceleration of the mass center. ac , is directed to the right. The moment of inertia is

35 1b-fta G

444 CHARTER 17 PI-ANAR 1CSNETtCS OF A RICID BODY: FORCE AND ACCELERAT1ON

The uniform slender pole shown in Fig. 17-23a has a mass of 100 kg. If the coefficients of static and kinetic friction hetween the end of the pole and the surface ara p, = 0.3, and j..tk = 0.25, respectively, determine the pole's arigular acceleration at the instant the 400-N horizontal (orce is applied.Thc pole is originally at rest.

S O L1J TI ON

Free-Body Diagram. Figure 17-23b,The path of motion of the mass center G wili be along an unknown curvcd path having a radius of curvature p, which is initially on a vertical lino. However, therc is no normal or y component of acceleration since the pole is originally atrest, i.e., vG = 0, so that (ac-,), = p = 0. We will assume the masscenter accelerates to the right and that the pole has a clockwise angular

9g1

uG acceleration of a. The unknowns ara ,ac, and a.

(1 )

(2 )

(3 )

x

400 Ti

Equation of Motion.

SF, = in(aG),; 400 N - FA = (100 kg)a6

+ T SP>, = m(aG)y;NA — 981 N = O

+21V/G = ka; FA(1.5 m) - (400 N)(1 m) =- 1+2(100 kg)(3 rn)I'la

I.5m

F.1

NAA fourth cquation is needed for a complete solution.

(b) Kinematics (No Slipping). With this assumption, point A acts as aFig. 17-23 "pivot" so that a is clockwise, then ac is directed to the right.

ar; = urm?: ac.; = (1,5 m} a (4)

Solving Eqs. 1 to 4 yields

NA = 981 N FA = 300 N

aG = 1 mis2 a = 0.667 rad/s2

The assumption of na slipping requires F, However,300 N 03(981 N) = 294 N and so the pole slips at A.

(Slipping). For this case Eq. 4 does nor apply. Instead the frictional equation FA = g*NA musi be used. Hence,

FA = 0.25NA (5)

Solving Eqs. ł to 3 and 5 simultaneously yieldsNA = 981 N FA = 245 N aG = 1.55 ni/.2 = -0.

428 raclis2 = 0.428 raclis2.5

40-0 N.5 m

(a)

17.5 EQUATIONS OF MOTION: GENERAL PLANE MOTION445

EXAMPLE

17.16

The uniform 50-kg bar in Fig.17-24a is heir, in the eqtrilibrium position by cords AC and RD. Determine the tension in RD and the angular acccicration of the bar immccliatcly after AC is cut.

SOLUTIONFree-Body Diagram. Fig, 17-24b. There are four unknowns. TE, (aG),, (aG),, and a.

( a )

Equations of Motion. = ni(aG),.; 0 = 50 kg (aG),

(c1c)x = O

+ T m(ac)y; T8 - 50(9.81)N = —50 kg (a,c)y(1)

1 IGa; TB(1.5 m) = 'm) (2)

.50(9.81) N

GB

1 . 5 m

Kinematics. Since the bar is at rest just after the cable is cut, then its(b)

angular velocity and the velocity of point B at this instant are equal to zero. Thus (as)„ = ViipRD = O. Therefore, a5 oniy has a tangential component, which is directed along the x axis, Fig. 17-24c. Applying the relative acceleration equation to points G and B,

aG = a8 + a X r GIB — 012ro8

-(aG)j = asi + (ak) X (-1.51) — OG (aG), = 01,3m

-(aG)j = agi - 1.5aj rGCI I i .5 in

Equating the i and j components of both sides of this equation,(c)

O = as Fig. 17-24

(aG), = 1.5a(3)

Solving Eqs. (1) through (3) yields

a = 4.905 rad/s?Ans.TE = 123 NAns.

(ac), = 7.36 m/s2

446 C H A R T E R 17 P L A N A R I C N E T f C . S O F A R I C I D B O D Y : F O R C E A N D A C C E L E R A T 1 O N


F17-13. The uniform 60-kg siender har is initially at rest on a smooth horizontal piane when the forces are applied. Determine the acceleration of the bar's mass center and the angular acceleration of the har at this instant,

F17-13

F17-14. The 100-kg cylinder rolIs without slipping on the horizontal piane. Determine the acceleration of its mass center and its angular acceleration.

F17-14

F17-15. The 20-kg wheel has a radius of gyration about its center O of k„ = 300 mm. When the wheel is subjected to the couple moment, it slips as it rolls, Determine the angular acceleration of the wheel and the acceleration of the wheel's center O. The coe[ficient of kinetic friction between the wheel and the piane is pk = 0.5.

F i l - 1 5

F17-16. The 20-kg sphere rolls down the inclined piane without slipping. Determine the angular acceleration of the spherc and the acceleration of its masa center.

F17-16

F17-17, The 200-kg spool has a radius of gyration about its mass center of kG = 300 mm. If the coupłe moment is applied to the spoci and the coefficient of kinetic friction hcłween the spool and the ground is ttk = 02, determine the angular acceleration of the spool. the aece!eration of G and the lension in the cable.

F17-18. The 12-kg siender rod is pinned /o a smali ratler A that slides freely along the słot. If the rod is released Erom rent at fJ =

0°, determine the angular acceleration of the rod and the acceleration of the miler irnmediately after the release.

F17 -18

20 N0.75 ml,

0.5 md

1.75m

14

BON

F17-17

17.5 EQUATIONS OF MOTrON: GENERAL PLANIE MOTION 447

E PROBLEIVIS

17-90. If the disk in Fig. 17-20 roiksiippffig, showthat when moments are summed about the instanianeous center of zero veiocity, /C, it is possible to use the moment equation SkIłe = f,ccx, where i‚c represenłs the moment of inertia of the disk calcutated about the instantaneous axis of zero v elocit y.

17-91. The 20-kg punching bag has a radius of gyrationabout its center of mass G of= 0.4 m. If it is initially atrest and is subjected to a horizontal fotce F = 30 N. determine the Miii ai angular acceleration of the bag and the tension in the supporting cable AB.

Prob. 17-91

17-92. The uniform 150-lb beam is initially at rest when the forces are appHed to the cables. Deterrnine the magnitude of the acceleration of the mass center and the angular acceleration cif the beani a( this instant.

17-93. The rocket has a weight of 20 000 lb, mass center at G. and radius of gyration about the mass center of kc = 2I ft when it is fired. Each of its two engines provides a thrust T = 50 000 lb. M a given instant engine A sudclenly fails ta operate. Dc łermine the angular acttleratioH of the rocket and the acceleration ot its nose R.

Prob. 17-93

17-94. The tire has a weight of 30 lb and a radius ofgyration of= 0.6 ft. If the coefficients of starte andkinetic friction between the tire and the piane are= 0.2and pk = 0.15, determine the tires angular acceleration as it rolls down the incli n e. Set O = lr.

17-95. The tire hasa weight of 30 Ib and a radius of gyration of ko = 0.6 ft. If the coefficients of stalic and kinetic friction between the tire and the piane arc g, = 0.2 and i.t.k = 0.15, deterrnine the maximum angle O of the inclinecl piane so that the tire roik without s]ipping.

= 100 lb

FB— 200 lb

60'

12 lE

Pu). 17-92

Prubs. 17-94/95

448 CHAPTER 17 PLANAR KfNETrOS OF A RIGID BODY: FORCE AND ACCELERATJON

417-96. The spool has a mass of 100 kg and a radius of gyration of kG = 0.3 m, If the coefficients of stalic andkinetic friction at A are = 0.2 and Ak = 0.15, respectiveły,determine the angular acceleration of thc spool if P = 50 N.

17-97, Solve Prob, 17-96 if the cord and form P -= 50 N are directed vertically upwards.

17-98. The spool has a mass of 100 kg and a radius of gyration k0 = 0.3 m. If the coefficients of static and kineticfriction at A are = 0.2 and gl, •.= 0.15, rcspectively,determine the angular acceleration of the spool if P = 600 N.

''17-1110. A uniform rod having a weight of 10 lb is pin supported at A from a roller which rides on a horizontal track. If the rod is originally at rest, and a horizontal force of F = 15 lb is appliecl to the miler. determine thc acceleration of the miler. Neglect the mass of the miler and its size d in the computations.

17-101. Solve Prob. 17-1_00 assuming that the roller at A is replaced by a slider błock having a negligibie mass. The eoefficient of kinetic friction between the biock and thetrack is= 0.2. Neglect the dimension d and the size ofthe block in the computations.

P

Probs. 17-96/97/98

17-99. The upper body of the crash dummy has a mass of 75 lb, a center of g,ravity at G. and a radius of gyration about G of kc -= 0.7 ft. By means of the seat pelt Ibis body segment is assurned to be pin-conneeted to the seat of the car at A. If a crash causes the car to decelerate at 50 ft/62, determine the angular velocity of the body when it bas rotated to A = Yr.

Prob. 17-99

Probs, 17-100/101

17-102. The 2-kg slender bar is supportcd by cord BC and then re1eased from Test at A. Determine the indial angular acceleration of the bar and the tension in the cord.

Prob. 17-1112

17.5 EQUATIONS OF MOTrON: GENERAL PLANIE MOTION 449

17-103. [f the truck acceIerałes at a constant rate of 6 m/s2, starting from rest,determine the initial angular acceleration of the 20-kg ladder. The ladder c.an be considered tss a uniform slender rod. The support at B is smooth.

Prob. 17-103

*17-104. 1f P = 30 lb, determine the angular acceleration of the 50-lb raller. Assume the roller to be a uniform cylinder and that no slipping occurs.

17-105. 1f the coefficient of static friction betwecnthe 50-0, miler and the ground is = 0.25, determine themaxinium furce P that can be applied to the handle, sn that rollcr rolls on the ground without slipping. Also. find thc angular acceleration of the ratler. Assume the ratler to be a uniform cylinder.

17-106. The spool has a mass of 500 kg and a radius ofgyralion = 130 m. 1t rests on the surface of a conveyorbell for which the coefficient of static friction is ji.s = 0.5 and the coefficient of kinetic friction is tek = 0.4. If the conveyor accelerates at ac = 1 m/s?, determine the initial tension in the wire and the angular acceleration of the spool.The spool is nriginally at resł.

17-1117. The spod bas a mass of 500 kg and a radius of gyration /co = 130 m. It rests on the surface of a conveyor beli for which the coefficient of static friction is µs = - 0.5. Determine the greatest acceleration ac of the conveyor so that the spool will not slip. Also, what are the initial tension in the wire and the angular acceleration of the spool? The spod is originafly at test.

*17-108. The semicircular disk having a mass elf 10 kg is 17rotating at w = 4 radfs at the instant = 60°. If thecoefficient of static friction nt A is s, = 0.5. determine if the disk slips at Ibis instant,

L'(

A

Probs. 17-106/107

Prob. 17-108

Probs. 17-104/1115

17-109. The 500-kg conerete culvert has a rnean radius of 0.5 m. If the truck lias an acceleration of 3 m/s2.. determine the culvert's angular acceleration. Assume that the culvert does not slip an thc truck hed, and negłeet its thkkness.

3m/s10.5m

Prob. 17-109

4 m

45a CHAPTER 17 PLANAR K[NETfCS OF A RICID BODY: FORCE AND ACCELERAT1ON

17-110. The 101b hoop or Thin ring is given an initiaI angular velocity of 6 rad/s when i l is placed on the surface.If the coefficient of kinetic friction between the hoop and thesurface is µk 0.3, determine the distance the hoop movesbetore it stops slipping.

= 6 rad is

Prob. 17-110

17-111. A long strip of paper fis wrapped finto two roik. each having a mass of 8 kg. Roli A is pin supported about its center whereas roli B is not centrally supported. If B is brought into contact with A and released from rest, determine the initiał tension in the paper between the rolla and the angular acceieration of each roli For the calcułation. assurne the rolla to be approximated by cylinders.

Prob. 17-111*17-112. The circular concrete culvert rolls with an angular velocity of w = 0.5 rad/s when the man is at the position shown. At this instant the center of granity of the culvert and the man is located at point G. and the radius of gyration obont G is kG = 3.5 ft. Determine the angular acceleration of the culvert. The combined weight of the culvert and the man is 500 lb. Assumc that the culvert rolis without slipping. and the man does not move within the culvert.

17-113. The uniform disk of mass m is rotating with an angular velocity of wf, when 11 is placed on the floor. Determine the lnitial angular acceleration of the disk and the acceleration of its masa center.The coefficient of kinetic friction between the disk and the floor is p..

17-114. The uniform disk of mass m is rotating with an angular velocity of wo when it is placed on the floor. Determine the time hefore it starta to roli without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the diak and the floor is pk,

410Probs. 17-1131114

17-115. The 16-lb bowling bali is cast horizontaliy anto a lanc such that indiany w = O and its masa center bas a velocity v = 8 ft/s. If the coefficient of kinetic friction between the lanc and the bali is pk = 0.12. determine the distance the bali trave.ls before it ralls without slipping. For the caleulałion, neglect the finger bolca in the bali and assurne the bali basa uniform density.

Prob. 17-112 Prob. 17-115

Prob. 17-117

17.5 EQUATIONS OF MCDT1ON: GENERAL PLANE MON:*451

*17-116.The uniform heam has a weight W. If it isoriginallyrest while being supported at A and B bycables, determine the tension in eable A if cable B suddenly fails.Assurne the heam is a slender rod.

17-119. The 30-kg uniform slender rod AB rests in the posiikm shown when the couple moment of M= 150 N.m is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

Prob.17-116

17-117. A cordwrapped around each of the twa 10-kgdisks. tf they are released from rest. determine the tension in the fixed cord D. Neglect the mass of the cord.

17-11$. The 500I3-1h beam is supported at A and when itis subjeeted to a force of 1000 lb as shown.

If the pin support al A suddenly fai ls, determine the beam's init ial angular rt on the

heam. rne that the bram is a slender rod sa acceleration and the forte of the rol ler supportFor the caleulation, assuthat its thickness can be neglected.

1000 lb

R`1* 2 ft -IProb. 17-

118

Prob. 17-119

ntal (orce P = 50 N*17-120 Thein th. T 30-kg slender rod AR restse positionshown when the horizois applied.Determine ihe initial angular aeceleration of the rod. Neglect the mass of the rollers.

A

13 m

B- 9 -

Prob. 17-120P = 50N

S 1r

2

452 CHAPTER 17 KANAR I.CNETrCS OF A RICID BODY: FORCE AND ACCELERAT1ON

CONCEPTUAL PROBLEMS

PI7-1. The truck is used to pull ihe heavy container.To be most effective at providing traction to the rear wheels at A, is it beat to keep thc container where it is or place it at the front of the trailer? Use appropriate numerical values to expfain your answer.

P17-3. How can you tell the driver is accelerating Ibis SUV? To explain your answer. draw the free-body and kinetie diagrams. Herc power is supplied to the rear wheels. Would the photo ;ciek the same if power were supplied ta the Front wheels? Will the accelerations be the same? Use appropriate numerical values ta explain your answers.

P 1 7 - 1P 1 7 - 3

P17-2. The tractor is about to tow the piane to the right. Is it possible for the driver to cause the front wheel of the piane to lift off the ground as be accelerates the tractor? Draw the free-body and kinetic diagrams and explain algebraica.11y (letters) if and how this might be possible.

P17-4. Here is sornething you shoul d not try at borne, at least not wilhout wearing a beim& Draw ihe free-body and kinetic diagrams and shaw what the rider nmst do to maintain trik position. Ust appropriate numerical values to explain your answer.

PI7-2 P17-4

CHAPTER REVVIEW 453

CHAPTER REVIEW

Moment of Inertia

The moment of inertia is a measure of ihe resistance of a body to a change in its angular velocity. h is defined by I =. f r'ż

clin and will be different for each axis about which it is computed.

Many bodies are composed of simple shapcs. If this is the casc, then tabuiar vatues of I can be used, such as the ones given on the inside bark cover of this book. To obtain the moment of inertia of a composite body about any specified axis, the moment of inertia of each part is determined about the axis and the resułis are added together. Doing this often requires use M the paraffel-axis theorem.

I = I G m d 2

Planar Equations of Motion

The equations M muflon define the transłałional. and rotational motion cif a rigid body, In order to account for alf of the ferma in [hese equations_ a free-body diagram should always accompany their application, and for somo problems, it may also be convenient to draw the kinetic diagram which shows niaG and fan.

2Fr = Pn(ac)i,

= ffir(aG)v

SMG. = O

Rectifinear translation

2F, = tn(aG),

rn(aG),

SMG = O

Curvilinear translation

SF„ = m(ac,),, tmo2r,G

SF, = on(20), =- maro

SM6.- = loa ar £M„ = 1-0,a

Rotafion About a Fixed Axis

= nr(aG),

m(ao),

SMG = kr-or ar =

General Piane Mulimi

Chapter

Rołer coasters must be able to coast over loops and through turns, and have enough energy to do so safely. Accurate calculation of this energy

must account for the size of the car as it moves along the track.

Planar Kineticsof a Rigid Body:Work and Energy

CHAPTER OBJECTIVES

■ To develop formulations for the kinetic energy of a body, and define the various vvays a „orce and couple do work.

▪To apply the principle of work and energy to solne rigid-body planar kinetic problems that involve kirce, velocity, and d isplacerr ent.

▪To show how the conservation of energy can be used to solne

rigid-body plener kinetic problems.

18.1 Kinetic Energy

In this chapter we will apply werk and energy mełbocis to solne planar motion problems involving furce, velocity, and displacement. But first it will be necessary to develop a means of obtaining the bodys kinetic energy when the body is subjected to translation. rotation about a fixed axis, ar gencral piane motion.

To do this we will consider the rigid budy shown in Fig. 18-1. which is represented here by a slab m oving in the inertial x-y reference piane. An arbitrary ith particie of the body, having a mass drn, is located a distance r frorn the arbitrary point P. If at the insrreti shown the particie has aveiocity vi, then the particle's kinetic energy is = v, Fig. 18-1

4 5 6 C H A P T E R 1 8 K A N A R N E T r C S O F A R I C I D B O D Y : W O R K A K D E K E R G Y

Fig. 18 1 (repeatd)

The kinetic energy of the entire body is determined by writing similar expressions for each particie of the body and integrating the results, i.e.,

1T = j r d m v ?

2 m

This equation may also be expressed in terms of the velocity of point P. If the body bas an angular veloeity w, then from Fig. 18-1 we have

Vp i- Vzfp

= (Vp)xi {Vp)yi wk x (xi + yj)

= l(vp).t Ryp)v

The square of the magnitude of vi is thus

vi • vl = = [(vp)„ — (012 + [(vp), + wy/2

(vp),2 2(y4.),my ± w'y' + (vp),2 + 2(vp)px 412X2

=71p — 2(Vp),ĆJy2(Vp),GOX w2r2

Substituting this into the cquation of kinetic cnergy yields

T = —i( — (vp),0)( jr y don) + (vp),..w( i x dIn) + wz( r2 dm)na 2 2

The first integral on the right represents the entire mass raz of the body. Since tirwr = fy dan and :un = J x dm, the secund and third integrals locate the body's center of mass G with respect to P. The last integral represents the body's moment of inertia fp, computed about the, z axis passing through point P. Thus.

T = in712, — p)rcui:m ± (v p),41..,Pr + j., i pco2 (18-1)

As a special case, if point P coincides with the mass center G of the body, then y = = O. and therefore

T = 1.nr();-; + i fGw2 (18-2)

Both terms on the right side are always positive, since pc and w are squared. The first term represents the translational kinetic energy, referenced from the mass center, and the seconcl term represents the body's rotational kinetic energy about the mass center.

18,1 KNEric -ENERGy 457

Translation. When a rigid body of mass m is subjected to either Wrectilinear ur curvilinear łranslation, Fig. 18-2, the kinetic energy due to rołation is zero, since w = O. The kinetic cnergy of the body is therefore

vG

•T =rrvq ; (18-3)

Rotation About a Fixed Axis. When a rigid body rotates about a fixed axis passing through point 0, Fig. 18-3, the body has both transladonal and rotational kinetic energy so that

T= + [Gw')(18-4)

The body's kinetic energy may also be formulated for this case by noting that ya = rGw, so that T ,= ,-1,(1 + inr,;-)co2 . By the parallel-axis theorem, the ten/1s inside the parentheses represent the moment of inertia /0 of the body about an axis perpendicular to the piane of motion and passing through point O. Hence,*

T = ź low2(18-5)

From the derivation, this equation will give the same resuit as Eq, 18-4, since it accounts for bodz the translational and rotational kinetic energies of the body,

General Piane Motion. When a rigid body is subjected to generał piane motion, Fig. 18-4, it has ara angular velocity w and its mass centerhas a velocityTherefore. the kinetic energy is

T 71, co.12(18-6)

This equation can also be expressed in terms of the body's motion about its insłantaneous center of zero velocity i,e.,

T = ;11 ic<4,2(18-7)

where is the moment of inertia of the body about its instantaneouscenter, The proof is similar to that of Eq, 18-5. (See Prob. 18-1.}

*The similarity befween this derivation and that uf IM° =should be noted. Also thesame result ean be obtanied direztly from Eq. 18-1 by seleciing point P at O, reali7(ng that co = O.

Translatien

Fig. 18-

2

Rutation About a Fixed Axis

Fig. 18-3

General Piane Marian

Fig. 18-4

458 CHAPTER 18 PLANAR </NETICS OF A IR 'GID BODY: WARK AND ENERGY

The wlał kinetic energy of this soil compactor consists of the kinetic energy of the body nr frame cif the maehine dtte to its translation. and the transiational and rotational kinetic energies of ihe roller and the wl-zeels due to Iheir generał piane morion. Here we exelude lhe additional kinetic energy deveioped by the moving paris of the engine and drive train.

System of Bodies. Because energy is a scalili- quantity, the total kinezie energy for a system of connecred rigid bodies is the sum of the kinetic energies of alt its moving parts, Depending on thc type of =hon, the kinetic energy of each body is found by applying Eq. 18-2 or the alternative forms mentionecl above.

18.2 The Work of a Force

Several types of forces are often encountered in planar kinetics probiems involving a rigid body. The work of each af these forces has Been presented in Sec. 14.1 and is listed below as a summary.

Work of a Variable Force. If an extcrnal furce F acts on a body, the work done by the furce when thc body mw/es along the path s, Fig. 18-5, is

U r = F•d r = F cos 0 d s

(18-8)

Fig. 18-6

Herc 0 is the angle betwecn the "tails" of thc forte and the differential displacement.The integration must accourrt for thc variation of the force's direction and magnitude.

Fig. 18-5

Work of a Constant Force, if art external lorce F, acts on a body. Fig. 18-6, and maintains a constant magnitude F, and constant direction 0, while thc body undcrgocs a translation s, thcn the above equation can be integrated, so that the work becomes

= (F, cos 6)s (18-9)

18.2 THE WORK OF A FORU 459

Work of a Weight. The weight of a body does work ordy when the body's center of mass G undergoes a vertical displacernent Ay. Tf this displacement is upward. Fig, 18-7, the work is negative, since the weight is oppositc to the displacemenł.

(18-10)G `żwi

Uw = —W Ay

Likewise, if the displacement is downward (—y) the work becomes Fig. 18-7positive. In butli cases the elevation change is considered to be smalt so that W, which is caused by gravitation. is constant.

Work of a Spring Force. If a linear elastil spring is attached to a body. the spring furce Fs = ks acting on the body does work when the spring either stretches or compresses front si to a firriher posiłion s2. In both cases the work will be negative since the displacentent of the bady is in the opposiłe direction to the furce, Fig. 18-8, The work is

= — (Iks] — 421(A)

kF.hQAAiSWITO

„,ł

Urnaretched position of spring, s -= O

where s21 si

Sz

Fig. 118-8Forces That Do No Work. There are same externai forces that do no work when the body is displaced. These forces act either at fixed poirns on the body, ur they have a direction perpendicular ło łheir dkplacement. Examples include the reactions at a pin support about which a body rotates, the norma! reaction acting on a body that moves along a fixed surface, and the weight of a body when the center of granity of the body moves in a horizontal piane, Fig. 18-9. A frictional furce Ff acting on a round body as it roik wiłhout slipping over a rough surface also does no work.* This is hecause, during any instam of nrne

Ff acts at a point on the body which has zero velocity (instantaneous center, K) and so the work done by the force on the point is zero. In other words, the point is not displaced in the direction of the force during this instant. Since F.s contacts successive points for only an instant, the work of F1 will be zero,

W

*The work done by a frictional furce when Ure body sfips is discussed in Sec. 14.3. Fig. 18-9

460 CHAPTER 18 PLANAR KNETICS OF A "RICH) BODY: WORK ANO ENERGY

1 8.3 The Work of a Couple Moment

Consider the body in Fig. 18-10a, which is subjected to a couple moment M = Fr. If the body undergoes a differential displaeernent, then the, wark dane by thc couple forces can be found by considering the displacement as the sum of a separatc translatiav plus rotation. When the body translates. the wark af each furce is produced anly by the component of displacernent ałong the line of action af the forces dsr. Fig. l 8-10b. Clearly the "positive" work of one force cancels the "negałive" work of the other. When the body undergoes a differential rotation d8 about Ehe arbitrary point 0, Fig. 18-10e, then each force undergoes a displacement dsH = (r/2) c/6` in the direction of the furce. Hence, the Wtul work dane is

r rciUm = F(--c/6) + F(.— d0) = (Fr)rlO

2= M (10

The work is posidve when M and ar0 have the same sense of direction and negarive if these veełors are in the opposite sense.

When the body rotates in the piane through a finite angle O measured

Traiisla t ion

(h)

in radians, from Bl to the wark of a couple moment is therefore

Um = M dOjf E (18-12)

F.6(1.4 O

11 the couple marnem M has a consianr magnirude, then

2 F

= — 9 [) ( 1 8 - 1 3 )Rotatiun

(c)

Fig.18-1Q

18.3 THE WARK OF A COUPLE MOMENT 461

The har shown in Fig. 18-11a has a mass cif 10 kg and is subjected to a couple moment of M = 50 N m and a force of P = 80 N, which is always applied perpendeular to the end of the bar. Also, the spring has an unstretched length of 0.5 m and remains in the vertical position due to the miler guide at B. Determine the tota' work done by all the forces acting on the bar when it has rotated downward from O -= to O = 90°.

SOLUTIONFirst the free-body diagram of the har is drawn in order to accou.nt for all the forces that act on it, Fig. 18-11b.

Weight W. since the weight 10(9.81) N = 98.1 N is displaced downward 1.5 m, the work is

Uw = 98.1 N(1.5 rn) -= 147.2 J

is the work positive?

Couple Moment M. The couple moment rotates through an angle M 6 -=ir /2 rud. Rence,

Um = 50 N- m(ir/2) = 78.5 J

Spring Force F,. When = the sprin g is stretched (0.75 m - 0.5 ni)= 0.25 m, and when O = 9()°, the stretch is (2 m + 0.75 - 0,5 m -=225 rn,Thus,

= - [:1,(30 Nim)(2.25 m)2 - 4(30 Nini)(0.25 m)21 = -75.0 J

By inspection the spring does negative work on the bar since F. acts in the opposite direction to displaccment. This checks with the result,

Force P. As the bar moves downward, the force is displaced through a distance of (r/2)(3 in) = 4.712 m. Thc work is posiłive. Why?

Up = 80 N(4.712 m) = 377.0 J

Pin Reactions. Forces Az and Ay do no work since they are not displaccd.

%tal Work. The work of ali the forces when the har is displaced is thus

= 147.2 J + 78.5 J - 75.0 J ± 377.0 J = 528 1 Ans.

P = $(1N

1.5 m

0.5 m

( b )

Fig. 18-11

(a )

462 CHAPTER 18 KANAR 1C NET fCS OF A RIGID BODY: WORK AKD EKERGY

1$.4 Principle of Work and Energy

By applying the principle of work and energy developed in Sec. 14,2 to each of the particies of a rigid body and adding the resuits atgebraieaUy, since energy is a scalar, the principle of work and cnergy for a rigid body becomes

7) YLFi-2 = T2

(18-14)

This equation states Chat the body's initial translational and rgtational kinetic energy, plus the wark done by alt the external forces and couple moments acting on the body as the body moves from its initial to its fina] position, is equal to the bodys finał translational and rotational kinetic energy. Note that the work of the body's interna/ forces does not have to be considered.These forees occur in equal but opposite colhnear pairs,so that whcn thc body moves, thc work of one tarce canceis that of its countcrpart. Purtherniore, since the body is rigid, no relative movement between thcsc forces occurs, so that no interna] wark is done.

When severai rigid bodies are pin connected, connected by inextensible cables, or in mesh with one another, Eq. 18-14 can be applied to the endre system of connected bodies, Fn alt these cases the internal forces. which hołd the various members together, do no work and hence are eliminated frorri thc analysis.

The wark of the torą ue ar moment developed by the chiving gears on the motory is iransformed into kinetic energy of rot mion of the Brum.

4 6 3 18.4 PRIN{IPLE OF WORK ANO -ENERGY


The principle af work and energy is used to solve kinetic problems that involve velacity, _rym., and dispłacemenl, since these terrns are involved in the formulation. For application, it is suggested that the following procedure be used.

Kinetic Energy (Kinematic Diagrams).

• Thc kinetic energy of a body is made up of twa parts. Kinetic energy of translation is referenced to the vcIocity of the mass center, T = Lmi, and kinetic energy of rotation is determined using the moment of inertia of the body about the mass center, T = fGrvZ. In the special case of rotation about a fixed axis (ar rotation about the /C), these twa kinetic energies are combined and can be expressed as T = 410.6r2, where /0 is the moment of inertia about the axis of rotation.

• Kinematic diagrams for velocity may be useful for determining ve;

and w ar for establishing a relationship between vG and w.*

Work (F~Bady Diagram),

• Draw a free—body diagram af the body when it is locatecl at an interrnediate point along the path in order to account for alf the forces and couplc momcnts which do work on the body as ił moves along the path.

•A tarce does work when it moves through a displacement in the direction of the tarce.

•Forces that are functions of displaccment must be integratcd to obtai n the wark. Graphically, the wark is equal to the arca undcr the forte—displacement curve.

•The work af a weight is the product af its magnitude and the vertical displacement, Uw = Wy. It is positive when the weight moves downwards.

• The work of a spring is of the form U, = where k is thespring stiffness and s is the strełch or compression of the spring.

•The wark of a coupie is the product af the couple moment and the angle in radians through which it rotates, U.44 = MO.

• Since algebraie £.1ddition of the wark terrns is required,it is important that the proper sign of each term be specified. Specifically, work as posidve when the kirce (couple, moment) is in the same direction as its displacement (rotation); otherwise, it is negative.

Principle af Work and Energy.

• Apply the principIe of work and energy. T, + 1111_2 = T2. Since this is a scalar cquation, it can be used to solve for oniy one unknown when it is applied to a single rigid body.

*A brief review of Secs. 16.5 to 16.7 may prove lielpfuł when solving proNems, since computations for kinefic erbem requir£ a kinernafic analysis of veloeity.

464 CHARTER 18 PLANAR r / N E T I C S O F A IR ' G I D B O D Y : W A R K A N D E N E R G Y

EXAMPLE

The 30-kg disk shown in Fig. 18-12 is pin supported at its center. Deterrnine the angle through which it most rotate to attain an angolar velocity af 2 rad/s startinę from rest. It is acted opon by a constans couple moment M = 5 N m. The spring is orginally unstretched and its cord wraps around the rim af the disk.

A4= 5 N-m

k= N i m

( a )SOLUTION

Kinetic Energy. Since the disk rotates about a fixed axis, and it is initially at rest, then

T, = O

= = 4, P;(30 kg){0.2 m)2] (2 rad/s)2 = 1.2 J

Work (Free-Body Diagram). As shown in Fig. 18-12b, the pin reactions O, and O, and the weight (294.3 N) do no work, sinice they are not displaced. The couple ‚nomem, having a constant magnituda. does positive work UM = M as The disk rotałes through a clockwiscangle af O rad, and the spring does ncgative wark U.$ = ks2.Principle M Wark and Energy.

{r t} + {łu,_2} = {r2} {

T, } + {MO - 1ks2} = {T,}

{o} + (5 N - m)0 - 1.2(IGNitn)f0(0.2 m)]2} = {1.2J}

- 0.20' + 58 - 1.2 = D

Solving this quadratic equation for the sma1lest positive ront,180°

= 0.2423 rad = 02423 rad( i r rad - 13.9'

294.3 NI- - • M = 5 N.rn

f l _ 2 1 - n h )

ĘOjp,

(h)

Fig. 18-12

18.4 PRINCIPLE CF WQRK ANO -ENERGY 465

EXAMPLE

18 .3

The wheel shown in Fig. 18-13a weighs 40 lb and has a radius °f gyration kra = 0.6 ft about its mass center G. "If it is subjected to a clockwise couple moment of 15 Ib • ft and roll from rent without

determine its angular velocity after its center G rnoves 0.5 ft. The spring has a stiffness k = 10 Ib/ft and is initially unstretched when the couple moment is appIied.

SOLUTION

Kinetic Energy (Kinematic Diagram). Since the wheel is initially at rest.

T t = O

The kinematic diagram of the wheel when it is in the finał position is shown in Fig. 18-1313. The final kinetic energy is dełermined from

= 42-11cĄ

1 [ 40 lb ( (0 ib ,6 ft)2 + 413 )(0.8 ft)21w,2,_

2 32.2 Ft/s- 32.2 ft,/s2

0.6211 (JA

Wark (Free—Body Diagram). As shown in Fig. 18-13c, only the spring kirce F, and the couple moment dn work. The norma' force does not move along itr line of action and the frictional force does no work-

, since the wheel does not slip as ił rulls.The work of F. is found using U. =ks2. Herc the work is ncgative

since F, is in the oppositc direction to displacement. Since the wheel does not slip when the center G moves 0.5 ft, then the wheel rotates

= sG i rG i i c = 0.5 f t /0.8 f t = 0.625 rad, Fig. 8-13b. Hence, thespring stretches s = Or = (0.625 rad)(1.6 ft) = 1 ft_


{Ti } + {su1-2} = (r2)

{Tj} + 1M8 {7.2}

{O} + {15 lb • ft(0.625 rad) — I O lbjft)(1 ft)2} =,{0.62I1 634 ft • lb }

=- 2,65 radis A.

k = 10 lbjft

( a )

( c )

Fig. 18-13

466 CHAPTER 18 KANAR ICNETfC.S OF A RICID BODY: WORK AKD EKERGY

EXAMPLE

The 700-kg pipe is equally suspended from the two tines cif the fork lift shown in the photo. It is undergoing a swinging motion such that when B = 30° it is momentarliy at rent. ❑etermine the norma] and frictional forces acłing on each tine which are needed to support the pipe at the instant 6 = O. Measurements ot the pipe and the, suspender are showri in Fig.18-14a. Neglect the mass of The suspender and the thickness of the pipe,

(a)

Fig. 18-14

SOLUTIONWe musi use the equations of mation to find the forces on the tincs since thcse forces do no work. l3cfore doing this, however, we will apply the principle of work and energy to determine the angular velocty of the pipe when O = 0°.

Kinetic Energy (Kinematic Diagram). Since the pipe is originally at rest, then

T 1 = O

The final kinetic energy may be computcd with reference to eiihcr thc fixed point O or the center of mass G. For the calculation we will consicler the pipe to be a thin ring so that iG—nar. If point G is considered. we have

-;a1(vG)':-■„Gw.;= -:E,(740 kg)[(0.4 m)f.v2i2 kg(0.15 m)2]w2

= 63.875w3

If point O is considered then the parallel-axis theorem must be used to determine /o, Hence,

-21foto:i 1[700 kg(0, 15 rri)2 + 700 kg(0.4 m)2Ie

63,875

18,4 PRINCIPLE OF WORK AND ENERDY467

Work (Free-Body Diagram). Fig. 18-14b. The narmal and frictional forces on the tines do no work since they do not move as the pipe swings.The weight does positive work since the weight moves downward through a vertical distance 9 y = 0.4 ni — 0.4 cos 30° m = 0.05359 m.


{T1} + {,1/1..2} = {T2}700(9.81)N

{13} + {700(9.81) N(0.05359 in)} = {63.875(03.} (b)

= 2.400 rad/s

Equations of Motion. Referring to the free-body and kinetic diagrams shown in Fig. 18-14c, and using the result for oh, we have

FJ = rn(aG),; FT = (700 kg)(aG),

m(aG)„; NT — 700(9.81) N -= (700 kg)(2.400 rad/s)2(0.4 m)I o = loa; O = [(700 kg)(0.15 m)2 + (700 kg)(0.4 m)2

]a

Since (aG)t = (0.4 then

0, (aG), =

FT = O

NT = 8.480 kN

There are two tines used to support the load, therefore

FT = O Ans.

8.480 kN 2 — 4.24 kN Ans,

NOTE: Duc to the swinging modem the tines arc subjccted to a greaier normal force than would be the case if the load were static, in which case NT = 700(9.81) N/2 = 3.43 kN.

Fig. 18-14

468 CHAPTER 18 KANAR K NETrCS OF A RICID BODY: WORK AKD EKERGY

EXAMPLE

1 8 . 5

The 10-kg rod shown in Fig. 18-15a is constrained so that its ends rnove alang the grooved slots. The rod is initially at rest when Q =O. Tf the slider block at R is acted upon by a horizontal furce P =50 N, A determine the angular velocity of the rod at the instant O = 45°.Ncglect friction and the mass of blocks A and B.

SOWTIONWhy can the, principle of work and energy be used ta solne this problem?

Kinetic Energy (Kinematic Diagrams). Twa kinematic cliagrams of the rod, when it is in the initia] position 1 and finał position 2, are shown in Fig. 18-15b. When the rod is in position 1, T1 = O since (vG), = wl = O. In position 2 the angular velocity is w, and the velocity of the mass center is (vG)2. Hence, the kinetic energy is

72 4/11.(11G):'Z 4fGwi

= 1(10 kg)(vG).2, + [142(10 kg)(0.8 rn)2 w?,

= 0.2667(w2)'

The twa unknowns (vG)2 and w, can be related from the instanłaneous center of zero velocity for the rod. Fig. 18-15h. It is scen that as A moves downward with a velocity (vA),, B moves horizontally to the Ieft with a velocity (vz,),, Knowing these directions, the IC is located as shown in the figure. Hence,

(trG)2 pcpcw2 = {0.4 tur]. 45° m)w2

= 0.4w,Therefore,

= 0.8w; + 0.2667w; = 1.0667w;

Of coursc, wc can also dctcrmine this result using T, = 4 bcw-Z.

Work (Free-Body Diagram). Fig. 18-15c, The norma' forces NA and Na do no work as the rod is displaced. Why? The 98.1-N weight is displaced a vcrtical distance of ay = (0_4 - 0,4 cos 45i m; whereas the 50-N furce moves a horizontaI distance of s = (0.8 sio 45°) ni. Both of these forccs do positive werk. Why?


{Ti} + {2.t_ = {TA{Ti} {1. f Ps} =

{O} + 198.1 N(0.4 rn - 0.4 cos in) + 50 N(0.8 sin 45° ni)}

= {1.0667 ł } Solving for wz gives

Fig. 18-15 (02 = 6.11 rad/s Ans:

(1'4.3)4 = ° W1 = O

(h )

0.4 m

45°

Q.4 m

98.1. N50 N 'f_11

(0.8 sio 45°) in

1

(0.4 cos 45') m

G

18.4 PRINCIPLE OF WORK ANO -ENERGY 469

• FUNDAMENTAL PROBLEMS

F 1 8 - 1

F18-2. The uniform 50-lb slender rod is subjected to a couple moment of A4 = 100 lb • ft. U the rod is at test when fi = 0°, deterrnine its angular velocity when O = 90°.

F18-2

F18-3. The uniform 50-kg siender rod is at rest in the position shown when P = 600 N is applied. Determinc the angular velocity of the rod when the rod reaches the vertical position.

F18-3

F18-4. 'nie 50-kg wheel is subjected to a force ot 50 N. If the wheel starts from rest and roik without sfipping, determine its angular velocity atter il has rotated 10 revolutions. The radius of gyration of the wheel about its rnass center O is ko = 0.3 m,

F 1 8 - 4

F18-5. If the uniform 30-kg slender rod starts from rest at the position shown, determine its angular velocity after it has rotated 4 revointions. The forces remain perpendicular to the rod.

3 0 N

. 0 .

5 m

0 . 3

m20 N

F 1 8 - 5

F18-6. The 20-kg wheel has a radius of gyration about its center O of kc; = 300 mm. When it is subjected to a couple moment of M = 50 N m, it rolls without slipping. Determine the angular velocily of the wheel artel- its center O has traveled through a distance of so = 20 m, starting from rest.

F 1 8 - 6

F18-1. The 80-kg wheel has a radius of gyration about its mass center O of ko = 400 mm. Determine its angular vdocity after il has rotated 20 revolutions st arting from rest.

0 . 6 m P = S O N

0 5111.5 m

470CHAPTER 18 PLANAR KfNETrCS OF A RICID BODY: WORK AND ENERGY

■ PROBLEMS

18-1. At a given instant the body of masa rit lias an angular velocity w and its masa center has a velocity v6. Show that its kinetic energy can be represented asT = where lrc is the moment of inertia of the bodydeterrnined about the instantaneous axis of zero vełocity. located a distancc rGlic from the masa center as shown.

Prob. 18-1

*18-4. The 50-kg flywheel bas a radius of gyration of= 200 mm about its center of masa. If it is subjected to a

torque of M = (901/2) N m, where H is in radians,determine its angular velocity when it has routed 5 revolutions. starting from rest.

Prob. 18-4

18-2. The wheel is mada from a 5-kg thin ring and two 2-kg slender rods. 1f the torsional spring attached to the wheel's center has a stiffness k = 2 N - m/rad, and the wheel is rotated until the torque M = 25 N • nr is developed. determine the maximum angular vełocity of the wheel if it is rereased from rest.

18-3. The wheel is made from a 5-kg lhin ring and Iwo 2-kg siender rody. If the łorsionat spring attached to the wheels center has a stiffness k = 2 N • mjrad, so that the torque on the center of the witce] is M = (20)N • m, whcre

is in radians, determine the maximum angular velocity of ihe wheel if ii is rota led two revoiutions and [hen released from rest.

Probs. 18-2/3

18-5. The spool has a masa of 60 kg and a radius of gyration kG = 0.3 m. If it is reicased from rest, determine how far its center descencis down the soi ooth piane before it attains an angular velocity of w = h rad/s. Neglect friction and the mass ❑f the card which is wound around the central core.

1R-6. Salw Prob. 18-5 if the coefficient of kinetic friction between the spool and piane at A is pk = 0.2.

Probs. 18-5/6

18.4 PRINCIPLE OF WORK AND -ENERGY 471

18-7. The double pulley consists of twa parts that arc attached to one another. It hasa weight ol 50 lb and a centroida) radius of gyration of ko = 0.6 ft and is turning with an angular velocity of 20 rad/s clockwise. Determine the kinetic energy of the system. Assume that neither cable slips on the pulley.

*18-8. The double pulley consists of two parts that are attached to one another, ft has a wcight of 50 lb and a centroida] radius of gyration of kt, = 0.6 ft and is turning with an angular velocity of 20 radis elockwise. Deterrnine the angular velocity of the pulley ht the instant the 20-lb weight moves 2 ft downward.

ProlK. 18-7/8

18-9. If the cable is subjected lo force of P = 300 N. and the spool starts from rest. determine its angular velocity after lis center of mass O has movcd 1.5 m. The mass of the spoi)] is 100 kg and fis radius of gyration about i ts center of mass is /co = 275 mm. Assume that the spool rolls without slipping.

400 mm

Prob. 18-9

18-10. The twa tugboats each exert a constant forte F on the ship. These forces are always directed perpendicular to the ship's centerline. It" the ship lias a mass m and a radius of gyration about its center of mass G of ko. determine the angular velocity of the ship after it turni 90'. The ship is origin alty at rest.

Prob. 18-10

18—li. At the instant shown. link AB has an angularvelocity = 2 rad /s. If each link is considered as a 18uniform slendcr bar with a weight of 0.5 lb/in.. determine the totak kinetic energy cif the system.

Prob. 18-11

30 Ib

472CH APT ER 18 KANAR KrNETrCS OF A RICID BODY: WORK AKD ENIERGY

*18-12. Determine the ve1ocity of the 50-kg cylinder after it bas descended a distance of 2 m.. Inilially, the system is at rest. The red hasa mass of 25 kg and a radius of gyration about its center of mass A of k.,, = 125 mm.

18-15. The 50-kg gear has a radius of gyration of 125 mm about ils center of mass O. TE gear rack B is stationary. wbite the 25-kg gear rack C is subjected to a horizontal forte of P 150 N, determine the speed of Cafter the gcar's center O has moved to the right a distanee of 0.3 m, starting from rest.

75 mm

Prob. 18-12

B

18-15

18-13, The wheel and the attached reel have a combined weight of 50 lb and a radius cif gyration about their center of

= 6 in. If pufie}, 8 attached to the motor is subjected to a torque of M = 40(2 — e-r114) ib' ft, where H is in radians, determine the velocitv of the 200-lb erate after it has moved upwards a disiance of 5 fi. starting from resi. Neglect the mass of pulley B.

18-14. The wheel and the attached red have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B thal is attached to the motor is subjected to a torque of M = 50 Ib • ft. determine the veloctty of the 200-lb erate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

4.5 in.

Probs, 18-13114

Gear B is rigidly attached to drum A and is supported by two smali rollers at H and D. Gear H is in mesh with gear C and is subjected to a torque of M = 50 N - m. Determine the angular velocity of the drutu after C bas rotated 10 revolutions, starting from resł. Gear B and the drum have 100 kg and a radius of gyration about their rotating axis of 250 mm. Gear C has a mass of 30 kg and a radius of gyration about its rotating axis of 125 mn-i.

Prob. 18-16

18.4 PRINCIPLE OF WORK AND -E NERGY 47 3

18-17. The center O of the thin ring of mass m is given an angular vełoetty of wEy. If the ring polis withoul slipping, determine its angular velocity after it lias traveled a distance of s down the piane. Negleet its thickness.

18-19. When H = G°, the assembly is held at Test, and the torsionai spring is untwisted.If the assernbly "is released and falls downward, determine its angular velocity at the instant

90°. Rod AB has a mass of 6 kg. and disk C has a mass of 9 kg.

Prob, 18-17 Prob. 18-19

18-18. If the end of the cord is subjected to a kirce of P = 751b, determine the speed of the 100-1b block C after P has moved a distance of 4 ft, starting from rest, Pulleys A and B We iden lica', each of which hasa weight of 10 ib and a radius of gyration of k 3 in. about its center of mass.

*18-20. If P = 200 N and the 15-kg uniform slender rod starts from Test at 6 = O°, determine the Toci's angular velocity al the instant just before H = 45°.

Prob. 18-18 Prob. 18-20

474 CHAPTER 18 KANAR KINETrCS OF A RICID BODY: WORK AKID EKERGY

18-21. A yo-yo has a weight of 0,3 Ib and a radius ot gyration kU = 0.06 ft. If it is released from rest. determine how far i[ musi descend in order to ai tain an angularvelocity 70 radfs. Neglect the mass of the string andassurne that the string is wound around the central peg such that the mewo radius at which it unravels is r = 0.02 ft.

18-23. The combined weight of the load and the platform is 200 lb. with the center of gravity localed at G. If a couple moment of M = 900 Ib • ft is applied to link AR, determine the angular velocity of linka AR and CD at the instant

= 60°. The system is at rest when H = 0°. Neglect the weight of the linka.

Prob. 18-21

18-21 If the 50-lb bucket is released from rest, determine its velocity after it has fallen a distance of 10 ft. The windłass A can be considere4.1 as a 30-lb cylinder. while the spokes are slender rod& each having a weight of 2 Ib. Neglect the pulley's weight.

Prob. 18-22

Prob. 18-23

*18-24. The tub of the mixer bas a weight of 70 th and aradius of gyration = 1.3 ft about its center of gr avlty. If aconstant torque M = 60 Ib - it is applied to the dumping wheel, determine the angular velocity cif the tub when it bas rotated O = 90°. Originally the tub is at rest when O =

18-25. The tub of the mixer has a weight of 74 lb and aradius of gyration .= 1.3 ft about its center of granity. ft aconstani torque M = 60 lb • ft is applied to the tub, determine its angular velocity when it bas rotated O = 45°. Originalły the tub is at rest when O = 0°,

Probs. 18-24/25

18.4 PRINCIPLE OF WORK AND -ENERGY 475

18-26. Twa wheels of negligible weight are rnounted at corners A and B of the reciangular 75-Ib plate. Tf the plate is released from rest at O = 90°, determine its angularvelocity at the instant just before łfT.

18-28. Be band winch is used to lift the 50-kg boad. Determine the wark required to rotale the handle five revolutions, starting and ending at rest. The gear at A lias a radius af 20 mm.

1 OO m m

Prob. 18-26

Prob. 18-28

18-27. The 100-lb bIock is transported a short distance by using twa cylindrical roilers.each having a weight af 35 Ib. If a horizontal furce P .= 25 lb is applied to thc bibek, determMe the block's speed atter it bas been displaced 2 ft to the left. Originally the block is at Test. No slipping occurs.

P ---- 25 ib

18-29. A matur supplies a constant torque ar twist of M = 120 Ib • ft to the Brum. If the drutu has a weight of 30 lb and a radius af gyration of ko = 0„8 Ft. determine the speed of the 15-Ib crate A afier it rises s = 4 ft storting from rest. Neglect the mass of the cord,

M = 120 lU•ft

18

s1

1.5 ft 1.5 ft

Prob. 18-27 Prob. 18-29

476CHAPTER 18 KANAR KrNETrOS OF A RICH) BODY: WORK AND EKERGY

18-30. Motor M exerts a constant force of P = 750 N on the rope. If the 100-kg post is at rest when Y = 0°, idetermine the angular velocity of the post at the instant d = 60°. Negleci the mass of the palle); and its sine, and consider the post as a slender rod.

Prob. 18-30

18-31. The uniform bar has a mass m and length 1. ff it is releasecl from rest when Y = 0°, determine its angular velocity as a function of the angle H hefore it slips.

*18-32. The uniform bar has a mass m and length 1. !f it is relcased from rest when O ---- 0°. deterrnine the angle O at which it first hegins to slip. The ceefficient of static frictionat O is = 0.3.

18-33. The two 2-kg gears A and B are attached lo the ends of a 3-kg slender bar. The gears rolł within the fixed ring gcar C, which hes in the horizonłal piane. If a 10-N- m tarque is applied to the center of the har as shown, determine the number of revolutions the bar m ust rot ate siarting from rest in order for it to have an angular velocity of [Dm, =

20 rad/s. For the calculation, assurne the gears can be approximated by thin disks. What is the result if the gears lie in the vertical piane?

Prob. 18-33

18-34. A bali of mass m and radius r is cast anto the hori7ontai surface such that it rolls without slipping. Deterrnine iis angular velocity al the instant H = 90°, if it has an initial speed of ur; as shown.

18-35. A bali of mass in and radius r is cast anto the horizontal surface such that it rolls without shpping. Determine the minimum speed .t)0 of its mass center G so that it rolls completely around the loop of radius R + r without leaving the track.

3

Probs. 18-31/32

Probs. 18-34/35

18.5 CONSERVATION OF -ENERGY 477

1$.5 Conservatlon of Energy

When a force system acting on a rigid body consists only of conservałiye Prus, the conservatłon of energy theorem can be used to solve a problem that otherwise would be solved using the principle of work and energy. This theorem is often easier to apply since the work of a conservativc fotce is independent of łhe pwh and depends only on the initiaI and finał positions of the body. I t was shown in Set. 14.5 that the work of a conservative furce can be expressed as the differcnce in the bodys potential cnergy m easured from an arbittarity seIecłed reference or datum.

Gravitational Potential Energy. since the total weight of a body can be considered concentrated at its center of gravity_ the gravirational potennal energy of the body is determined by knowing the height of the body's center of granity above or below a horizontal datum.

Vg = WyG(18-15)

Here the potential energy is posiaue when yG is positive upward, since the weight has the ability to do posithm mirk when the body moves back to the datum, Fig. 18-16. Likewisc, if G is located betow the datum (—yG), the gravitatonal potential energy is negarive,sincc the weight does negative work when the body rcturns to the datum.

Gravitational potential energy

Fig. 18-16

Elastk Potential Energy. The [orce devcloped by an elastic spring is also a conservałive furce. The eiastic potemMI energy which a spring irnparts to an attached body when the spring is stretched or compressed from an initial undeformed position (s = 0) ta a fina] position s, Fig. 18-17, is

V, = (18-16)

In the deformed position, the spring furce acting on the body always has the ability for doing positive wark when the spring retu rns back to its original undefonned position (see Set. 14.5).

Unstretchedpusi.tion uf

spring, S = O 5

Elastic potential energy

Fig. 18-17

47S CHAPTER 18 KANAR 1CNETrCS OF A RICID BODY: WORK AKD EKERGY

Conservation of Energy. In generał, if a body is subjected to both gravitational and elastic forces, the total putential energy can be expressed as a potential function represen Led as the algebraie sum

V V g 1 1 , (18-17)

Here measurement of V depends upon the location of the body with respect to the selected datum.

Reali7ing that the work of conservative forces can be written as adifference in their paten fial energies, i.e., _.2),„ VI - , Eq. 14-16,we can rewrite the pi-inciple of work and energy for a rigid body as

+ + T2 + 1/2 (18-18)

Herc (1/./1_,),„„K„, rcpresents the work of the nonconservativc forces such as frietion. if this term fis zera, then

+ V, T2 + V-2 (18-19)

The torsional springs located ai the top o( the garage door wind up as the ebor is lowered. When thc door is raised, the potential energy stored in the springs is [hen transferred into gravitationa] potential energy of the doors weight, thereby rnaking it easy to opera.

This equation is referred to as the conservałion of mechanical energy. It stater that the sum of the potential and kinetic energies of the body remains constant virhen the body moves from one position to another. It also appfics to a system of smooth, pin-cannected rigid bodies, bodies conneeted by inextensible cords, and bodies in mesh with other bodies. In all these cases the forces acting at the points of contact are elitninated from the analysis, since they =ur in equal but opposite collinear pairs and each pair of forces tnoves through an equal distance when the system undergoes a displacemen t.

It is imp(,->rtant to remember that only probłeMS involving conservative furce systems can be solved by using E. 18-19, As stated in Sec. 14.5, friction or othcr drag-resistant forces, which depcnd on velocity or accelcration, arc nonconservative. The work of such forces fis transformed int° thermai energy used to beat up the surfaces of contact, and consequently this energy is dissipated finto the surroundings and niay not be recovered. Therefore, problemy involving frictional forces can be solved by using either the principle of work and energy written in łhe form of Eq. 18-18, if it applies, or the equations of motion.

4 7 9 18.5 CONSERVATION OF 1=NERGY

•-•cedure for Analysis

The conseiwation of energy equation is used to solne problems involving velucity, thspigłcement, and cumervative forte systems. For application i t is suggested that the folIowing procedure be used.

Potential Energy

• Draw Iwo diagrams showing the body located at its initiai and final positions along the path.

• If the center of granity, G, is subjected to a verricvl displacernent, establish a fixed horizontal datum from which to measure the body's gravitatlonal potential energy yg.

• Data pertaining to thc clevation yG of thc body's center of granity from the datum and thc extension or compression of any conneeting springs can be determined from the problem geometry and listed on the two diagrams_

• The potentlaI energy is determined from V = Vg + V,. Herc Vs

= WyG, which can be positive or negative, and V, = which is always positive.

Kinetic Energy.

• The kinetic energy of thc body consists of two parts, namely translational kinetic energy, T = ,L,m1;(2;, and rotational kinetic energy, T = -11-Gw2.

• Kinematic diagrams fur velocity may useful for establishing arelationship between% and w.


• Apply the conservation of energy equation T1 + VI = Tr + V2.

480 CHAPTER 18 KANAR KfNETrCS OF A RICiD BODY: WORK ANID EKERGY

0 2 ni

1 8 . 6

The 10-kg rad AR shown in Fig. 18-18a is canfined so that its endy move in the horizontał and vertical sloty. The spring has a stiffness of k = 800 N/rn and is unstretched when O = 0°. Determine the angułar velocity of AR when O = 0°, if the rod is released from rest when k - 84QNIm O = 30°. Neglect the mass of the slidcr blocks.

SOLUTION

Potential Energy. The two diagrarns of the rod. when it is located at its initial and finał posiłions,are shown in Fig. 18-I Sb.Thc daturo, used to measure the gravitational potential encrgy, is pIaceci in line with the rod when O = O.

When the rod is in position 1, the center of gravity G is located be!~ the dumni so its gravitatianal potential energy is negative. Furthermore, (positive) elast-ic potential energy is stored in the spring. since it is stretched a distance of s i = (0.4 sin 30°) ni. Thus,

Dalom YE = -Wy +

si (0A sin 30°) m = -(98.1 N)(0,2 sio 30° m) + :t,(800 Nim)(0,4 sin 30° ninr)'`= 6.19 ł

When the rod is in position 2, the potential energy of the rod is zero, since the center cif gravhy G is located at the latum. and the spring is unstretched,si= 0. Th us,

Sz =V2 = O

Kinetic Energy. The rod is released from rest from position 1, thus (v0), = w

E = 0, and sa

T I = D

I n p o s i i i o n 2 , t h e a n g u l a r v e l o c i t y i s t o , a n d t h e r o d ' s m a s s c e n t e r h a s a v e l o c i t y o f ( v 0 ) 2 . T h u s .

(b) sn(v(;), + l /(-(.4A

= 3(10 kg)(.%)2 + i[h(10 kg)(0.4 m12

Using kinerrwtics-. (vG), can be related to itai as shown in Fig, 18-18c. At the instant co-nsktered, the instantaneous center of zero velocity (K) for the rod is at point A; hcncc, (%), = (rcd/c)o,(, = (0.2 m)zo2. Substituting into the above expression and simplifying (or using -/fccoD, we gol

T2 = D.2667w


{T1} + {1/1} {7.2} + {y,}

{o} {6,19 J} = {0.2667wi} + foł

= 4.82 rad/s

0.2 m

(a)

W

Fig. 18-18

18.5 CONSERVATION OF ENERGY 481

The wheel shown in Fig. 18-19a has a weight of 30 lb and a radius of gyration of kc = 0.6 ft. It is attached to a spring which bas a stiffness k = z lb/ft and an unstretched length of 1 ft. If the diak is released from rest in the position shown and rolls without slipping, detcrmine its angular veloeity at the instant G moves 3 ft to the left.

SOLIJTION

Potential Energy. TWo diagrams of the whecl, when it at the initial and finał positions, are shown in Fig. 18-19b, A gravitationai datum is not needed here since the weight is not displaced verticalIy. From the problem geometry the spring is stretched si = V32 + 4^ — I) = 4 ft in the initial position, and spring s2 = (4 — 1) = 3 ft in the fina] position. Hence, the positive spring potential energy is

VI = ksi = 4(2 lbift)(4 ft)2 = 16 ft • lb

= ks; = Ibift)(3 ft)2 = 9 ftlb

Kinetic Energy. The diak is releascd from rest and so (vc)i = O. cni = O. Therefore,

T I = O

Since the instantanems center of zero velocity is at the grouncl, Fig. 18-19c, we have

1 ,=

- [ ( 3 0 t h

(0.6 ft).-> 30 lb )(0.7.5 ft)lo.i22 [ ftłs-'

32.2 ft/s2 J

= 0,4297wZ

Conservation °f Energy.

jT[} j1/1 = {yz}

{0} + (16 ftlb) = (0.4297W;21 + {9 ft• lb}

4,04 rad/s nv.

NOTE: If the principle of work and cnergy were used to solut ihis

problem, thcn the work of the spring would have to be determined by

considering both the change in magnitude and direction of the spring

forte.

C1.12

E i D

(e)

Fig.

482 CHAPTER 18 KANAR KfNETrCS OF A RICID BODY: WORK AKD EKERGY

EXAMPLE

( b )

Fig. 18-20

The 10-kg homogeneous disk shown in Fig. 18-20a is attached to a uniform 5-kg rod AB. If the assembly is released from rest when O =60°. determine the angular vetocity of the rod when O = 0°. Assume that the disk rolls without nlipping. Neglect friction along the guide and the mass of the coilar at B.

SOLUTION

Potential Energy. Tom diagrams for the rod and disk, when dwy are located at thcir initial and finał positions,are shown in Fig. 18-20b. For convenience thc datom passes through point A.

When the system is in position 1, only the rod's weight has positive patential energy. Thus,

V1 = Wryi = (49.05 N)(0.3 sio 60° m) = 12.74

When the system is in position 2, both the weight of the rod and the weight of the diak have zero potential energy. Why? Thus,

= O

Kinetic Erwrgy. Since thc wtłuc system is at rest at the initial position,

T i =

In thc fina position the rod bas an angular vełocity (cu,.)2 and its mass center has a velocity (vG)2. Fig. 18-20c. Since the rod is fu/1y ex(ended in this position, the disk is momentarily at rest, so (to,i)2 = O and itv,4)2 = O. For the rod (vG), can be related to (w,.)2 from the instantaneous center of zero velocity, which is Iocated at point A. Fig. 18-20c. Hence, (ya), = rGlidwp.)2or (tia), = 0.3(wr)2. Thus,

1 i 1T2 = .3-mr(VG)E; + idcor>'1' + -2" mnr(V4).'i

= 2(5 kg)[(0.3 n1)(0JA2i2 + [12 kg)(0,6 rn)21(Q,A tO + O 0.3(0)A2


{Ti} {yi} = {T2} + {172}

{o} + {12:74 J} = 0.3(w,.)::;} {O}

(c,4)2 = 6.52 rad/s Ang

NO'TE: We can also determine the fina' kinetic energy of the rod using T2 =

48318.5 CONSERVATION OF f N ERG Y•FUNDAMENTAL PROBLEMS

F18-7. If the 30-kg disk is released from rest when H = 0°, determine its angular velocity when o = 90°

F18-10. The 30-kg rod is released from rest when H = Determine the angular velocity of the rod when O = 90°. The spring is unsireiched when O = 0°.

F18-8. The 50-kg red hasa radius ol g yration aboui lis center O of k, = 300 mm. JE il is released from rest, determine its angular velocity when its center O has traveled 6 m down the smoołh indined piane.

F18-8

F18-9. Thc 60-kg rod OA is released from rest when = 0°, Determine its angular velocity when N = 45°, The spring

remains vertieal during the morion and is unstretched when O = i)°.

1 8-9

F18-10F18-11. The 30-kg rod is released from rest when O = 45°, Determine the angular velocity of the rod when ł) = 0°, The spring is unstretched when O = 45°.

Ź

1.5 m

k=3Uf)Nfm

F18-11

F18-12. The 20-kg rod is released from rest when O = 0°. Determine its angular velocity when O = 90°. Thc spring bas an unstretched length of 0.5 rn.

2 m

F18-12

F18-7

2 rn

484 CHAPTER 18 PLANAR KfNETrOS OF A LICH) BODY: WORK AND EKERGY

PROBLEMS

*18-36. Al the Instant chowu, the 50-1b bar rotates clockwise at 2 rad/s. The spring attached to its end always remains vertical due to the miler guide at C. If the spring bas an unstretched lenglh of 2 ft and a sliffness of k = 6 tb/ft, determine the angutar velocity of the bar the instant it has rotated 30° clockwisc.

18-37. At the instant s kown, the 50-lb bar rot ates clockwise at 2 rad/s. The spring attached to its end always remains vertical due to the roller ęuide at C. ff the spring bas an unstrełched lengih of 2 ft and a stiffness of k = 12 lb/ft, determine the migle O. measured from the horizontal, to which the bar rotates before it momentarity stops.

C

4 ft

Probs. 18-36137

18-38. Tlic spool has a mass of 50 kg and a radius of gyration ko = 0.280 m. if the 20-kg block A is released from rest, determine the distance the block musi fali in order for the spool to have mi anguIar velocity w = 5 rad/s. Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

18-39. The spool has a mass of 50 kg and a radius of gyration kQ= 0.280 m. 1f the 20-kg block A is released from rest, determine the velocity of the block when it descends 0.5 m.

Probs. 18-38/39

*18-40. An automobile tire has a mass of 7 kg and radius of gyration kc) = 0,3 m. If it is released from rest at A on the incline, determine its angolar velocity when it reaches the horizontal piane. The tire rolls without słipping.

13

Prob. 18-40

18-41. The system consists of a 2[1•1b disk A, 4-Ib slender rod 8C, and a l-lb smooth collar C, If the disk rolls without slipping. determine the velocity of lite collar al the instant the rod becornes horizontal. i.e.. H = fr. The system is released from Test when O = 45°,

18-12. The system consisłs of a 24.1-Ib disk A, 4-1b slender rod BC., and a 1-lb smooth collar C. If the disk notis without slipping, determine the velocity of the collar at the instant =

30°, The system is released from rest when O = 45°.

Probs. 18-41/42

r3

2 radis

6 ft

1$.5 CONSERVATION OF -ENERGY485

18-43. The door is made from one piece, whose sides mc-ue along the horizontal and vertical iracks_ If ihe door is in the open position, O = 0°. and then released, determin e the speed at which its end A strikes the stop at C. Assume the door is a 180-lb thin plate having a width of 10 ft.

Prob. 18-43

18-44. Determine the speed of the 30-kg cylinder aftcr it has descended a distance ot 2 m, starting from rest. Gear A hasa masa of 10 kg and a radius of gyration of 125 mm about ils center of mass. Gear B and drem C have a combined mass of 30 kg and a radius of gyration about their center of mass of 150 mm,

Prob. 18-44

18-45. The disk A is pinned at O and weighs 15 lb. A 1-fl rod weighing 2 lb and a 1-ft-diamełer sphere weighing 10 lb ara welded to the disk, as shown. If the spring is orginally stretched I ft and the sphere is released from the position shown, determine the angular velocity of the disk when it has rotated

18-46. The disk A is pinned at O and weighs 15 lb. A 1-ft rod weighing 2 ib and a I-tt-diamater sphere weighing 10 lb are welded la the disk. rts shown. II the spring is originally stretched I ft and the sphere is released from ihe position shown. dctermine the angular velocity of the diak when it łtas rotated 45'.

Probs.18-45/46

18-47. At the instant the spring beeomes undcformcd, the center ot the 4fl-kg disk has a speed of 4 m/s. From this point determine the distance d the disk moves down the piane before mesmeotariły słoppiog. The disk rolls withqut sl ippin g.

Proli. 18-47

486CHAPTER 18 PLANAR KfNETrOS OF A RICH) BODY: WORK AND EKERGY

18-51. A spring having a stiffness of k = 300 Nim is attached to the end of the 15-kg rod, and it is unstretched when O = 0°. li the rod is released from rest when O = 00. detcrmine iks angular veloeity at the instant H = 30% The motion is in the vertical piane.

*18-48. A ehain that bas a negligible mass is draped over the sprocket which hasa mass of 2 kg and a radius of gyration of k = 50mm. 1f the 4-kg biock A is released from rest in the position s = l rn, dcterminc the angular velocity of the sprocket at the instant s = 2m.

18-49. SoIve Prob. 18-48 if ihe ehain has a mass of 0.8 kg/m. For the calculation neglect the portion of the ehain that wraps over the sprocket,

[00 mm- ,

s = I m

Probs, 18-18/49

18-50. 'The compound disk pulley consists of a hub and attached outer firn, If it has a mass of 3 kg and a radius of gyration ke = 45 mm. deterrnine the speed of błock A a fter A descends 0.2 m from resl. Blocks A and B each have a masa of 2 kg. Neglect the mass of the cords.

30 mm

11

LJProb. 18-50

k= 300 NA-n

Prob. 18-51

*18-52. The two mars arc released from resl at the position O. Determine their angular velocities at the instant thcy beeome horizontal. Neglect the mass of the rolter M C. Each bar bas a mass m and length L.

18-53. The Iwo bars are released from rest at the poSitiOn O =Determine their angular velocittes at the instantthcy bccorne horizontal.Negleet the mass cif the ratler at C. iracki bar lias a mass rn and length L.

Pro Lis. 18-52/53

18.5 CONSMATION OF -ENERGY487

18-54. 1f the 250-lb biock is released from Test when the spring is unstretched, determine the velocity cif the błock after it lias descended 5 ft. The drum bas a weight of 50 łb and a radius of gyration of 40 = 0.5 ft ahout "its center of mass O.

1856. 1f the chain is released from test from the position shown, determine the angolar velocity of the pulley after the end B has risen 2 fl.The policy hasa weight of 50 lb and a radius of gyration of 0.375 ft about its axis, The choin weighs 6 Ibift.

4 I I

6 f t

Prob. 18-54

Prob. 18-56

18-55. The 6-kg rod ABC:1s corxnected to the 3-kg rod CD. If the system is released from rest when O' = T. determine the angurar velocity of rod ABC at the instant it bccomes horizontal.

18-57. 1f the gear is released from rest. determine its angolar velocity atter its center of gravsty O has descended a distance of 4 ft. The gear bas a weight of 100 lb and a radius of gyration about its center of gravity of k = 0.75 ft. 1 8

Prob. 18-55 Prob. 18-57

488CHAPTER 18 PLANAR KrNETrOS OF A RICH) BODY: WORK AND EKERGY

18-58. When the slender 10-kg bar AB is horizontaI it is at Test and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the bar is momentardy stopped when it has rotated clockwise

18-59. When the slender 10-kg bar AB is horizontaI il is al Test and the spring is unstretched. Determine the stiffness k of the spring so that the motion of the har is momentarily stopped when it has rotated clockwise 45°,

18-61. A uniform ladder having a weight of 30 ibis released from rest -when it is in the vertical milion. Ii ii is allowed to fali freely, determine the angle O at which the bortom end A starty to slide to the right of A. For die ealculation, wsunie the ladder to be a slender rod and neg1ect fric6on at A.

Pruł,. 18-61

*18-60. 1f the 40-kg gear B is released from rest at O determine the angu1ar velocity of the 20-kg gear A at theinstant U = The raduj of gyration of gears A and Bab-out their respective centers of mass are kA = 125 mm and k8 =- 175 mm. The auter gear ring P is fixed.

18-62. The 50-Ih wheel lias a Tad os cif gyration abouł its center of granity G of kL = 0.7 ri. If it rolls without delermine its angolar velocity when it has rotated clockwise 90' from the position shown. The spring AB has a stiffnessk1.20 ibift and an unstretchcd lcngth of 0.5 ft. The wheelis released from rest.

Prob. 18—b0 Proh. 18-62

ft

0.5 ft

I ft

"1.11:wh-1I 1 5 m _I

Probs. 18-58/59

18.5 CONSERVATION OF -ENERGY489

18-63. Tle uniform window shade A B has a total weight of 0.4 lb. When it is released, it winds up around the spring-foaded core O. Motion is caused by a spring within the core, which is coiled so that i: exerts a torque M = 0.3(10-3)0 lb • ft.. where is in radians, on the core. If the shade is released from rest, determine the angular velocity of the core at the instant the shade is eompletely roned np, i.e., after 12 revolutions, Whcn Ibis occurs, the, spring becomes uncoiled and the radius of gyratig.wi of the shade about the axle at O is ko = 0.9 in. Nok: The elastic potential energy of the torsional spring is Wp = 402, whereM = H and k = 0,3(10-3) lb • flirad.

M

B

- L

Prob. 18-45318-64. The motion cif the uniform 80-lb garagc door is guided at its ends by the track. Determine the requircdinitiai stretch in the spring when the door is open, O = sothat when it falls freely it come,s to rest when it just reaches the fully elosed position, O -= 90°, Assume the door can be treated as a thin plate, and there is a spring and pulley system on each of the IWO sides of the door.

18-65. The motion of the uniform 80-lb garage door is guided at its ends by the track. If it is released from rest at

=

determine the doors angular velocity at the instantO = 30. The spring is odginany stretched 1 [l when the door is held open, O -= O. Assume the door can be treated as a thin plate, and there is a spring and pullłey system on each of the Iwo sides of the door.

Prohs. 18-64165

18-66. The end A of the garage door A B travels along the horizontal track, and the end of member BC is attached to a spring at C. If the spring is odginany unstrctchcd, determine the stiffness k so that when the door falls downward from TeSi in the position shown, it will have zero angular velocity the moment it closes, i.e., when it and BC becorne vertical. Neglect the mass cif member BC and assurne the door is Chin plate having a weight of 200 lb and a width and height of 12 ft. There is a similar cormection and spring on the other side cif the door.

12 ft

Prob. 18-66

18-67. Determine the stiffness k of the torsional spring at A. 48 so that if the bars are released from rest when O = T, bar A B has an angular velocity of 0.5 rad is at the closed position, O = 90°. The spring is uneoiled when O = 0°.The bars have a masa per unit length of l O kg/rn.

*18-68. The torsional spring at A has a stiffriess ofk = 900 N • ni/rad and is uncoiled when O = Deterrninethe angular velocity of the bars. A B and BC,when O = T, ifthey are released from rest at the closcci position. O90°.The bars have a mass per unit length of 10 kg/m.

f '

Probs. 18-67168

490 CHAPTER 18 P L A N A R 1 0 E N E T f G S O F A R I C I D B O D Y : W O R K A K D E K E R G Y

CONCEPTUAL PROBLEMS

P18-1. The bicycle and rider start from rest at the top of the hill. Show how lo determine the speed of the lider when he freeły coasts down the hill. Use appropriate dimensions of the whcels, and the mass of tlic rider,frame and wheels of the bicycle to explain your resulis.

P 1 8 - 1

P18-2. Twa torsional springs, M = WP. are used to assist in °petting and closing the hood of this truck. Assuming the springs are uncoiled lH = 0°) when the hood is openecl. determine the stiffness k (N. m/ rad) of eaeh spring so that the hood can easily be lifted.i.e..practicaIly no kirce applied to it, when it is closed in the onlixked position. Use appropriate numerical values to cxplain your result.

P18-3. The operation of this garage door is assisted using iwo springs A B and side members BCD, wind' are pinned at C. Assuming the springs are unstrekhed when the door is in the horizontal (opon) position and ABCD is vertical, determine each spring stiffness k so that when the door falls to the vertical (closed) position, it will słowly come to a stop. Use appropriate numerical values to explain your result.

P 1 8 - 3

P18-4. Detemiine the connterweight of A needed to halance the weight of the bridge deck when H = (r, Show that this weight will maintain equilibrium of the deck by considering the potential energy of the system when the deck is in the arbitrary position fl. Both the deck and AB arc horizonial when H = O°. Neglect the weights of the other members. Use appropriate n umerical values Cti explain this result.

P18-2 P18-4

CHAPTER R wiEw 49• CHAPTER REVI EW

Kinetie Energy

The kinetic energy of a HO body that undergoes planar motion can be referenced to its mass center. h includes a scalar sum of its translational and rotational kineric energies.

vr;S —11 .

CiTran_slatinn

TranshIlion

Rotation About a Fixed Axis

T =4mzi 41,c02

ar

T = low2

General Piane Molion

T = 1 my /Gw2

or

T = i cw2

Rotation About a F xtri A.xis

General Piane MQtion

492CHAPTER 18 PLANAR iCFNETECS OF A RICID BODY: WORK AND ENIERGY

Work of a Force and a Couple Moment

A furce does wark when ił undergoes a displacement ch in the direction of the forte. In particular, the frictional and norma forces that act on a cylinder or any circular bady that rolls withoirt s fipping will do no wark. since the normal furce does not undergo displacement and the frictional forte acts on successive points on the surface of The body.

U F = i F w s R a l s

Uw = —WAy

Weight

k F.. f i n o w w f f i c

Unstretched position of spring. s = O

IU = 2 s -

Spring

14, = (F, co:i 0)s

Constant rurce

az

U m = M d Or . ;

M= M(02 —

Constant rrkagniTude

CHAPTER REVIEW493

Principle ofWork and Energy

Problems that involve veIocity, forte, and displaeement ean hc solved using the principle of work and energy.The kinetic energy is ihe sum of both its rotational and translational parts. For application. a free-body diagram Omoki be drawn in order lo accoun t for the work of alt of the forces and couple moments that act on the body as it moves along the path.

Conservation v# Enemy

If a rigid body is subjected only to conservative forces, (hen the conservation-of-energy equation can he used ta solve the problem. This equation requires that the sum of the potential and kinetic energies of the body remain ihe same at any Iwo points along the path.

The potential energy is the sum of the body's gravitational and elastic potential energie& The gravitational potential energy will be positive if the body's center of granity is located above a datum. If it is below the daturo, then it will be negative, The elastic potential energy is always positive. regardiess if the spring is stretched or compressed.

Tt + =

T, + Vi = -I- Vi

where V = Vx + VP

Vs = Wyc

+yoDałum

ł Y u

Unstretehed pontion of spring—s =VR = -Wyk

V, = ks2

Gravitational potential energyElastic potential energy

Chapter 19

The irripulse that this tugboat imparts to this shup will cause it to tum in a manner that can be preckteci by applying the principles of

irnpuise and rnornentum,

Planar Kinetics of aRigid Body: Impulseand Momentum

CHAPTER OBJECTIVES

•Ta develop formulations for the linear and angular momentom of a body.

•To apply the princr.)les of linear and angular irnpulse and momentum to solne rigid-body planar kinetic problerns that involve korce, velocity, and tirne.

•To cliscuss application of the conservation of momentom_

•To analyze the mechanics of eccentric impast.

19.1 Linear and Angular Momentum

in this chapter we wili use the prineiplcs of linear and angolar impulse and ~meritum to solne problems involving force, velocity, and finie as relatcd to the planar motion of a rigid body. Bcforc doing this, wc will first formałize the methods for obtaining a bodys linear and angular momentum_ assuming łhe body is symmetrie with respect to an inertial x-y reference piane.

Linear Momentum. The linear momentum of a rigid body is determined by summing vecłorially the linear momenta of all the particIesof the body,L = Sincevi = invG (see See. 152) we cenalso write

L = rinic (19-1)

This equation st ates that the bodys linear momentum is a vector quantiły having a rnagnitude myci, which is cornmonly measured in units of kg - mis nr sług ft/s and a direerion defined by vG the velocity cif the body's mass center.

496CHAPTER 19 I"LANAR KfNETrQS OF A RICH) BODY: IMPULSE ANO MOMENTUM

(a)

x

( b )

Fig. 19-1

Angular Momentum. Consider the body in Fig, 19-la, which is subjected to generał piane motion. At the instant shown, the arbitrary point P has a known velocity Vp, and the body has an angular vciocity Therefore the velocity of the ith particie of the body is

= V p + V i p r r = Yp 4- toxr

The angular momentum of this particie about point P is equal to the "moment" of the particic's linear mornenturn about P ,Fig, 19-1a. Thus,

(HA = r X m iv i

Exprcssing yi in terms of Vp and using Cartesian vectors, we have

(.11Aik = ini(xl + }j) X [(VAi + (VAyj ł ark X (xi + yi)1 (11p)i = -

miy(vp), + rnix(vp), + mgor2

Letting ‚31 —>din and integrating over the entire mass m of the body, we obtain

H p = - ( i m a d m ) ( v p ) , ± U x d i n ) ( v p ) ) , + ( ] r d m ) w

Here Hp represcnts the angular momentum of the body about an axis (the z axis) perpendicular to the piane of motion that passcs through point P. Since tra = f y dm and ,Tcln = Jx dm, the integrals far the first and seconci terms on the right are used ta łacate the bodys center of mass G with respect to P, Fig. 19-1 b, Also, the lasł integraf represents the bodys moment of inertia about point P Thus,

Hp =/pw(19-2)

This equation reduces to a simp]er form if P coincides with the mass center G for the body,* in which case ź = = D. Hence,

HG = 7G.3(19-3)

It also reduces to the same simple form, Hn = 1Pm, if point P is a fixcl poini (see Ey. 19-9) nr the vdocity of P is directed ałong the line PC.

19.1 LINEAR ANO ANGULAR MCMENTUM 497

Ifere the angolar mDmenretm of rhe body about G is equal w the producł of the moment of inertia of the body abour an ax.is passing through G and the body angolar velocity. Realize that He is a vector quantity having a magniłude fow, whieh is corntnonly measured in units of kg m2is nr sług ft2is, and a direction defined by w, which is a lways perpendcular to the piane of motion.

Equation 19-2 can also be rewritten in terms of the x and y components of the velocity of the body's mass center. (vG), and (vG),, and the body'smoment of incrtia Since G is located at eoordinates then by theparallel-axis theorem, „p = „G + m(12 + 5,2). Substituting into Eq. 19-2 and rearranging terms, we have

Np = 5;n[—(vp), + y.y.d] + _Tunavp)., ± rrJ + Iorr (19-4)

From the kinematic diagram of Fig. 19-1b, vG can be expressed in terms of vp as

vG = vp + w X

(vG),i + (vG)yi = (vp),i (vp)yj wk X Gi + 5j)

Carrying nut thc cross product and equating the respective i and j components yields the twa scalar equations

(vp),, —

(vG), (vp),

Substituting these results into Eq. 19-4 yields

+ Wp = —5.-"nl(190)x :1C.M(VG)y IW (19-5)

As shown in Fig. 19-1c, this rsolr indleaws !har when the angolar momentom of che body is compoted abouł point P, ił is eqtavalent w themoutent of (be lirTear mornenmmor ds components Myc), andm(vG),, about P plus thc angolar manienium oruoi.Using these results, we wili now consider three types of rnotion.

x

Body MOERC nturn diagram

(e)

Fig. 19-1

Y

498 CHAPTER 19 PLANAR KINETICS OF A IR 'GID BODY: IMPULSE AND MOMENTUM

Tra[Watiun(a)

Rotation abulii a fixed axis(b)

Fig. 19-2

Translation. When a rigid body is subjected to either rectiiinear or curvilinear łransiałion, Fig. 19-2a, then rx = O and its mass center has a velocity of vG = v, Henee, the linear momentum, and the angular momentum about G, become

L = rrly

K G "" O

If the angular momentum is computed about same other point A, the "moment" of the linear momentum L must be found about the point. Since d is the "moment arm" as shown in Fig. 19-2a, then in aceordance witli Eq. 19-5, H A = (4)(MVG)).

Rotation About a Fixed Axis. When a rigid body is roładng about a fixed axis, Fig. 9-2b, the linear momentum, and the angular momentum about G, are

L = i r a v r ,

l i G ! G w (19-7)

is sometimes convenient w cnmpute the angular momentum about point O. Noting that L (or vG) is always perpendieldar to rG, we have.

Na = /Gai rG(riTvG)(19-8)

Since = this equation can be written as 110(IG + mr7(-3)6).

Using the parallel-axis theorem,*

(19-6)

Ho -=- łow

(19-9)

For the calculation. then, either Eq, 19-8 or 19-9 can be used.

Tbe simi!arity between this derivation and that of Eq. 17-16 (1,M0 = Ipa) and Eq_ 12-5 (T = .4,10ab2) should be noted. Also hole that (be same result can be obtainedfront Eq, t9-2 by selecting point Paf 0, realizing that ćvo), = (yeA,0,

19.1 LINEAR ANO ANGIJLAR MOMENTUM 499

.---_,-- .,,.. ` L = m v G

\A

Generał piane motion

(c)

Fig. 19-2

Generai Piane Motion. When a rigid body is subjecłed to general piane motion. Fig. 19-2c, the linear momentum, and the angular momentum about G, become

LI J =

(19-10)

'f the angular momentum is computed about point A, Fig. 19-2c, it is necessary to include the moment of L and FIG about this point. In this casc,

= /Gw (d)(in vG)

Here d is the moment arrn, as shown in the figure.As a special case, if point A is the instantaneous center of zero velocity

then, Iike Eq. 19-9, we can write the above equation in simplified form as

H I c (19-11)

where is the moment of iztertia of the body about the LC,(See Prob. 19-1)

As the pendułum swinp downward, its angular momentum about point O can be determined by cornputing the moment °f /Gw and rnyu about O. "This is 110 = /Gw + (nroc>ri. Since pa = r»d, ihen 110 = /Gw ł on{wci)d = (IG md2}w = Tow.

J500 CHAPTER 19 1"I_ANAR KfNETrQS OF A RICH) BODY: IMPULSE ANO MOMENTUM

At a given instant the 5-kg sIender bar has the motion shown in Fig. 19-3a. Determine its angular momentum about point G and about the /C ai this Instant.

( a )

SOLUTION

Bar. The bar undergocs generał piane rnotion. The (C. is established in Fig. 19-3b, so that

2 m/s= = 0.5774 radis

4 m cos 30°vG = (0.5774 rad/s)(2 m) = 1.155 m/s

Thus,

/C 4 m cos 30'

30'

,2 m

(C -01-1(;= 1G-w= [(5 kg)(4 m)2](0.5774 rad/s) = 3.85 kg • m2/s

Adding liry and the, moment of invG about the (C yields

2 m/sA (C +) /gwirt;(-0 + d(inVG)

= [71.--z(5 kg)(4 01)1(0,5774 rad/s) + (2 ni)(5 kg)(1 .155

= 15.4 kg • m2/s)

We can also use

(C ÷) //ew

= (5 kg)(4 m)2 + (5 kg)(2 m)21(0.5774 rad/s)

= 15.4 kg • rn2is

(b)

Fig. 19-3

19.2 PRINCIPLE OF 1ty1PULSE AND MOMENTUM 50 1

19.2 Principle of Impulse and Momentum

Like the case for particie motion, the principie of impulse and momentum for a rigid body can be developed by cornbining the equation of motion with kinematics. The resulting equation will yield a direct solation to problems involving furce, uetocity, and tiule.

Principle of Linear Impulse and Momentum. The equation of translational motion for a rigid body can be written as SF = maa = tn(dvG, dt). Since the mass of the body is constant,

d= t

(1nvG)

'GW]

IniliaF mumentum diagram

(a)

Multipiying botki sides by dt and integrating from t = ti, lik = ("v6)1 to

t = t 2 , v c = ( v G ) . , y i e l d s „

d r f „ r , d t +i l = in(VG)2 - f"("VG)1

f ' }'r,

This equation is referred to as the principle af linear impaise and momenrum. It states that the sum of aIl the impulses created by the external furce system which acłs on the, body during the time interval ri to

is equal to the change in the Iinear momentum of the body during this time interval, Fig. 19-4.

Principle of Angular Impulse and Momentum. If the body has general piane motion then !MG = ica = fc(dwidt). Since the moment of inertia is constant,

IMG= -(Gw)dt

Iv[ultiplying bulb sides by dr and integrating from r = ti, w - wi to t = ł,. w = w2 gives

S fr . M G = - / G t o i ( 1 9 - 1 2)

t , I n a s i T n i l a r m a n n e r , f o r ~ d u t a a b y m a f i x e d a x i s p a s s i n g t h r o u g h

point 0, Eq. 17-16 (S3/0 = 1`602/) when integrated becomes

Impulsediagram

( b )

F3 dr

Mo dt = 10w2 -log] (19-13)

Equations 19-12 and 19-13 are referred to as the prmeiple of angular impahe and MO mentem. Both equations state that the sum of the angular irnpulses acting on the body during the time interval ri to t, is equal to the change in the body's angular momentum during this time interval.

Final momen(um diagram

(c)

Fig. 19-4

S02 CHAPTER 19 PLANAR KfNETrCS OF A RICID BODY: iMPULSE AND MOMENTUM

Inilial

morncntum diagram

( a )

j:Mi dr,(„, G

W(r2— tt)

f Fzdr

f F3 dt tiMprJ1Se-diagram

(b)

To summarize łhese concepłs, if motion occurs in the x-y piane. the following ihree senior equarions can be written to describe the planar ~un of the body.

(19-14)

The tcrms in these equations can be shown graphically by drawing a set of impulse and momentum diagrams for the body. Fig. 19-4. Not e that the linear momentum yrzvG is applied at the body's mass center, Figs. 19-4a and 19-4c; whereas the angular momentum /Gw is a Irce vector, and

herefore, like a couple moment, it can be applied at any point on the body. When the impulse diagram is construct ed, Fig. 19-4b, the, forces F and moment M vary with time. and are indicated by the integrals. Howevcr, if F and M are constani integration of the impulscs yiclds F(r2 - /t) and M(r2 - Ii), respectiveiy. Such is the case for the body's weight W. Fig. 19-4b.

Equations 19-14 can also be applied to an entire system of connected bodies rather than to each body separately. This eliminates che need to include interaction impulses which occur at che connections sutce they are iniernal to the system. The resultant equations may be written in symbolic form as

J rin(vc„,), + .£ F, cit = pn(vG,),

If

ni(vG,)1 + S 1, Fy d! = nr(vG_„)2i

rz.

LITG do [ +MG di = 1G,Lo2

• 1 L

i

2 momen rum xi

syst. linear +momentum tI

cyst. linear) impulse ),(1_2)

syst_ linear

impulse )1„0-21,

( syst, linear

Xmomentu

msyst. linear .1'rnornenturri ,-2

syst, briear

„syst. augular2d(÷ ( syst. angularmornentum o t impulse 0(1_,)

( syst, angularmornen Lum f 02

11

FinałrncmiuMurri diagram

Fig. 19-4 (repeated)

(19-15)

As indicated by the third equation, the system's angular momentum and angular impulse must be computed with respect to the .same reference point O for all the bodies of the system.

5 0 319.2 PRINCIPLE OF i MPUISE AND MOMENTUM

Procedure For Analysis

Impulse and momentum principles are used to solve kinetic problems Chat involve velocity, furce, and sitne since these terms are involved in the formulation.

Free-Body Diagram.• Establish the x, y, z inertia] frame of reference and draw the freebody

diagram in order to account for all the forces and couple moments that produce impufines on the body.

• The direction and sense of the initial and fina] velocity of the body's mass center, vG, and the body's angular velocity rs should be e.stablished.lf any of these rnotions fis unknown,assurne that the sense °f lis components is in the direction of the positive inertial uvorclinates.

• Compute the moment of inertia IU or to.

• As an alternative procedure, draw the impuise and momentum diagrams for the body or system of bodies. Eaeh of these diagrams repre,sents an outlined shape of the body which graphically accounts for the data required for Bach of the dirce terms in Eqs. 19-14 or N-15, Fig, 19-4. These diagrams are particulariy helpful in order to visualize the "moment" terma used in the principle of angular impulse and momentum, if application is about the .1C or another point other than the body's masa center G or a fixed point O.

Principle of Impulse and Momentum.

• Appiy the three scalar equations of impulse and momentum,

• The angular momentum of a rigid body rotating about a tixed axis is the moment ot mvG plus /Gw about the axis. This is equal to Ho = /0w, where f0 is the moment of inertia of the body about the axis.

• A II the forces acting on the, body's free-body diagram will create an impulse; however, some of these forces will do no work.

• Forces that are functions of time musi be integrated to obtain the impulse.

• The principle of angular impulse and momentum is often used to eliminate unknown impulsive forces that are parali& or pass through a cornmon axis, since the moment of these forces is zero about this axis.

Kinematics.

• If more than three equations are needed for a complete solution, it may be possible to relate the velocity of the body's mass center to the body's angular velocity using khncrnatiea. If the motion appears to be complicated, kinematic (velocity) diagrams rnay be helpful in obtaining the necessary relation.

504 CHAPrEa 19 KANAR KfNETrGS OF A RIGID BODY: IMPULSE AND MOMENTUM

E X A M P L E

The 20-Ib disk shown in Fig, 19-5a is acted upon by a canstant couple moment of 4113 ft and a force of 10 lb which is applied to a eord wrapped around its periphery. Determine the angular vełocity of the disk two seconds after storting from rest. Also, whal are the force components of reaction at the pin?

SOLUT1ONSince angular velocity, furce, and time are involved in the problems, we will apply the principles of impulse and momentum to the solution.

(a)F= 10 lb Free-Body Diagram. Fig, 19-5b, The disk's mass center does not

move: however, the loadnig causes the disk to rotate clockwise.The moment of inertia of the disk about its fixed axis of rotation iS /A =

ntr- = 2 ( 32 20 ftlb/s-

) ( 0 . 7 5 f t ] . 2 = 0 . 1 7 4 7 s l u g • f t ' - - . 2

Principle cif Impulse and Momentum.

4 lb • ił)

mi./2A)J15. f '2

F, di -= m(vAi2

fr

rri(vA,)1 4-/.. P",d1 = in(vA,)2i

O + 4,(2 s) -- O

20 Ib

0.75 ft

A,

10 Ib O + A,(2 s) — 201b(2 s) — 10 lb(2 s) -= OJAWI + f MA cit •=

J

<b)

Fig. 14-5

O + 41b • ft(2 s) + [10 ]b[21)1([]_75 ft) = 0.1747(D2

Solving these equations yields

A, = O ,-1 f S.

Ay = 30 lhARS.

= 132 rad/s) A In

19.2 PRINC1PLE DF IMPULSE AND MOMENTUM 505

The 100-kg spool shown in Fig. I9-6a has a radius of gyration kG = 0.35 m. A rabie is wrapped around the central hub of the spool, and a horizontal furce having a variable magnitude of P = + 10) N is applied, wherc ł is in seconds. 1f the spool is initially at rest, determinc its angular velocity in 5 s. Assume that the spool roIls without slipping at A.

t a

Y I r

X

98! P = (t + 1()) N

0.4 m ?;

0.75 Mti

A

(h)

Fig. 19-6

( a )

SOLUTION

Free-Body Diagram. From the free-body diagram, Fig. 19-61), the variable fotce P will cause the friction force FA to be variable, and thus the impulses created by both P and FA must be determined by integration.Force P causes the mass center tu have a velocity vG to the right. and so the spool lias a clockwise angular velocity w.

Principle of Impulse and Momentum. A direct solution for w can be obtained by applying the principia of angular impulse and momentum about point A, the /C, in order to eliminate the unknown friction impulse.

<C+) JAWI f MA aft = łA,w2(,) Li

55 r+ I0) N dt1(0.75 m + 0A- m) = [100 kg (0.35 m)2 + (100 kg)(0.75 m)21w2

O

625(1_15) = 68.5w2

= 1.05 rad/s) Ans.

NOTE: Try solving this problem by applying the principia of impulse and momentum about G and using the principle of linear impulse and nionnenłum in the x direction.

506 CHAPTER 19 KANAR Ki NETrCS OF A RIGIO BODY: IMPULSE AND MOMENTUM

The cylinder B, shown in Fig. 19-7a has a mass of 6 kg. h is attached to a cord which is wrapped around the periphery of a 20-kg disk that has a moment of inertia Ii = 0.40 kg • m2. lf the cylinder is inirially moving downward with a speed of 2 m/s, determine its speed in 3 s. Neglect the mass of the cord in the caleulation.

— 2 mis

( a )

V R

T

58.86 N

Fig. 19-7

(b)

SOLUTION I

Free-Body Diagram. The free-body diagrams of the cylinder and disk are shown in Fig. 19-7b.A11 the forces are =staw since the weight of the cylinder causes the motion. The downward motion of the cylinder, vB, causes w of the disk to be c]ockwise.

Principle of Impulse and Momentum. We can eliminate A, and A,. from the analysis by applying the principle of angular impulse and momentum about point A. He-nce

Disk

(C +) lewi + f MA& = irAtoz

0.40 kg m2(wi) + T(3 s)(0.2 m) = (0.40 kg m2)w2

Cylinder(±t) "g(Vp)i S f F,th = nzlAva);-2

—6 kg(2 in/s) + T(3 s) — 58.86 N(3 s) —6 kg(v8)2

Kinematics. Since w = vidr, then wi = (2 mis)/(0.2 m) = 10 radis and w, = (vB)2/02 m = 5(vA2. Substtuting and soIving the equat1ons sim ultaneously for (va)7 yields

(v8)2 13.0 m/s .1,

5 0 719.2 PRINCIPLE OF i MPUISE AND MOMENTUM

O 4.frcg • rn2(1-iiihti

0 , 2 m

.q5:-EW3+ l A/ —OP-

0.2 ril tok,. (3 ,), '‚

,---'7

6 kg(2 m /s) 58.86 N(3 s) (c)

6 kg(v 8).2

S O L U T I O N I I

Impulse and Momentum Diagrams. Wc can obtain (vB), chrectly by considering thc system consisting of the cylinder, the cord, and thc disk. The impulse and momentum diagratns have Been drawn to c]arify application of the principle of angular impulse and momentum about point A, Fig. 19-7c.

Principia of Angular irnoulso and Momentum. RealiTing that = 10 rad/s and w, = 5(y8),, we have

(ć +)(=syst. angular+syst, angular(syst. angularmomentum ,i iimpulseAt I-2) momentum A2

(6 kg)(2 m/5)(0.2 ni) + (0.40 kg • rn2)( I 0 rad/s) + (58.86 N)(3 s)(0.2 m) = (

6 kg)(v8)2(0.2 m) Jr (0.40 kg m2)[5(vB)zl

(y8)2 = 13.0 m/s 1 Ans.

Fig. 19-7

508 CHAPTER 19 PLANAR KINETICS OF A IRIGIO BODY: IMPULSE AND MOMENTUM

EXAMPLE

(a)

The Charpy impact test is used in materials testing to detertnine the energy absorption characteristics of a materia] during impact. The test is performed using the penduIum shown in Fig. 19-8/1, which has a mass mass center at G, and a radius of gyration kc ahout G, Deterrnine the distance rp from the pin at A to the point P where the impact with the, specimen S should occur so that the horizontal furce at the pin A is csscntially zcro during the impact. For the calculation, assumc thc spoci= absorbs all thc pcndulum"s k inetic energy gaincd during thc tint c it falls and thereby stops the pendulum from swinging when O =

SOLUTION

Free-Body Diagram. As shown on the free-body diagram, Fig. 19-8b, the conditions of the problem require the harizontal furce at A to be zero. Just before impact, the pendulurn has a clockwise angular velocity wI, and the mass center of the pendulurn is moving to the Ieft at (vG)1 =rwa.

Principie of Impulse and Momentum. We will apply the principle of angular impulse, and momentum about point A. Thus,

4 i

60 1 + MA dt = 1,1w2

'vii — (I F di)rp = O m(vG)]

+ 2 J F dt = m(vc)2 —m(7-all

) + ji F dt = O

Eliminating the impulse f F dr and substituting fA + nar'- yields

ntr2jw,i in(7-r.fh )p = O

Factoring aut inwi and solving for rp, we obtain

rp = ± An.r

NOTE: Point P, so defined, is called the center pf percussion. By placing the striking point at P, the furce developed at the pin wili be minimized. Many sports rackets, cluhs, etc, are designed so that collision with the objecł heing struck occurs at the center of percussion. M a consequence, na "sting" or little sensation occurs in the hand of the płayer, (Also see Probs. 17-66 and 19— l .)

A,

A, - « A

rn

F

(b) .

Fig. 19-8

19.2 PRINCIPLE OF i MPUISE ANO MOMENTUM509• FUNDAMENTAL PROBLEMS

F19—L The 60-kg wheel has a radius of gyration about its center O of ko = 300 mm. If it is subjected to a couple moment of M = (3t2) N ni. where r is in seconds, deterrnine the angular velocity of the wheel when t = 4 s, stariing from rest.

F 1 9 - 1

F19-2. The 300-kg whcel bas a radius of gyration about its mass center O of ko = 400 mm. If the wheel is subjected to a couple moment of M = 300 N - m. deterrnine its angular velocity 6 s after it starts from rest and no slipping occurs. Also, dctcrminc the friction forcc that thc ground appiies to the wheel,

F 1 9 - 2

F19-3. If rod CM of negligible mass is subjected to the couple -moment M = 9 N m, deterrninc the angular velocity of the r0-kg inner gear r = 5 s after it starts from rest. The gear hasa radius of gyration about its mass center of kA = 100 mm. and it rolls on the fixed aster gear. Morion occurs in the horizontal piane.

F 1 9 - 3

F19-4, Gears A and 8 of mass 10 kg and 50 kg have raduj of gyration about tkeir respective mass centers of

= 80 mm and kfi. = 150 mm. If gear A is subjected to the couple marnem M = !O N • nt when it is at rest, determine the angular velocity af gear B when t = 5 s.

0,2 m

F 1 9 - 4

50-kg spool is subjected to a horizontal furce af P = ISO N. If the spool rolls without slipping, determinc its angular velocity 3 s after i# starts from rest. The radius of gyratkin of the spool altom its center of mass is kc = 175 mm.

P 150N

F19-5

F19-6. The reel lias a weight of 150 lb and a radius of gyration about its center of gravity °f kG = 1.2.5 ft. If ii is subjected to a torque of M = 25 lb - ft, and starts from rest when the torque is applied, dc łermine its anguIar velocity in 3 seconds. The coefficient of kinetic friction between the reel and the horizontal piane isk = 0.15.

l II_

F 1 9 - 6

51 0CHAPTER 19 PLANAR ICNETfC.S OF A RICiD BODY: IMPULSE AND MOMENTUM

■ PROBLEMS

19-1. The rigid body (slab) hasa mass m and rotates with an angular velocity w about an axis passing through the fixed point O. Show that the momenta of ali the particles composing the body can be represented by a single vector having a magnitude mt3G and acting through point P, called the center of percussion, which Iies at a distance rp/G =kufrrm

ja from the mass center G. Het-e 1(6 is the radius of gyration of the body, compuied about an axis perpendicular to the piane of motion and passing through G.

❑ c ;

Prob. 19-1

19-2. At a given instant, the body bas a linear momentumL = myc and an angular momentum-= /GW computedabout its mass center. Shaw that the angular moment um cif the body computed about the instantaneous center of zero velscity IC can he expressed as Iłże = ficw, where 1,,c represents the bodys moment of inertia computed about the instantaneous axis of zero vek-K:ity. As shown, the IC is located at a distance tctie away from the mass center G.

19-3. Show that if a slab is rolaling about a Iixed axis perpendicular to the slab and passing through its mass center G, the angular momentum is the same when computed about any other point P.

€19-1. The cakle is subjectcd to a furce of P (Wt2) lb, where r is in seconds. Determine the angular ueloeity of the spod 3 s after P is apphed, slarting from test. The spod has a weight of 150 Ib and a radius of gyration of 1.25 ft about its center of granity.

Prob. 19-3

Prob. 19-2 Prob. 19-4

19.2 PRiNciPLE OF IMPULSE AND MOMENTUM511

19-5. The impact wrench consists of a slender 1-kg rod which is 580 mm long. and cyhndrical end we-ights at A and B that each have a diameter of 20 run and a mass

❑f 1 kg.This assembly is free to tulu about the handle and socket, which are attached to the lug nut on the wheel of a car, 1f the rod iti3 is given att angolar velocity of 4 rad is and it strikes the bracket C on the handle without rebounding, determine the angular impulse impartcd ta the lug nut.

Prob. 19-5

19-7. The airplane is traveling -in a straight line with a speed of 300 km/h, when ihe engines A and B produce athrust of T A = 40 kN and T820 kN. respectively.Determine the angolar velocity of the airpłanc in r =5 s, The piane has a mass of 200 Mg, its center of mass is located at G. and its radius ot gyration about G is k0= 15 m.

Prob. 19-7

19-6. The space capsule hasa mass af 1200 kg and amoment of inertia FG900 kg • m2 about an axis passingthrough G and clirected perpendicular to the page. If it is traveling forward with a speed 1.10 = 800 m/s and executes a t urn by means of iwo jets, which provide a constant thrusi of 400 N for 0.3 s, determine the capsules angular velocity just after the jets arc turncd off.

19-8. The a.ssemb1y weighs 10 lb and bas a radius of gyration kG = 0.6 fł about its center of mass G. The kinetic energy of the assembly is 31 ft • lb when it is in the position shown.If it rolla counterelockwise on the surfacc without slipping. determine its linear momentum at [bis instant.

T=4.0 N

Prob. 19-6

Lf 1

1 ft

Prob. 19—S

S 1 2 C H A P T E R 1 9 K A N A R K I N E T r O S O F A R I G I D B O D Y : I M P U L S E A N O M O M E N T U M

19-9. The wheel having a masa of 100 kg and a radius of gyration about the z ais of k, = 300 mm, rests on the smooth horizont al piane. If the beli issubjected to a lorce of P= 200 N. determine the, angular vdocity of the wheel and the speed of its center of mass O. three seconds after the force k appIied.

19-11. The 30-kg reel is mounted on the 20-kg cart. If the cable wrapped around the inner hub of the red is subjectedto a force of 50 N. determine the velocity of the cart andthe angular velocity of the red when ł -= 4 s. The radius of gyration of the red about its center of mass O is k0 = 250 mm. Neglect the size of the smali wheels.

P = 204 N

Prob. 19-9

19-10. The 30-kg gear A has a radius of gyration about its center of mass O of 'co = 125 mm. If the 20-kg gear rack B is subected to a force of P = 200 N, determine the tlnie required for the gear to obtain an angular veiocity of 20 rad/s, starłing from rest. The contact surface between the gear rack and the horizontal piane is smooth.

200B N

1'r01]. 19-10

Prob. 19-11

19-12. The spool h as a weight of 75 lig and a radius ofgyration 1.20 tt. If the bibek B weighs 60 lb, and aforce P = 25 lh is applied to the cord, determine the speed of the block in 5 s starting from rest. Neglect the mass trf the cord.

Prob. 19-12

19.2 PRINCIPLE DF [MPULSE ANO MOMENTUM513

19-13. The slender rod hasa mass m and is suspended al its end A by a cord. If the rod reeeives a horizontai blow giving ił an imputse I at its bodom B,determLne the location y of the point P about which the rod appears to rntate during the Impact.

1915. The assembly shown consists of a 10-kg rod AB and a 20-kg circular disk C. If it is subjected to a torque of M = (202) N•rn, where t is it in seconds. determine its angular velocity when t = 3 s. When t = O the assembly is rotating at co/ = 1-61t) rad/s.

Prob. 19--13

v

Prob. 19-15

19-14. If the hall lias a weight Wand radius r and is thrown onto a rough surface with a velocity vsl parallel to the surface, determine ihe amount of backspin, edo, it musi be given so that it stops spinning at the same instant that its forward velocity is zero. It is not neeessary to know the coefficient of friction at A for the calcuIałion.

19-16. The frame of a tandem drum roller lias a weight of 4000 lb e xeluding the twa roIlers.Each roller lias a weight of 1500 Ib and a radius of gyration about its axle of 125 ft. If a torque of M = 300 lb•ft is supplied to the rear roller A. cleterrnine the speed of the drum roller 10 s later, starting from rest.

dl

A

Prob. 19-14 Prob. 19-16

514 CHAPTER 19 PLANAR KINETICS OF A RIGID BODY: IMPULSE ANO MOMENTUM

19-17. A motor transmits a tamile of M = 0.05 N • m ło the center of gear A. Determine the angular velocity of each of the three (equal) smaller gears in 2 s sla_Tting from Test. The smaller gears (B) are pinned at their centers. and the masses and centroida! radli of gyration of the gears are given in the figure.

19-18. The man pulls the rope off the reel with a consłanł force of 8 lb in the direction shown, If the reel bas a weight of 250 lb and radius of gyration kG = 0.8 ft about the trunnion (pin) at A, deterinine the angular velocity of the red in 3 s starting from rest. Negleet friction and the weight of rope that is removed.

Prob, 19-18

19-19. The double pulley consists of twn wheels which are attached to one another and tum at the same race. The pulley has a mass of 15 kg and a radius of gyration 40 -= I 10 mrn, If the bIock at A has a mass of 40 kg, determine the speed of the Noc! in 3 s after a constans force P = 2 kN is applied to the rope wrapped around the inner hub of the pulley. The błock is originaIly at rest. Neg I ect the mass of the !wc.

Prob. 19-19

19-20. The cable is subjected to a furce of P= 20 Ib. and the spocił roik up the rai] without slipping. Determine the angular velocity of the spool in 5 s. słarting from rest, The spool has a weight of 100 lb and a radius of gyration about łts center of gravity O of kej = 0.75 ft.

Prob. 19-20

rn,., = 0.8 kg

= 31 mm

mh = 0.3 kg

kh = 15 mm

Prob. 19-17

192 PRINCIPLE OF i MPUISE AND MOMENTUM515

19-21. The inner hub of the wheel rests on the horizontal track. it does not slip at A, determine the speed of the 10-lb block in 2 s after the bIock is released from rest. The wheel has a weight of 30 lb and a radius of gyration kG = 1.30 ft. Negleet the mass of the pulley and card.

19-23. The 100-kg red bas a radius of gyration about its center of mass G of kG = 200 mm. ff the cable Bis subjected to a force ol P = 300 N, determine the linie required For the reel to obtain an angular velocity of 20 rad/s. The coefficient of kinctic friction between the reei and the piane is Juk = 0.15.

Prob. 19-23

Prob. 19-21

19-22. The 1.25-1b tennis racket bas a center of granity at G and a radius of gyration about G of ku, = 0.625 ft. Determine the position P where the bali most be hit so chat no sting is Pelt by the hand holding the racket, i.e.. the horizontal force exerted by the racket on the hand is zero.

19-24. The 30-kg gear is subjected to a lorce of P= (200 N. where t is in seconds. Deterrnine the angular velocity of the gear at r= 4 s, storting from Test. Gear rack B is fixed to the horizontal piane, and the gears radius of gyration about its mass center O is ko = 125 ram.

19

Prob. 19-22

P = (20!) N

f l r u - S f t r J Y f i n n

Prob. 19-24

516 CHAPTER 19 PLANAR KfNETrCS OF A RIGIO BODY: IMPULSE AND MOMENTUM

19-25. The doubIe pulley oonsists of two whecis which arb attached to one another and turn at the same rate.The pulley has a mass of 30 kg and a radius of gyralion k0 = 250 mm. If two men A and B grab the suspended ropes and step off the iedges at the same time, determine their speeds in 4 s starting from rest.The men A and B have a mass of 60 kg and 70 kg, respectively. Assume they do not move relałive to the rope durMg the rnotion. Ncglect the mass of the rope,

Prob. 19-25

19-26. If the shaft is subjected to a torque of M = (I 5/2} N. rn, where r is in seeonds, determine the angular velocity of the assembly when r = 3 s, starting from rest. Rods AB and Be each have a mass of 9 kg.

1.9-27. The square plate has a mass m and is suspended at i ts corner A by a card. If il receives a horizontar impulse I at corner B. determine the location y of the point P about which the plate appears to rotate during the Impact.

Prob. 19-27

*19-28. The crate has a mass mc. Deterrnine the constant speed UQ it acquires as it =yes down the conveyor, The rollers each have a radius of r, mass m, and arc spaced d apart, AFate chat friction causes bach miler to rotate when the crate comes in contact with it,

Prob. 19-26 P rob. 19-28

19.3 CoNsERvATioN OF MOMENTUM 517

1 9 3 Conservation of MomentumConservation of Linear Momentum. If the sum of alt the linear impteLres acting on a system of connected rigid hodies is zero in a specific direction, then the linear momentum of the system is constanł, or conserved in this direction, that is,

syst. linear 2 syst. linear- ' rnomentuinh f ( momentum,i)

7

(19-16)

This equation is referred to as the conservarhm of Linear momentum.Without inducing appreciable errors in the calculations. it may be

possible to apply Eq. 19-16 in a specified direction for which the linear impulses are smali or nonimpuisive. Specifically, nonimpulsive forces occur when smali forces act over very sport periods of time. rypical cxamples include the furce of a slightly deformcd spring. the initial contact furce with soft gro un d, and in some cases the wcigh t of the body.

Conservation of Angular Momentum. The angular momentum of a system of connected rigid I-aodies is conserved about the system's center of mass G. or a fixed point O. when the sum of alt the angular impulses about thesc points is 7cto or appreciably smali (nonimpulsive). The third of Eqs, 19-15 ]hen becorries

( ._.., syst. angular2.i . ł syst. angularmomentum ol momentum 02

(19-17)

This equation is referred to as the conservadon of angular momerazon. In the case of a single rigid body, Eq. 19-17 applied to point G becornes (iGm)[ = orGw).... For example, consider a swimmer who executes a somersault after jumping off a diving board. By tucking his arms and lcgs in closc to his ch est. he decreases his bocly's moment of inc rti a and thus inereases his angular vclocity (f6w must be constant). If he straightens out just beforc entcring the water, his body's moment of inertia is increased, and so his angular velocity decreases. Since the weight of his body creates a linear irnpulse during the time of modon, this exa-mple also illustrates how the angułar momentum of a body can be conserved and yet the lineał- momentum is nut. Such cases occur whenever the external forces creating the linear impulse pass through either the center of mass of the body or a fixed axis of rotation.

J`18 CHAPTER 19 KANAR Ki NETrCS OF A RIGiO BODY: IMPULSE AND MOMENTUM

The conservation of linear nr angolar momentom should be applied using the following prncedure.

Free-Body Diagram.

• Establish the x, y inertial frame of reference and draw thc frec-body diagram for the body or system of bodies during the Orne of impact. From this diagram ciassify each of the applied forces as being either "Impu1sive" or "nonimpulsive."

▪By inspection of the free-body diagram, the conservation of frnnur

mornewurn applies in a given direetion when rw external impulsivc forces act on the body or system in that direction: whcreas the conservation of angular momenturn applics about a fixed point O or at the mass center G of a body ar system of bodies when all the external i triputsive forces acting on the body ar system create zero moment (ar zera angular impulse) about O or G.

• As an alternative procedura. draw thc impulse and momentum diagrams for the body ar system of bodies, Thesc diagrams arc particularly heipful in order to visualize the "moment" terms used in the conservation of angular momentum equation, when it has Been decided that angolar momenta are to be computed about a point other than the body's mass center G.

Conservation of Momentum.

• Apply the conservation of linear ar angular mornentunn in the appropriate direetions.

Kin ernati cs.

• If the motion appears to be compbcated, kinematic (velocity) diagrams may be helpful in obłaining the necessary kinematic relations.

19.3 CONSERVAT ION OF MO MENTUM519

EXAMPLE

simplifying yields(vG), = 0.8921(%)1(1)

Conservation of Energy. In order to roll over the obstruction, the wheel must pass pasition 3 shown in Fig. 19-9c. Hence, if (L5G)2 [or (nG)] is to he a minimum, it is necessary that the kinetic energy of the wheel at position 2 be equal to the potential energy at position 3. Placing the datum through the center of gravity. as shown in the figure, and applying the conscrvation of energy cquation, we have

{Tz}1/,2} = {7.3}{1/3}

{-I( I 0 kg)(vG)-.--(0.156 kg rn2),(0.;){o} ={o}{(98.1 N)(0.03 m)}

Suhstituting col = 5(?)).., and Eq. 1 into this equa.tion, and solving,(v0)] = 0,729 miisAns_

*This principle does not apply during impact. since energy śs iosł during the collision. However. 'ust after impact, as in Fig. 19-9c, it can be used.

( b )

18.114

D

(c)

Fig. 19-9

The 10-kg wheel shown in Fig. 19-9a lias a moment d nenia fc = 0.156 kg - m2. Assuming that the wheel does not slip or rebound, deternrine the minimum velocity va it must have to just roll over the obstruction at A.

SOLLITIONimpuise and Momentum Diagrarns. Since na slipping or rebounding occurs, the wheel essentially pivots about point A during contact. This condition is shown in Fig. 19-9b, which indicałes, respecłiveiy, the momentum of the wheel pist before impacr, the irnpulses given to the wheel during impacr, and the momentum of the wheel jusr afrer impact. Only two impulses (forces) act on the wheel. By cornparison, the force at A is much greater than that of the weight, and since the time ot impact isvery sport, the weight can be considered nonimpulsive. Thc impulsive force F at A has both ars unknown magnitudc and an unknown direction O. Ta eliminatc, this kirce trom the analysis, note that angular momentum about A is essentially conserved since (98.1 At)d = 0.

Conservation of Angular Momentum. With reference to Fig. 19-9b.

+)(J4)1 = (HA)2r'in(vG)1i,;(01 = rm(vG)2Irc-p2

(0.2 m — 0.03 m)(10 k,g)(vG)[ + (0.156 kg • m2)((-01) = (0.2 n-1)(10 kg)(vG)2 + (0.156 kg - 1112)(<02)

Kinematics. Since no slipping occurs, in gencraI w = v<Vr = vc/0.2 m = 5yG. Substituting this into the abovc equation and

---- (0 2 — 0.03) in

5 2 OCHAPTER 19 KANAR ICNETfC.S OF A RICID BODY: IMPULSE AND MOMENTUM

EXAMPLE

1 : 1 3 5 1 1 .

The 5-kg slender rod shown in Fig. 1.9-10a is pinned at O and is

initially at rent. If a 4-g bullet is fired finto the rod with a velocity of 41430 m/s, as shown in the figure, determine the angular velocity of the

075mrod just after the bullet becomes e,mbedded in it.

SOLUTION

025 mImpulse and Momentum Diagrams. The impulse which the bulletexerts on the rod can be eliminated from the analysis, and the angular vclocity of the rod jus! after impact can be determinecl by considcring the bullet and rod as a single system.To clarify the principles involved. the impulse and momentum diagrams are shown in Fig. 19-10b. The momentum diagrams are drawn just before and jusy after Impact. During impact, the bullet and rod exert equal but opposde interna! impulses at A. As shown on the impulse diagram, the impulses that are exte,rnal to the system are due, to the re,actions at O and the weights of the bullet and rod. Since the time of impact. dt. is very sport, the rod moves only a slight amount, and so the "rnoments" of the wcight impulses about point O arc essentially zero. Therefore angular momentum is conservcd about this point,

O,. Ar

O, At

0.5 in

ms(vii)i 0.75m 49.05 d10.75m : : ( 0 2 G fuR(vG)2

~. IP B (V B)20.0392 At

(b)

Conservation of Angular Momentum. From Fig. 19-10b. we have

(C+) 1(K0)1 = i(No)2

rna(vŃ3) cos 30°(0.75 m) = nia(vn)0.75 m} -1-rn(vG)2(0.5 m) +(0.(X)4 kg)(400 coh. 30° mis)(0.75 nn) =

(0.004 kg)(%)2(0.75 m) + (5 kg)(vG)?,(0.5 ni) + [5(5 kg)( 1 in)".'](.02 (1)OF

Fig. 19-10

1.039 = 0.003(vB)2 2.50(vG)2 + 0.4167w-,

Kinematics. Since the rod is pinned at O. from Fig. 19-10c we have

(vG)2 = (0.5 in)to-, (vR)2 = (0.75 in)co, Substituting finto Eq. 1 and solving yiclds

(02 = 0.623 nidfs

( a )

19.4 ECCENTRC IMPACT521

*19.4 Eccentric Impact

The concepts involving central and oblique impact of particles were presented in Sec. 15.4. We will now expand this treatment and diseuss the eccentric impact of two bodies. Eccentric impact occurs when the line connecting the mm centers of the two bodies does n« coincide with the line of impact. This łype of impact often occurs when onc or both of the bodies are constrained to rotate about a fixed axis. Consicler. for example. the collision at C berween the two bodies A and B, shown in Fig.19-11ś7. It is assumed that just before collision B is rotating counterclockwise with an angular velocity (w8)1, and the velocity of the comact point C Iocated on A is (uA), . K-inematic diagrams for both bodies just before collision are shown in Fig. 19-1 I b. Provided the bodies are smooth, the impulsive forces they exert on each other are direcied along the line of impact, Hence, the component of velocity of point Cm body R, which is directed along the- line of im pact, is (ve) I = (e.s,,) I r, Fig.19-11b.I,ikcwise, on body A the cum ponent of velocity (u,4)1 along the line of impact is (v,4)t. In order for a collision to occur,(vA)1

During the impact an equal but opposite impulsive forte P is exerted between the bodies which deforrns their shapes at the point of eontact. The resulting impulse i.s shown on the impulse diagrams for both bodies. Fig. 19-11e, Note that the impuLsive Lirce at point C on the rotating body creates impulsive pin reactions at O. On these diagrams it is assumed that the impact creates forces which arc much larger Chan the nonimpuIsive weights of the bodies, which are not shown. When the deformation at point C is a maximum, C on both the bodies movcs with a common velocity V along the line of impact, Fig. 19-I1 d. A period of restitution then occurs in which the bodies tend to regain their original shapes. The restitution phase creates an equal but opposite impulsive forte R acting between the bodies as shown on the impulse diagram, Fig. 19-11e, After restitution the bodies move apart such that point C on body B has a velocity (v8), and point C on body Ahas a velocity Fig. 19-11f, where (yzę),(VA

In general, a problem involving the impact of two bodies requires deterraining thc two rinknowris (v,1)2 and (1.,R).-,, assuming (y,k)1 and (N), arc known (or can be dctermined using kincmatics. encrgy methods, the equations of morion, etc.). To solve such problems, twa equations must be written.The frrst equarion generally Involves application of the consetvation of angalar rracvmenrEtrra to the two badies. In the cale of both bodies A and B, we can state that angular momentum is conserved about point O since the impulses at C are internal to the system and the impulses at O create zero mmcm (or zero angular impulse) about O. The second equation can be obtaincd using the definition of the coefficient of restiturion, e, which is a rafio of the restitution impulse to the deformaiion impulsc.

L ine o f i m p a c t

B

PI a n,: ut impact(a)

Fig. 19-11

Herc is an example of ecceni.ric impact"When [hese lincs coincide, central impacł occurs and the problem can be anałyzed asoceurring between this bowl i ng balicliscussed in Sen. i5.4. and pin.

522 CHAPrER 19 KANAR 1.CNETfC.5 OF A RICID BODY: IMPULSE AND MOMENTUM

t5O,.d1

i-0. o, d,

Pdt

Velocitybefore mllisiort

(b)

Defoririation impulse

(c)

Restitution irnpuise

(e)

cif Rdr

Velocilv at maximum deforrnation

(d)

tt4)2

Velocityafter co I lision

Is is important to realize, however, Chat this arialysis has only a very limiłed appiicadon in engineering, because values of e for this case have bem found to be highly sensiiive to the małerial, geometry, and ihe vedrycity of each of the colliding bodies: To establish a useful form of the coefficient of restitution equation we must first apply the principle of angular impulse and momentum about point O to bodies B and A separately, Combining the rcsuits, we then obtain the neccssary cquation. Proceeding in this manner, the principle of impulse and momentum applicd to body B from the time }ust before the collision to the instant of maximum defonnation, Figs. 19-11b,19-1 lc, and 19-11d, becomes

(C -F) Id(w8)[ + r P dr low (19-18)

Here i is the moment of inertia of body B about point O. Similarly, applying the principle of angular impulse and momentum from the Instant of maximum deformation to the time jurt after the Impact, Fi&s.19-11d, 19-11e, and 19-11f, yields

(C +) locv + r j R dt = /0(0)8)2 (19-19)

Solving Eqs, 19-18 and 19-19 for f P dr and f R di, respectively. and formulating e, wc have

f Rdr r(WB)2 - rw (vB)2 - v

e ` i P rh - r(w 8)1 - (N)[

( t j

Fig. 19-11 (cont.)

19.4 ECCENTRC livn,ACT523

In the same manner, we cari write an equation which relates the magnitudes of velocity (vA)i and (N)2 of body A. The result is

— (va)2 ,

4)1 — v

Cornkrifflng the above two equatiorts by eliminating the common velocity yields the des;.red result, i.e.,

(vri)-). (vA}-

(vA)1 (v.13)1(19-20)

This equation is identical to Eq. 15-11. which was derived for the central impact between iwo partieles. It stater that the coefficient of restitution is equal to the ratio of the relative velocity of separation of the points of contact (C) just after impact to the relative velocity at which the points approach one another just before impact. In deriving this equation, we assumed that the points of contact for bot]] bodies move up and to the right boi h before and after impact. If rnotion of any one of the contacting points occurs down and to the left, the velocity of this point should be considered a negative quantity in Eq. 19-20.

D arMg impact the eolu= ol marsy highway signs arc intended to break ola ot their supports and easily coilapse at ti-seirjoints.Th+S is Shown by the slotted connections at their bac and the brak at the cołumn's midsection.

524 CHAPTER 19 I"LANAR KfNETrOS OF A RICH) BODY: IMPULSE ANO MOMENTUM

19.11

(2)

1.5 f t

0 )

Au-r

1.5 ft

The 10-lb slender rod is suspended from the pin at A, Fig. 19-12a, If a 2-ib bali B is thrown at the rod and strikes its center with a velocity of 30 ft/s, determine the angular veloci ty of the rod just after impast, The coefficien t of restitution is e = 0.4.

SOLUTIONCo nservation of Angula r Momentum. Consider the bali and rod as a system. Fig. 19-12b.Angular momentum is conserved about point A since the impulsive forte between the rod and bali is internal. Also, the weIghłs of the bali and rod are nonimpulsive. Noting the directions of the velocities of the, bali and rod jurt after impast as shown on the, kmematic diagram. Fig. 19-12c. we require

( + ) (II A)1 = (11„)-_,

Since (vG)2 = 1.5w, then2.795 = 0.093 I 7(?,,B)2 f 0.9317m,(1)

Coefficient of Restitution. With referentce to Fig. 19-12c, we have

Ans..

12.0 = 1.5w:_. — (v6)2 (2)

Solving Eqs. 1 and 2, yields(vB)2 = —6.52 ft is = 6.52 ftis <-

w2 = 3.65 rad/s

30 I i /s

(c)Fig. 19-12

n-i(vir?)1(1.5 ft) = Pn(v a),-,(1 .5 ft) + nik(vG):.(1.5 ft) + ./Gw2

( 2 l b 32.2 flis2)(3e) ft/s)(1-5 ft) = ( 2 lb \\

32.2 kis2)(4)2(1.5 ft)

10 Ib 1- 1 ( 10 lb }i32.2 ft/s2i(vG)2(1.5 ft) + 1.12 ks 32.2 ft/s213 ftHw2

(vG)2 (vB), (1.5 11)(02 (vB)2N — 0 . 4 —

(.v8)1 — (%)i 30 ft/s —

"L 's ff

2 I b

19.4 ECCENTRC IMPACT525

E PROBLEIVIS

19-29. A man has a moment of inertia 1, about the z axis. He is originany at resi and standing on a smal] platform which can tum freely. If be is handed a wheel which is rotating at w and bas a moment of inertia J about its spinning axis, determine bis angular velocity if (a) he holds the wheel upright as shown, (b) turris the wheel out, O = 90°. and (c) turns the wheel downward, = 180°. Neglect the effect of holding the wheel a distance, d away Erom the z axis.

Prob. 19-2919-30. Twe wheels A and B have masses mA and mi?, and radii of gyration about their central vertica/ axes of kA and

respectiveły. If they are freely rotating in the same direction at wA and 018 about the same verdeal axis. determine their cornmon angular velocity after they are brought finto contact and slipping between Chero stops.

19-31. A 150-lb man kaps off the circular platform wilk a velocity ofarnf~= 5 fr/s,relative to the platform. D etermine the angular velocity of the platform afterwards. Initially the man and platform are at rest. The platform weighs 300 łb and can be treated as a uniform circular disk.

Prob. 19-31

19-32. The space satcllite bas a mass of 125 kg and a moment of inertia Ż. = 0.940 kg • m2, excluding the, four solar paneis A, B, C. and D. Bach solar panel hasa mass of 20 kg and can be approximated as a Chin plate. If the satellite is griginally spinning about the z axis at a constantrace= 0.5 rad/s when H = 90°, determine the rate ofspin if alt the panels are raised and reach the upward posiłion, O = 0°, at the same instant.

Prob. 19-32

19-33. The 80-kg man is holding Iwo dumbbells while standing on a turntable of negligible mass, which tunis freely about a verłical axis, When his arras are fully extended, the turntable is rotating with an angular velocity of 0.5 rev/s. Deterrnine ihe angular velocity of the man when he retracts his arms to łhe position shown. When bis arras arc fully extended, approximate each arm as a uniform 6-kg rod hav-

ing a length of 650 mm, and bis body as a 68-kg solid cylinder of 400-mm diameter. With bis arms in the 19 rctracted position, assume (ho man is an 80-kg solid cylinder of 450-mm diameter. Each dumbbell consists of Iwo 5-kg spheres of negligible size,

Prok 19-33

5 2 6 C H A P T E R 1 9 P L A N A R K [ N E T r C S O F A R I C I D B O D Y : I M P U L S E A N D M O M E N T U M

19-34. The 75-kg gymnast lets go of the horizontal bar in a fully stretched position A, rotating with an angular velocity of (dA = 3 rad/s. Estimate his angular velocity when he assumes a łucked position £I. Assumc the gymnast at positions A and B as a uniform slender rod and a uniform circular disk. respecfively.

*19-36. The 5-11) rod AB supports the 3-lb disk at its end.If the diak is given an angolar velocity [91, = 8 rad/s while the rad is held stationary and then released, determine the angular velocity cif the rod after the disk has stopped spinning relative to the rod due to trictional resistance at the bearing A. Iviotion is in the horkonłał piane. Neglect friction at the fixed bearing B.

754:31-nm 5 rad/s

Prob. 19-34

19-35. The 2-kg rod ACB supports the two 4-kg disks at its cuda. If hoth disks ara given a elockwise angular veiocityKOI = i = 5 rad/s while the rod is held stationary andthen released, determine the angular veiOcity of ihe rod after both disks have stopped spinning relative to the rod due to frictional resistance at the pins A and B_ Tvfotion is in the horizol-frał plano. Neglect friction at pin C.

19-37. The pendulum consists of a 5-1b slender rod AB and a 10-lb wooden block. A projectite weighing 0.2 lb is fired into the center of the bkick with a velocity of 1000 ft/s, If the pendulum is initially at rest, and the projeeftle embeds itself into the hlock. determine the angular velocity of the pendulum just after the impact.

2 f i

v. It>1101= ,

I f i

; - - 1 f t -

Prob. 19-37

0,4

0.15m(w8 )1 [7.15

0.75in

PrI■13.19-35

Prob. 19-36

19.4 ECCENTRC iMPACT527

19-38. The 20-kg cylinder Ais free to slide along rod BC. When the cylinder is at x = 0, the 50-kg circular disk D is rotating with an angular velocity of 5 rad /s, If the cylinder is given a slight puste. determine the angular velocity of the disk when the cylinder strikes B at x = 600 mm. Neglect the mass of the brackets and the smoot h rod.

19-40. The uniform rod assembly rotates with an angular velocity of r,r() on the smooth horizontal piane just before the hook strikes ihe peg P without rebound. Determine the angular velocity of the assembly immediately after the impact. Each rod has a mass of m.

Prob. 19-40

Prob. 19-38

19-39. The slender har of mass m pivots at support A when it is released from rest in the vertied position. When iii falls and rotates 90°, pin C will strike support B, and the pin at A will leave its support. Dełermine the angular velocity of the har immediałcly after the impact. Assume the pin at

will not rebound,

19-41. A thin disk of rnass m lias an angular velocity cdi while rotating on a smoodi surface. Deterrnine its new angular velocity ;ust after the hauk at Its edge strikes the peg P and the disk starts W rotate abk-att P withoutrebounding.19

Prob, 19-39 Prob. 19-41

528CHAPTER 19 PLANAR KINETrCS OF A RIGIO BODY: IMPULSE AND MOMENTUM

19-42. The vertical shaft fis rotating with an angular velocity of 3 rad/s when H = 0".1f a furce Fis applied to thecollar so that N = delermine the angular velocity of theshaft. Also, find the work dane by furce F. Neglect the mam of rods Gil and EF and the collars I and J. The rods AR and CD each have a mass of 10 kg.

19-44. A 7s bullet having a velocity of 800 m/s 3s fired finto the edge of ihe 5-kg disk as shown. Determine the angular velocity of the disk just after the bullet becomes embedded in it. Also. calculate how far the disk will swing until it stops.The disk is ariginaliy at rest,

0.3 m03 m

0.3 rri0.:+ ITI

E1JO I M

/1

Prob. 19-42

Prob. 19-44

19-43. The mass center of the 3-1b hall has a velocity of (v6)1 = 6 ftis when it stdkes the end of the smooth 5-Ih slender bar which fis al test. Determine the angolar veIocity of the bar ubaw the z axis just after irnpact if e = 0.8,

0.5 ft

Prob. 19-43

19-45. The IO-lb block is sliding on the smooth surlace when thc eorner D hity a stop block S. Deterrninc the minimum velocity v the block should have which would allow it to tip over on its side and land /n the position shown. Neglect the size of S. Hiw: During impact consider the weight of the block to be nonimpolsive.

AB

13

Prob. 19-45

V

H - 1 C

E n

A Srl

19.4 ECCENTRC iMPACT 529

19-46. The two disks each weigh _I 0 lb, If they are rekased from rest when O = 34°, determine O after they collide and rebound from each other. The eoefficient oI restitution is e --- 0.75. When O = T. the disks hang so that they just touch one another.

2 f i

Prob, 19-46

19-47. The pendulurn consists of a 10-łb solid bali and 4-1b rod.If it is released from rest when Odetermine the angle

after the bal] strikes the wali, rehounds, and the pendulurn swings up to ihe point ot momentary rest. Take e = 0.6. 0.3 ft.

Prob. 19-47

*1.9-48. The 4-1b rod AB is hanging in the vertiwl position. A 2-lb block,sliding.011 a smooth horizontal surface with a veloeity of 12 ft/s, strikes the rod at its end R. Determine the velocity of the blask immediately after the colłision, The coeftciern of restitwtion between the block and the rod at B is e =

121:lits

Prob. 19-48

19-49. The hammer conssts ol a 19-kg solid cylinder C and 6-kg uniform sic nder rod AR. If the hammer is released from rest when O = 90n and strikes the 30-kg hlock D when O = 0°, determine the velocity of blosk 19 and the angular velocity of the harnmer inunediately after the impaet. The coeffident of restitution loetween the hammer and the block is e = 0.6,

Prob. 19-49

19-511. The 6-lb slencler rod Ali is originally at rest. suspended in the vertical position. A 1-lb bali is thrown at the rod with a velocity v = 50 ft/s and strikes the rod at C. Determine the angular ve1ocity of the rod just after the impact. Take e = 0.7 and d = 2 ft,

CO

v = 50 tt/s

BProb. 19-50

0_3 ft

2f[

3 f t d

3 F1

19

5 3 0 C H A P T E R 1 9 P L A N A R K f N E T r C S O F A R I G I O B O D Y : I M P U L S E A N D M O M E N T U M

19-51. The solid bali of mass m is dropped with a velocity vi anto the edge of the rough step. If it rebounds horizontalły off the step with a velocity v,. deterrnine the angle 0 at which contact occurs. Assume no slipping when the hall strikes the step. The coefficient of resfitution is e,

Prob. 19-5l

19-52. The wheel has a mass of 50 kg and a radius of gyration of 125 mm about its center of mass G. Determine the minimum value of the angular velocity col of the wheel, sa that it słrikes the step at A without rebounding and then roi is over it without slipping.

19-53, The wheel has a masa of 50 kg and a radius of gyration of 125 mm aboui its center of mass G. If it ro!ls without slipping with an angular velocity cif <dl = 5 rad/s hefore it strik es the step at A, determine its angular vclodty after it roIls over the step. The wheel does not lose contact with the step when it strikes it.

Probs. 19-52(53

19-54. The disk has a mass m and radius r. If it strikes the step without rebounding, deterrnine the largest angular velocity Ó,1., the chsk can have and not Iose contact with the step.

P rob. 19-54

19-55. A solid bali with a mass m is thrown on the ground such that at the instant of contact it bas an angular vclocty f.di and velocity components (vG), I and (vG),1 as shown. [f the ground is rough so no slipping occurs. deterrnine the cornponents of the velciciły of its mass center just after impast. The coefficient of restitution is e.

Prob. 19-55

'19-56. The pendulum consists of a 10-1b sphere and 4-lb rod, If it is reLeased from rest when .0 = 90°, determine

the angle H of rebound after the sphere strikes the fluor. Take = 0.8.

Prob. 19-56

19.4 ECCENTRC IMPACT 531


P19-1. The soil compactor rnoves forward at constant vclocity by suppłying power to the rear wheels. Use appropriate numerical data for the wheel, roller, and body and calculate the angular momentum of this system about point A at the ground, point B on the rear axle, and point G, the center of granity for the system.

P19-1

P19-2. The swing bridge opens and closes by turning using a motor located under the center of the deck at A that applies a torque M to the bridge. If the bridge was supported at its end B, would the same torque open the bridge at the same time, or would it open slower or faster? Explain your answer using numerical values and an impulse and momentum analysis. Also, what are the benefits of making the bridge have the variable &pih as shown?

P 1 9 - 2

P19-3. Why is ił necessary to have the tail blade B on the lielicopter that spins perpendicular to the spin of the main blade A? Explain your answer using umerical vałues and an impulse and momentom analysis.

P 1 9 - 3

P19-4. The amusement park ride consists of two gondolas A and B, and counterweights C and D that swing in opposite dircetions. Using realistic climensions and mass, calculate the angular momentum of Ibis system for any angular position of the gondolas. Explain through analysis why il is a good idea to design ibis system to have counterweights with each gondola.

P 1 9 - 4

532CHAPTER 19 PLANAR ZENETFGS OF A RICiD BODY: IMPULSE AND MOMENTUM• CHAPTER REVIEW

Linear and Angular Momentum

The linear and angułar momentom of a rigid body can be referenced to its rnass center G.

f the angular momentom is to be determined about an axis other than the one passing through the mass center, then the angular momentum is detcrmined hy summing vector HG and the moment of vector L abouł this axis.

L = nivGL = niv<i L = ma0

Ku = O H0 = 1FGW H0=fcw

HA = («G)d HA = [Gw {rnvo}d

Principle of Impuke and Momentom

The prineiples of I near and angukr impulse and momentum are useki to solve problem~ that in voIve (orce, veloci ty, and tiule. Before applying these equations, it is important to establish the x, y, z incrtial coordinate system. The free-body diagram for the body should also be drawri in order to accouni for alt °f the forees and couple moments that produce impulses on the body.

tn<vG,), + 2 fF, dr -=

+ I Fy dr = pi(vG,.)2

j .rr 2

!GW' +MG 111 = kW:,

TranslatIon

Rotation about a fixed avk

AGeneral piane motion

C HAPTER R EV1EW 533

Conservation of Momentum

Provided the sum of the linear impulses acting on a system of connected rigid bodies is zero in a particular direcłion, then the Linear momentom for the system is conserve d i#1 th s (tirection. Con5ervatign of angular rnomentum occurs if the impulses pass through an axis ar are parallel to it. Momentum is also conserved if the externał forces are smali and thereby create nonimpuNive forces on the system. A free-body diagram should accompany any application in order to classify the forces as impulsive or nonimpulsive and to deterrnine an axis about which the angular momentom may be conserved.

Eceentric Impact

If the line of impact does not coincide with the line connecting the masa centers of two eoilidmg bodics, then eccentric impact will occur. Tf the motion of the bodies just after the impact is to be dełermined. Ulen it is necessary to consider a conservation of momentom equation for the system and use the coefficien t of resiitution eyualion.

▪syst. ł inew(.,, syst. linear

E'rnorn C11 LIM1 ) i — .r momentom),

( ▪ syst. angularmomentum )0 j

momentum )0,syst. angolar

(v B)2 (vA)2e -

(1)R}1 — ft)i

ReviewPlanar Kinematicsand Kinetics of aRigid Body

Having presented the various topics in planar kinematics and kinetics in Chapters 1.6 through 19, we will now surnmarize thcse principles and provide an opportunity for applying them to the solution of various types °f problems.

Kinematics. Here we are interested M studying the geometry of motion, widiom concern for the forces which cause the motion. Before solving a planar kinernatic,s problem, it is first necessary to classify the mofion as being either rectilinear or curvilinear transiation, rotation about a fixed axis, or general piane motion. In particular, problems involving genua] piane motion can be solved cither with reference to a fixed axis (absolute motion analysis) or using translating or rotating frames of reference (relative motion analysis). The choice generally depends upon the type of constraints and the probiern's geometry. In all cases, applicałion of the necessary equations can be clarified by drawing a kinernatic diagram. Remember that the veloeity of a point is always tangent' to its path of motion, and the aceeleration of a point can have components in the n—t directions when the path is curved.

Transiation. When the body moves with rectilinear or curvilinear translation, ali the points on the body have the same morion.

//B = VA aR aA

Rotation About a Fixed Axis. Angular Motion.Variobk Angukr Acceleratiun. Provided a nnathernatical reIationship is given between any iwo of the four variables 8, w, Ci , and r, then a third variabic can be determined by solving one of the following equations which reIate all three vartables.

dO dww =, a a (18 =. w dw

di dr

Conmant Angular Aceeleration. The following equations apply when it is absolutely ceriain that the angular aeceleration is constant,

8 = 80 ł worw = wo £1,2 = ra.? — Op)

REVłEW 2 PLANAR KNEMATICS AND K1NETICS OF A RFGID BODY535

Motion of Point P. GUC w and a have been determined. then the circular motion of point P can be speeifie4.1 using the foilowing scaiar ar vector equations.

W r 1 1 = = X r

a,= a„= a=a X r— 6J2r

Generał P1ane Motion—Reiative-Motion Analysis. Recall that when transiałing axes are placed at the "bose point” A. the relative motion of point B with respect to A is shnply eirenlar motion of B abort! A. The following equations apply to two poinłs A and B located on the same rigid body.

Vs = VA VsłA •-•• VA X rs7A

aB = aA aB/A — oA + a X rB/A w2rBm

Rotałing and translating axes are often used to analyze the motion af rigid bodies which are connected together hy collars or slider blocks.

37.8 - v, + + (v ,1„...)„:

aB = aA + 1). X rs/A + 1 x (n x rp/A) + (.17.8,,A).yz +

Kinetics. To analyze the forccs which cause the motion we musi use the principles af kinctics. When applying the necessary equations, it is important to firsł establish the inertial coordInate system and define the posiłive directions of the axes. The directions should be the same as those selected when writing any equations of kinematics if simultaneous sointion of equations becomes necessary.

Equations of Motion. 'These equations are used to determine acceIerated mations or forces causing the motion. If used to determine posiłion, velocity, ar tirne of motion, then kinematics will have to be considered to complete the solu.tion. Before applying the equations of ~bon, always draw a free-body diagram in order to identify all the forces acting on the body. Also, establish the directions af the acceleration of thc mass center and thc angular accelcration of thc body, (A kinetie diagram may also be drawn in order to represent inoG

and 10a graphically. This diagram is particularly convenient for resoIving mac into components and for identifying the terms in the moment sum .':(Jit,,)p.)

536 REviEW 2 PLANAR KNEMATICS AND KINETICS OF A RIGID BODY

Tle three equations of motion are

In(aG),

Q F y = ~ O ,

SMG = Ic;cg or £/417, = S(yfi)e

In particular, if the body is rołałing about a ftxed axis, moments may also be summed about point O on the axis, in which case

'Mo = = ioa

Work and Energy. Tle ~atm of wark and energy rs uwd to solve prohlems involving force, velocity, and di.splaeemeeny. Before applying this equation, always draw a free-body diagram of the body in order to identify the forces wWch da work. Recall that the kinetic energy of the body is duc to translational motiori of the mass center. vG, and ratational motion of the body, w.

T1 + S Li 1_2

wherc

T = źrravł Gw2

Lir. = J F cos O ds(variable (arce}

F, cos 0(s2 — si) (constant furce)

—W 41,3: (weight)

LTs = — -W2) (spring)

MO (constant couple moment)

If the forces acting on the body arc conservałive forces, then apply the conservation of energy equation. This equation is easier to use than the equation of work and energy, lince it applies only at rwo poinłs on the path and does nor require calcu]ation of the work done by a force as the body rnoves along the pach.

T1 -F = T2 + V,

where V = V i- V, and

Wy (graviłational potential energy)

Vy = -14ks2 (eIastic pOleiiiiiii energy)

REviEw 2 PLANAR KNEMATICS AND KINETICS OF A RUD BODY 537

Impulse and Momentum. The principles of linear and angular impulse and mornemum are used to salce problemy involving furce, velocity, and time. Before applying thc equations, draw a free-body diagram in order to identify ali the forces which cause linear and angular impulses on the body. Also. establish the directions of the velocity of the mass center and the angular velocity of the body jusa before and jurt atter the impulses are applied. (As an aiłernative procedure, the impulse and momentum diagrams may accompany the solution in order to graphicaIly account for the, tems in the equations. Thcsc diagrams arc particularly advantageous when computing the angular impulses and angular momenta about a point other Chan the, body's mass center.)

In(vG)1 "2, f F dr = m(vG),

(HG)t -F S f MG dr = (HG),

O l "

(Ho)1 X." Modt = (H0)2

Conservation of Momentum. !f nonimpuisive forces or no irnpulsive forces act on the body in a particular direction, ur if the motions of several bodies arc involved in the problem, ibm consider app]ying the conscrvation of linear or angular momentum for the solution. Investigation of the free-body diagram (or the impulse diagram) will aid in determining the directions along which the impuisive forces are zero, or axes about which the impulsive forces create zero angular impulse. For these cases,

in(vG)i = 1n(vG)2

(H0)1 = (H0)2

The problems that follow involve application of ali the above concepts. They are presented in rundom order so that practice may be gained at identifying the various types cif problems and developing the skills necessary for their solution,

538REMEW 2 PLANAR KINEMATICS AND K1NETICS OF A RIGID BODY

REVIEW PROBLEMS

R2-1. Blocks A and B weigh 50 and 10 !b. respectively. If

15 P = 100 lb, determine the normal force cxerted by block A on block R. Neglect friction and the weights

of the pulleys, cord. and bars of the triangular h ame,

Prob. R2-1

R2-3. The truck carries the 800-lb crate which bas a center of grovity at G,. Determine the largest acceleratiun of the truck so tkał the crate will not slip or tip on the truck bed, The cocfficient of static friction between the crate and the truck is g, -= 0.6.

15 fłfl

3 ftG,

a

Prob. R2-3

R2-1. The spool has a weight of 30 lb and a radius of gyration ko = 0.Ó5 ft. If a forte of 40 lb is applied to the cord at A. determine the angular velocity of the spool in = 3 s starting from rest. Neglect the mass of the pulley and cord.

R2-2. The handcart has a mass of 200 kg and center of mass at G. Determine the norma] reacłions at mdi of the wheels at A and /I if a fotce P = .50N is applied to the handle. Neglect the mass and rolling resistance of the wheels.

Prob. R2-2

R2-5. Solne Prob. R2-4 if a 40-lb block is suspended from the cord at A. rather than applying the 40-lb furce.

Probs. R2-415

40 Ib

REVIEW 2 PLANAR KINEMATICS AND K1NETICS OF A RUD BODY539

R2-6. The uniform place weighs 40 lb and is supported by a roller at A. If a horizonial kirce F = 70 Ib is suddeniy appbed lo The roller. deterrnine the acceleralion of the center of the roller at the instant the furce is applied. The place has a moment of inertia ahout its center of mass of /G = 0.414 sług - ft2. Neglect the weigh t of the roller.

R2-8-. The double pendalurn consists of two rods. Rod A B has a constant angular yelocity of 3 rad /s, and rod BC has a constans angular vetociły of 2 radl s. Both of (hese absolute TFI

Gi 1ORS are measured counterclockwise. Determine the velocily and acceleration of point C at the instam shown.

2 f t

Prob. R2-6

R2-7. The center of the palicy is being lifted yertically with an acceleration of 4 m/62 at the instant it has a yelocity of 2 m/:;. If thc cakle does not slip on thc pulley's surface, determine the accelerations of the cylinder B and point C on the palicy.

ciA -= 411-0.2

1 vA = 2 mis.

Prob. R2-7

Prob. R2-8

R2-9. The link OA is pinned at O and rotates because of the sliding action of rod R aiong the borizontal grooye. If R starta front rest when N = tr and lias a constant acceleration

= 6.0 mints2 to the right, determine the angular velocity and angular acceleration of OA when t = 2 s.

Prob. R2~9

54a REMEW 2 PL4NAR KNEMATICS AND KINETICS OF A RIGID BODY

R2-10. The drive wheel A has a constant angular velociły of co,k. At a pdrlicular instant, the radius of rope wound on each wheel is as shown. If the rope has a thickness T, determine łhe angular acceleration of wheel B,

*R2-12. If the bali bas a weight of 15 lb and is thrown ()rito a rough surface so that its Center bas a velocily of 6 ft/s parailel to the surface, determine the amount of backspin, the bali must be given so that it stops spinning at the same instant that its forward vehcity is zero. R "is not necessary to know t h e coeffici cni of kinetic friction alki for the calcu' ation.

Ą

Prob. R2-10Prob. R2-12

R2-11. Thc dresser has a weight of 80 lh and is pusbcd along the floor. If the coefficient of static friction at A and is t.a.s =. 0.3 and the coefficient of kinetic friction is fik = 0.2, determine the smallest horizontal furce P needed to cause morion. If this force is increased slightiv. determine the acceleration of the dresser. Also, whal are the hormaI reactions at A and B when it begins to move?

R2-13. The dragster has a mass of 1500 kg and a center of mass at G. If the coefficient of kinetic frictian between therear whcels and the pavernent is = 0.6, detcrmine if it ispossible for the driver to lift the front wheels, A, off the ground wbite the rear wheels are slipping. If so. what aceeleration is necessary to do this? Neglect the mass of the whcels and assume that the front wheels arc frec to roll.

R2-14, The dragster has a mass of 1500 kg and a center of masa at G. If no sIipping occurs, determine the frietion force F which must be applied to car* of the rear wheels B in order to develop an accełeration c = 6 m/s'-. What ara the norrnaI reactions of each wheel on the ground? Neglect the masa of the wheels and assume ihai ihe front wheels are free to roli.

P

Prob. R2-11

Lr

2.5 Ił

• 4.

BL

4 li

Prob& R2-13/14

REVIEW 2 PLANAR KLNEMAT1CS AND KJNETICS OF A Rio BODY541

R2-15. ft the operator drives the pedals at20 revimin, and then begins an angular acceleraiion of 30 revimin2, determine the angular velocity of the flywheel F when s = 3 s. Note that thc pedal arm is fixed connected to the chain wheei A. which in turn drives the sheave i3 using the fixed connected clutch gear D. The beli wraps around ilie sheave then drives the pulle y E and fixed-connected flywheel.*R2._16. If the operator initially drives the pedals al 12 revirnin, and then begins an angular aceeleration of 8 revimin2, determine the angular velocity af the flywheel F after the pedal arm bas rotated 2 revolutions. Note that the pedal arm is fixed connected to the chama wheel A. which in turn drives the sheave B using the fixed-connected clutch gear D. The beli wraps around Ule sheave then drives the pulley ki and fixed-connected flywheel.

= 125 min rs = 175 mm

nhnh ••• 10 mm Probs. R2-15f16

R2-17. The drutu lias a mass of 50 kg and a radius of gyrationabout the pin at O of = 0.23 m. Starting from rest, thesuspended 15-kg błock Bis allowed to fali 3 ni withoul apptying, the brake ACD. Determine the speed of the błock at this instant. If the coefficient of kinetic friction at thc brake pad Cis = 0.5, determine the furce P ihat musi be applied at thebrake handle which will then stop the błock after it descends another 3 m. Neglect the thickness of the handle.

R2-1S. The drurn has a mass of 50 kg and a radius of gyration about the pin at O af k = 0,23 m. If the /5-kg block is moving downward at 3 m/s, and a furce of P = 100 N is applied to the brake arm, deierrnine how far the błock descends from the instant the brake is applied untii it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is µk = 0.5.

P robs. R2-17/18

R2-19. The I .6-Mg car shown has Been "raked" by increasing the height h = 0.2 m of -its center of mass. This was done by raising the springs on the rcar axle. If the coefficient of statie friction between the rear wheels and theground is = 0.3, show that the car can accelerate slightlyfaster (lian its counterpart for which h = 0. Neglect the mass of the wheels and driver wid assume the front wheels at B aro free to roli while the rear wheels slip.

Prob. R2-19

*R2-20. The disk isrotating at a consta nt rato = 4 radjs, and as it falls freeiy, its center bas an acceieration of 32.2 ft/s2. Deten-nine the acceIeration of point A on the rim of the disk at the instant shown.

R2-21. The disk is rotating at a constant rate W = 4 rad/s, and as ii falls freely, its center has an aceeieration of 32.2 ft/s2. Determine the aceeleration of point B on the rim cif the disk at the instant shown.

= 4 rad 11,Probs. R2-20/21

R2-22. The board resis on the surface of Iwo drums. Al the instant shown, it lias an aceeleration of 0.5 rn/s2 to the right, whiłe at the same instant points on the outer rim of each dram have an aceeleration with a rnagnitude of 3 m/s2. If the board does not slip on the drums, determine iis speed due to the motion.

a= 03 mis'

Prob. R2-22

542 REMEW 2 PLANAR KINEMATICS AND KJNETICS OF A RIGIR BODY

R2-23. A 20-Irg roll of paper, originally at rest, is

pin-supported at its ends to bracket AB. The roli rests against wali for which the coefficient of kinetic fricłion at C is ge 03. If a force of 40 N is applicd uniformly to the end of the sheet, deterrnine the initial anguiar acceleration of the roli and the tension in the bracket as the paper unwraps. For the calculałion, treat the roll as a cylinder.

Prob. R2-23

-R2-24. Ai the instant shown, link AB has an arigniar vełocity rgAB = 2 rad is and an angular acceleration

6 razi/s2, Determine the aceeleration of the piu at C and the angolar acceieration of link GB at this instant. when (P = 60°.

Prob. R2-24

R2-25. The truck has a weight of 8000 lb and center of grani ty at G1, Ił carries the 800-lb crate. which has a center of gravity at G,. Deierrnine the normal reaction at each of ils four tires if it accelerates at a =13.5 ft/s2-. Also, what is the frietinnal force acting betwcen the cratc and the truck, and between each of the rear tires and the wad? Assume that power is delivered only to the rear tires. The front tires nre free to roli. Neglect the mass of the tires. The crate does not slip nr kip on the truck,

Prob. R2-25

R2-26. The 15-lb cylinder is initially at rest on a 5-113 plate. If a couple moment M = 40 lb Ct is applied tn the cylinder, determine the angutar acceieration of the cylinder and th e linie needed for the end B of the plute to travel 3 ft and strike the wali. Assume the cylinder does not slip on the plate, and neglect the mass of the wilers unizier the plate.

Prob. R2-26

RFVfW 2 PLANAR KkNEMATICS AND KINETICS OF A RFGio BODY 543

R2-27. At the instant shown. twa forces ad on the 30-lb slender rod which is pinned at O. Deterrnine the magnitude of furce F and the initial angular acceleration of the rod so that the horizontal rcaction which the pin exeds on the rod is 5 lb directed to the right.

20 Ib

Ptob. R2-27

R2-30. The wheeibarrow and its contents have a mass of 40 kg and a mass center at G, excluding the wheel. The whecl has a mass of 4 kg and a radius of gyration

=17.120 m. If the wheelbarrow is released from rest from the position shown, determine its speed after it travels 4 m down the indii-ie. The coefficieni of kinetic friction between the incline and A is feA = 0.3. The wheels roli R2 without slipping at D.

Prob. R2-30

*R2-28. The 20-lb solid bali is cast on the Floor such that it has a backspin w = 15 radis and its center bas an initial horizontal velocity aG = 20 ft/s. If the coefficient of kinetic friction between the fluor and the bali is pA = 0.3, determine the distance it lravels before it stops spinning.

R2-29. Deterrnine the backspin w which shonłd be given to the 20-lb bali so that when its center is given an initial horizontal velocity a = 20 flis it stops spinning and translating al the same instant. The coefficient cif kinetic friction is µA = 0.3.

VG = 20 ft/s

A

Probs. R2-2W29

R2-31. Al the given instant rnember AB has the angolar rnotions shnwn. Determine the velocity and acceieration of the slidcr block C at this Instant.

Prob. 1{2-3I

544 REMEW 2 PLANAR KrNEMATICS AND K1NETICS OF A RIGID BODY

*R2-32. The spool and wire wrapped around its core have a mass cxf 20 kg and a centroida) radius of gyration iro = 250 mm. If the coeffclent of kinetic friction at theground is = 0. ł. determine the angular acccicration ofthe spool when ihe 30-N • m couple moment is applied.

R

Proh. R2-32

R2-33. The car has a mass of 1.50 Mg and a mass center at G. Determine the maximum aceeleration ii can have if (a) power is supplied only to the rear wheels, (b) power is suppIied ony to the front wheels, Negłect the mass of the wheels in the calculation, and assumc that the wheels that do not receive power are free to roli. Also. assume that slipping of the powered wheels occurs, where the coefficient of kinetic friction is lak = 0.3.

R2-34, The tire has a mass ot 9 kg and a radius of gyration -= 225 mm. If it is released from rest and rolls down the piane

without slipping, determine the speed of its center O when r = 3 s.

Prob. R2-34

R2-35. The bar hasa mass rn and length is releasedfrom rest from the position U .= 30°. determine its angular acceleration and the horizontal and vertical cornponents of reaction al the pin O.

''R2-36. The pendulum consists of a 30-lb sphere and a 10-Ib slender rod. Computc the reaction at the pśn O inst after the cord A 13 "is cut,

Proh, R2-36

R2-37. Spoci B is at rest and spooI A is rotating at 5 radfs when the siack in the cord connecting thern is token up. If the cord does not stretch, determine the angular velocity of each spoci śmmediately afler the cord is .jerked light. The spools A and B have weights and radli of gyrationWA = 30 1h, = 0,8 ft and Wg = 15 ]b, kg = 0,6 ft,respectively,

6 r a d / s

Prob. R2-37

Prob, R2-33

Pmh. R2-35

REVIEW 2 l''LANAR KINEMATICS AND K1NETICS OF A RUD BODY 545

1112-38. The rod is bent finta the shape of a sine eurve and fis forced to rotate aboul the y asis by connecting the spinele S to a motor. If the rod starts front rest in the position shown and a motor drives it for a short time with an angular acceleration a = (1.5er)radfs2, where r is in seconds, determine the magnitudes of the angular velocity and angolar disptacement of the rod when r = 3 s. Locate the point on the rod which has thc greatest velocity and aceeleration, and compute the magnitudes of the velocity and aceeleration of Ibis point when r = 3 s. 'The curvedcfining the rod "is 0,25 sin(./ry, where thc argument forthe sine is given in radians when y is in meters.

Prob. R2-38

R2-39. The scaffold S is ralsecl by moving the roller at A toward the pil at R, If A is approaching B with a speed of 15 ft/s. determine the speed at which the platform rises as a function of O. The 4-ft links are pin connected at their midpoint.

Prob. R2-39

*R2-40. The pendulum of the Cliarpy impast maehine has a mass of 50 kg and a radius of gyration of ka = 1,75 m, If it fis released from Test when B = O. determine iłs angular velocity jut before it strikes the specimen S, B = 900.

Prob. R2-40

R2-41. The gear rack lias a mass of 6 kg, and the gears cacki have a mass of 4 kg and a radius of gyration k = 3fl mm at [heir centers.lf the rack is originally moving downward at 2 mis, when s = U. determine the speed of the rack when s = 600 rnm,The gears dre Erce lo Wrn aboul (heir centers, A and B.

Prob. R2-41

546 REMEW 2 PLANAR KNEMATICS AND K1NETICS OF A RIGID BODY

R2-42. A 7-kg automobile tire is reieased frorn Test at A on the incline and «As wilhout slipping to point B. where it then travels in free flight.Determine the maximum height h the tire attains. The radius of gyration of the tire ahout its mam center is ka -= 0.3 m.

R2-44. The uniform connecłing rod BC has a mam of 3 kg and is pin-connected at its end points. Deterrnine the vertical forces which the pins exert cm the ends B and C ofthe rod at the instant (a) d = and (b) 8 = 94°. The crankAB is iurningwiłh a constant anguffir velocity wAR = 5 radts,

700 mm

(Iq k2.:\00 mm

4.1....!H

Prob. R2-44

1h

Prob. R2-42R2-45. if bar A8 lias an angułar velocity wAB = bradfs, determine the velocity of the slider błock C at the instant shown.

R2-43. The two 3-lb rods EF and HI are fixed (welded) to the link AC at E. Dełermine the internal axial force shear force E,., and moment ME. which the bar AC exerts on FE at F if at the instant N = 30' link /1.3 has an angular velocity w = 5 radis and an angular acceleration

= raci/s2 as shown.

Prob. R2-45

f t

Prob. R2-43

R2-46, The drwa of mass m, radius r. and radius of gyration ka roils along an inclined piane for which thecoefficient of static friction is If the drum is relcascd fromrest, determine the maximum angle N for the incline 50 that it polis without

Prob. R2-46

RFVfW 2 PLANAR KkNEMATICS AND KINETICS OF A RGIO ROPY 547

R2-47. Determine the velocity and acceleration of rod R For any angle H of cum C if the carn rotates with a constani angular velocity ra. The pin connection al O does not cause an interference with the motion of A on C.

Prlilr. R2-47

*R2-48. When the crank on the Chinese windlass is turning. the rope on shaft unwinds wh ile that cm shaft B winds np. Determine the speed at which the błock lowers if the crank is turning with an angular velocity w = 4 rad/s. What is the angular velocity of the policy at C? The rope seg-nients on each side of the pulley are both parailel and vertical, and thc rope does not slip on the pulley.

R2-49. The semicircular disk hasa mass of 50 kg and is released from rest from the posilion shown.The coefficients of stalic and kinetic friction between the disk and the beamarc o., = 0.5 and= 0.3, respectively. Determine theinitial reactions at the pin A and miler R, used to support the beam. Negiect the mass of the beam for the calculation.

R2-50. The sernicircular diak hasa masa of 50 kg and is released from rest from the position shown.The coefficients of static and kinetic friction between thc disk and the heamare = 0.2 and juk = 0.1, respectively. Determine theinitial reactions at the pin A and roller B used to support the heam, Neglect thc masa of the beam for the calculation.

Probs. R2-49/50

R2-51. The hoisting gear A has an initial angular velocity of 60 rad/s and a constard deceleration of 1 radis2. Determine the velocity and deceleration of the block which is being hoisted by the hub on gear B when r = 3 s.

1=I

Prob. R2-48 Prob. R2-51

Design of industrial robots requires knowing the kinernatics of theirthree-dimensional motions.

Chi ter 20r A .

o's

4

Three-DimensionalKinematics of aRigid Body

CHAPTER OBJECTIVES

el To analyze the kinematies of a body subjected to rotation about a fixed point and general piane motion.

MI To provide a relative-motion analysis of a rigid body using translating and rotating axes.

20.1 Rotation About a Fixed Point

When a rigid body rotates about a fixed point, the distance r from the point to a particie located on the body is the same for any pasifianof the body. Thus, the path of motion for the particie iks on the surface af a sphere having a radius r and centcred at the fixed point. Since moden along this path occurs only from a senes of rotations mada during a finite time interval,we will first develop a familiarity with same of the properties of rot ation al di sp lacemen ts.

The boom can rotate up and down, and because it is hinged al a point on the verlical axis about which it turns. it is subjeełed te rotation about fixed point.

550CHAPTER 20 TI łREE-D[MENSIONAL KINEMATICS OF A RIGID BODY

Eulers Theorern. Euler's theorem states that two "component" rotations about different axel passing through a point are equivalcnt to a single resultant rotation about an axis passing through the point. If more than two rotations arc applied, they can be combincd into pairs, and cach pair can be further reduced and combined into one rotation.

Flnite Rotations. If cornponent rotations used in Euler's theorcm arc finite, it is important that the order in which they are applied be rnaintained. To show this, consider the two finite rotations Oi + applicd to the block in Fig. 20-1a. Each rotation bas a magnitude of 90' and a direction defined by the right-band rule, as indicated by the arrow. The final position of the block is shown at the right. When these two rotations are applied in the order 02 + 01, as shown in Fig. 2t-Ib. the final position of the block is not the same as it is in Fig. 20-1a. Beeause firtiłe rotations do not obcy the commutative law of addition (01 + B, g B, + (Pt}, they cannot be classtfied as vectors. If smaller, yet finite, rotations bad been used ta illustrate this point. e.g., 10° instead of 90', the fina"' position of the błock aftcr cach combination of rotations would also bc different; however, in this case, the differencc is only a smali amount.

Oz =

-1-)Lvvy

x

z

20

I j

70111 V

X

x

(h)

Fig. 20-1

20.1 ROTATION ABOUIT A FIXED POINT 551

Infinitesimal Rotations. When defining the angular motions of body subjected to three-dimensional motion, ody rotations which arc infiniłesimally smali will be considered. Such rotations can be classified as vectors, since they can be added vecłorially in any manner. To show this. for purposes of simplicity let us consider the rigid body itself to be a sphere which is allowed to rotate about its central fixed point 0, Fig. 20-2a. If we impose two infinitesimal rotations 40 + 402 on the body, it is seen that point P rnoves along the path d91 x r + 40, X r and endy up at Pr Had the two successive rotations occurred in the order da, 40 , then the resultant displacements of P would have been d02 X r d0 x r. Since the vector cross product obeys the distributive law, by comparison (de[ + 40-,) X r = (40, + d@,} x r. Herc infinitesimal rotations 40 arc vectors, since these quantities have both a magnitude and direction for which the order of (vector) addition is not important, i.e., ciói + dó2 = c/O, + dOt. As a resuit, as shown in Fig. 20-2a, the two "component" rotations 40i

and d02 are equivalent to a single resultant rotation 40 = 401 + 402, a consequence of Euler's t henrem.

Angular Velocity. If the body is subjected to ara angular rotation 40 about a fixed point, the angular velocity of the body is defined by the time derivative.

w = (20-1)

The line specifying the direction of w, which is coliinear with 40, is referred to as the insłantaneotts axis of rotation, Fig. 20-2b. In general, this axis changes direction during each instant of time. Since 40 is a vector quantity, so too is w, and it fonows from vector addition that if the body is subjected to two component angular motions, wl = Ói and w2 =the resultant angular velociły is w = wE +

Angular Acceleration. The body's angular acceleration is determined fratra the time derivative of its angular velocity,i.e.,

= sil I (20-2)

For motion about a fixed point, a must account for a changc in both the magnitude and direction cif w, so that. in generał, a is not dircctcd along the instantaneous axis of rotation, Fig. 20-3.

As the direction of the instantaneous axis of rotation (or the line ofaction ofchanges in space, che locus of the axis generates a fixed spacecone, Fig. 20-4, If the change in the direction of this axis is viewed with respect to the rotating body, the locus of the axis generates a budy ,cotte.

d Ow

Instantaneous axis of rotation

4A)

_ dO

( b )

Fig. 20-2

Fig. 20-3

Instalitannous axis ()f rutatiun

S52 C H A P T E R 20 T H R E E - D I M E N S I O N A L K I N E M A T I C S O F A R I G I D B O D Y

Instantaneous axis of rotation

insiaknaneous axis ofrotation

At any given instant, these cones meet along the instantaneous axis of rotation, and when the body is in motion, the body cone appears to roli either on the inside er the outside surface of the fixed space cone. Provided the paths defined by the open ends of the cones are describecl by the head cif the w vector, then most act tangent to these paths at any given instant, since the tirne rate of change of w is equal to cv. Fig. 20-4_

Td illustrate this concept, consider the disk in Fig. 20-5a that spins about the rod at to„ while the rod and disk precess about the vertical axis at wp. The resultant angular velocity of the disk is therefore w = ca, + wp. Since both point O and the contact point P have zero velocity, then both w and the instantaneous axis of rotation are along OP. Therefore, as the disk rołałes, this axis appears to move along the surface of the fixed space cone shown in Fig. 20-5b. If the axis is observecl from the rotating disk, the axis then appears to rnove on the surface of the body cone. At any instant, though, these two cones meet each other along the axis OP. If cv has a constant magnitude, then a indicates on 1y the change in the direction af which is tangent to thc cones at the tip of w as shown in Fig. 20-5b.

Vei0City. Once ca is specified, the veiocity of any point on a body rotating about a fixed point can be determined using the same rnethods as for a body rotating about a fixed axis. Hence, by the cross product,

v=wxr (20-3)

Herc r defines the position af thc point measured from the fixed point 0, Fig. 20-3.

Acceleration. If w and oć are known at a given instant, the acceleration of a point can be obłained from the time derivative of Eq,20-3.which yields

( a )

a = a X r - F r o X ( c o X r ) (20-4)

1"- Space cone Rudy cone

*20.2 The Time Derivative of a Vector Measured from Either a Fixed or Translating-Rotating System

{b)

Fig. 20-5

Instantancous N

axis of rotationIn many types of problems involving the muflon of a body about a fixed point, the angular velocity w is specified in terms of its components,Then, if the angular acceleration a of such a body is ta be determined, it is often easier to cum pule the time derivative of w using a coordinate system that bas a rotation defined by one ar more cif the components of w. For example, in the case of the disk in Fig. 20-5a, wherc w = ws + wp, the x, y, z axes can be given an angular velocity of wp. For this reason, and for other uses Iater, an equation will now be derived, which relates the time derivative cif any vector A defined from a translating-rotating reference to its time derivative defined from a fixed reference.

20.2 THE TIME DERIVATIVE OF A VECTOR MEASURED FROM EITHER A FIXED DR TRANSLATkNG-RCTATING SYSTEM553

Consider the x, y, z axes of the moving frarne of reference to be rotating with an angular velocity SZ, which is measured from the fixed X, y, Z axes, Fg. 20-6a. In the following discussion, it wili be convenient to express vector A in terrns of its i, j, k components, which define the directions of the moving axes. Hence,

A = A v j A zk

Tri general, the time derivative of A must account for the change in both its magnitude and direction, However, i# this derivative is taken with respect to rhe moving frame of reference, only the change in the magnitudes of the components of A must be accounted for, since thc directions of the components do not change with respect to the moving reference. Hence,

= ki ± ± (20-5)

When the tlnie derivative of A is taken wich respect w the fixed franie of reference, the directions d i, j, and k change only on account of the rotation U of the axes and not their translation. Hence, in generał,

z

Y

( a )

A = ± + Ayj + A,k

The time derivatives of the unit vectors will now be considered. For example, i = difdt represents only the change in the direcrion of i withrespect to time, since i always bas a magnitude of 1 unit. As shown in r i al time r I dr

Fig. 20-6b, the change,di,is tangent to the path described by the arrowhead diof i as i swings due to the rotation n. Accounting for butli the magnitude i at time tand direction of di, we can thercfore define i using the cross produci,

(b)i = .1-1. x i. In general, then

Fig, 20-6

These formulations were also developed in Sec. 16.8. regarding planar motion of the axes. Substituting these results int° the above equation and using Eq. 20-5 yields

i = 1 / X i j = i 1 X j k = 1 1 X k

2 0

=(A) . , , , ,+nxA (20-6)

This result is important, and will be used throughout Sec. 20.4 and Chapter 21.11 states that thc time derivativc of any vector A as observed from the fixed X, Y, Z franie of reference isequal to the time rate of change of A as observed from the x, y, z translating-rotating frame of reference. Eq. 20-5. plus 11 X A, the change of A caused by the rotation of the x, y, z franie. As a result, Eq. 20-6 should always be used whenever produces a change in the direction of A as scen from the X, X Z reference. If this change does not occur, i.e., n = 0, then A = (k),)„, and so the time rate of change of A as observed from both coordinate systems will be the same.

554 CHAPTER 20 THREE-DtMENSIONAL KINEMATICS OF A RICIE) BODY

2 0 . 1

The disk shown in Fig. 20-7 spins about its axle with a eonstant angular velocity ra, = 3 rad/s, while the horizontal platform on which the disk is mounted rotates about the vertical axis at a constant rato wp

= 1 rad/s. Determine the angular acceleration of the disk and the velocity and acceleration of point A on the diak when it is in the position shown.

Z . z

Fig. 20-7

50 LUTIONPoint O represents a fixed point of rotation for the disk if one cunsiders a hypothetical extension of the disk tu this point. To detcmine the velocity and acceleration ()t. point A, it is first necessary to determine the angular vclocity w and angular acceleration a of the disk, since thesc vcctors arc used in Eqs. 20-3 and 20-4.

Angular Velocrty. The angular velocity, which is measured from X, Y, Z, is siniply the vector addition of its two component motions. Thus.

w = -= 3j - 1k} radis

20.2 THE TIME DERIVATIVE OF A VECTOR MEASURED FROM EITHER A FIXED OR TRANSLATkNO-ROTATING SYSTEM 555

Angular Acceleration. Since the magnitude of w is constant, only a change in -its direction, as scen from the fixed reference. creates the angular accelerałion a of the disk. One way to obtain a is to compute the tinie derivative of each of lfze rwo components of w using Eq. 20-6. At the instant shown in Fig, 20-7. imagine the fixed X, Y, Z and a rotating x, y, z framc to be coincident. If the rotating x, y, z framc is chosen to have an angular velocity of S2 = wr, = {-1k rad/s, then ws

will always be directed along the y (not Y) axis, and the time rato ofchange of w, as seen from x, y, z is = O (the magnitude

and direction of w5 is constant). Thus,

w s = (W,)„, + cop X 6;5 = + (-1k) X (3j) = {31} rad/s'

By the sanie choice of axes rotation, = wp, or cycu with U = a,

the tinie derivative (60)„y, = 0, since wp has a constant magnitude and direction with respect to x, y, z. Hence,

The angular acceleration of the disk is łherefore

= Wt = cifs + Wł, = (M} radis2Ans.

Velocity and Acceleration. Since w and oz have now Been cletermined, the velocity and aceeleration of point A can be found using Eqs. 20-3 and 20-4. Realia.ing that 1-.4 = 1 j + 0.25k ) m, Fig. 20-7, we have

= w X xy = (3j - 1k) X (1J + 0,25k) {1,751} m/s Ans.

a A = a X r A + X ( w r A . )

= (3i) x (lj + 0.25k) + (3j - lk) x [Oj lk) x (lj + 0.25k)]

= { -2.50j - 2.25k) MiS2 Ans.

5 5 6 C H A P T E R 2 0 T H R E E - D t M E N S I O N A L K I N E M A T I C S O F A R I C I E ) B O D Y

20.2

At the instant O = 60°. the gyrotop in Fig. 20-8 has three components of angular motion directed as shown and having magnitudes defined as:

Spin: w, = 10 rad/s, increasing at the rate of 6 rad/s2

Nufation: w„ = 3 rad ./ s, increasing at the rate of 2 rad/s-2

Precession: w1, = 5 rad/s, increasing at the, rate of 4 rad/s2

Determine the angular velocity and angular acceleration of the top,

Z. Z

5 radis= 4 ractis2'

Always in---- oZ directiun

t o „ = 3 r a d ! \ - \ \ , = 2

r a d / s 2 - Always in x—y planu

Fig. 20-8

X. x

SOLUTION

Angular Velocity. The top rotates about the fixed point O. if the fixed and rotating framus aru coincident at the instant shown, then the angular velocity can be cxpressed in terms of i, j. k componcnts, with reference to the x, y, z frame; i.e.,

w = + co, ,;in Hj + (wp + co„ eos 0)k

w.v= IQ rad/s = -3i + 10 sin 60-j + (5 + 10 cos 60°)k= 6 radls2 = ł -31 4- 8.66j + lok I rad/s

Angular Acceleration. As in the solution of Example 20.1, the angular acceleration ar will be determined by investigating separately the time rate of change each of the angular velocay componems as observed from the fixed X, Y, Z reference.We will choose an Ił for the

Y , v- x, y, z reference so that the component of w being considered is viewed as

having a ~stara direction when observed from x, y, Z-Careful examination of the rnation of the top reveals Chat cus

has a runstant directiuri relative to x, y, z if these axes rotate at + co Thus,

= + (co„, + wp) X w,= (6 sim + 6 cos 6010 + (-3i + 5k) X (10 sin 601 + 10 cos 60°k)

= -43.301 + 20.20j - 22.98k } rad/s2

Since co„ always łies in the fixed X-Y płatw, this vector has a consfanłdirection if the motion is viewed from axes x, y, z having a rotation of

= wf, (not -= w, + wp). Thus,

= + wp X co„ = -2i + (5k) X (-3i) = -2i - 15j ł raclis'-'

Finally, the component wpis always directed along the Z axis so that herc it is not necessary to think of x, y, z as rotating. i.e., 4= 0. Expressing the data in terms of the i, j, k components, we therefore have

<Uj, = (cUp),y, + O X cup = 14k } raclis2

Thus, the angular acceleration of the top is

a= (;)„, ci>„ + <Uj, = 1-45.31 + 5.20j - 19.0k j- radis2 .1h%

20.3 GENERAŁ MoTiaNi557

20.3 General Motion

Shawn in Fig. 20-9 is a rigid body subjected to general motion in three dimensions far which the angular veIocity is w and the angular acceleration is a. If point A has a known motion of vA and aA, the mation of any other point B can be determined by using a relative-rnotion analysis. In this seetian a trangating coordinette system will be used to (Winę the rciative ~lon, and in the next scetian a refereriee that is both ratating and translating will be considercd.

If the arigin of thc translating coordinatc system x, y, z= 0) islocated at the "base point" A. then, at the instant shown, the mation of the bady can be regarded as the sum °f an insłantaneous translatian of the body having a motion of vA, and aA, and a rotation of the body ahout an insłantaneous axis passing through point A, Since the bady is rigid, the triatlon of point 73 measured by an observer located at Ais dicrefarc the same as the rotation of the body abym a fixed point. This relative malim occurs abaut the instariłaneaus axis of rotation and is defincd by vom = w X ro, Eq. 20-3, and aB,A = nt X N/A + w X (co X tufy), Eq. 20-4. For translating axes, thc relative mations arc related to absalute

!"nstantaneausaxis of rotation

mations by vB = vA + vBiAthat the absolute veloeity from the equations

and aB = aA aBiA , Eqs. 16-15 and

and acceleration of point B can be

16-17, so determined

A

;ki

VB V A + l a l X r B i A j (20-7) A . ) X

Fig. 2.0-9

and

ao - aA .ex x rs/A + w x (ov x r8/A) (20-8)

These twa equations are essentially the same as ta those describing the general piane motion of a rigid body, Eqs. 16-16 and 16-18. However, difficulty in applicałion arises for three-dimensional motion, because now measures the change in brah the magnitucle and direction of w.

5 5 8 CHAPrER 20 THREE-DiMENSIoNA>_ KINEMATICS OF A IŚiGIp BODY

20 .3

(b )

Fig. Z(I-10

If the collar at C in Fig. 20-10a rnoves towards B with a speed cd 3 rl-LN deterraine the velocity of the collar at D and the angolar velocity °f the bar at the instant shown, The bar is connected to the col lars at its end points by ball-and-socket joints.

SOLUTIONBar CD is subjected to generał median, Why? The velocity of point D on the bar can be related to the velocity of point C by the equation

VD = Vc + WP X rD/c

The fixed and translating frames of reference are assumed to coincide at thc instant considcred, Fig, 20-10b. We have

vn = k vc = {3j} m/srEvc { li ± 2j — 0,5k} ni = W?, i + Wyj + wzk

Substituting juto the abave equation wc gct

i j k

—vi) k = 3j + wx £0,..

1 2 — 0 . 5

Expanding and equating the respective i, j, k components yie14.1s—0.5wy — = O

( 1 )

0.5w„ + + 3 = O ( 2 )

2(ox — 1(o, -F vn = O( 3 )

These equations cantain four unknowns." A fourth equation can be

written if the direetion of w is specified. In particular, any component ofacting along the bar's axis has na effeet on moving the co1lars,This is

because the bar isfree to rokne about its axis, Therefore, ii w is speciiied as acting perpendicular ło the axis of thc bar, then w must have a uniq ue ma gnitude to sa tis fy the aboye equations.Perpendicularity is guaranteed pravided the dat product of w and rDic is zero (see Eq. C-14 of Appendix C). Hence,

w • rDic =+ w,j + w,k) (li + 2j — 0.5k) = Ow„ + 204 — 0.5w, = O(4)

Solving Eqs. 1 through 4 simul t ancously yields

= —4.86 rad/s wy = 2.29 rad is = —0.571 rad/s Ans.

vp= 12.0 m/s Ans.

'Although [bis is the case, the magnitude of vij can be obtained. For example, solve Eqs. 1 and 2 for ek,. and e), in terrrts of o.J., and substitute into Eq. 3. It will be noted that will cancel oni, which will allow a solulion for y0.

20.3 GENERAL. M OTION 559

EPROBLEMS

20-1. At a given Instant, the satellite disb has an angular motion w, = 6 rad/s and W, = 3 rad/s2 about the z axis. At this same instant H -= 25°, the angular motion about the x axis is w, = 2 rad/s. and 6, = 1.5 rad/s2.Determine the velocity and acceleration of the signal horn A at this Instant.

Prob. 20-1

20-2. Gears A and B arc fixed. while gem C and D are free to rotate about the shaft S. If the shaft turns about the z axis at a constant rate of w, = 4 rad/s, determine the angular velocity and angular acceleration of gear C.

Prob. 20-2

20-3. The ladder of the lite truck rotates around the z axis with an angular velocity w, = 0.15 rad/s, which is increasing at 0.8 rad/s2. At the same instant it is rotating upward at aconstant rate = 0.6 rad/s. Determine the velocity andacceleration of point A located at the top of the ladder at this Instant.

*20-4. 'The ladder of the fire truck rotates around the z axis with an angular velocity of w, = 0.15 rad/s, which is increasing at 0.2 rad/s2. At the same instant it is rotating upwards at w, = 0.6 rad/s while increasing at 0.4 rad/s2. Determine the velocity and acceleration of point A located at the top of the ladder at this instant.

x

Probs. 20-314

20-5. Gcar B is connected to thc rotating shaft, while thc plate gear A is fixed.tf the shaft is turning at a constant rateof = 10 rad/s about the z axis. determine the magnitudesof the angular velocity and the angular acceleration of gear B. Also, determine the magnitudes of the velocity and 20 acceleration of point P.

z

H200 mm.___

10 rad/s4-.,

Prob. 20-5

560 CHAPTER 20 THREE-DIMENSIONAL KINEMATICS DF A RIGID BODY

20-6. Gear A is fixed while gear B is free to miale on theshaft S. If the shaft is turning about the z axis at = 5 rad/s,wbite increasing at 2 rad/s2, determine the velocity and acceleration of point P at the instant shown. The face of gear B lies in a vertical piane.

*20-8. The cone rolls without slipping such that at the instant shown w. = 4 ntd/s and W, = 3 rad/s2. Determine the velocity and acceleration of point A at this instant.

20-9. The cone rolls without slipping such that at the instant shown w, = 4 rad/s and W, = 3 rad/s2. Determine the velocity and acccleration of point Bat this instant.

= 4 rad/s= 3 rad/s7

2 0

p

Prob. 20-6

20-7. At a given instant, the antenna bas an angular motion w, = 3 rad/s and W, = 2 rad/s'' about the z axis. At this same instant O = 30", the angular motion about the x axis is w2 = 1.5 rad/s, and W, = 4 rad/s2. Determine the velocity and acccleration of the signal horn A at this instant. The distance from O to A is d = 3 ft.

Prob. 20-7

Probs. 20-8/9

20-10. At the instant when O =- 90°. the satellite's body is rotating with an angular velocity of w, = 15 rad/s and angular acceleration of wi = 3 rad/s2. Simultaneously. the solar panels rotate with an angolar velocity of w2 = 6 rad/s and angular acceleration of w, = 1.3 rad/s2. Determine the velocity and accelcration of point B on thc solar panel at this instant.

20-11. At the instant when O = 90', the satellite's body travels in the x direction with a velocity of vo = {5001} m/s and acceleration of ao = (50i) m/s2. Simultaneously. the body also rotates with an angolar velocity of w, = 15 rad/s and angular acceleration of W, = 3 rad/s2. At the same time. thc solar panels rotate with an angolar velocity cif w2 = 6 rad/s and angular acceleration of w2 = 1.5 rad/s2 Determine the velocity and acceleration of point B on the solar panel.

Probs. 20-10/11

20.3 GENERAL MoTON561

'20-12. The disk is free to rotate on the shaft S. If the shaft is turning about the z axis at wa = 2 radfs. while increasing at 8 rad/s2, cletermine the velocity and aceeleration of point A at the instant shown.

8 rarl/s.

2 radj's

Prob. 20-12

20-13. The disk spins ahout the arm with an angular velocity of a,,s = 8 raclis, which is increasing at a constantrate of = 3 radis2 al the instant shown. If the shaftrotates with a constant angular velocity of wjr, -= 6 rad/s, detcrrnine the velocity and aceeleration of point A located on the rim ot the disk at this instant.

20-14. The wheel is spinning about shaft A8 with an angular velocity of w, = 10 rudis, which is increasing at aconstant rate of = 6 rad/s2, while the frame precessesahout the z axis with an angolar velocity = 12 rad/s,which żs increasing at a constant raie of= 3 rad/52.Determine the velocity and aceeleration of point C located on the rim of the wheel at this instant.

20-15. At the instant shown, the tawer cranc rotates about the z axis with an angular velocity rai = 0.25 rad/s. which is increasing at 0.6 radis2. The boom 0/1 rotates downward with an angular velocity w, -= 0.4 rad/s, which is increasing at 0.8 rad/s.l. Determinc the velocity and accelcration of point A located at the end of the boom at this instant.

rup = tU raclis z6 rac1/52

0.15 m

ren = 12 rad is ryp = 3 rad/s'

Prob. 20-14

20

Prob. 20-15Prob. 20-13

562 CHAPTER 20 THREE-DtMENSIONAL KINEMATICS OF A RICIE] BODY

20-16. If the top gear 8 rotates al a constant race ol w, determine the angular velocity of gear A. which is free to rotate about the shaft and rolls on the bottorn fixed gear C.

Prob. 20-16

20-17. When 0 = 0°, the radar disk rotałes about the y axis with an angular velocity of 9 = 2 radfs.increasing at a constant race ol d = I.5 radl s2. &mul tarienusly. the disk olsu precesses about the z axis with an angular velocity of raP = 5 rad/s., increasing at a constant rafo of rsp -= 3 radis2, Determine the velocity and aceeleration of the reeeiver A al this instant.

20-18. Gear A is fixed to the crankshaft S, while gear C is fixed. Gear B and the propeller arc free to rotate. The crankshaft is turning at 80 radis about its axis. Determine the magnitudes of the angular velocity of the propeller and the angular acceleration of gear B.

P rob. 20-18

20-19, Shaft HO is conneeted ta a b0-and-socket joint at B. and a beveled gear A is attached lo its other end.The gear is in mesh with a fixed gear C. If the shaft and gear A are spinning wich a constant angular velocity CEJ = 8 radfs, determine the angular velocity and angular accelerasion of gear A.

20

Prob. 20-19

—20 ft rap = S rad/s3 rad/s'

2 rad/s 0= 1.5 rad /s'

Prob. 20-17

20.3 GENERAŁ M oTioN563

620-20. Gear B is driven by a motor rnounted on turntable C. If gear A is held fixed, and the motor shaft rotates with aconstant angular velocity of w,. 30 radis, dełermine theangular velocity and angular acceleration of gear B.

20-21. Gear is driven by a motor mounted on turntable C. If gear A and the motor shaft rotate with constant angularspeeds of {I0k) radfs and wy. I30j) rad/s. respectively.clełermine the angular wiochy and angular acceleration of gear B.

Probs. 20-20/2120-22. The crane boom OA rotates about ihe z axis with a

constant angular velocity of 0.15 radis. while it isrołating downward with a const ant angular velocity of = 0.2

rad/s. Determine the velocity and accelexation of point A located ihe end of the boom at the instant shown.

Prob. 20-22

20-23. The differential of an automobile allows the Iwo re

rear wheels to rotale at di rfent speeds when the automobile łravels along a curve. For operation, the rear axles are attached to the whecls at one end and have beveled gears A and R on their other ends. The differential case D is placed over ihe left axle but can rotate about C independent of the axle. The case supports a pinion gear E on a shaf t. which meshes with gears A and B. Finally, a ring gear G is fixed to the dłfferen tial case so that the case rotates with the ring gear when the atter is driven by the drive pinion H.This gear.like the diffc ren tial casc. is free to rotate about the left whecl axle,1f the drive pinion is turning at -= 100 radis and thepinion gear is spinning about its shaft at wE -= 30 rad/s, determine the angular velocity, and m, of each axle.

Prob. 20-23*20-24. The truncated cone rotates about the z axis al a constant rate w = 0.4 rad/s without slipping on the borizontal piane. Determine the velocity and acceleranon of point A on the cone.

z

Prob. 20-24

= 30 rad/s )

0_1

Y

15 m

T o l e f t j w h e e l

50 mm

k J'1171ir

40 mm

01

From motor' / /

• I

C7:(14- r

20

564 CHAPTEP 2 0 THR-D[MENSIONAL KINEMAT1CS OF A RIGID Boby

2 0

20-25. Disk A rotates al a cc}nstanl angular velocity of 10 rad/s. If rod BC is joined to the disk and a collar by ba11- and-socket joints, deterrninc the velocity of eol lar B at the instant shown. Also, what is the rod's angular velocity oBj;(- if it is direeted perpendicular to the axis cif the rod?

r.

E

300 in ni

Prob. 20-25

20-26. If the rod is attached with ball-and-socket joints to smooth colIars A and R at its end points, determine the speed of Bat the instant shov.m if A ismovingdownward at a constant speed of vA = 8 ftis. Mso. determine the angular velocity of the rod if it is directed perpendicular to the axis of the rod.

20-27. If the collar at A is moving downward with an acceleration a4 = { —5k } ft/s2, at the instant its speed is

= 8 ft/s, detemin.e the acceleration ot the collar at B at ibis instant.

Probs. 20-26/27

920-28. If wheei C rotates with a conslnm angular velocity of wc = 10 rad/s. determine ihe velocity of ihe collar al B when rod A$ is in the position shown.

20-29. At the instant rod AB is in the posiłion shown wheel C rołates with an angular velocity of coc = 10 rad/s and has an angular acceleration of rxr. = 1.5 rad/s2. Determine the acceleration of collar B at this instant.

20-30. If wheel D rotates with an angular velocity of win = 6 radis,determine the angular velocity of the foilower link BC al the inslanl shok•n.The link rotates about the z axis ał z = 2 ft.

2 fi

O

Prob. 20-30

z

300 mm4100 mm

200 mm

600 rum

- ' > ' -

Probs. 20-28/29

li

200 mm

A100;nm

= 10 radisA

500 mrn

20.3 GENERAŁ M OTI ON565

20-31. Rod AB is attached to the rotating arm using bali-and-socket joint& If AC is rotating wich a constant angular velocity of 8 rad/s about the pin at C. determine the angular velocity of hnk BD at the Instant shown.

*20-32. Rod AB is attached to the rolating arm using ball-and-socket joints II AC is rotating about point C with an angular velocity of 8 rad/ s and bas an angular acceleration of «Ac = 6k} radfs2attlteinstantsNawtt.determtnetheangular velocity and anguIar aeceleralion of link BD at ibis instant.

Probs. 20-31/32

20-33. Rod AB is attached to collars at its cnds by bai!-and-soeket joints. If collar A moves upward with a velocity of vA = (8k} ftis.determine the angular velocity of the rod and the speed of collar B at the instant shown. Assume that the rod's angular velocity is directed perpendieular to the rod.

20-34. Rod A8 is attached to collars at its ends by hall-and-socket joints. If collar A moves upward with an acceleration of aA -= 4k } ft/s2, determ i ne the angular acceleration of rod AB and the magnitude of acceleration of collar B. Assunne that the rod's angular acceleration is directed perpendicular to the rod, and use the result of Prob. 20-33 for osm,.

Probs. 20-33/34

20-35. Solne Prob. 20-25 if the conneetion at B consists of a pin as shown in the figure belowi rather than a bal I-and-socket joint. Mm: The constraint allows rota ton cif the rod both about bar DE (j direction) and about the axis of the pin {n direct-ion). Since there is no rotational component in the u direction. i.e.. perpendicular to n and j where

= j X n, an additional equation for soiution C2I1 Lic obtained from w • u = 0. The vector n is in the same direction as ra,c X rD/c.

Prob. 20-35

*20-36. The rod assembly is supported at B by a ball-and-socket joint and at A hy a clevis. If the collar at B moves in the x—z piane with a speed v,s = 5 ft/s, determine the velocity of points A and C on the rod assembly at the instant shown. Hint.• See Prob. 20-35.

Prob. 20-36

2 0

566 CHAPTER 20 T H R E E - D t M E N S I O N A L K I N E M A T I C S O F A R I C I E ) B O D Y

*20.4 Relatiye-Motion Analysis Using Translating and Rotating Axes

The most generał way to analyze the three-dimensional mation of a rigid body requires the use of x, y, z axes that both translate and rotate relative to a sccond frame X, Y, Z, This analysis also provides a mcans to detennine the motions of twa points A and B Iocated on separate members of a mechanism, and the relative motion of one particie witki respect to another when one Gr both parłicles are moving along curved padu.

As shown in Fig. 20-11. the locatiorts of points A and B are specified relative to the X, Y, Z frame of reference by position vectors rA and r8. The base point A represents the origin of the x, y, z coordinate system, which is translating and rotating with respect to X, Y, 7.. At the instant

considered, the velocity and acceleration of point A are vA and ki, and the angular velocity and angular acccieration of the x, y, z axes arc ft

and = dl-1/dt. All these vectors are measured with respect to the X, Y, Z frame of reference, alłhough they can be expressed in Cartesian

component form along either set of axes.

2 0

Fig. 20-11

56 720.4 RELATIvE-MonoN ANALYSIS USING TRANSLATING AND ROTATING AXES

Position. If the position of "B with respect to A" is specified by the relative-position vector rstA, Fig. 20-11, then, by vector addition,

Ta = rA + rB/A (20-9)

whe re

ra = position of B

rA position of the origin A

rBiA = position of "B with respect ta k'

Velocity. The velocity of point B measured from X, Y, Z can be determined by taking the time derivative of Eq. 20-9,

T/9 = TA -I-

The first two terms represent vB and vA. The last term must be evaluated by applying Eq. 20-6, since TB/A is measured with respect to a rotating reference. Hence,

i'D/A = (i-B/A)zyz + a X rE/A = (BA-1).xyz fl X rs/A (20-10)

Therefore,

V g = v A n X r „ , , ( v , , , , ) , y , (20-11)

where

vB = velocity of fś

vA = velocity of the origin A of the x, y, z frame of reference

(vBfA). = velocity of "B with respect to A" as measurcA:1 by an obscrver attached to the rotating x, y, z frame of reference

Sl = angular velocity of the x, y, z frame of reference r

B/A = position of "B with respect to A"

568 C H A P T E R 20 T H R E E - D t M E N S I O N A L K I N E M A T I C S O F A R I C I E ] B O D Y

Acceieration. The acceleration of point B measured from X. Y, Z is determined by taking the time derivative of Eq. 20-11.

di"+„1, + ii X TB/A + ii X im + dt(VNA)x-v.:

The time derivatives defined in the first and second terms represent aB and aA, respectively. The fourth term can be evaivated using Eq. 20-10, and the last term is evaluated by applying Eq, 20-6, which yields

cit (Vs/A/kyz = (1PB./A)xy: + a X (V ElA).xyz -+ 11 x (v.s/A)_ryz

Here (aBm),,, is the acceleration of B with respect ta A measured from x, y, z. Substituting this result and Eq. 20-10 into the above equation and simpiifying, we have

aB = aA. + x rB,,,A 11 (1.1 x ram) + 211 x (vB/.4), (aBIA),y,

(20-12)

whe re

Compiittedspaal motion of the concrete bucket B nccurs due to ihe rotation of the boom about the Z axis, motion of the caniage A along the boom. and exlension and swinging of the cakle AB. A transJating-rotating x, y, z coordinate system can be establIshed on thc carriage, and a relative-motion analysis can [hen be apptied to dudy this motion.

aB = acceleration of B

aA = acceleration d the origin A of the, x, y, z frame of reference

(a ,ISIA)xv:_. relative accelcration and relative veiocity ofwith respect to A" as measured by art observer attached to the rotating x, y, z frame of reference

ii, Sl = angular acceleration and angular velocity of the x, y, z frame of reference

11-BiA position of "B with respect to A'

Equations 20-11 and 20-12 ara identical to those used in See. 16.8 for analyzing relative piane motion,'' In that case, however, application is simplified since ri and ii have a constant direeriort which is always perpendicular to the piane of rnoton. For three-dirnensional motion, must be computed by using Eq. 20-6, since depcncls on thc change inbotki the m agnitude and direetion of

*kafar to Ser. 16.8 for an interprełation of the terma.

5 6 920.4 RELATIVE-MOTION ANALYSIS USING TRANSLATING AND ROTATING AXES


Three-dimensional motion of particles ar rigid bodies can be analyzed with Eqs, 20-11 and 20-12 by using the following procedure.

Coordinate Axes,

• Select the iocation and orientation of the X, Y. Z and x, y, z coordinate axes. Most often solutions can be easily obtained if at the instant considered:

(1) the origins are comcidera

(2) the axes are collincar

(3) the axes are parallel

• If several componews of anguIar velocity are invoived in a problem, the calculations will be reduced if the x, y, z axes are selected such that only one component of angular velocity is observed with respect to this frame (fl„,) and the frame rotates wich (1 defined by the other components'of angular vełoci ty.

Kinematic Equations.•After the origin of the rnoving reference, A, is defined and the rnoving

point I3 is specified, Eqs. 20-11 and 20-12 should then be written in symbolic form as

VA Ił X rBIA (VB/A)Ay.:

a& = + ił X rB/A + S2 X (n X rRIA) + 2n x (y,„,,)a.„ (abiA)„).,

• If r, and ił appear to change direction when observed from the fixed X, Y, Z reference then use a set of primed reference axes, x', y', z' having a rotation W = 53. Equation 20-6 is then used to determine ił and the motion v, and a, of the origin of the moving x, y, z axes.

• If rBiA and n„, appear to change direction as obscrved rum x, y, z, then use a set ci double-primed reference axes x", y", z"

having and apply Eq, 20-6 to determinc xa and the reiative motion (v,r?/„k), and (aB,,,,),,„.

• After the final forms of ił, vA, aA, ń„„ (v8/,),,, and (aBiA),, are obtained, numerical problem data can be substituted and the kinematic terms evaluated. The components of all these vecłors can be selected either aiong the X, Y, Z ar along the x, y, z axes, The choice is arbitraly,provided a consistent set of unit vectors is used.

5 7 0 CHAPTER 20 THREE-DtMENSIONAL KINEMATICS OF A RIOID BODY

A motor and attached rod AR have the angular rnotions shown in Fig. 20-12. A Miar C on the rod is locałed 0.25 m from A and is moving downward along the rod with a velocity of 3 m/s and an acceleration of 2 m/s. [kici-mit-1c the velocity and acceleration of C at this instant.

SOLUTION

Coord i nate Axes.The origin of the fixed X, Y, Z reference is chosen at the center of the platform, and the arigin of the moving x, y, z franie at point A, Fig. 20-12. Since the collar is subjected to two components nf angular Triotion, wp and wm, it will be viewed as having an angular velocity ofSixy,, wm in x, y, z. Therefore, the x, y, z axes will be attached to theplatform so that n •

0.25 rn

Y. v'

v . y

wn = 5 rad/s =

2 rad/s2

Z z '

2X, x, x'. x" 8

wAr 3 rad/s = ł rad /s2

Fig. 20-12

20.4 RELATivE-MorioN ANALYSIS US1NG TRANSLATiNG AND ROTATING AXES571

Kinematic Equations. Equations 20-11 and 20-12, applied to points C and A, hecome

ve = VA + n X re/A (Vc/A),y,

ar = aA + .1). X rci„ + I/ X (U X rcm) + 211 X (vc1A)3,), + (ac/)A:v:-

Motion of A. Here rA changes direction relative to X, Y, Z. To (ind the time derivatives of r, we will use a set of x', y', z' axes coincidentwith the X. Y, Z axes that rotate at II' = Thus.

= wr = {5k } rad/s (U does not change direction relative to X, Y, Z.) Śt

= Wp = {2k } rad/s2

rA = 21} m

v/4 = + top X O + 5k x 21 = { 10j } m/s

a, = = + wp x (r,)x,, ,z,] + x r, + wr X j ' A

= [ O ± 0 ] + 2k x 21 + 5k x 10 j = { —501 + 4 j} in / s 2

Motion of C with Respect to A. Here rciA changes direction relative to x, y, z, and so to find its time derivatives use a set of x", Y",z" axes that rotate at IV = = wM. Thus,

= {3i } rad/s (1-1,y, does not change direction relative to x, y, z.)= = {11} rad/s2

re/A = { —0.25k } ni

(VCIA)xyz (i.CIA)Ły":"WM X rc/A

= —3k + [3i X (-0.25k)] = {0.75j — 3k } m/s

81(.."/A).■yr_ = (.1:ClAayr [0CIA)xyz' WM X (ii-CIA)x"y"z") WMrCłAWM(i'CIA)xy:[-2k + 3i X (-3k)] + (li) X (-0.25k) + (33) X (0.75j — 3k) { I8.25j

+ 0.25k} m/s2

Motion of C.

vr — vA + U X rciA + (vc/A)„y.:

— 10j + {5k X (-0.25k)] + (0.75j — 3k)

{10.75j — 3k } m/s Ans.

al = a, + 11 x rci, + S/ X (11 X rcy,) + 21/ X (vc/A)xyz + (ac/A)xyz

= (-501 + 4j) + [2k X (-0.25k)] + 5k X [5k X (-0.25k)] + 2[

5k x (0.75j — 3k)] + (18.25j + 0.25k)

= —57.51 + 22.25j + 0.25k } m/ s` Ans.

572 C H A P T E R 2 0 T H R E E - D I M E N S I O N A L K I N E M A T I C S O F A R I G I D B O D Y

EXAMPLE

20 .5

z, z, z', Z"A

= 1.5 rad/s2,

X, x, - Ll= 4 ra.'„S

Y, y'

The pendulum shown in Fig. 20-13 consists of two rods; AR is pin supported at A and swings only in the Y-Z piane, whereas a hearing at B allows the attached rod RD to spin about rod AR. At a given instant, the rods have the angular motions shown. Also, a collar C, located 0.2 m from R, has a velocity of 3 m/s and an acceleration of 2 rn/s2 along the rod. Determine the velocity and acceleration of the collar at this instant.

0.5 m

w2 = 5 rad/s ł

U#2 = 6 rad/s?

Fig. 20-13

SOLUTION ICoordinate Axes. The origin of the fixed xy, Z frame will be placed

3m /s at A. Motion of the collar is convenicntly observed from B,so the origin."-1.» 2 m /s2 of the x, y, z framc is located at this point. We wił' choose Sit = co, and

n =- Kinematic Equations.

vc = v8 + IZ x r" + (vc/B),a, = aft + U x rcfs + (fi x re/8) + 2U x (vc/B)„, + (ac75.)„...

Motion of R. To find the time derivatives of rB let the x', y', z' axes rotate with I/ = w . Then= cdi = {4i} rad/s = 61 = {1.5i} rad/s2

rB = { -0.5k } m

vs = iR = + w, x rB = 0 + 4i x (-0.5k) = {2j} m/s

aB = = + to, X + Wi x rg Wi X rB

= + 0] + 1.51 X (-0.5k) + 4i X 2j = {0.75j + 8k} m/s`

Motion of C with Respect to a To find the time derivatives af rcts relative to x, y, z, let the x", y', z" axesrotate with 11„z = co2.Then11.„„ = w2 = {5k } rad/s = iw-, = { -6k} rad/s2

rem = 0,2j } m

= (icm)xyz, = + co, X Tc /R = 3j + 5k x 0.2j = {-li + 3j} m/s

1.acys>ty: = (YL7B)Ay.: = [(Vc/B).1-y"z- ÷ "2 X (rC/B)i-y'z'l Iri)2 X re/B + '02 x "18)

.xyz

= (2j + 5k X 3j) + (-6k X 0.2j) + [5k X (-li + 3j)] = { -28.8i - 3j }

mis'

Motion of C.

Ve = VB + n X ric/s + (v cvB)xy, = 2j + 4i X 0.2j + (- Ii + 3j)

=- {-li + 5j + 0.8k} m/s Ans..ac = 118 + S2 X rem + X (11 X rc/B) + 21/ X (vcis)„., + (ac/8),„

= (0.75j + 8k) + (1.51 X 0.2j) + [41 X (41 X 0.21)] + 2j4i x (-

li + 3j)] + (-28.8i - 3j)

= { -28.81 - 5.45j + 323k } m/s2 Ans.

20.4 RELATIVE-MOTION ANALYSIS USING TRANSLATING AND ROTATING AXES 573

SOLUTION II

Coordinate Axes. Herc wc will let the x, y, z axes rotate at

= ce i l + w, = 4 i + 5k} rad/s

Then = O.

Mation cd B. From the constraints of the problem o.b does not changc direction relative to X, Y, Z; however. the direction of w2 is changed by wi. Thus, to obtain ś i consider x', yr, z' axcs coincidentwith the X, Y, Z axes at A, so that = , . Then taking the derivativeof the components of n,

+ 62 --- + + + url x w21

-= [1.5i + 0] + [-6k + 4i X 5k] = {1.51 — 20j — 6k} radis2

Also, wi changes the direction af rB so that the time derivatives of r8 can be found using the prirned axes defined above. Hence,

vB + (.01 X rB

= O + 41 X (-0 .5k) = {2j} m/s

ilB = t 8 = + Cell X cel i X rB +Gil X '8

= [O + 0] + 1.51 X (-0.5k) + 41 X 2j = {0.75j + 8k } m/s2

Motion of C with Re%pect to B.

= O

= O

r GIB = {0.2j} m(v c18).k.. {3.1} m/s(21c/B)„ = {2j } m/s2

Motion of C.

vc = vI? + U X rcps INCiłdty:

= 2j + [(41 + 5k) X (0.2j)] + 3j

= {—li + 5j + 0.8k} m/s A n s.

ac = 28 + + x (n x re/B) 211 X (vc,$),, + (ac,,,$)xyz

= (0.75j + 8k) + [(1.51 — 211j — 6k) X (02j)]

+ (41 -i- 5k) X [(41 --F 5k) X 0.2j] + 2[(41 -I- 5k) X 3j] 2j

= { —28.81 — 5.45j + 32.3k} m/52 Arfa'

574 CHAPTER 20 THREE-DtMENSIONAL KINEMATICS OF A RICIE) BODY

■ PROBLEMS

20-37. Solvc Example 203 such that the x, y, z axes move with eurvilinear translation. Sl = O in which case the collar appears to have both an angular velocity

= w, w2 and radial molton.

20-38. Solve Example 20.5 by fixing x, y, z axes to rod BDso that fl = w,. In this case the collar appears only tomove radially outward along BD; hence 14, = 0.

20-39. Al the instant H = 60'_ the lelescopic boom AB of the construction lift is rotating with a constant angular velocity akont the z axis of wi = 03 rad/s and about the pin at A with a constant angular speed of w2 = 0,25 rad/s. SirnuIlaneousiy, lhe boom is exlending with a velocity of 1.5 ft/s, and it bas an acceleration of 0.5 ft/s2.both measured relative to the canstruction lift. Dctermine the velocity and acceleration of point B located al the end of the boom al this instant.

*20-40. Attheinstantft = 60°. thetonst,ructionlirtisr4tsting about the z axis with an angular velocity of w = 0.5 rad/s and an angular acceleration of W I = 0.25 rad/s2 while the telescopic boom AB rotaies about the pin at A with an angular velocity of w, = 0.25 rad/s and angular acceleration of

= 0.1 rad/s'. Simultanecrusly, the b00111 is cxtending with a velocity of 1,5 ft/s. and ii has an acceleration of 05 ft/s2,both measured relative to the frame. Deterrnine the velocity and acceleration of point B located at the end of the boom at this Instant.

2 0

Probs. 20-39/40

20-41. At a given instant. rod BD is rotating about the y axis with an angular velocity w BD = 2 rad/s and an angular acceleration ć = 5 rad/ s2. Also, when 9 = hO° /lnic AC is rotating downward such that B = 2 rad/s and

= rad/s2. Detcrmine the velocity and acceleration of point A on the link at this instant.

Prob. 20-41

20-42. Al the instant H = 30°. the frame of lhe crane and the boom AB rotate with a constant angular velocity of = 1.5 radis and w, = 0.5 rad/s. respectively. Determine the

velocity and acceleration of point B at this instant.

20-43. At the instant 8 31:1°. the frame of the orane isrotating with an angular velocity af w1 = 1.5 raci/s and anguIar acceleration of wl = 0.5 rad/s'-.while the boom AB rotates with an angular velocity of wa = 0.5 rad/s and angular acceleration of w, = 0.25 rad/s2. Deterniine the velocity and acceleration of point B at this instant.

Probs. 20-42/43

20.4 RELATIVE-MOTION ANALYSIS USING TRANSLATING AND ROTATING AXES 575

*20-44. At the instant shown, the boom is rotating about the z axis with mi angolar velocity roj = 2 rad /s and angular acceleration i;.11 = 0.8 rad/s2. M [bis same instant the swivei is rotating at ar = 3 rad is when cd-r = 2 rad/s2. both measured relative to the boom. Oeterrnine the velocity and acceleration of point P on the pipe at Ibis instant.

20-45. During the instant shown the france of thc X-ray camera is rotating about the vertical axis at cd, = 5 rudis and ty= = 2 rad/s2. Relat1ve to the frame the arm is rotating ai air,j = 2 radis and (;),] = 1 rad/s2. Determine the velocity and acceleration of the center of the camera C at this instant,

Prob. 20-45

20-46. The boom A B of the crane is rotaling about the z axis with an angular velocity w= = 0.751-ad/s. which is increasing at m, = 2 rad/s2. At the same instant, H = 60° and the boom is rotating upward at a constant rato

= 0.5 radis2. Determine Lhe velocity and acceleration of the lip B of Lite boom at Ibis instant.

*20-47. The boom AIS of the crane is rotating about thez axis wich an angular velocity of = 0.75 rad/s, which isincreasLng at = 2 radis2. Al the same instant, H = 60°and the boom is rotating upward at h - U.S rad/s2, which is increasing at 6 = 0.75 radis2. Determine the velocity and acceleration of the lip B of the boom at this instant.

20-48. At the instant shown, the motor rotates about the z axis with an angular Mocny of ml = 3 radis and angular acceleration of tul -= 1.5 rad/s2. Simultaneously, shaft DA rotates with an angular wiochy of 0,12 = 6 rad/s, and angular acceleration of at2

= 3 rad/s2, and collar C sHdes along rod A with a vehreity and acceleration of 6 m is and 3 m ł Det ermLn ethe velocity and acceleration of collar Cat this instant.20

z

Pruł). 20-48

£02 = 3 rad/s(;.! 2 radia` "

= 2 radfsj), = 0.8 rad/s2

ti

4

Prob. 20-44

0.5 rad /52

cd, = 0.75 rad /s = 2 radfs2

5 ft Y

Probs. 20-46/47

576CHAPTER 20 THREE-DIMENSIONAL KINEMATICS OF A RIGID BODY

2 0

20-49. The motor rotates about the z axis with a constant angular velocity of w1 = 3 rad/s. Simultancously, haft OA rotates with a constant angular velocity of w, = 6 rad/s. Also, collar C slides along rod A13 with a velocity and acceleration of 6 m/s and 3 m/s2. Determine the velocity and acceleration of collar C at the instant shown.

Prob. 20-49

20-511. At the instant shown, the arm OA of the conveyor beli is rotating about the z axis with a constant angular velocity w, = 6 rad/s, while at the same Instant the arm is rotating upward at a constant rate w, = 4 rad/s. If the conveyor is running at a constant rale r = 5 ft/s, determine the velocity and acceleration of the package Pal the instant shown. Neglect the size of the package.

20-51. At the instant shown, the arm OA of the conveyor beli is rotating about the z axis with a constant angular velocity

= 6 rad/s, while at the same instant the ann is rotating upward at a constant rate w = 4 rad/s. If the conveyor is running at a rate r = 5 ft/s, which is increasing at = 8 ft/s2, determine the velocity and acceleration of the package P at the instant shown. Neglect the size of the package.

Probs. 20-50/51

*20-52. The boom AB of the locomotive crane is rotating about the z axis with an angular velocity wi = 0.5 rad/s, which is increasing at W, = 3 rad/s2. At this same instant, O = 30° and thc boom is rotating upward at a constant rate of t? = 3 rad/s. Determine the velocity and acceleration of the tip B of the boom at this instant.

20-53. The locomotive crane is traveling to the right at 2 m/s and has an acceleration of 1.5 m/s2, while the boom is rotating about the z axis with an angular velocity w, = 0.5 rad/s, which is increasing at W, = 3 rad/s2. At this same instant,

= 30° and thc boom is rotating upward at a constant rate= 3 rad/s. Determine the velocity and acceleration of the

tip B of the boom at this instant.

Probs. 20-52/53

20-54. The robot shown has four degrees of rotationa] freedom, namely, arm OA rotates about the x and z axes, arm AB rotates about the x axis. and CB rotates about the y axis. Al the Instant shown. w, = 1.5 rad/s. w, = I rad/s2. w3 = 3 rad/s, ir3 = 0.5 rad/s2, w, = 6 rad/s. w, = 3 rad/s2, and w, = w, = 0. If the robot does not transłate, i.e., v = a = 0, determine the

velocity and acceleration of point C at this instant.

Prob. 20-54

CHAPTER REVIEW 577

CHARTER REVIEW

+ n X rif/A +

= aA + ttŻ K rym -F fk x ((t x r,v,) + 2x (v,914),,

instantancous axis of rotalion

V p = w x r

a p = a X r ÷ o i X ( w X r )

VB = ym -F w X r8/A

a,,aA + ar X r,vA r_tp X (tei X r37A)

Relative Motion Analysis Using Translating and Roiating Axes

The motion of two points A and B on a body, a series of connected bodies, or each point located on two different paths, can be related using a relative motion analysis with rotating and translating axes at A.

When applying the equations, to find vs

and ak, it is important tn accnunt for both the magnitude and directional ehanges of r,4, r$/A. il, and £1,,, when taking lheirtime derivatives to finda4, (1713./A

{aKłA),,,, fl, andTo do this properly,one must use Eq. 2Ó-6.

Rotation About a Fixed Point

When a body rotałes abc:Fut a fixed point 0, then points on the body follow a path that lies on the surfaee of a sphere centered at 0,

Since the angular acceleration is a time rate of change in the angular velociły, then it is nccessary to account for both the magnitude and direcflonal changes of w when findingits time derivative. To do this, the angular velocity is often specified in terms of its component motions. such that the di.rection of some of [hese components will remain constant relative to rotating x. y, z axes. If this is the casc, t h en the time derivative relative tn the fixed axis can be determined using A = (A), + Ii X A.

Once w and cz are know], the velocity and acceleration of any point P in the body tan then be determined.

General Motion

If the body undergoes general motion, then the motion of a point B on the body can be related to the motion of another point A using a relative motion analysis, with translating axes artached to A.

Chapter 21

The forces acting on each of these motorcycles can be determined using the equations of mcition as discussed in this chapter.

Three-Dimensional Kineticsof a Rigid Body

CHAPTER OBJECTIVES

■ To introduce the methods for finding the moments of inertia and products of inertia of a body about vanous axes.

■ To show how to apply the principles of work and energy and dimensional motion.

■ To develop and apply the equations of motion in three dimensions.

■ To study gyroscopic and torque-free motion.

*21 .1 Moments and Products of Inertia

When studying the planar kinetics of a body, it was necessary to introduce the moment of inertia IG, which was computed about an axis perpendicular to the piane of motion and passing through the body's mass center G. For the kinetic analysis of three-dimensional motion it will sometimes be necessary to ca lculate six inertial quantities. These terms, called the moments and products uf inertia, describe in a particular way the distribution uf mass for a body relative to a given courdinate system that has a spccificd orientation and point of

580 CHAPTER 21 THREE-DtmENsICNm_ KINErics o.F A RIGID BeDY

Moment of Inertia. Corisider the rigid body shown in Fig. 21-1. The moment oj. Merita for a differentiał element dm of the budy about any one of the thrce coordinate axcs is defined as the product of the mass d the element and thc squarc of thc shortest distance from the axisto the element. For example, as noted in the figure, rx 1)2 + z2. sothat the mass moment of inertia ot the element about the x axis is

= rk dm = (y + z2) dm

The moment of inertia f„ fur the body can be determined by integrating this expressiart ovex the entire mass of the body. Hence. for each uf the axes, we can wriłe

Fig. 21-1

.4= jp1( ts?cfiri= (y2 + Z2) dm

my y =

(x2 z2) din r.";:dim

Imr2din (x2 + y2) dm

(21-1)

Herc it is scen that thc moment of inertia is always a positive (mamiły, since it is the summation of the product af thc mass dm, which is always posirive, and the distances squared.

Product of Inertia. The product of inertia fora differential element dm with respect to a set of Iwo orthogonal planes is defined as the product of the mass of the element and the perpendicuiar (or shortest) distances from the planes to the element. For example, this distance is x to the y—z piane and it is y to the x—z piane, Fig. Z1-1. The product of inertia dł,• for the element is therefore

= xy dm

Notc also that dl,.x = dl,. By integrating over the entire mass, the products of inertia af the body with respect to each combination of planes can be expressed as

i = ł - = T c y d m "Y Yx

m

j r . i

M 4 , = 1 , _ , = x z d m

(21-2)

21.1 1VIOMENTS AND PRODUCTS Of INERTIA 581

(u) (b)

Fig. 21-2

Unlike the moment of inertia, which is always positive, the product of inertia may be positive, negative, ar zero. The result depends on the algebraic signs of the twa defining coordinates, which vary indepe-ndentiy Erom one another, In particular, if either one ar both of the orthogonal planes arc planes of synnnełry for the mass, the product of inertia with respeet to these planes will be zero. In such cases, elements of mass will occur in pairs located on each side of the pianc of symmetry. On one side of the piane the product of inertia for the element will be positive, while on the other side the product of inertia of the corresponding element will be negative, the sum łherefore yielding zero. Examples of this are shown in Fig, 21-2. In the first case, Fig. 21-2a, the y-z piane is a piane of symn-ietry, and hence I, = I,. -= O. Calculation of 4., will yield a positive result, since all elements of mass are located using only positive y and z coordinates, For the cylinder, wita the coordinate axes łocated as shown in Fig. 21-21), the x-z and y-z planes are both planes of syrnmetry. Thus,

= 0-

Parallel-A)ds and Parallel-Plane Theorems, The techniques of integration used to determine the moment of inertia of a body were described in Sec, 17.1, Also discussed were methods to determine the moment of inertia of a composite body, i,e., a body that is composcd of simpler segments, as tabulated on the inside bacik cover. In both of these cases the parallel-axis theorem is often used for the calculałions. This theorern, which was developed in Sec. 171 allows us to transfer the moment of inertia of a body from an axis passing through its mass center G to a parallel axis passing through some ather point. If G bas coordinatcs xG,

yG, ZG defined with respect to the x, y, z axes, Fig. 21-3, then the parallel-axis equations used to calculate the moments °f inertia about the x, y, z axes are

z,2G)

fy., = (/),..y,)G + .a.a(x:6" +

izz = 1nOcć +

(21-3) _vG

Fig. 21-3

582 CHAPTER 21 THRE-D[IvIENSIONAL KINETICS OF A RiG1t3 BODY

The products of inertia of a composite body arc computcd in the same manner as the body's moments of inertia. Here. however. the parallet-plarie łheoremis important. This theorem is used to transfer the products of inertia of the body with respect to a set of three orthogonal planes passing through the body's mass center to a cotTesponding set of three parallel planes passing through some other point O. Defining the perpendicuiar distanees between the planes as x0, y and .zG. Fig. 21-3. the parailel-piane equations can be written as

Y

xG Ixy = (4,3,,)0 + truc-37G

ly, = (iy,,,)G + rnyGza,

+ inzGxG,

Y G

(21-4)

Fig. 21-3 irepeaied)

The dyrzarrzics of the space shutt[e while il orbiis the. earth can be predicted °n b, ii ils moments and products of inertia are known relative to its mass center.

Thc derivation of these formulas is similar to that given for the paralleb axis equation, See. 171

inertia Tensor. The inertial properties oś a body are therefore compietely characterized by nine terms, six of which are independent of one another. This set of terms is defined using Eqs. 21-1 and 21-2 and can be written as

(I

JCX —i.ICV —1,i'.7 kk4.1' IYZ

—4.„ — lzyIZZ

This array is called an inertia łensor. It has a unique set of values for a body when it is deterniined for each location of the origin O and orientation of the coordinate axes.

In genem], for point O we can specify a unique axes inelination for which the products of inertia for the body are zero when computed with respect to these axes. When this is done, the inertia tensor is said to be "diagonalized" and may be written in the simplifiecl farm

( t , O O

Oi ) a i ,

Here Ix =Ixx,= , and I. = I. are termed the principal moments of Merita for the body, which a-re computed with respect to the prineip£11 axes of inertia. Of these three principa I moments of inertia, one will be a maximum and another a minimum of the body's moment of inertia.

The negaEive signs are herc asa consequence uf the development of angolar momentum, Eqs. 21-10.

21,1 MOMENTS AND PRODUCTS OF INERTIA583

The mathematical determination of the clirections of principal axes of inertia will not be discussed here (see Prob. 21-22). However. there are many cases in which the principal axes can be determined by inspection. From the previous cliscussin it was noted that if the coordinate axes are oriented such that two of the three orthogonal planes containing the axes are planes of symmetry for the body, then all the products of inertia fur the body are zero with respect to these coordinate planes, and hence these coordinate axcs are principal axes of inertia. For example. the x, y, z axes shown in Fig. 21-2b represent the principal axcs of inertia for the cylinder at point O.

Moment of Inertia About an Arbitrary Axis. Consider the body shown in Fig. 21-4, where the nine elements of the Media tensor have bccri dctermincd with respect to the x, y, z axes having an origin at O. Herc we wisk to dctermine the moment of inertia of the body about the Oa axis, which has a direction defined by the -unit vector u,. By definition for = f b2 dm, where b is the perpendietdar distartee from dm to 0a. If the position of d z is located using r, then b = r sio El, which represents the magndaede of the cross product u„ X r. Hence, the moment of inertia can be expressed as

fo„ = f (u,, X r) I 2dm = J (u, X r) • (tr X rkim• m

Prov ided q , = u, i + u , j u ,k and r = x i yj zk , t hen u„ X r =(urz + 04x - u,z)j (uj uAk. After substituting andperforming the dot-product operation:the moment of inertia is

fo, -= f Ruyz - uzy)2 + (ug - 14„4 ,2 (gxY uyx)2idmm

u;c ty2 + z2).dm + (z2 + x2)dm + 14;2 (x2 + y2) din

‚11

turu, j ry din - yz chn 224,14,_ j zx durrm FTP

Rccognizing the integrais to be the nriorricnts and products of inertia of the body, Eqs. 21-1 and 21-2, we have

dm

b = r sinr

rr

Fig. 21-4

10„ = + 1„uy2 + fz,u,2

(2I--5)

Thus, if the inertia tensor is specified for the x, y, z axes, the moment of inertia of the body ahout the inehned Oa axis can be found. For thecalculałion, the direction cosinesu,,u,cł the axes must be determined.These tenns specify the cosines of the coordinate direction angies a, 113, y macie between the positive Oa axis and the positive x, y, z axes, respectively (sce Appendix B).

584 CHAPTER 21 THREE-DtMENSIONAL KINETICS OF A RIGID BODY

EXAMPLE

0 . m 4 k

V

Determine the moment of inertia of the kent rod shown in Fig, 21-5a about the Aa axis. The mass of each of the łhree segments is given in the figure.

SOLUTIONBcforc appIying Eq. 21-5, it i5 first necessary to determinc the momcnts and products of inertia of the rod with respect to the x, y, z axes. This is done using the formuła for the moment of inertia of a slender rod.I = and the parallel-axis and parallel-plane theorems, Eqs. 21-3and 21-4. Dividing the rod finto three parts and locating the mass center of each segment, Fig, 21-5b, we have

I„ = [1(2)(0,2)2 + 2(0.1)2i + [0 + 2(0.2)2]

++ 4((0.2)7 + (0.2)7)1 = 0.480 kg • m2

= [(2)(0.2)2 + 2(0.1)2] + [h(2)(0.2)2 + 20-0.1)2 + (0.2)2)]

+ [0 + 4((-0.2)2 + (0.2)2)] = 0.453 kg • m2

iz, = [1:1 + + [-142(2)(0.2)2 + 2( -0,1)21 + [1.(4)(0.4)2 +

4{(-0.2)2 + (0.2)2) I ={}.400 kg •

m2 = [0 + O] + [0 + O] + [0 + 4( - 0.2)(0,2)1 = -0, I60 kg • m2

/,, = [0 + O] + [0 + Oj + 0 + 4(0.2)(0.2)] = 0.160 kg • rn2

2 kg(-0.1,0,02)

f„ = [0 + O] + [0 + 2(0.2)(-0.1)] 4-

[0 + 4(0.2)( -0.2)1 = -0.200 kg • m2

The Aa axis is defined by the unit vector

rD -0.21 + 0.4j + 0.2k= = -0.4081 + 0.816j + 0.408k14, -

ro "\,/ (-0.2)2 + (0.4)2 + (02)2

B

CDThus,

u,. = -0.408 u) = 0.816 u, = 0.408

- 1 1 _ 3

c\4 kg( -G2.02 .02)

2 kg) Substituting these results anto Eq. 21-5 yields

Y

x(b) I A w —fxxuź-F I vyu+2 + ł24vuxuv24.7.44vu:21;:truzi€x

Fig. 21-5 - 0.480( - 0.408)2 + (0.453)(0.8 I 6)2 + 0.400(0.408)2

- 2( - 0.I60)( -0.408)(0S16) - 2(0.160)(0.816)(0.408)

- 2( -0.200)(0 ,408)( - 0.408)

- 0.169 kg m2 A i7.V.

21.1 MOMENTS AND PRODUCTS OF iNERTIA 585

■ PROBLEIVIS

21-1. Show that the suni of the moments of inertia of abody, Iiz + 4, L. is independent of the oricntation of thex, y, z axes and ibus depends only on the locatton of the origin.

21-2. Determine the moment of inertia of the cone with respeet to a vertleal axis passing through the cone's center of mass. What is the moment of inertia about a parallel axis y' that passing through the diameter of the base of the cone? The cone has a masa m.

I n

Prob. 21-2

21-3. Determine the moments of inertia i, and I,. of the paraboloid c.lf revalulion.t7le mass of the paraboloid is

*21-4. Determine the radli of gyrałion k, and k, for the solid formed by revolving the shaded area aboul the y axis. The density °f the materiał is p.

- 4 f t

Prob. 21-4

21-s. Determine by direct integration the produci ofinertia for the homogeneous prism. The density of themateriał is p. Express the result in terms of the total mass m of the prism.

21-6. Determine by direct in tegration the produci of inertia I. for the homogeneous prism. The density of the materiał is p. Express the resuit in terms of the total mass m of the prism.

4 l't

.21

Prob. 21-3 Probs. 21-5/6


21-7. Determine the produci of inertia of the objectformed by revolving the shaded arca about the line x = S ft. Express the result in terms of the dcnsity of thc materiał, p.

Determine the moment of inertia /). of the objeet formed by revolvmg the shaded area about the hne x = 5 ft. Express the rewii in terms of the densitv °f the materia'. p.

21-10. Determine the mass moment of inertia of the homogeneous block with respect tt its centroida] x' axis. The mass of the błock is m.

L3 f t

2 fi

Probs. 21-704

Y

Prob. 21-10

21-9. Determine thc elements of the inertia tensor for thc cube with respect to the x, y, t coordinate system. The mass of the cube is nr.

21-11. Determine the moment of inertia of the cylinder with respect to the a—a axłs of the cylinder.The cylinder has a mass m.

21

Prob. 21-9 Proh..21—i1

21

21.1 IVIomENT5 AND PRODUCTS Of iNERTIA 587

*21-12. Detennine the moment of inerh a of the composite plate assembly. The plater have a specific weight of 6 lbfift2.

2143. Determine the produci of inertia l,„ of the composite plate assembly.The plates have a weight of 6 ihift2.

z

*21-16. The hem rod has a mass of 4 kg/m. Determine the moment of 'nenia of the rod about ihe Da axla.

Probs. 21-12/13

21-14. f)e te rrn in e the products of in ertia 1,.,, f„,andlxZ,of the Chin plate. The maierial hasa density per unit area of 50 kg/rnz.

Prob. 21-14

21-15. Determine the products of inertia , 4, and /„ of ihe solid. The materiał is steel.whch hasa specific weight of 490 ibift3.

21-17. The bont rod bas a wcight of 1,5 Ibift. Locate the center of gravity G{x, } ant] determine the principai moments of inertia 1,,, 4., and 1, of the rod with respect tothe axes.

OS ft __ 0.4 m

Prob. 21-17

03 ft .0.5 ft

Prob. 21-16

0.6 ta

Prob. 21-15

588CHAPTEP 21 THRE-D[MENSIONAL KINETICS OF A RIOJD BODY

21-18. Deterrnine the mornents of inertia about the x, y, z axes of the rod assembiy. The rods have a mass cif 0.75 kg/m.

21-20. The assembly consists of a 15-lb plale A. 40-lb pl ate 8, and four 7-1b rods.Determine the moments of inertia of the assembly with respect to the principal x, y. z axes,

z

Prob. 21-20

21-19. Determine the moment of inerłia of the composite body about the au axis. The cylinder we3ghs 20 lb. and each hemisphere weighs 10 Ib.

21-21. Deteamine the moment of inertia cif the rod-and-t hin-ring assembly about the z axis. The rods and ring have a mass per unit length of 2 kg/m.

21

P rob. 21-19

Y

Prob. 21-21

Prob. 21-18

21.2 ANGUL4R MeMENTLIM 589

29.2 Angular Momentum

In this section we will develop the necessary equations used to deterrnine the angular momentum of a rigid body about an arbiłrary point. These equations will providc a mcans for devcloping both the prineple of impulsu and momentum and the equations of rotational motion for a rigid body.

Consider the rigid body in Fig. 21-6, which has a mass m and center °f mass at G. The X, Y, Z coordinate system represents an inertial frame °f reference, and hence, its axel aru fixed or transIate with a constant vełoci ty.The angular momentum as measured from this reference will be determined relative to the arbitrary point A. The position vectors tq and pą aru drawn from the origin of coordinates to point A and from A to the ith particie of the body. If the particie's mass is mi, the angular momentum about point A is

= PA X mivi

where vi represents the particle's velocity measured from the X, Y, Z coordinate system. If the body has an angular velocity ca at the instant considered, v, nicy be related to the velocity of A by applying Eq. 20-7, i.e.,

V i = V A + (d X p .4

Thus,

=- PA X ini(VA ± cz) X PA)

= fPĄ►ni) X 17,1 + Na X (w X p.1)ira i

Summing thc rnomcnts of ail thc partielcs of the body requires anintegration, Siwe din, we have

HA = PAdm) X vA LPA X (£t) X pA)din (21--6)

Inc rtia I coordinate system

Fig. 21-6

5 9 0 C H A P T E R 2 1 T H R E E - D t M E N S I O N A L K I N E T I C S O F A R I G I D B O D Y

z

Fi‚wd Point

(n)Center at Mass

(b)

Y

Y

zz

Fig. 21-7

Fixed Point O. If A becomes a fixeci point O in the body, Fig.21 —7a, then v = D and Eq. 21-6 reduces to

Ho = po X ((Lo X po)dm (21-7)

Center of Mass G. If A is located at the center of mass Gaf the body. Fig. 21-7b, then ,,p,1 df = O and

pc X (w X pc)dinrrr

(21-8)

HG

(;

Y

Arbitrary Point

(c)

Arbitrary Point A. In general, A can be a point other than O ar G, Fig. 21-7c, in which case Eq. 21-6 may nevertheless be simplified to the following form (see Prob. 21-23),

= PGM X nivG HG (21-9)

Herc the angular momentum cansists af twa parts —the mantent af the linear momentum myc, of the body about point A added (vcctorially) to the angular momentum HG. Equation 21-9 can also be used to determinc the angular momentum of the body about a fixed point O. The resu]ts, of course, will be the same as those found using the more convenient Eq. 21-7.

Rectangular Components of H. To m ake practical use af Eqs. 21-7 through 21-9, the angular momentum must be expresscd in terms of its scalar components. For this purpose, it is convenient to

21.2 ANGU LAR M M ENTUM 591

choose a second set of x, y, z axes having an arbitrary oric n talion rclativeto the X. Z axes, Fig. 21-7, and for a gcncral formulation. noto thatEqs. 21-7 and 21—S are botki of the form

H= ifpX (ct i X p)dm

Expressing W, p. and w in terrris of x, y. z corripcinents. we have

+ H, k = f (xi + 3/J zk) x Rw, i + wyj ćik

X ( x i + y j + z k ) l d m

Expanding the cross products and combining terms yieldsfi,i + U [

FI}, + I i ,k —, w, (y z2)din — w,, J xy dn — wzxz dtn if / 1 m * /

+ [—tor faxy drn + wy j (x2 + z2)(bri — co, J yz dmij

+ rvx zx drr — yz drn + ł (x2 + y2)dni jkrff

Equating the respective i, j, k components and recognizing Chat the integrals represent the moments and products of inertia, we obtain

N,.-ILTWA..— [1,34)V 1x.5.(-0

H v = —IvaLla Jrzgov — v:£0.:= — lzyw,

(21-10)

These equations can be sim pIified furthcr if the x, y, z coordinate axes are oriented such that they become principal axes of inertia for the body at the point. When these axes are used, the products of inertia1`,„ = f„, = = 0, and ił the principal moments of inertia about the x, y,z' axes are represented as Ix. =-= ,ry,, and = the threecomponents of angular momentum become

k = 11, = = 1,6Z] ( 2 1 4 1 )

5 9 2 CHAPTER 21 THREE-DtmENsioNAL KINErics OF A RIGID BODY

The morion of the astronaul is controfied by use of smali directional jets attached W his nr her space suit. The impulses these jets provide musi be carefully specified in order to preyent tumbhng and less of orientation

Principle of Impulse and Momentum. Now that the formulation of the angular momentum for a body has Been developed, the principle of impulse and mornenłum, as discussed in Sec. 19.2, can be used to solne kinetic problems which involveforce, velocity, and time. For this case. the following two vector equations are availabłe:

t,rn(vdil +F dr = m(vG),

r,

(Ha) L + f Mo dt = (Ho)2

In three dimensions each vector term can be represented by three scalar components. and therefore a totał of six scalar equations can be written. Three equations relate the linear impulse and momentum in the x, y, z direcfions, and the other three equations relate the body's angular impulse and momentum about the x, y, z axes. Before applying Eqs. 21-12 and 21-13 to the solution of problems, the materia] in Secs,19,2 and 193 s hould be reviewed.

(21-12)

(21-13)

21 .3 Kinetic Energy

Inertial coordina te system

Fig. 21-8

In order to appły the principle of wark and energy to solne problems involving general rigid body mk-stion, it is first necessary to farmulate expressions for the kinetic energy of the body. To do this, consider the rigid body shown in Fig. 21-8, which has a mass ni and center of mass at G. The kinetic energy of the ith particie of the body having a mass mi and vclocity vi, measured reIative to the inertial X, Y, Z frame of reference, is

4 vi)

Provided the velocity Uf an arbitrary point A in the body is known,vF

can be related to vA by the equation v, = vA + w X pA, where w is the angular velocity of the body, measured from the g, y, z coordinate system, and pĄ is a position vector cxtending from A to i. Usiag this expression. the kinetic energy for the particie can be written as

= - irn(vA + w X pA)•(vA + w X pA)

I= — 2(vA - vA)mi + vp • (w X p,l)mi + 2fw X PA) - (w X pA)mi

The kinetic energy for the entire body is obtained by summing the kinetic energies of all the particles of the body. This requires an integration. Since ms

—› dm, we get

T = 2 in(v.4-vA) + vA-(0) X PAdrn) + i (0.3 PA)' (co X pA)don

21.3 KNETiC fNERGY593

The ]ast ten)] on the right can be rewritten using the vector identitya x b•c = a-Its x c, where a = b = pA , and c = w X pĄ. The finalresult is

J T = rri(vA • vA) -H v, • (W X pA(tn)

+ 4 W • ffiA Z (w X pA)dm (21-14)

This equation is rarely used hecause of the computations involving the integrals. Simplification occurs, however.if the reference point A is either a fixed point tir the center &mas&

Fixed Point O. If A is a fixed poru O in the body, Fig. 21-7a, then vA = 0, and using Eq. 21-7, we can express Eq. 21-14 as

T = -.1,6) • Ho

If the x. y, z axes reprcsent the principal axes of incrtia for the body. then = u)., i + wyj + u), k and Ho = fxwk i + 1.3:cuj + 1,w, k. Substituting

int° the abovc equation and performing the dot-product operations yields

T 4 ir,0„.,,2 4iywy2 ± 44(0_2 (21-15)

Center of Mass G. If A is located at the center of muss Gaf the body, Fig. 21-7b, then Jpq rinz = O and, using Eq. 21-8, wc can write Eq. 21-14 as

In a manner similar to that for a fixed point, the last term on the right sicie may be represented in scalar form, in which case

+ 2Ix + 2t3p),= + (21-16)

Here it is scen Chat the kinetic energy consists Gf two parts; nameiy, the translational kinetic energy of the mass center, 3ntv6, and the body's rołational kinetic energy.

Principle of Work and Energy. Having formulated the kinetic cncrgy for a body, the principle of work and energy can be applied to solve kinetics problems which involve furce, veloeity, and displacement. For this case only one scalar equation can be written for each body, namely,

+ =Tz (21-17)

Before applying this equation, the material in Chapter 18 shouid be reviewed,


The rod in Fig. 21-9a bas a weight per unit length of 1.5 Ib/ft, Deterrnine its angular veIocity just after the end A falls onto the bank at E. The hook provides a permanent conneetion for the rod due to

t ftthe spring-lock mechanism S. Just before striking the hook the rod is faIling downward with a speed (vG)/ = 10 ft/s.7-70m7The principIe of impuIse and momentum will be used since impaet occurs. SOLUTION

A

0333 ftImpulse and Momentum Diagrams. Fig. 21-9h. During the short03 fl 7 time ❑t, the impulsive force F acting at A changes the momentum of

the rod. (The impulse ereated by the rods weight W during łhis time is smali compared to f F dt, so that it can be neglected, i.e„ the weight is a nonimpulsive force,) 1-lence. the angular moment urn of the rod is conserved about point A since the moment of f F dz about A is zero. Conservation of Angular Momentum. Equation 21-9 ni ust be used to find the angular momentum of the rod, since A does not bccomc a fixed point unii] after the impulsive interaetion with the hook. Thus, with reference to Fig. 21-9b, (ROI = (HA )2, °r

rri./A x m(va)i rG/A x Myc), (11(;)2. (1)

From Fig. 21-9a, rc,b, =- j -0.667i + 0.5j} Ft. Furthermore, the primcdaxcs arc principal axcs ofinertia for the rod bccausel,,,,, =1,,,, = = 0.Hencc, from Eqs. 21-11, (Fic), = rw, i + /yrwyj + k, Thc principalmornents of inertia are Iz = 0.0272 shig • 02, ty, = 0.0155 sług • ft2-, = 0.

0427 slug ft' (see Prob. 21-17). Substituting int° Eq. 1, we have

( -0.6671 + o,5JD x [ (—)(- 10k) = ( -0.6671 + 0.5j) X [ ()( -vG)2k1

+ 0.0272w, i + 0,0155w, j + (1.0427w, k

Expanding and cquating the respectivc i, j, k components yields

-0.699 = -0.0699(vG).2 + 0.0272w,

-0.932 = -0.0932(vG)i + 0.0155w, (3)

G = 0.0427w, (4)Kinematics. There ara four unknowns in the above equations: however, another equation may be obtained by relating w to (vG),using kinmaatics. Since = U (Eq. 4) and after impast the rod rotatesabout the fixed point A, Eq. 20-3 can be applied, in which case (vG) = w

xrG/A or-(73G),k w,,,j) x (-0.667i + 0.5j)—(vG), = 0.5w, + 0.667w, (5)

Solving Eqs. 2,3 and 5 simultaneously yields

Fig. 21-9 {V = -8.41k} ftis w = -4.091 -- 9.55j} rad/s

x '

( a )

W A r O

( b )

( 2 )

21.3 KNETiC ENERGY 595

A 5-N • m torque is applied to the vertical shaft CD shown in Fig. 21-10a, which allows the 10-kg gear A to tern freely about CE. Assuming that gear A starts from rent, determine the angular velocity of CD after it bas turned twa revolutions,NegIect the mass of shaft CD and axle C F and assumc that gear A can be approximated by a thin disk, Gcar B is fixed.

5OLUTIONThe principle o( work and energy may be used for the solution. Why?

Work. If shaft CD. the axle CE, and gear A are considercci as a system of connected bodies, only the applied torque M does work. For tworevolutions of CD, this work is = (5 N • rn)(=kr rad) = 62.83 L

Kinetic Energy. Since the gear 1s initially at rent, its initial kinetic energy is zero. A kinernatic diagram for the gear is shown in Fig. 21-10b, If the angular velocity of CD is taken as weD, then the angular velocityof gear A is rac!, + weE, The gear may be i magined as a portionof a massless extended body which is rotating about the fixed point C. The instantaneous axis of rotation for this body is along line CH, bccausc butli points C and H on the body (gear) have zero velocity and must therefore lie on this axis. This requires that the components rxiscp and £(),-E be related by the equation wcD/O.I m = wcE/0.3 m ur wcE = 3wCp. ThILS,

—(-1-/CEJ WCD k = 36)CD + WeDk (1)

The x, y, z axes in Fig. 21-10a represcnt principal axes of inertża at C for the gear. Since point C is a fixed point of rotation, Eq. 21-15 may be applied to determine the kinetic energy. i.e..

T '= -121„.(f4 42l (2)

Using the paralIel-axis theorem, the moments of inertia of the gear about point C are as follows;

= .!(10 kg)(0.1 ni)2 = 0.05 kg -

= Ir = kg)(0.1 m)2 + 10 kg(0.3 rn)2 = 0.925 kg

Sinice (dx = —3(0c-D, = U. = WIE,, Eq. 2 becomes

Ty = -•:,(0.05)(-303cD)2 ÷ O + 5[10,925)((-1>CD)2 = 0.075w:2h)

Principle of Work and Energy. Applying the principle, of work and energy, we obtain

TI + =

O + 62.83 = 0.6875c,2.D

wen = 9.56 rad/s A/1.5.

Tnstantancous z axis of rotation

"J'enW.7

x

ea )

Fig. 21-10

5 9 6 C H A P T E R 2 1 T H R E E - D t M E N S I O N A L K I N E T I C S O F A R I G I O B O D Y• PROBLEMS

21-22. u a body contains no planes of symrnetry, the principal moments of inertia can be determined mathematically.To shGw how this is done, consider the rigid body which is spinning with an angular velocity (o, directed dong one of its principal axes of inertia, lf the principal moment of inertia about lhis axis is I. the angular m ornen ium can be expressed as H = 7w = luxri + koj + Iw, k. The components of 11 may also be expressed by Eqs. 21-10, where the inertia tensor is assumed to be known. Equ.ate the i, j, and k components of boli] expressions for H andconsider Q.■_„and 0.), to be unknown. The solution cif thesethree equations is obtained proyidcd the determinant of the coeffieents is zero. Show that Ibis determinant, when expanded, yields the tubie equation

13 - + 13,. + 4,)12+ + /3„1„ + 122/, - -1,'.z- 1",)/

- (Ix.,[11),IZZ 21XN 4J:X- 13,31;7 IzA) = 0

The three positive roots of I, obtained from the solution of this equation, represeni the principal inoments of inertia Ii., and I,.

21-23. Shaw that if thc angular momentom of a body is deterrnined with respect to an arbitrary point A. alert HA can be expressed by Eq. 21-9. This requires substituting PA = pc, + PG.IA lato Eq. 21-b and expanding. noting that fp,) dm = O by definition of thc mass center and vc, = v j+ w x pG/A

Y

Prob. 21-23

*21-24. The 15-kg circuiar disk spins about its axle with a cons tan t angular velocity of wi = 14 rad/s. Stmultaneously, the yoke is rotating with a constant angular velocity of

= 5 rad/s. Determine the angular momentum of the disk about 115 center of mass O. and its kinetic energy.

Z

z

21

5 rad s

Prob. 21-22 Prob. 21-24

21.3 KNUJ-C ENERGY 597

21-25. The C(}310 has a mass m and roik withoui slipping on the conicał surface so that it has an angular velocity about the vertleal axis of rxl. Determine the kinetic energy of the cone due to This motion.

Prob. 21-25

21-28. The space capsule hasa mass of 5 Mg and the radii of gyration are k, = k, = 1.30 m and k,. = 0.45 m."1f il travels with a velocity= {400j ± 20(ik) m/s,compute its angular velocity just after it is struck by a meteoroid having a mass of 0.80 kg and a velocityv, = {-3001 + 200.115014} m/s. Assume thal themeteoroid embeds itself into the capsule at point A and that the capsule initially has no angular velocity,

Prob. 21-28

21-26. The circular disk has a weight ❑f 15 lb and is mounted on the shaft AB at an angle of 45 with the horizon tal. Determine the angular velocity of the shafi when ,r = 3 s if a constans torque M = 2 lb • ft is applied to the shaft. The shaft is originally spinning at w = radis when the torque is applied,

21-27. The circular disk hasa weight of 15 lb and is mounted on the shaft AD at an angle of 45° with the horizontal. Detennine the angular veloeity of the shaft when r = 2 s if a torque M = (4e" ) lb - ft, where r is in seconds, is applied to the shaft. The shaft is originalły spinning at w, = 8 rad/s when the torque is applied.

21-29. The 2-kg gear A roik on the fixed plate gear C, Determine the angular velocity of rod OB about the z axis after it rotates one revolution about the z axis. starting from rest. The rod is acted upon by the constans moment M = .5 N • m, Neglect the mass of rod OB. Assume that gear A is a uniform diak having a radius of 100 mm.

21

Prob. 21-29Prob& 21-26/27

598CHAPTER 21 THREE-DtMENSIONAL KINETICS OF A RIGIO BODY

21-3O. The rod weighs 3 Ih/ft and is suspended from parafie] cords al A and B. If the rod has an angular velocity of 2 rad/s about the axis at the instant shown. determine haw high the center ()f the rod rises at the Instant the rod momen tarily stops swinging.

Prob. 21-30

21-31. Rod /LB lias a weight of 61h and is attached to Iwo sm ooth collars al i ts ends by bolI-and-socket joints.If collar A is nioving downward wich a speed of 8 ft/s when = 3 ft, determine the speed of A al the instant z = 0. The spring bas an unstretched length of 2 ft. Neglect the mass of the collars, Assume the angular velocity of rod AB is perpendicular to its axis.

Prob. 21-31

*21--32. The 5-kg &udar disk spins about AB willi a constant angular velocity of wl = 15 rad/s, Sim ultane ously. the shaft lo which arm OAB is rigidly atłached,rowes with a constant angular velocity of w = 6 radis. Determine the angular momentom of ihe disk about point 0, and its kinetic energy.

Prob. 21-32

21-33. The 20-kg spherc rotates ahout the axle with a constant angolar veloeity of w, = 60 rad/s. If shaft AR is subjecied to a torgue of M = 50 N m, causing it to rotaie, determine the value ol wn after the shaft has turned 90° from the position shown. Initially, wf, .= 0. Neglect the mass of arm CDE.

Prob. 21-33

= 5U tki.m

21.3 KNETiC -ENERGY599

21-34. The 200-kg satellite bas -its center of mass at point G. Ib radli of gyration abulil the z. x, y axes are kz. =

300 mm, kz = k,.• = 500 mm. respectiveiy. At the instant shown. the satellite rotates about the x, y, and z axes with the angular welocity shown, and its center of mass G has u welocity of vG = —250i + 200,1 4- 120k m/s. Determine the angular momentum of the satellite about point A at this Instant.

21-35. The 200-kg satellite has its center of mass at point G. Ilsradii of gyralion about the z ,x ,y axes are kz, = 300 mm, k_,. = k,. = 500 mm, respectivel y, At the instant shown, the satellite rotates ahout the x, y, and z axes with the angular vełocity shown, and its center of mass G bas a welocity of w.G = —250i + 200J + 120k 1 m/s. Determine the kinetic energy of the satellite ał this instant.

Z. z'= 1250 radis

elf

21-37. The circular place has a weight uf 1t1 lb and a diameter of 1.5 ft. If ii fis released from rest and falls horizontally 2.5 ft anto the hook at S. whieh provides a permanent connection, determine the wiochy of the mass center of the plate jus) atter the connecdon with lhe hook is m ade.

Prob. 21-37

= 600 radw,r = 300 radia

v

Probs. 21-34/35

21-36. The 15-kg rectangular plate is free to rotate about the y axis because of the bearing supports at A and B. When the plate is balanced in the vertical piane. a 3-g bitnet is fired finto it. perpendicular to its surface, with a welocity

=

—20001 } m/s. Compute the angular velocity of the plate at the instant it has rotated 180, If the buliet strikes torner D with the same welocity w, instead of al C, does the angular welocity remain the same? Why or why not?

Prob. 21-345

21-38. The I0-kg disk Mis on the horizontal pIane without slipping. Detennine che magnitude of its angular momentum when it is spinning about the y axis at 2 radis.

21-39. If arm OA fis subjeeted to a torquc of M = S N • m. deterrnine the spin angular welocity of the 10-kg disk after the arm has turned 2 rew, starting from fest. The disk rolla on the horizontal piane without slipping, Neglect the mass of che arm.

1 - 1 1

Probs. 21-38/39

6 0 0 CHAPTER 21 THREE-DtMENSIONAL KINETICS OF A RIGID BODY

*21.4 Equations of Motion

Having becorne familiar wich the techniques used to descrihe both the inertial properties and the angular momentum of a body, we can now write the equations which describe the, morion of the body in their most usefuI forms.

Equations of Translational Motion. The frartslatu7nal morion of a body is defined in terms of the acceleration of the body's mass center, which is measured from an inertial K, Y, Z reference, The equatioi of łranslational motion for the body can be written in vector form as

= mau (21-18)

ar by the three scalar equations

F, = m (aG),F, = m(aG),

x. = rn(aG),(21-1.9)

Here, x = F, ż + 5f:j + Fzk represents thc sum of all the external forces acting on the body.

Equations of Rotationa I Motion. Tr: Sec, 15,6. we developed Eq. 15-17. namely.

=I0(21-20)

Fig, 21-11

'nur Lia! cuurdi nate system

which stater that the sum of the moments of all the external forces acting on a system of particies (contained in a rigid body) about a fixed point O is eq uaI to the time rate of change of thc total angular momentom cif the body about point O. When moments of the external forces acting on the particles are summed about the system's masy center G, one again obtains the same simple form of Eq. 21-20, relating the momentsumniation to the angu]ar momentum LIG. To shaw this, considerthe system of particles in Fig. 21-11, where X, Y, Z represents an inertial frame of reference and the i, y, z axes, with origin at G, iransiate with respect to this frame. In generał, G is acceierating, so by definiłi.on the translating frame is not an inertial reference. The angolar momentom of the ith particie with respect to this frame is. however,

(111)c = rjG X +91(vic

where ri/G and v,fG represent the position and velocity of the ith particie with respect to G. Taking the time derivative we have

k/G X ri/G X mii'WG

21.4 EQUATIONS CP MoTioN ó01

By definition, vi/G _rrfc. Thus, the first term on the right sicie is zero since the cross product of the same vectors is zero. Also, aipr; = so that

{lho (ruo X m ial-IG)

Similar expressions can be written for the other particles of the body. When the results are summed, we get

= X(riła x inai/G)

Here 1=1G is the time rate of change of the total angular momentum of the body computed about point G.

The relative aeceleration for the ith particie is defined by the equationa,/G -=- ai - where ai and k represent, respectively, the accelerationsof the ith particie and point G measured with respect to the inertia[ frarne of reference. Substituting and expanding, using the distributive pruperty of the vector cross product, yields

= (rila (1,miriło) X a°

By definition of the mass center, the sum (Smiri/G) = (2,Prii)i" is equal to zero, since the position vector T reiative to G is zero. Hence, the last term in the above equation is zero. Using the equation of motion, the product miar can be replaced by the resultant exrernal forte acting on the ith particie. Denoting lk = (ri/G X Fi), the fina] result can be written as

sivIG (21-21)

The ratatio-nal equation of motion for the body will now be cleveloped from either Eq. 21-20 (..31- 21-21. In this regard, the scaiar cornponents of the angular mumentum HO nr 1-10 are defined by Eqs. 21-10 nr, if principal axes of inertia are used either at point O nr G, by Eqs. 21-11. if three cumponents arę computed about x, y, z axes Chat are rołałing with an angular velocity n that is different from the body's angular ve.locity <0, then the time derivative H = dilldr, as used in Eqs. 21-20 and 21-21, musi account for the rotation of the x, y, z axes as measured from the inertial X, Y, Z axes. This requires application of Eq. 20-6, in which case Eqs. 21-20 and 21-21 become

i Mo = (Ho)xyz + S X 1-1,9Nfe (1=1°)xyz Si x 113 (21-22)

Here (14),y, is thc time rate of changc of H measurcd from the x, y, z reference.

There are three ways in which one can define the motion of the x, y, z axes. Obviously, motion of this reference should be chosen so that it will yi.eld the simplest set of moment equations for the solułion of a particular problem.

602CHAPTER 21 THREE-DtMENSIONAL KINETICS OF A RIGID RCDY

x, y, z Axes Having Motion — O. If the body bas general

m o t i o n , t h e x , y , z a x e s c a n b e c h o s e n w i t h o r i g i n a t G , s u c h t h a t t h e a x e s o n l y

f r a n s / w e r e l a t i v e t o t h e i n e r ł i a l X , Y , Z f r a m e o f r e f e r e n c e . D o i n g

t h i s s i m p l i f i e s E q . 2 1 - 2 2 , s i n c e = O . H o w e v e r , t h e b o d y m a y h a v e a

r o t a t i o n w a b o u t t h e s e a x e s , a n d t h e r e f o r e t h e m o m e n t s a n d p r o d u c t s o f i n e r t i a

o f t h e b o d y w o u l d h a v e t o b e e x p r e s s e d a s f u n d i o n s o f t i r n e . I n m o s t

c a s e s t h i s w o u l d b e a d i f f i c u l t t a s k , s o t h a t s u c h a c h o i c e o f a x e s h a s r e s t r i c t e d

a p p l i c a t i o n .

x, y, z Axes Having Motion = The x, y, z axes can be

c h o s e n s u c h t h a t t h e y a r e f z x e d i n a n d m o v e w i d z f i l e b o d y . T h e

m o m e n t s a n d p r o d u c t s o f i n e r t i a o f t h e b o d y r e l a t i v e t o t h e s e a x e s w i l l t h e n b eL.onsfri.nt during the moton. Since = Eqs. 21-22 become

ZM° = ± X H02.MG (HAy, LO X HG(21-23)

W e c a n e x p r e s s e a c h o f t h e s e v e c t o r e q u a t i o n s a s t h r e e s c a l a r e q u a t i o n s

u s i n g E q s . 2 1 - 1 0 . N e g l e c t i n g t h e s u b s c r i p t s O a n d G y i e l d s

-= /:„.(;), — (,/yy — —fxy({JY—

- V_((1). (14) ZA:((;‚.: (')X(1‚Y)

SMy yyĆkliv —1„)&0 zW x — ł yz(CU — wxwy) (21-24)- /w,(<0? — (ob — 1.(6).

!M, = f„io, — — /3,y)w,w3, — — w")- ./zy(w;`, — w?) — + w.,wx)yr.

I f t h e x , y , z a x e s a r e c h o s e n a s p r b a c i p a l a x e . , ( ) f t r z e m a , t h e p r o d u c t s o f

i n e r t i a a r e z e r o , e t c . , a n d t h e a b o v e e q u a t i o n s h e c o m e

EMx = — (f, — 4)wyw..„

. 4 f y , = t y d r y — ( f , —

= — (Ix —

( 2 1 - 2 5 )

T h i s s e t o f e q u a t i o n s i s k n o w n h i s t o r i c a l l y a s t h e Euler equat ions o f rnoł ion , named af te r the Swiss mathernatic ian Leonhard Euler , who f i r s t developed them, They appły ordy f o r m o m e n t s s u m m e d about e i ther po in t O or G,

21.4 EQUATIONS OF MOTION 603

When applying these equations it should be reałized that ćox, wy., represent the time derivatives of the magnitudes of the x. y, z components of w as observed from x, y, z. To deterniine these components, it is firstnecessary to find toy, w, when the x, y, z axes are oriented in a genem!position and then sake the time clerivałive of the magnitucle of these components, i.e., (6),. However, since the x, y, z axes are rotating at

= w, then from Eq". 20-6, it shouI d be noted that W = (6),„ + w X w. Sinice w X w = O. then w = (W),„. This important result in" dicałes that the time derivative of to with resPeeł to the fixed X, Y, Z axes, that is can alsu be used to obtain (W),)„,. Gencrally this is the easiest way to determinc the resuit, See Example 21.5.

x, y, z Axes Having Motion S w. To simplify thecalculations for the time derivative of w, it is (Men cunvenien t to choose the x, y, z axes having an angular veIocity 11 which is different from the angular vcłocity w of the body, This is parłicularly suitable for the analysis of spinning tops and gyroscopes which are symmetrkyzi about their spinning axes.* When this is the case, the moments and products of inertia remain constant about the axis of spin.

Equations 21-22 are applicable for such a set of axes. Each of these twn vector equations can be reduced to a set of three scalar equations which arc derived in a manner similar to Eqs. 21-25,t

[

..A4-x - LiTx(,?, - ivn,(0, + ir.„11,,w, .119, = /,,,c;.i,„ — /,11„go, + 411,tox ../Vf„, = i,to, — IxD,:to„ + 1,n,,,,I

(21-26)

Here [1,, represent the x, y, z components of ft, measured fromthe inertial 'frame of reference, and io„, rias, Wz musi be determined relative to the.; y, z axes that have the rotatinn SZ. See Example 21.6.

Any one of these sets of moment equations. Eqs. 21-24, 21-25, or 21-26, represents a sedes of three first-order nonlinear differential equations. These equations are "coupled," since the angular-velocity components are present in ali the terms. Success in determining the so1ution for a particular problem therefore depends upon what is unknown in these equations. Difficulty certainly arises when one atte-

mpts tu sulve for the unknown components of to when the external moments are functions of time. Further complications can arise if the moment equations are coupled to the three scalar equations uf translational motion, Eqs. 21-19. This can happcn because cif the existencc uf kincmatic constraints which relate the rotation cif the body to the translation of its mass center. as in the case of a hoop which rolls

'A detailed discussiun of such devices is given in Sec. 2ł .5. See Prob. 2 ł —42.

6 0 4CHAPTER 21 THREE-DtMENSIONAL K INETICS OF A R IGID BODY

without slipping, Problems that requirc thc simultancous solution of differcntial equations arc generally solved using numerka] mcthods with the aid of a compułer. In many engineering problems, however, we are given information about the motion of the body and are required to determine the applied momenty acting an the body. Most of these probiems have direct solutions, so that there is no need to resort to computer teehniques.


Problems involving the three-dimensional motion of a rigid body can be soivect using the following procedure.

Free-Body Diagram.

• Draw a free-body diagram of the body at thc instant considered and specify the x, y, z coordinate system. lite origin of this reference must be located either at the body's mass center G, or at point O, considered fixed in an inertial reference frame and located either in the body nr on a rnassless extension of the body.

• Unknown reactive force components can be shown having a positive senne ()f direction.

• Dcpending on the naturo of the problem, decide what type of rotational rnotion S2 the x, y, z coordinatc system should have,

i.e., = 0, fZ = w, or w. When choosing, keep in mindthat the moment equations are simplified when the axes move in such a manner Chat they represent principal axes of inertia for the body at ali times.

• Compute the necessary momenłs and products of inertia for the body relative to the x, y, z axes.

Kinematics.•Determine the x, y, z componcnts of the body's angular velocity and find

the time derivatives of w.

• Nate that if = w, then w = Therefore we can eitherfind the time derivative cif w with respect to the X, Y, Z axes, w.

and then detennine its components rur, 6y, or we can find thecomponents of w along the x, y, z axes, when the axes are oriented in a generał position, and then take the time derivative of the m agnitudes of these components,

Equations of ivlotion.

•Apply either the two vector equations 21-18 and 21-22 or the six scalar ~piment equations appropriate for the x, y, z coordinate axes chosen for the problem.

21.4 EQUATIONS OF MOTION 605

EXAMPLE

T h e g e a r s h o w n i n F i g . 2 1 - 1 2 a h a s a m a s s o f 1 0 k g a n d i s m o u n t e d a t

a n a n g l e o f 1 0 ° w i t h t h e r o t a t i n g s h a f t h a v i n g n e g l i g i b l e n i a s s . I f

I - = 0 . 1 k g • m 2 , I , = 0 . 0 5 k g • m 2 , a n d t h e s h a f t i s r o t a t i n g w i t h a

c o n s t a n t a n g u l a r v e l o c i t y o f w = 3 0 r a d / s . d e t e r m i n e t h e c o m p o n e n t s o f

r e a c t i o n t h a t t h e C h r u s t b e a r i n g A a n d j o u r n a l b e a r i n g B e x e r t o n t h e s h a f t a t t h e i n s t a n t s h o w n .

S O L U T I O N

F r e e - B o d y D i a g r a m . F i g . 2 1 - 1 2 b . T h e o r i g i n o f t h e x , y , z c o o r d i n a t e

s y s t e m i s l o c a t e d a t t h e g e a r ' s c e n t e r o f m a s s G , w h i c h i s a l s o a f i x e d

p o i n t . T h e a x e s a r e f i x e d i n a n d r o t a t e w i t h t h e g e a r s o t h a t t h e s e a x e s

w i l l t h e n a l w a y s r e p r e s e n t t h e p r i n c i p a l a x e s o f i n e r t i a f o r t h e g e a r . H e n c e = w .

K i n e m a t i c s . A s s h o w n i n F i g . 2 1 - 1 2 c , t h e a n g u l a r v e l o c i t y w o f t h e g e a r

i s c o n s t a n t i n m a g n i t u d e a n d i s a l w a y s d i r e c t e d a l o n g t h e a x i s o f t h e

s h a f t AR. S i n c e t h i s v e c t o r i s m e a s u r e d f r o m t h e X, Y , Z i n e r t i a l

f r a m e o f r e f e r e n c e , f o r a n y p o s i t i o n o f t h e x , y , z a x e s ,

wx = O = -30 sin 10° = 30 cos 10°

T h e s e c o m p o n e n t s r e m a i n c o n s t a n t f o r a n y g e n e r a l o r i e n t a t i o n o f t h e

x , y , z a x e s , a n d s o 6 , = c a y = 6 , = O . A l s o n o t e t h a t s i n c e S Z = t h e n w

= ( 6 ) , „ . „ . T h e r e f o r c , w e c a n f i n d t h e s e t i m c d e r i v a t i v e s r e l a t i v c t o

t h e X , Y , Z a x e s . I n t h i s r c g a r d w h a s a c o n s t a n t m a g n i t u d e a n d

direction (-4-2) since w = 0, and so mT = my = = 0. Furthermore,

since G is a fixed point, (aG)„. = (aG), = (aG), = 0.

Equations of Motion. Applying Eqs. 21-25 (1). = yields

SM, = I x .w x - (1, - 4)(a :w-_ .

-(A y)(0.2) (By)(0.25) = O - (0.05 - 0.1)(-30 sin 11:0(30 cos Kr)

-0.2A y + 0.25By = -7.70 (1)

1314%, = 46'y 4)0-1-(ox

A x ( 0 . 2 ) c o s 1 0 ° - B x ( 0 . 2 5 ) c o s 1 0 ° = 0 - O

rr A x = 1.25Bx (2)

= (ix - iy)Wa W v

A x ( 0 , 2 ) s i n 1 0 ° - B x ( 0 . 2 5 ) s i n 1 0 ° = O - O

A x = 1.25Bx (check)

A p p l y i n g E q s . 2 1 - 1 9 , w e h a v e

Fx = m(ac)x; Ax + Bx = O (3)Fy = m(ac)y; Ay +By 9$.1 = O (4)

Fz = (aG)z; A z = O

Solving Eqs. 1 through 4 simultaneously gives

Ax = B x = O Ay -= 7 1 . 6 N B) = 2 6 . 5 N

Y

(a

)

(b )

(

c )

Fig. 21-12

606 CHAPTER 21 THREE-DtME11510NAL KINETICS OF A RIGID BODY

EXAMPLE

Err

G

h )

Z . z

(c)

Fig. 21-Ą 3

Y. Y', Y

X . X ' . X —i,

The airplane shown in Fig. 21-13a is in the process of making a steadyhririzonral turn at the rate of During this motion, the propeller isspinning at the rate of w„. If the, propeller lias two blades. determine the mome-nts which the propeller shaft exerts on the propeller at the instant the blades are in the vertical position. For simplicity. assume the blades to bc a uniform slendcr bar having a moment of inertia about an axis perpendicular to the blades passing through the center of the bar, and having zero moment of inertia about a Iongituclinal axis.

( a )

SOLUTION

Free-Body Diagram. Fig. 21-13b. The reacłions of the connecting shaft on the propeller are indicated by the resultants FR and MR. (The propeller's weight is assumed to be negligible.) The x, y, z axes will be taken fixed to the propeller, since these axes always represent theprincipal axes of inertia for the propeller.Thus. = w. The momentsof inertia I., and fy arc equal (/„ = fy = 1) and Jr.„ = 0.

Kinematics. The angular velocity of the propeller observed from the X, Y, Z axes, coincident with the x, y, z axes, Fig. 21-13c, is

= w, = w, i + wr k, so that the x, y, z components of w are

w . , = w y w = 0 w . = c u p

Since ł/ = w, then cis = (6)„,. To find w, which is the timc derivativc with respect to the fixed X, Y, Z axes, we can use Eq. 20-6 since w changes direction relative to x, y, Z. The time rate of changeof each of these components w = relative to the X, Y. Z axescan be obtained by introducing a third coordinate system x', y', 7',which Kas an angular velocity = wp and fis coincident with theX, y Z axes at the instant shown. Thus

21.4 EQUATIONS OF MOTfON607

= (r.;J),, ± wp x w

= (ids)s' + wp X (ws + wp)

= O+ O + Np X W5 + f i p X Wp

=" O ± O + Wp k X co,1 4- O = wp.co,j

Sinice the X, Y, Z axes are coincident with the x, y, z axes at the instant shown, the componcnts of w along x, y, z are therefore

= O = urpw, w ,=

These same results can also be determined by direct calculation of (6

), however, this will invoive a bit more work. To do this, it will be neeessary to view the propeller (or the x, y, z axes) in some generał posinon such as shown in Fig. 21-13d. Herc the piane has turned through an angle 1 (phi) and the propeller bas turned through an angle tp (psi) relative to the piane. Notice trat wp is always directed along the fixed Z axis and w, follows the x axis. Thus the general components of w are

w„ = wj, sin rj.; = wp cos

Since w, and wp are constant, the tirfIC derivatves of these components become

Z

X

( d )

Fig. 21-13

= O w1 = wp eos —w sin 1.f., P

But d) = rfr = Cr and ii( = rt,. at the instant considered.Thus,

W, == Ow, — wp

= O= wpw, tvz=awhich are the, same results as those obtained

previously. Equations of Motion. Using Eqs. 21-25, we have

= (1„ 4)(.0„,w, = 1(0) — — 0)(0)wp

Ms = O ANS.

1Ty Wp£0, ) ( O i)iwpws

114y = 2/wpw, A

Mw = — (1:x — = 0(0) — (1 — i)w,(0)

M, = O Ans,


EXAMPLE

(b)

Fig. 21-14

3 radfs

The 10-kg flywheeI (or thin disk) shown in Fig. 21-14a rotates (spins) about the shaft at a constant angular velocity of ws = 6 rd.clis. At the sanie time, the shaft rotates (preeessng) about the bearing at A with 111 UnguIar velocity of wp = 3 rad/s. If A is a thrust bearing and B is a journal bearing, determine the components of furce reaction at each of these supports due to the motion.

SOLUTION I

Free-Body Diagram. Fig, 21-14b. The origin of the x, y, z coordinate system is located at the center of mass G of the flywheel. Here we will Id t hese coordinates have an angular velocity of 11 = wp = ł 3k } rad/s. Althouigh the whcc] spins reiative to these axes, the moments of inertia remain constant,* i.e.,

fx = I = 4(10 kg)(0.2 m)2 = 0.1 kg m2

= O kg)(0.2 m)2 = 0.2 kg -m2

Kinematies. From the coincident inertial X, X Z frame of reference. Fig. 21—I4c. the flywheel has an angular velocity of = {6j + 3k} rad/s, so that

w, = 0 wy= 6 radis = 3 raclis

The time derivativc of w musi be determined relative to the x, y, z axes. In this case both wp and w, do not change iheir magnitude or clirection, and so

ti).„ = O £i),. = O <;,)., = O

Equations of Motion. Applying Eqs. 21-26 (n w) yields

= 4W„. — /„.11,w, +

—40.5) + 8,(0.5) = 0 — (0.2)(3)(6) + O = —3.6‚11, = Ivio, — + i.„1/,(0„

O = O — 0 + O

5,14, = — 1„11,(0,, + 411"

ri,,(03) — ,(0,5) = O — O + O

*This wuuld not be [rue for the propeller in Example 21.5.

21.4 EQUATION5 OF MOTION 6 O 9

Applying Eqs. 21-19, we have

/Fx = ni(ac,)x; A, -H B, = O

Y F y = ni(at)y; A, = —10(0.5)(3)2

5Fz = rn(ac)z; — 10(9.81) = O

So/ving these equations, we obtain

= O Ay, = —45.0 N , = 52.6 N Ans,B, = O = 45.4 N Ans.

NOTE: 1f the precession rir had not oceurred, the z component of forte at A and 13 would be equal to 49,05 N. In this case, however, the difference in these componcnts is causcd by the "gyroscopie moment" created whenevcr a spinning body precesses about another axis. We will study this effect in detail in the next section.

SOLUTION IIThis example can a]so be soIved using Eulers equations of mołion, Eqs. 21-25. In this case S = w = {6j + 31(-} rad/s, and the linie derivative (i.›),y, can be conveniently obtained with reference to the fixed X. Y, Z axel since 6 = (6),y,... This calculation can be performed by ehoosing x', v', z' axes to have an angular velocity of fl = osi„Fig, 21-14e, so that X. ►,

+ wp X w = O + 3k X (6j + 3k) =- {-18i} rad/s2

—18 rad/s = O cu, = 0

The ni‹mtent equations then become

M . — —

—,4,(03) -H 13 ,(0 .5) = 0.1(-18) — (0,2 — 01)(6)(3) = —3.6 S My = ły (;.>y1▪,, )0) z (-040 = —'it = ł:t], — ▪ — iy)ay.03.

A,(0.5) — B = O — O

The solution then proceeds as before.

3 1.ad.s

L . z .

— 6 rad js

(c)

Fig. 21-14

610 CHAPTER 21 THREE-DWENSIONAL KINETICS OF A RIGID BODY• PROBLEMS

*21-40. Derive the scalar form of the rotation ul equationof motion about the x axis if w and the moments andproducts of incrtia of the body arc noz constanz with respect to time.

21-41. Derive the scalar form of the rotational equation of motion about the x axis if SL # w and the rnoments and products of inertia of the body arc constanz with respeet to łime,

21-42. Derive the Euler equalions of motion for ,S1 w.i.e.. Eqs.

21-43. The 4-1b bar rests along the smooth corners cif an open box, At the Instant shown, the box bas a velocity v = {3j} ftis and an acceleration a = ( —6,1) ft/s2. Deiermine the x, y, z COMponents of force which the corners exert on the bar.

*21-44. The uniform plate has a masa of M = 2 kg and is given a rotahon of w = 4 radis about its hearings at A and H. If a = 0.2 ni and c = 0.3 rn, determine the vertical reactions at the instant shown. Use the x, y, z axel shown and nota

( y r i c r e c 2 - d 2 )that f , -

1 2 C 2 a 2

21-45. If the shaft AB is rotałing with a constant angular velocity of w = 30 rad/s, determine the X, y. Z componenłs of reaction at the thrust bearing A and journal bearing B at the instant shown, The g.fisk has a weight of 15 lb. Negleet the weight cif the shaft AR.

Prob. 21-45

21-46. The 40-kg flywheel (disk) is mounted 20 mm off its true center at G. If the shafi is rotaling al a consian i speed rv =- 8 radfs, determine the maximum reactions exerted on the jawna' hearings at A and B.

21-41, The 40-kg flywheer (diak) is mounted 20 mm off its true center at C. If the shafi is rotating al a constim i speed 8 rad/s. determine the minimum reactions exerted

on the journal hearings at A and B during the motion,

2 ft

----------2 tx

Prob. 21-43

2 1

Prob, 21-44 Probs. 21-46/47

21.4 EQUATIONS OF MOTION611

21-48. The man sits on a swivel chair wliich is rotating with a constant angular velocity of 3 rad /s. He holds the uniform 5-lb rod AR horizontal. He suddenly gives it an angular aeceleration cif 2 rad/s:-. measured relative to blm, as shown. Determine the required tarce and moment components al the gTip, A. necessary to do ibis. Establish axes al the rods center of mass G, with +z upward, and +y directed along the axis of the rod towards A.

3 rud./.

Prob. 21-48

21-49. The 5-kg rod AB is supported by a rotating ano. The support at A is a jawna bearing, which develops reactions normal to the rod. The support at B is a thrust bearing, which develops reactions both norma! to the rod and along the axis of the rod. Neglecting Eriction. determine the x, y, z components of reaction at these supports when the kanw rotates with a constant angular vctocity of

= 10 radfs.

21-50. The rod assembly is supported by a ball-and-socket joint at Card a journal bearing at D, which develops only x and y furce reactions. The rods have a mass of 0.75 kg/m. Determine the angular aceeleration of the rods and the components of reaction at che supports at the instant w -= 8 radia as shown.

Prob. 21-50

21-51. The uniform hatch door, having a mass of 15 kg and a mass center at G. is supported in the horizontal piane by hearings at A and B. If a vertical lorce F = 300 N is applied to the door as shown. determine the components of reaction at the hearings and ihe angular acceleration °f the door. The bearing at A will resist a component of furce in the y direction, whoreas the bearing at B will not. For the ealcuiation. assume the door to be a thin Male and neglect the sile of each bearing. The door is originaily at rest,

= 10 radis

Prob. 21-49

100 mm200

200 m m

30 mm

G

_150 mm. 150 mm

30 mm3'

Prob. 21-51

61 2 CHRPTER 21 THREE-D r M E N 5 I O N A L K I N E T I C S O F A R I G I O B O D Y

*21-52. The conical pendulum consists of a bar of mass m and Iength L that is supported by the pin at its end A. If the pin śs subjected to a rotation ru, determine the angle B that the har makes with the verticat as it rotates.

Prob. 21-52

21-53. The car travels around the curved road of radius psuch that itr mass center has a constant speed Write theequations of rotational morion with respect to the x, y, z axes. ASSUMe that the can six moments and products of inertia with respect to these axes are known.

21-54. The rod asseinNy is supported by journal hearings al A and B. which develop only x and z force reactions on the shaft. If the shaft AB is rotating in the direction shown at w = { -5j} radjs, determine the reactions at the hearings when the assembly is in the position shown. Also, what is the shaft's angular acceleration? The mass of each rod is 1.5 kg/m.

21-55. The 20-kg sphere is rotating with a cons tan l angular speed of w, = 150 rad/s about axle CI], which is mounted on the circular ring.The ring rotates about shaft AB with aconstant angular speed of wg = 50 rad is. shaft AB issupported by a chrust bearing at A and a journal bearing at B, deterniine the X, Y, Z components of reaction at these hearings at the instant shown. Neglect the mass of the ring and shaft.

300 min500 mm 400 rrirn

300 mm500 mm

Prob. 21-54

Prob. 21-53 Prob. 21-55

21.4 EQUATIONS OF MOTION613

X21-56. The rod assembly bas a weight of 5 1h/ft, 1t is supported at B by a smooth jouriial bearing. which develops x and y force reactions, and at A by a smooth thrust bearing, which dcvelops x, y. and z furce reactions, If a 50-lb • ft torque is applied along rod A a. delermine the components of reaction at the hearings when the assembly has an angular veloci ty w -= 10 rad/s at the instant shown.

21-57. The blades of a wind turbine spin about the shaft S with a constant angular speed of wy, while the franie precesses about the vertical axis with a constant angular speed cif wp. Deterrnine the x, y, and z components of moment that the shaft exerts on the blades as a function of A. Consider each blade as a slender rod of mass m and length L

Prob. 21-57

21-58. The cylinder has a mass af 30 kg and is mountcd on art axle that is supported by bearings at A and R. If the axleis turning at w = rad/s, delermine the verticatcomponents af furce Reling at the bearings at this instant.

Prob. 21-58

21-59. The Chin rod hasa mass of 0.8 kg and a total length of 150 mm. It is rotating about its midpoint at a constant rate H = 6 rad/s, while the tubie la which its axle A is fastened is rotating at 2 rad/s. Determine the x, y, z moment components which the axle exerts on the rod when the rod is in any position A.

2 U-

Prob. 21-56

Prob. 21-59

61 4 CHAPTER 21 THREE-DtME11510NAL KINETICS OF A RIGID Beln>

Precession

(b)

*21.5 Gyroscopic Motion

in this section we will develop the equations defining the motion of a body (top) which is s yrometrical with respect to an axis and rotating about a fixed point. These equatłons also apply to the motion of a particutarly in teresting device, the gyroscope.

The bodys motion will be analyzed using Euler angles Ęb, 0, ib(phi, theta. psi). To illustrate how they define the position of a body, consider the top shown in Fig. 21-15a. To define its final position, Fig. 21-15d, a second set of x, y, z axes is fixed in the top. Starting with the X, Y, Z and x, y, z axes in coincidence, Fig. 21-15a, the final position of the top can be detcrmincd using the following thrce steps:

1 . Rota te the top about the Z (or z) axis through an angle(O 4 < 27r), Fig. 21-15b.

2. Rotate the top about the x axis through an angle O (0 O Fig. 21-15c.

3. Rotate the top about the z axis through an angle (O gr < 2.7r) to obtain the final position, Fig. 21-15d,

The sequence of these three angles, 4, B, then 4,, must be maintained,since finite rotations are nut veciors (see Fig. 2D-1). Although this is the case,the differential rotations cle, and dtpr are vectors. and thus the angular

Y velocity w of the top can be expresscd in terms of the timc derivatives ofthe Euler angles.The angular-veIocity components (), 0, and 4i are known as the preeession, ruitation, and spin, respectively.

x

y

Z, z

x

Y, y

21

Spiu rfi

(d )

Fig. 21-15

21.5 G YROSCOA M OTrON 61 5

Their positive direcłions are shown in Fig. 21-16. It is scen that these vectors are not alI perpenclicular to one another; however, w of the top can stili be expressed in terrns of these three components.

Since the body (top) is symmetric with respect to the z ar spin axis, there is no need to attach the x, y, z axes to the top since the inertial properties of the top will remain constant with respect to this fratrieduring the motion. Therefore= + (a,„ , Fig. 2I-16. Lence. theangular velocity of the body is

rr = cax i + wyj + toz k= 9i +sin 6)j + cos O + iji)k (21-27)

And thc angular velocity of the axes is

= + 1 1 , , j + , f i k

= t3i + (4: sin 0)j +cos 0)k (21-28)

Have the x, y, z axes represent principal axes of inertia for thc top, and sothe momcnts of inertia will be represented as i„ = = ,/ and l:, =Since .11 <A, Eqs. 21-26 are uscd to establish the rotational cquations ofmotion. Substituting into these equations the respective angular-velocity components clefined by Eqs. 21-27 and 21-28, their corresponding time derivatives, and the moment of inertia components, yields

SM, = cbz sin O cos B) 4- f, sin 0(q1 cos O +

S. k i y sin O + 2riO cos O) - "ki cos 6 + iii) (21-29)

= t:(;/;; 4, cos O - (kt sin 0)

Each moment summation applies only at the fixed point O ar the center of mass G of the body. Since the equations represent a couplecl set of nonlinear second-order differential equations, in generał a closed-formsolution may not be obtained. Instead, the Euler anglesand q maybe obtained graphically as functions of finie using numerical analysis and compułer techniques.

A special case, however. does exist for which simplification of Eqs. 21-29 is possible. Cornmonly referred to as steady precession, it occurs whcn the nutation anglc O. precession fb, and spin all remain =stara. Equations 21-29 then reduce to the form

zL

x

Fig. 21-16

-/r/fl2 sin O cos 6 + f. sin 0(4, cos O + (21-30)

616 CHAPTER 21 THREE-DtMENSIONAL KINETICS OF A RIGIO BODY

Equation 21-30 can be further simplified by noting that, from Eq. 21-27, = cos O + i, so that

= —./4)2 sin O Las O + f,,cb (sin 0)w,

o r

£M i = sin 8(14), his cos 0) (21-31)

It is interesting to note what cffects the spin jr has on the marnem about the x axis. To show this. consider the spinning rotor in Fig. 21-17. Here O = 90°, in which case Eq. 21-30 reduces to the form

=

o r

1113, = Iznymz (21-32)

=

z _

Fig. 21-17

Frani the figure it can be scen that 11). and t». act along their respective positive axes and thercfore are mutually perpcndicular. Instinctively. one would expect the rotor to fali down under the influence of gravity!However, this is not the case at all, provicled the product iscorrectly chosen to counterbalance the moment 1/14, = WrG of the rotor's weight about O. This unusual phenomenon of rigid-body motian is Oft011 referted ta as the gyroscopic effeet,

21.5 GYROSCOPZ MOTrON 61 7

Perhaps a more intriguing demonstration of the gyroscopic effect comes from studying the action of a gyroscope, frequently referred to as a gyro. A gyro is a rotor which spins at a nery high rate about its axis of symmetry. This race of spin is considerably greater than its precessional rata of rotation about the vertical axis. Hance, for alt practical purposes, the angular momentum of the gyro can be assurned directed along its axis of spin. Thus, for the gyro rotor shown in Fig, 21-18, o), >> n,, and the magnitude of the angular momentum about point 0, as deterMined from Eqs.21-11, reduces to thc form Ho = f.r.o,. Since butli the magnitude and direction of Ho arc eonsłan t as observed from x, y, z, direct applicationof Eq. 21-22 yiclds

x

y Z

SZr

— z. Y11Q

sm. nv x Hol

(2 I -33) Fig. 21-18

Using the right-hand role appIied to the cross product, it can be scenthat always swings Ho (or co,) toward the sense of ni,. In effect, thechange in direction of the gyro's angular momentum, dI7E0, is equivalent to the angular impulse catised by the gyro's weight about 0, i.e.,d1-1p = M,. dt, Eq. 21-20. Also, since Ho = /zw, and 41, , and Hoara mutually perpendicular, Eq. 21-33 reduces to Eq. 21-32.

When a gyro is mounted in gimbal rings, Fig, 21-19. it becornes free of external muments applied to its base. Thus, in theory, its angular momentum H will never prccess but, instead, maintain its sanic fixcd oricntation along the axis of spin when thc base is rotated, This typa of gyroscope is called a free gyro and is useful as a gyrocompass when the spin axis of the gyro is directed north. In reality, the gimbal mechanism is never completeły free of friction, so such a device is useful cały for the local navigation of ships and aircraft, The gyroscopic effect is also useful as a means of stabilizing butli the rolling motion of ships at sen and the trajectories of missiles and projectiles. Furthermore, this effect is of significant importance in the design of shafts and hearings for rotors which ara subjcctcd to forced precessions.

Bearings

GiniMs--411r I '

The spinning of the gyro within the frame of this boy gyroscope produccs angular

omen turn Ho. whieh is nhanging Liirection as the frame precesses aun aboui the vertical axis. The gyroscupe will net fali down since the moment °f its weight W ahoul the support is halance3 by the change in the direction of Ho.

Gyro

Fig. 21-19

618 CHAPTFR 21 THRE-DlIvIENSIONAl_ KINETICS OF A RiCIP Bony

21.7

(a)

Fig. 21-20

The top shown in Fig. 21-20a hal a mass of 0.5 kg and is precessingabout the verticai axis at a constant angle of O = If it spins withan angular velocity wY = 100 rad/s, determine the precession cdp. ASsUrne that the axial and transverse moments of inertia of the top are 0.45(10-3) kg • m2

and 1.20(10-3) kg • m2, respectively, mcasurcd with respect to the fixed point O.

Z

Y

( b )

SOLUTIO NEquation 21-30 will be used for the solution since the motion is steady precession. As shown on the free-body diagram, Fig. 21-20b, the coordinatc axcs arc established in the usual manner, that is, with the positive z axis in thc direction of spin, the positivc Z axis in the direction of precession, and the positive x axis in the direction of the moment ł M, (refer to Fig. 21-16). Thus,

= -$2 sin e cos f1 + INd.) cos +

4.905 N(0.05 m) sin 60° = -(1.20( I 0-3) kg • m2(k'l sin (:os

+ 10.45(10-3) kg - nr21(ji sin 60'((.") cos 60' + 100 rad /s)

o r120.0c.P 654.0 = 0 (1)

Solving this quadratic equation for the precession gives

= 114 rad/s (high precession) frr

and= 5.72 rad/s (low precession) Any

NOTE: In reality, low precession of the top would generally be observed, since high precession would require a Iarger kinetic energy,

21,5 GYROSCOPLC MOTłON619

SOLUTIONThe free-body diagram of the assembly is shown in Fig. 21-21b. The origin for both the x, y, z and X, Y, Z coordinate systems k located at the fixed point O. hi the conventional sense, the Z axis is chosen along the axis of precession, and the z axis is along the axis ()f spin, so that O = 90Q, Since the precession is swady, Eq, 21-32 can be used for the sołution.

Substituting the required data gives(9.81 N) (0.2 m) - (19.62 N)s = [1(1 kg)(0.05 m)210.5 rad/s(-70 rad/s) s =

0.102 m = 102 mm

The €--kg disk shown in Fig. 21-21a spins about its axis with a constant angular velocity r.,k) = 70 radis. The block at B has a mass of 2 kg. and by adjusting its position s one can change the precession of the disk about its supporting pivot at O while the shaft remains h orizontal. Deterrnine the position s that will enable the disk to have a constant precession wi, = 4.5 rad/s about the pivot. Negted the weight of the shaft,

:p = 0.5 rad/s

idf) = 70 rad /s

(a)

Z , y

9.81 N (b)

Fig. 21-21

6 2 0 CHAPTFR 21 THRE-DlIvIENSIONAL KINETICS OF A RICIP BenY

21.6 Torque-Free Motion

When the only external force acting on a body is caused by gravity, the general motion of the body is referred to as łorque-free morion. This type of motion is characteristic of planets, artificial satellites, and projectiłes-provided air friction is neglected.

In order to clescribe the characteristics of this motion, the distriblution of the bodys mass will be assumed axisvnimerric. The satellite shown in Fig. 21-22 is an example of such a body, wherc the z axis rcpresents an axis of symmetry. The origin of the x, y, z coordinates is located at the

mass center G, such that = and I, = = i. Since gravity is theonly external force present, the summation of moments about the mass center is zero. From Eq. 21-21, this requires the angular momentum of the body to be constant, i.e.,

HU = constant

At the instant considered, it will be assumcd that the incrtial frame of reference is oriented so that the pusitive Z axis is directed along HU and the y axis lies in the piane formcd by the z and Z axes, Fig, 21-22_ The Euler angle formed between Z and z is .0, and therefore, with this choice of axes the angular momentum can be expressed as

= lfGSin Flj + 1--/G cos e k

Furthermore, using Eqs. 21-11, we have

HG ho, i + [01.), + ico, k

Equating the respective i, j, and k components of the above t-wo equations yields

' . 2 1

Fig. 21-22

6 2 121.6 TORQue-FREE MorioN

HG sin B H G coswx-Q Wy tek

ł(21-34)

ar

H G sin O Ikcos— k (21-35)j +

In a similar mann«, equating the respeetive i, j, k curnpunents of Eq. 21-27 to those of Eq. 2 I -34, we obtain

= u

si.n e MG sin

(i) cos o + HG cos

ft

Solving, we get

= conslanlHG

f - _" HG cos O

/.

(21-36)

Thus, for torque-free motion of an axisymmetrical body, the angle O

formed between the angular-momentum vector and the spin of the body remains constant. Furthermore, the angular momentum EIG, precession and spin ii, for the body remain constant at ali times during the motion.

Eliminating HG from the second and third of Eqs. 21-36 yelds the following relation between the spin and precession:

(21 -37)ł, cos 8

622 CHAPTFR 21 THRE-DlIvIENSIONAl_ KINETICS OF A RICIP BenY

Axis of Z precession.

Instantaneonsaxis of rotalion

zMis of spin

Rody cone

Iustantaneons Zaxis of rotation

Axis ofprecession

< J, (b)

Fig. 21-23

Thesc t wo components of angular motion can be studied by usług the body and space cone models in troduced in Sec. 20.1. The space cone defining the precession is fixed from rotating, since the precession has a fixed direction, while the outer surface of the body cone rolls on the space cone's outer surface. Try to imagine this motk' in Fig. 21-23a. The interior angle of each cone is chosen such that the resultant angular velocity tif the body is directed along the line of contact of the twa cones. This line of contact represents the instantaneous axis of rotation for the body cone, and hence the angular velocity of both the body cone and the body must be directed along this line. Sincc the spin is a funetion of the moments of inertia I and of the body, Eq. 21-36, the cone model in Fig. 21-23a is satisfactory for describing the motion. provkledTorque-free motion which meets these requirements is called regular precession. If I < I., the spin is negative and the precession positive. This motion is represented by the satellite motion shown in Fig. 21-238(f < The cone model can again be used ta represent the motion:however, ta preserve the correct vector addition of spin and precession to obtain the angular velocity w, the inside surface of the body cone must roll on the outside surface of the (fixed) space cone. This motion is referred to as retrograde precession.

Saiciflies arc dum givcn a spin beforc they arc launched. If (heir angular momentum is not collinear with 11-te axis oi spin, they will exhibit precession. In the photo nn theIcCi,rcgular precession will occar słnceand in ihephoto on ihe right. retrogradeprecession wilE oceur sine 1 <

2 1

21.6 ToncuE-FizrE MOTION 623

EXAMPLE

BEL

The motion of a football is observed using a slow-motion projector. From the filie, the spin of the football is seen to be directed 30° from the horizontal, as shown in Fig. 21-24a. Also, the football is precessing about the vertical axis at a rate (j) = 3 radfs. If the, ratio of the axial to transverse moments of inertia of the football is 3, measured with respect to the center of mass. determine the magnitude of the football's spin and its angular vełocity. Ncglcct thc effect of air resistance.

I.3 rad/s

3 0 °

(a)(b)

Fig. 21-24

SOLUTIONSince the weight of the football is the only forte acting, the motion is torque-free. In the conventional senne, if the z axis is establishe,d along the axis of spin and the Z axis alung the, precession axis, as shown in Fig.21-24b, then thc angle O = 60°. A pplyi ng Eq. 21-37, thc spin is

I - _ i= um O = 113 (3) cos

3= 3 ra.dis

Using Eqs. 21-34, where = {.0 (Eq. 21-36), we have

= O

HG sin O 37 sirt 60°= 2.60 rad/s

1-1G cos 9 3T cos 60'w. - T = 1 f - 430 Tadis

3

ru = N7{(0,-)2 + (my )2 (w)2

= V/(0)2 + (2,60)2 + (4.50)2 =

5,20 rad/s

6 O

Thus ,

Ans:

i ? k .

624 CHAPTFR 21 THRE-DLMENSIONA1_ K INETICS OF A R IGID Bon) ,

PROBLEMS

21-60. Show thał the angular velociły of a body, in terms of Euler angles 4. H. and 4 ,, can be expressed asw =sin łk sin tkr + rlcosgryi -L (;,12 sir O cos — sin (tr)i

(A6 'ms where i. j, and k are directed along the x, y,z axes as shown in Fig. 21-15d.

21-61. A Chin rod is i-niłial]y coincident with the Z axis when it is given dirce rotations defined by the Euler angles

= 30°, H = 45°, and IP = 60°. If these rotations are gJven in the order stałed, delermine ihe coordinate direction angles a, p, y of the axis of the rod with respect to the X. Y. and Z axes. Are these directions the same for any order of the rotations? Why?

21-62. The turbine on a ship has a mass of 400 kg and is mounted on bearings A and B as shown, Its center of mass is at G. its radius ot gyration as k, =11,3 m, and k, = k,. = 0.5 m, If it is spinning at 200 rad/s, determine the vertieal reactions at the hearings when the ship undergoes each of the following motions: (a) ralling,

0.2 rad/s, (b) turning, m, = 0.8 radis, (c) pitching. c4i3 = 1.4 rad/s.

f e >

Prob. 21-62

21-63. The 10-kg disk spins about axle AB at a constant rate of w, = 100 md/s. If the supporting arm precesses about the vertical axis at a constant rate of wr, = 5 rad/s. determine ihe interna] moment at O caused only by the gyroscopic actian.

Prob. 21-63

''21-64. The 10-kg disk spins abaut axle A$ at a constant rate of w, = 250 rad/s. and B = 30°. Determine the ratc cif precess-

ton of arm OA. Neglect the mass of arm OA axte and the circular ring D.

21-65. When OA precesses at a constant rate of car = 5 rad/s, when O = 90°, determine the required spin of Inc 10-kg disk C, Neglect the mass of arm CM, axle A A, and the circular ring D.

Probs. 21-64/65

200 rad/ "N\ %NI' 111

A - • ; .>%

:Z roi t.3 mf• 21

0.8 rn--

ldi ,t

21,6 TORCIU-FREE MoTiow625

21-66. The car travels at a constan t speed of t‚c = 100 krn/h around the horizontal curve having a radian of 80 m. Tf each wheel has a mass of 16 kg, a radius of gyration k0 = 300 mm about its spinning axis, and a radius of 400 mm, determine the difference bełween the norma forces of the rear wheeis, caused by the gyroscopic effect. The distanee between the wheeis is 1.30 m.

21-68. The top consists of a thin disk chat bas a weight of $ lb and a radius of 0.3 ft.The rod hasa negligible mass and a lensth of 0.5 ft. If the top is spinning wich an angular vełocity w, = 300 radis, determine the steady-state precessionai angular vełocity w,, of the rod when 8 = 40°.

21-69. Solne Prob. 21-68 when 8 =

Prob. 21-66

21-67. A wheel of rnass rn and radius r rolla with constant spin w about a circular path having a radius a. If the angIe of inclination is 0, determine the rate of precession. Treat the wheel as a thin ring.. No slipping occurs.

Probs. 21-68169

21-70. The top hasa mass of 90 g, a center of mass at G, and a radius of gyration k = 18 mm about its axis of symmetry. About any transversc axis acting through point O the radius of gyration is k, = - 35 mm. 1f the top is connected to a bali-and-socket joint at O and the precessiou is run = 0,5 racliN,determine the spin

2 1

Prob, 21-67 Prob. 21-70

626 CHAPTFR 21 THRE-DLMENSIONAl_ K INETICS OF A R IC IP Ben',"

2 1

21-71.. The 1-lb top lias a center of gravity at point G. If it spins about its axis of symmetry and precesses about thevertical axis at constant raies of = 60 radfs andwp = !O radis, respectively, deterrnine the steady state angłe O. The radius of gyration of the top about the z axis is

= I in.. and about the x and y axes it is k, = = 4 in.

*21-72. Whilc thc rocket is in frec flight, it has a spin of 3 radjs and precesses about an axis rneasured 10° grom the axis of spin. If the ratio of the axial tcti transverse moments of inertia of the rocket is If /5, computed about axes which pass through the mass center G, determine the angle which the resultant angular velocity makes with the spin axis. Construct the body and space cones used lo describe the motion. Is the precession regular or retrograde?

Prob. 21-72

21-73. The 0.2-kg football is thrown with a spin w, = 35 rad/s. 1f thc angle H is measured as 60', de.termine the precession about the Z axis. The radius of gyration abulii the spin axis is k, = 0.05 nn, and about a transverse axis it is k, = 0.2 m.

Prob. 21-73

21-74. The projectile shawn is subjected to torque-free motion. The transverse and axial moments of inertia arc 1 and f.„, respectively. If O represents the angle between the precessional axis Z and the axis of symmetry z. and f3 is the angle between the angular velocity w and the z axis, shaw that p and O arc related by Ule equation tan O = (1/1,) tan p.

Prob. 21-74

. . . —

W p = ' G r a d / s . . 1 - - ' ' 'ruq= titl rad7s

3 in

Prob. 21-71

21.6 TORQUE-FREE MOTłON627

21-75. The 4-kg disk is thrown with a spin = 6 rad/s. Ifthe angle 8 is measured as 160°, determine the precession about the Z axis.

125 nim

21-78. The projectile preeesses about the Z axis at a constant rate of = 15 rad/s when it leaves the barrel of a gun. Dełermine its spin and the magnitude of its angular momentum HG.Tite projedne basa mass of 1.5 kg and radli of gyration about its axis of symmetry (z axis) and abulii its transverse axes <x and y axes) of k, = 65 mm and k, = k, = 1.25 mm, respectiveły.

Prob. 21-75

21-76. The rocket has a mass of 4 Mg and radli of gyration k, = 0.85 m and k, = 2.3 m. It is initiatly spinning about the z axis at wz = 0.05 radis when a meteoroid M strikes it ut A and creates an impulse I = {3001 N • s. Determine the axis of precession after the irnpact.

21-77. The football lias a mass of 450 g and radli of gyration about its axis of symmetry (z axis) and its transverse axes (x ar y axis) of k„ = 30 mm and k, = k, = 50 mm, respectively If the football hos an angular momentum of HG 0,02 kg • m= is, determine its precession and spin . Also. find the angle /3 that the angular velocity vector makes wich the z axis.

1.16= 0.02 kg • ralsZ

Prob. 21-77

Prob. 21-78

21-79. The space capsulc has a mass of 3.2 Mg, and about axes passing through the mass center G the mial and transverse radli of gyration are kI = 0.90 m and ks = 1.85 m, respectively. If it is spinning at w,. = 0.8 revis, determine its angular momentum. Precession occurs about the Z axis.

2 1

Prob. 21-79

628CHAPTER 21 THREE-DIMENSIONAL KINETICS OF A RiGID ROBY

CHAPTER REVIEW

Moments and Products of Inertia

A body has six components of inertia for any specified x, y. z axes. Three of these are mornents of inertia about each of theaxes, i,„ 1,2, and threc arc products ofMe dias e ach defined from Iwo orthogon Iplanes, 1,, i If either one ar both ofthese planes are planes of sy, rnmetry, then the product af incrtia with respect to these planes will be zero.

The moments and products of inertia can be detcrmined by direeł integration ur by using tabulated values. ff these quanitides are to be deterrained with respect to axes or planes that do not pass through the mass center, then parallel-axis and parallel-piane theorems must be used.

Provided the six components of inertia arc known, thcn the moment of inertia about any axis can be determined using the inertia iransfc>rmadon equation.

Principal Moment of lnertiu

At any point on ar off the body, the x, y, z axes can be oriented so that the products of incrtia will be xero. The resutting morneMsof inertia are calfed the principal mornents of inerda,one of which will be a maximum and the other a minimum.

Principie of Impulse and Momentum

The anguiar momentom fora body can be determined about any arbitrary point A.Once the linear and anguar momentum for the body have been formulated, then the principle of impulse and momentum cen be used to solne probierns that involve furce, vełocity_ and time,

Principie af Work and Energy

The kinetic energy far a body is usually determined relative to a fixed point or the body's mass center.

1_, = j rź dm = j(Y2- + z..2) dm j_ry dm

= r- dm = (x-1 + z2) din ft, = f = j yz dm

= j r dm = f (x2 + y2) dm= = xz dm

ło. = l eź+vo,2. 1„u,2 — 24.,J.ru„. — 24".m, —

( 1 . , O U) 0Q O f z

t;m(yc)i + f wh ero f i

F dr = m(vo), (H0)1 A- S N10 dr = (Hoy2Ho = po X (w X po) dm

1

nr

Center cif Mass= p,„ x mv, +

Arbirrary Point

T = + 4 4[012 + f.0):2 T = .mv;7') + + +

Fixe d Point Center of Mass

H = .1" 1 r xvw I ;guzr

Fixed Point O Hy = lyZW:

w pe X ( X pul = — 1„ro, + 1„td,1 1 G i

CHAPTER RVIEVV 629

These formulałions can be used with the principle of work and energy to solve problem that involve force. velocity, and displacement.

Equations °f Motion

There are three s.calar equations of translafionał motion for a rigid body that moves in three dimensions.

The three sealar cquations of rotational motion depend aport the motion of the x, y. z reference. Most often. [hese axes are oriented so that they are principal axes of inertia. If the axes are fixed in and move with the body sn that n = w, then the equations are referred to as the Euler cquations of ~jon,

A free-body diagram shookl a]ways accompany the application of the cquations of motion,

Gyroscopie Muflon

The angular motion of a gyroscope is bess described using the three Euler angtes O, 0, and tis. The angular veloeity cornponents are called the precession the nutation O. and the spin

11Ó = O and .r:b and tir arc eon stant, then the motion is referred to as steady precession.

It is the spin of a gyro rotor that is responsible for holding a rotor from failing downward, and histead causing it to precess about a verticai axis. This phenomenon is called the gyroscopic effect.

T, + Uł = T,

Fx = nr(aG}_r2.F,. = m(aG),£F, = nr(a;(;),— (1, —

— — 1,}w,

= 1,41, — ▪ —

f t =m, =faxy /Azw,= I▪„nzw,

- 1,11", +

2114, sin O cos d + 1,1:P sin 0(4; cos O + ,1,r)

= 1 3 , M = O

Toni ue-Free Muflon

A body Chat is only subjected to a gravitational force will have no rnoments on it about its mass center. and so the motion is dcscribcd as torque-free mafio n. The angolar momentum for the body aboul its mass center will rernaM constani. This causes the body to have both a spin and a precession. The motion depends upon the magnilude of the moment of inertia of a symmetric body about the spin axis, f:, versus that about a perpendicular axis, f,

O eonstant

C =

.f-1_

[kr — 1-1,3 cosi f,

Chapter 22

The analysis of vibrations plays an important role in the study of tne behaviorof structures subjected to earthquakes.

Vibrations

CHAPTER OBJECTIVES•To discuss undamped one-degree-of-freedom vibration of a

rigid body using the equation af motion and energy methods.

•To study the a natysis of undamped forced vibration and viscous damped forced vibration.

*22.1 Undamped Free Vibration

A vibration is the oseillating motion of a body ar system of connected bodies displaced from a position of equiltbrium, In general, there are two types of vibration, free and forced. Free vibration occurs when the motion is maintained by gravitational or elastie restoring forces, such as the swinging motion of a pendulum or the vibration af an elastie rod. Foreed vibration is caused by an external periodic ar intennittent force applied to the system. Both of these types of vibration can either be damped nr undamped. Undunxped vibrations exciude frictional effects in the analysis, Snice in reality botki interna) and external frictional forces arepresent,the motion of all vibrating bodies is actually damped.

632 CHAPTFR 22 ViRRATioNs

Equilibrium The simplest type of vibrating motion is undamped frec vibration,position represented by the block and spring model shown in Fig. 22-1a,

xVibrat ing motion occurs when the błock is re leased from a displaced posi t ion x so that the spr ing pul ls on the block. The block wil l a t ta in a veloci ty such that i t wi l I proceed to rnove ciut of equi l ibr ium when x = O. and provided f i le support ing surface is smooth, the błock wil l (a)oscillate back and forth,

The time-dependent path of motion of the błock can be determined by applying thc equation of motion to the błock when it is in the displaced

W =mg position x. The frce-body diagram is shown in Fig. 22-lb. The elastic

F = k x 1

(b) !F, = max; -

Fig. 22-1Note that the acceleration is proportional to the block's displacentent. Motion described in this manncr is callcd simpte harmonie motion. R earranging the terms into a "standard form" gives

ać + co2„x = O (22-1)

The constant w„ is called the natura.' frequeney, and in this case

(22-2)

T Equation 22-1 can aiso be obtaines.i by considering the block to be suspended so that the displacement y is measured from the biock's

Equilibriurn equilibrinm payirion, Fig. 22-2a. When the błock is in cquilibrium, the

rpositionspring exerts an upward force of F = W = mg on the błock. Hence.when the błock is displaced a distance y downward from this

posit ion, the magnitude of the spring force is F = W + ky, Fig. 22-2b. Appłying the equation of modem gives

= ma,,; -W - ky + W =

F = W ÷ ky or

4-1 + y = O

(b) which is the same form as Eq. 22-1 and (,)„ is dełined by Eq. 22-2.

Fig. 22-2

22-1 UNDAMPF6 FftE Vil3RATłOhil63 3

Equation 22-1 is a homogcneous, sccond-order, linear, differential equation with constant coefficients, It can be shown, using the methods of differential equations, that the general solution is

x = A sin cod + B cos itud (22-3)

Here A and B represent two constants of integration, The błock velocity and acceleration are determined by taking successive time derivatives, which yie1els

= x = Awn cns w„t - Bw„ sio orni (22-4)

a = sin co„t - Bw cos (22-5)

When Eqs. 22-3 and 22-5 are substituted into Eq. 22-1, the differential equation will be satisfiecl, showing that Eq. 22-3 is indeed the solution to Eq. 22-1.

The integration constants in Eq. 22-3 are generally determined from the initial conditions of the problem. For example, suppose that the block in Fig. 22-la has heen displaced a distance xi to the right from its equilibrium position and given an initial (positive) veiocity vi dircctcd to the right. Substituting x = x, when f = O into Eq. 22-3 yiclds B = xi. And since = vi when r = 0, using Eq. 22--4 we obtain A = vi iwf,. If thcsc values arc substiłuted into Eq. 22-3, the equation dcscribing the motion becomes

vXsin <0,,f + X ł cos a,t

WI2

(22-6)

Equation 22-3 may also be expressed in terms of simple sinusoida' m otion. To show this, let

A = C cos c¢ (22-7)

and

B = C sin d (22-8)

where C and 4) are now constants to be determined in place of A and B. Substituting into Eq. 22-3 yields

x C cci»; w„i + C sin r¢ cos

And sincc sin(0 + = shi O cos 49 + cos 8 sin then

x = C sin(w,,t 49) (22-9)

If this equation is plotted on an x versus CuT„r axis, the graph shown in Fig. 22-3 is obtained. The maximum dispkcement of the błock from its

634 CHAPTFR 22 ViFuRATioNs

w„f

Fig. 22-3

equilibrium position is defined as the amplitude of vibration. From either the figure or Eq. 22-9 the amplitude is C. The angle tii is called the phase anglesince it represents the amount by which the curve is displaced from the origin when r = O. We can relate these two constants to A and B using Eqs, 22-7 wid 22-8. Squaring and adding these two equations, the amplitude becomes

C = + 8 (22-10)

If Eq. 22-8 is divided by Eq. 22-7, the phase angle is then

rt) (22-11)

Note that the sine curve, Eq. 22-9, completes one cyrle in t ime t = T (tan) when (1 )„z = 2/r, or

2.7r'T (22-12)

This time interva I is called a period. Fig, 22-3, Usil-1g Ey, 22-2, the period can also be reprcsented as

T = 2 7 r —797

k (22-13)

Fina Ily. the frequency fis defined as the number of cycles completed per unit of time,which is the reciprocal of the period; that is,

w„f - T 2 7 r (22 -14)

o r

,

I \ i /T: 27r /r /

( 2 2 - 1 5 )

The frequency is expressed in cyclesis, This ratio of units is called a hertz (117.),where 1 Hz = 1 cycleis = 27r rad/s,

When a body or system of connected bodies is given an initial displacement from its equilibrium position and releascd, it will vibrate with the natura' frequency. w,,. Providcd the system has a single dcgree of frecdom, that is, it rcquircs only one coordinate to specify completely the position of the system at any time, then the vibrating motion will have the same characteristics as the simple harmonie motion of the bIock and spring just presentecl. Consequently, the motion is described by a differential equation of the same "standard form" as Eq. 22-1, i.e,,

+ cox = O (22-16)

Herce, if the natural frequency wn is known, the period of vibration T,

frequency f, and other vibrating characteristics can be established using Eqs. 22-3 through 22-15.

22.1 LINDAMPF6 FREZ ViRRATłON

ortant Points

• Free vibration nceurs when the motion is maintained by gravitational nr elastic restoring forces.

• Thc ampli tude is the maximum elisplaccment of thc body.

• The period is the time required to complete one cycle.

• Thc frequcuey is the number of eycles completed per unit of time, where I Hz = 1 cycleis,

• Only one position coordinate is needed ta describe the location of a one-degree-of-freedom system,


As in the case of the block and spring, the natura] frequcncy w„ of a body nr system of conneetecl boclies having a single degree of freeduni can be determined using the following procedure;

Free-Body Diagram,

• Draw the free-body diagram of the body when the body is displaced a small anlmnł from its equilibrium position.

• Locate the body with respect to its equilibrium position by using an appropriate inertiai coordinate q. The acceleration of the body's mass center aG or the body's angular acceleration « should have an assurned sense of direction which is in the posirive direciłyn of the position coordinate.

• If the rotational equation of motion SA,1‚. = /(.44.k)p is to be used, thcn it may be heneficial to also draw thc kinetie diagram since it graphically aceounts for the components im(aG),, ni(aG)i., and /G«, and thereby m akes it convenient for visualizing the terms needed in the moment sum ł (44,1)p,

Equation of Motion.•Appły the equation of motion to relate the elastic or gravitational restoring

forces and coupie moments acting on the body to the hody's accelerated motion.

Kinematics.

• Using kinematics, express the body's accelerated motion in terms of the second time derivative of the position coordinate,

• Substitute thc rcsult into thc equation of motion and determine w„ by rcarranging the terms so that the resulting equation is in the "standard form," 9 + w„2ri = O.

6 3 5

636 CHAPTER 22 V I B R A T I O N S

EXAMPLE

( a )

t

W = mg

(1,)

Fig. 22-4

Determine the period of oscillation for the simple penduIum shown in Fig. 22-4a. The bob hasa mass m and is attached to a card of length i. Neglect the sfze of the bob.

SOLUTIONFree-Body Diagram. Motion of the system will be related to the position coordinate (q =) 49, Fig. 22-417, When the bob is displaced by a smali angle 0, the resturing fony acting on the bob is created by the tangential componen t of its weight. gyrrq si n 0, Furthermore, aż acts in the direction of increasing s- (m 0).

Equation of Motion. Applying the equation of motion in the iarkgrendal direction,sincc it involves the rcstoring furce, yields

- F, = tria,; -mg sin 9 = ma,

Kinematics. ur = d2skit2 = 5. Furthermore, s can be related to 0 by the cquation s = 10, so that a, = id. Hence, Eq. 1 reduces to

-sin 0 = O (2)

The solution of this equałion involves the use of an elliptic inłegral. For smali displacemews, however, sin O ----- 0, in which case

( 3 )

Comparing this equation with Eq. 22-16 + wx - 0), it is scen thatio„ = From P4. 22-12, the period of thne required for the bob tomake one complete swing is therefore

27r= — =

This interesfing result, origfnally discovered by Galileo Galilei through expertment. indicates that the period depends only on the, length of the cord and not on the mass of the pendelum bob or the angle 0.

NOTt: The solution cif Eq. 3 is given by Eq. 22-3, wherc = Vg/1and O is substituted for x. Like the block and spring, the constants A and B in this problem can be determined if, for example, one knows the displacement and velocity of the bab at a given instant.

22.1 UNDAMPED FftE VlBRATrONb37

The 10-kg recłangułar plate shown in Fig. 22-5a is suspended at its center from a rod having a torsional stiffness k = 1.5 N • m/rad. Determine the natural period of vibration of the plate whe-n it is given small angular displacement d in the piane of the plate.

b = 0.3 ma = .2 m

SOLUTION

Free-Body Diagram. Fig. 22-5b. Since the plate is displaccd in its own piane, the torsional reworing moment created by the rod is M = kO. This moment acts in the direction opposite to the angular dispiacement O. The angular acceleration 9 acts in the direc-tion of posiłive O.Equation of Motion.

= loa; — k0 = 100

n r

U+—k0=0 loSince this equation is in the "standard form." the natural frequency is = v

Irito.From the table on the inside back cover, the moment of inertia of

the plate about an axis coincident with the rod is fo = b2 ) .HeTICe,

1lo = —(

1210 kg)[ (0.2 m)2 + (0.3 in)2-1 = 0.1083 kg m2

The natural period of vibration is therefore,T =

27r \ITo 1

2 . 7 r — 2 7 r — 1 . 6 9 s

.5 083 --

T = W k Hw r r 4 ~ =

W

(b)

Fig. 22-5

EXAMPLE

638 CHAPTER 22 V I B R A T I O N S

EXAMPLE

[. —200 mm- —5 k . —5

c-

( a )

The kent rod shown in Fig. 22-6a has a negligible mass and supports a 5-kg collar at -its end. If the rod is in the equilibrium position shown. deterrnine the natural period of vibration for the system,

SOLUTION

Free-Body and Kinetic Diagrams. Fig. 22-6b. Here the rod is elisplaced by a smali angle O from the equilibrium position. Since

100 mm the spring is subjected to an initial compression of x„ for equilibrium,

`1 then when the displacemenł x x, the spring exerts a force of F, =kx kx„ on the rod.To obtain the "standard form," Eq. 22-16, Sa,k = 400 N /m must act upward, which is in accordance with positive O dispłacement.

h 0_2 m - - 1

49.05 N 0.1' m

kx

5 2 ,

O 2 m

( h

= -0.2o

x = 0 . 1 0

( c )

Fig. 22-6

Equation ot Motion. Moments will be summed aboui point B to eliminate the unknown reaction at this point. Since 6 is smali,

C-F£MB = £(-14.)8;

kx(0,1 m) – kx„(0.1 m) + 49.05 N(0.2 m) = –(5 kg)ay (0.2 m)

The seeond term on the left side, –kx„(10. I m), represcnts the mornen t created by the spring force which is necessary to hołd the coliar in

i.e., at x = 0. Since ibis moment is cqual and oppositc to thc moment 49.05 N(0.2 m) created by the weight of the collar, thcse twa termy cancel in the above equation, so that

kx(0.1) = –5,23,(0.2) (1)

Kinernatics. The dcformation d the spring and the position of thc collar can be related to the angle 0, Fig. 22-6c. Since O is smali, x = (0.1 m)0 and y = (0.2 m)0. Therefore, a, = y = t1.2B. Substituting finto Eq. 1 yields

400(0.10)0,1 = –5(0.20)0.2

Rewriting this equation in the "standard form" gives

d + 200 = O

Compared with + f.o;x = O (Eq. 22-16), we have

= 20 w„ = 4,47 rad,R

The natural period of vibration is therefore2-tr 27r— 1.4 sr_o 4.47= C) An.s-.

22.1 UNDAMPED FREE Vł13RAPON639

EXAMPLE

A 10-Ib block is suspended from a card that passes over a 15-lb disk, as shnwn in Fig. 22-7a. The spring has a stiffness k = 200 113/ft. Determine the natural period of vibration for the system.

k= 200 Ibift

t5 Ib

10 Ib

(a) (b)

SOLUT1ON

Free-Body and Kinetic Diagrams. Fig. 22-7b. The system consists of the, disk, wbici undergoes a rotation defined by the angle 0, and theblock, which translates by an amount s. The vector acts inthe clirection ofpositive O, and consequently acts downward in thedircction of positive s.

quation of Motion. Summing moments about point O to eliminatethe reactions realizing that to = -!mr2, yields

C+ ł Mo = I (.4ti)0;

10 112(0.75 ft) — F,(0.75 ft)

1 ( s- 2 32.12 fl 5 lb 0 lb,)(0.75 ft)2 ( 32.12 fl is-

+ (2(0.75 ft) (1)i )

Kinematics. As shown on the kinematic diagram in Fig. 22-7c, a smalt positive displacernen t 6 of the disk causes the błock to lower by an amount s = 0.750; hence. a,, -= 's' = 0.75U. When O = O°, the spring lorce rcquired for equilibrium of the disk is 10 lb, acting to the right. For position 0, the spring forte fis F, = (200 1bift)(0.750 ft) + 10 lb. Substituting these results finto Eq. 1 and simplifying yields

+ 3680 = O

Hence, (.0,„2 = 368 w, = 19.18 rad/s

Therefore, the natural period of vibration is9 2ir

T = 17 = 19 18 — 0,328 s A n s ,

640 CHAPTFR 22 VIRRATioNs• PROBL E MS

22-1. A spring lias a stiffness of 600 Nim, If a 4-kg block is attached to the spring. pushed 50 mm above its equilibrium posiiion. and relesed from rest, determine the equation which describes the biock's molton. Assume that positive dispiacement fis measured downward.

22-2. When a 2-kg błock is suspended from a spring, the spring is stretchcd a distance of 40 mm. Determine the frequency and the period of vibration for a 0.5-kg block attached to the same spring.

22-3. A spring is stretched 200 mm by a 15-kg block. If the błock is displaced 100 mm downward from its equilibrium position and given a downward vclocity of 035 m/s. determine ehe equation which describes the motion, What is the phasc angŁe? Assume that positive displacement fis downward.

22-4. When a 20-1b weight is suspended from a spring, the spring is stretched a distance of 4 in. Determine the natural frequency and the period of vibration for a 10-lb weight attached to the same spring.

22-5. When a 3-kg block fis suspended grom a spring, the spring is stretched a distance of 60 mm. Determine the natural frequency and the period of vibration for a 0.2-kg błock attached to the same spring.

22_g. A 6-lb weight is suspended from a spring having a stiffness k = 3 lb/in, If the weight is given an upward velocity of 20 ftis when it is 2 in. above lis equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.

22-9. A 3-kg block is suspended from a spring having a stiffness of k = 200 N/rn. if the block is pushed 50 mm upward fr‹.->m its equilibrium position and then released from rest. determine the equation that describes the motion. What aro the amplitude and the natura] frequency of the vibration? Assume that positive displacetnent is downward.

22-10. Determine the frequency of vibration for the błock. The springs aro odginany compressed J,.

Prob. 22-11

22-6. An 8-kg block fis suspended from a spring having a stiffness k -= 80 Nim, If the block is given an upward velocity of 0.4 m/s when it is 90 mm above its equilibrium position. determine the equation which describes the motion and the maximum upward displacement cif the block measured from the equilibrium position, Assume that positive dispiacementis measured down ward.

22-7. A 2-lb weight is suspended from a spring having a stiffness k = 2 1h/in. If the weight is pushed i in. upward from its equilibrium position and then released from test. determinc the equation which describes the motion, What is the amplitude and the natura' frequency of the vibra lian?

22-11. The sernicircufar disk weighs 20 lb, Determine the natural period of vibration if it is displaced a smali amount and released.

Prob. 22-11

22.1 UNDAMPE D FREE Vi BRATra N 641

*22-12. The uniform beatu is supported at its ends by two springs A and B. each having the same stiffness k.When nothing is supported on the bean-Lit has a period of vertical vibration of 0.83 5.1f a 50-kg mass is placed at its center, the period of verfleal vibration is 1.52 s. Cornpute the stiffness of each spring and ihe mass of the beatu.

22-14. The conneeting rod is supported by a knife edge atA and the period of vibration is measured as = 3.38 s. Itis then rernoved and rotated 180w so that ii is supported by the knife edge at B. In tłiis case the period of vibration is measured as 78 = 3.96 s. Determine the location d of the center °f gravity G. and compute the radius of gyration kc.

Prob. 22-12

Pruli. 22-14

22-13. The body of arbitrary shape bas a mass m. mass center at G, and a radius of gyration about G of kc. If it is displaced a slight amanta O from its equilibrium position and released,determine the natural period ()f vibration,

22-15. The thin hoop of mass m is supported by a knife-edge, Deterrnine the natural period of vibration for smali amplitudes °f swing.

2 2

Prob. 22-13 Prob. 22-15

642CHAPTFR 22 VIRRATION5

22-16. A block of mass m is suspended front two springs having a stiffness of k1 and k2, arranged a) parallel lo each other, and b) as a senes. Determine the equivalent stiffness of a single spring with thc same oscillation charaoeristics and the period ot (ascii lation for each case.

22-17. The 15-kg block is suspended from lwa springs having a different stiffness and arranged a) parallel to each other. and b) as a sedes. If the natural periods of oscillation of the parallel system and senes system wre observecl to be 0.5 s and 1.5 respectively, determine the spring, stiffnesses k1 and k2.

22-19, The 50-kg błock is suspencled from the 10-kg pulley that bas a radius of gyration about its center of mass of 125 mm. 1f the błock is given a smali vertical displacernen t and then released. determine the natural frequeney of osci I I ation,

Prob. 22-19

22

( b ) Probs. 22-16/17

22-18. The p‹.3inter on a metronome supporls a 0.4-lb slider A, which is positioned at a fixed distance from the pivot O of thc pointer. When the pointer is clisplaceci, a torsional spring at O exerts a restoring torque on the pointer having a magnitude M = (1.20) ib • ft, where represents the angle of displacement from the vertical, measured in radian& Detcrmine the natural period of vibration when the pointer s clisplaced a smali amount and released. Neglect the mass of the pointer.

k = 1.2 Ib ft /rad Prob. 22-18

22-20. A uniform board is supported on Iwo wheels which rotate in oppositc directions at a constant angolar speed, 1f the coefficient of kinetic fricłion between the wheels and board is p, determine the frequency of vibration at the board if it is displaced slightly, a distance x from the midpoint between the wheels. and released.

d

Prob. 22-20

22.1 LINDAMPF6 FREZ ViRRATłON643

22-21. If the 20-kg hlock is given a downward velocity of 6 m/s at its equilibrium position. deterrnine the equation !hal describes the ampiitude of the block's oscillation.

22-23. The 50-lb spool is attached to iwo springs. If the spool is displaced a smali arnount and released, determine the natura' period of vibration. The radius of gyration ofthe spoOt is kG1.5 ft, The spool roik without slipping.

Prob. 22-21

Prob. 22-23

22-22. The bar has a Iength i and mass m. It is supported at its ends by rollers of negiigiblc mass. If it is given a smali displacement and released. determine Ehe natural frequency of vibration.

*22-24. The can has a mass of m and is attached to twosprings, each ha\.Ąng a stiffneS9 of kl == k, unstretchedlength of io, and a streiched length of I when the cart is in the equilibrium position, If the cart is displaced a distance of .r = ..ro such that both springs remain in tension (xo < f — iU1, determine the natural frequency of oscillation.

22-25. The cert hasa mass of iv and is attached to two springs, each having a stiffness of ki and k2, respectively. If both springs are unstretched when the cert is in the equilibriurn position shown, detcrmine the natural frequency of oscilłati on ,

2 2

Prob. 22-22Prxbs. 22-24/25

644 CHAPTER 22 VIBRATIONS

22-26. A tływheel of mass r►a, which has a radius of gyration about its center of mass of ko, is suspended from a circular shaft that has a łorsional resistance of M = CO. If the flywheel is given a smali angular displacernent of O and released. determine the natural period of oscillation.

Prob. 22-26

22-27. If a block D of negligible sine and of mass m is attachcd at C, and the bell cran k of mass Mis given a small angular dispiacement of H, the natural period of oscillation is Ti. When D is removed, the natural period of oscillation is r,. Determine the bell crank's radius of gyration about its center of mass, lin B, and the spring's stiffncss k. The spring is unstretched at S = O'', and the motion occurs in the horizontal plan e.

22-28. The platform AB when empty has a mass of 400 kg, center of masa at G1, and natural period of oscillation

-= 2.38 s. If a car. having a masa of 1.2 Mg and center of masa at G. is placed on the, platform, the natural period of oscillation be corries r, = 3.16 s. Determine the moment of inertia of the car about an axis passing through G,.

22-29, A wheel of mass m is suspended from thrcc cqual-length cords. When it is given a smali angular displacement of fii about the z axis and released, it is observed that the period of oscillation 15 T. Determine t/te radius of gyration of the wheel about the z axis.

Prob. 22-28

22

Prob. 22-27 Prob. 22-29

22_2 ENERGY METHOOS645

*22.2 Energy Methods

The simple harmonie motion of a body, discussed in the previous section, is due Only to gravitational and elastic restoring forces acting on the body. Since these forces are conservative, it is also passible w use the conservation at energy equaton to obtain the body's natura] frequency ar period of vi bration. To show how to do this, consider again the b]ock and spring model in Fig, 22-8. When the block is displaccd x from the

position, the kinetie energy is T = 4InV"'2 = rni=7- and thepotential energy is V = Since energy is conserved, it is necessary that

T + V -= coristant

-1/r/i7'1 + kx2 = constant (22-17)

The differential equation describing the accelerated motiun of the biock can be obtained by difprentiating this equation with respect to time;

+ k x i c O

ż(tnr+= O

Since the velocity ż fis not aiways zero in a vlblating system,

+ os,z,x = O co„ = kfrrr

which is the same as Eq. 22-1.If the conservation cif energy equation is written for a system af conrrectcd bodies', the natural frequency ar the equation of motion can also be determinecl by time differenłiation. It is not necessary to dismember the system to accoun t for the internat forces beeaue they do no work.

Fig. 22-8

Equilibrium position

x


The suspension of a railroad car consists ol a set of springs which are mounted between the franie of the car and ihe wheel truck. This will give the car a natural frequency of vibraton which can be deterrnined.


The natura] frequency ton of a body or system of connected boclies can be determinecl by applying the conservation of energy equation using the following procedure.

Energy Equation.

• Draw lhe body when it is displaced by a smali amarant from its

equilibrium position and define the locałion of the body from itsequilibrium position by an appropriate position coordinate q.

• Formulate the conservałion of energy for the body, T+ V = constant, in terms of the position coordinate.

▪In general, the kinetic energy musi account for both the bodys translationat and rotatianai motion. T = +• The potential energy is the sum of the graviłational and elasticpotential energies of the body, V = Vg V., Eq. 18-17. In

particufar, Vb should be measured from a daturn for which q = O (equilibrium position),

Ti me Derivative.

• Take the time derivative of the energy equation using the chain rule of calcutus and factor aut the eoni= terms. The resulting differential equation represents the equation of morion for the system. The natural frequency of ton is obtained after rearranging the terms in the "standard form," 4 + w2,7 = O.

222 ENERGY METHODS647

EXAMPLE

The Chin hoop shown in Fig. 22-9a is supported by the peg at O.Determine the natura) period of oscillation for smali amptitudes ofoswing. The hoop has a mass

rn. SOLUTION

Energy Equation. A diagram of the hoop when it is displacedsmali amount (qO from the equilibriurn position is shown inFig. 22-9b, Using the table on the inside back cover and the parallelaxis theorem to determine f, thc kinetie energy is

T = kpw,2, = mr2P92 mr2Ó2

If a horizontal datum is placed through point 0, then in the displaced position, the potential energy is

V = - fng(r cos 0)

The total energy in the system is

("a)

O D a t =

r cos ri

T + V = mr2-Ó2 - mgr cos O

Time Derivative.

mr'(29)6 + mgr(sin 0)Ó = O

mrÓ(2r6 4, g sin O) = O

Since 9 is not ałways equal to zero, from the terms in parentheses,

(b)

Fig. 22-9

f1+sir19=I1 2rFur Smalt angic 0,sin O O.

+ 6 = O 2r

co„ = \11-

2rso that

2-w7' ==

(-Lin

648 CHAPTER 22 VIRRATIONS

Ans .21T 2.7r

T == = 1.57 sto„ 4

98.1 N(b)

Fig. 22-10

A 10-kg block is suspended from a cord wrapped around a 5-kg disk. as shown in Fig. 22-10a. 1f the spring has a stiffness k = 200 N/m, determine the natura' period of vibration for the system.

SOLUT ION

Energy Equation. A diagram cif the block and disk when they are displaced by respective amounts s and 0 from the equilibrium position is shown in Fig. 22-10b. Since s = W.15 ru)0, then vb = ś --- {0.15 m)8. Thus, the kinetic energy of the system is

T = mbvi, + I ow3= -1(10 kg)[(0.15 + kg){0.15 m)21{0)2

= 0.1406(0)2

Estahlishing the darum at the equilibrium position of the błock and realizing that the spring stretches s,, for equilibrium, the potential energy is

V = + s)-2- - Ws

= 42(200 Nirri)[ss, + {0.15 m)0]2 - 98.1 N[(0.15 m)0]

The total energy for the system is łherefore.T + V = 0,1406(0)2 + 100(s81 + 0.150)2 - 14,7150 Time

Derivative.

0.28125(0)9 + + 0.150)0.150 - 14,720 = O

Since s,, = 98.1/200 = 0.4905 m, the above equation reduces to the "standard form"

B+ 160 = O

sra that

oi„ 11.716 = 4 rad/s

Thus.

64922.2 ENERGY M ET HODS•PRQBLENIS

Prob. 22-32 Prob. 22-35

22-30. Determine the differential equation of motion of ihe 3-kg błock when it is displaced slightly and releasedThe surface is stricroth and the springs are originally unstretehed.

Prob. 22-30

22-31, Determine the natural period of vibration of the pcndulum, Considcr the twa rods to be siender.each having a weight of 8 its/ft.

I ft-A

Prob. 22-31

22-32. The uniform rod af mass m is supported by a pin at A and a spring ni B, If the end ir3 is given a smali downward displacerneni and released, determine the natura' period of vibration.

22-33. The 7-kg disk is pin connected al its niidpoint. Determine the natura' period of vibration of the disk if the springs have sufficient tension in thern to prevent the cord from slipping on the disk as h oseillates, Nrnty Assume Chat the inilial stretch in erach spring is 80. This term will cancel aut after taking the tiule derivative af the energy equation.

Prob. 22-33

22-34. The machine has a rnass m and is uniformly supported by foru springs, Bach having a stiffness k, Determine the natural period of vertical vibration,

Prob. 22-3422-35. Determinc the natura] period of vibration of the 3-kg sphere.Neglect the mass of the rod and the size of the sphere,

= 500 Nim

M m

i ft

k = 50.0 Wrnr>

L300 mm—.1. 300 rnm—,

22k

6 5 0 C H A P T E R 22 V I B R A T I O N S

*22-36. The slender rod hasa mass m and is pinned at its 22-38. If the disk has a mass of 8 kg, determineend O. When it is vertical, the springs are unstretched. the natural frequency of vżbration.The springs are odginanyDetermine the natural period of vihration, unstretched.

2 2

Prob. 22-36

22-37. Determine the natura] frequency of vibration of the 2G-1b disk. Assurne the disk does not slip on the inelined surf ace.

Prob. 22-37

Prob. 22-38

22-39. The sernicircular disk bas a mass m and radius r, and it rolls without slipping in the semicircular trough. Deterrnine the natural period of vibration of the disk if il isdispiaced slightly and released. IQ= mr2.

Prob. 22-39

*22-40. The gear of mass m has a radius of gyration about its center of mass O of ko. The springs have stiffnesscs of k] and k

2. respectiveły. and both springs are unstretched when ihe gear is in an equilibrium position. If the gear is given a smali angular displacement of O and released, determine its natural period of osedlation.

Prob. 22-40

22.3 UNDAMPED FORCED V[sRArioN ó51

*22 .3 U n d a m p e d F o r c e d V i b r a t i o n

Undamped foreed vibration is considered to be one of the most important types of vibrałing mot'« in engineering. lts principles can be used to describe the motion of many types at nriachines and structures.

Periodic Force. The błock and spring shown in Fig. 22-11a provide a convenient model which represents the vibrałonal characteristics of system subjectcd to a periodic force F = Foin wor. This furce has an arnplitude of Fo and a forcing frequency w0. The free-body diagram forthe block when it is displaced a distance x is shown M Fig. 22-11b.Apply-ing the equation motion, we have Equilibrium

poci ton2F, = ma„; F0 sin wpt - kx

o rk .

-x = F0 sm courin m

This equation is a nonhomogeneous second-order differential equation. The generał solution consists of a complementary solution, xc, plus a particular solution.

The complementary solution is determined by setting the term on the right sicie of Eq. 22-18 equal to zero and solving the resulting homogeneous equation. The solution is defined by Eq. 22-9.1.e..

x, = C sin(w,,r cf)) (22-19)

where co„ is the natural frequency, =Vic7rn, Eq. 22-2.

Since the motion is periodic, the pimictdar solution, of Eq. 22-18 can be determined by assuming a soIution of the form

xP -= X sin «Jot (22-20)

where X is a constant. Taking the second timc derivative and substituting into Eq. 22-18 yicids

k F0

-X4 sin wet + -sn (X sin wet} —sm aictM

Factoring out sin war and solving for X gives

Foikx — 2 sin mot

P — (WO/ wrr)

Sbaket łabies provide Foreed vibration and arę used to separate oul granular materiał&

1111111g.(.449Z/-4 /d„ t

(a)

W = mg

F Fo sm No

N = W (b)

Fig. 22-11

2 2

(22-18)

( k i n i ) - - ( u i v i c , . % ) 2

Substitut-ing into Eq. 22-20, we obtain the particular solution

Faim FoikX - (22-21)

(22-22)

6 5 2 CHAPTER 22 VIBRATIONS

The general soludon is łherefore the sum of twa sine functions having clifferent frequencies.

F o i k x = x, + xp = C sin(clint + 4)) + ,sin &Jor (22-23)

J — (tooło4p)-The complemeniary solunon x, defines the free vibration, which depends on the naturai frequency to„ =\,/.1/ and the constants C and (f). The purdcufar solution xt, describes the forced vibration of the block causedby the applied foree F F0 sin cer. Sincc alt vibrating systems ara subjectto fritiżon, the frcc vibration, , will in timc dampen aut. For this reason the frec vibration is referred to as transient, and the forced vibration is called sready-słałe, since it is the only vibration that remains.

From Eq. 22-21 it is scen that the amplitude of forced or steady-state vibration depends on the frequency rado wo/wn. If the tnagnification factor MF is defined as the ratio of the amplitude of steady-state vibration. X, to the static deflection, Fo/k., which wouid be produced by the amplituda of the, periodic furce FQ, then, from Eq. 22-21

The soi' compactor operates by foroed vibration developed by an interna' motor, It is importnnt that the forcing frequency not be ciose la the natura' frequency, of vibration of the compactor.which can be determined when the motor is iturneduffipffierwiseresonance will occur and the machine will become uncontrollable.

22.3 UNDAMPED FORCED VIERATION 1553

X 1 MF =-

Foł k I - (wo/4)2

This equation is graphed in Fig. 22-12. Note that if the force ar displacement is appIied with a frcquency close to the naturaI frequencyof the system. i,c., I, the amplitude of vibration of the błockbecomes extremely large. This occurs because the force F is applied to the block so that it always follows the motion of the block.This condition is called resunarace, and in practice, resonating vibrations can cause trernendous stress and rapid failure of parts,*

periodic Support Displacernent. Forced vi bra łions can also anse from the periodic exeitation of the support of a system.The model shown in Fig. 22-13a represents the periodic vibration of a block which is caused by harmonie movement S = Si} sin cooi of the support. The hec-body diagram for the block in this case is shown in Fig. 22-13b. The displacement 3 of the support is measured from the point of zero displacement, i.e., when the radial line OA coincides wich OB. Therefore.generał deformałion of the spring is- 30 sin coos). Applying theequation of morion yields

+ F = IWA.; -k(Jc So sio wąt) = rnrFig. 22-12

( 2 2 - 2 4 )3

2

O

— 2

(aio 77 mp)

(Z)

o r

k k50X + X =—sin CJfll

m nc

By comparison, this equation is kientical to the form of Eq. 22-18, provided F0 is replaced by k5o. If this substitution is made into the solutions defined by Eqs. 22-21 to 22-23, the results are appropriate for describing the motion of the block when suhjected to the support displacement S = So sin war,

A

6 = 61, sin

Equilibrium pos'tion

( a )

(22-25)wił

W = mg

— Sc sin coor)._

N = W 22

*A swing has a natura' -period of vibralion, 2.3 delermined in Example 22.1. If sorneone (b)pushes an the swing onty when it reaehes iks highest point, negleeting drag or windresistanee, resonanee will oceur sinee the natural and forcing frequendes are the same_ Fig. 22-13


EXAMPLE

The instrument shown in Fig. 22-14 is rigidly attached to a platform P. which in tam is supportecl by foru springs, each having a stiffness k = 800 N/rn. f f the Floor is subjected to a vertical displacement

=

i0 sin(80 mm, where t is M seconds. determine the amplituda of steady-state vibration. What is the frequency of the tloor vibration rcquired to causc rcsonance? The instrument and platform have a tołal mass of 20 kg.

Fig. 22-14

SOLUTIONThe natural frequency is

„fr = .\14(800 Nim)— — 12.65 rad isg20 k

The arnplitude of steady-state vibration is found using Eq. 22-21.with 00 replacing Fo.

X — 10 — 16.7 mmI — (woilw,,)2 1 — [(8 radis)/(12.65 rad/s)]2

Resonance will occur when the amplitude of vibration X caused by the fluor displacement approaches infinity. This requires

= co„ = 12.6 radisAns.

22.4 Vlscous DAmPEe FREE ViBRATrON 655

* 22.4 Viscous Damped Free Vibration

The vibration analysis considered thus far has not included the effects of friction or damping in the system, and as a resulł, the solutions obtained are only in close agreement with the actual motion. Since all vibrations die out in time, the presence of damping forces shauld be included in the analysis.

In many cases damping is attributed to the resistance created by the substance, such as water, oil, or air, in which the system vibrates. Provided the body moves slowly through this substance, the resistance to motion is directly proportional to the bodys speed. The type of furce developed under these conditions is called a viscous damping force. The magnitude of this force is expressed by an equation of the form

F = cie (22-26)

where the constant c is called the coefficient of viscous damping and has units of N • sim or lb • sift.

The vibrating motion of a bady or system having viscous damping can bc characterized by the błock and spring shown in Fig. 22-15a.Thc cffect of damping is proyidcd by the dashpot connected to the błock on the right side. Damping occurs when the piston P moves to the right or left wiłhin the enclosed cylinder. The cylinder contains a fluid, and the motion of the piston is retarded since the fluid musi Baw around or through a smali hole in the piston. The dashpot is assumed to have a coefficient of viscous damping c,

H the block is displaced a distance x tram its equilibrium position, the resulting fnree-body diagram is silowi, in Fig. 22-1513,13oth the spring anddamping furce opposc the forward motion of the block, so that applyingEquilibriurn

pasitiunthe equation of motion yicldsFs = ; Ck = I k

ar (a )

rtix + cx + kx = O (22-27)

This linear, second-order, homogeneous, differential equation has a solution of the form

= eA W = mg

ł

N W

(b)

Fig. 22-45

where e is the base ai' the natural logari ibm and A (lambda) is a constant. The vałue of A can be obtained by substituting this solution and its time dcrivatives into Eq, 22-27. which yiclds

nOt2e'l' + = O

a r

eAr{razu? + cA ± k) = O

656 CHAPTER 22 VIBRATIONS Since e?`" can never be zero, a solution is possible provided

/nhl. + ch + k = 0

Hence, by the quadrałie form ula, the twa values of A are

c 2 k —+2rn"\i'f(2m— m

(22-28)kA2 = \/(_)

2rrt 2ony I7J

The generał solution of Eq. 22-27 is therefore a combination of exponentials which invoives both of these roots,Therc arc three possible combinations of A[ and A, which musz be considered, Before discussing these combinations. however. we will first define the critical damping coefficient Cr as the value of c which makes the radical in Eqs. 22-28 cq ua] to zero; i.e.,

— c < A 2 =2rri rld

or

er = 2m j— 2ttown

on(22-29)

Ove rda m pecl System. When e e,, the roots h] and )12 arc both real. The generał solution of Eq. 22-27 can then ho writtcn as

x -- AcAll Be`21 (22-30)

Motion corresponding to this solution is nonvibrating. The effect of damping is so strong that when the block is dispiaced and relcased, it simply creeps hack to its original position without oscillating.The system is said to be overdamped.

Critica I ly Dam ped System. If c = Cr, then h, = A, = —c,/2mThis situation is known as erident damping, since it represents a condition where e has the smallest value necessary to cause the system to be

22 nonvibrating. Using the methods of differential equations, it can be shownthat the solut'« to Eq. 22-27 for critical damping is

x = (A + BtleTt (22-31)

22.4 Vlscous DAMPEJD FREZ ViRRATłON657

Underdamped System. Most often c ci., in which case the system is referred to as i.mderdamped. In this case the roots h i and A, arc complex numbers. and it can be shown that the generał s()Iution of Eq. 22-27 can be written as

x = D[e-(c12"( sin(wdt (22 -32 )

where D and 4, are constanfs generally deterrnined from the initial condi tions taf the problem. The constant wd is called the damped nawal frequency of the system, It lias a value of

k

(22-33)

where the rafio is called the damping faetor.Thc graph of Eq. 22-32 is shown in Fig. 22-16. Thc initial limit of

motion. D, diminishes with each cycle of vibration, since motion is confined within the bounds af the exponential curve. Using the damped natura' frequency .ffid, the period of damped vibration can be written as

2irT d = (22-34)

Since wd < wn, Eq. 22-33, the period of damped vibration, rd, will be greater than Chat of free vibration, T = 21r/w„.

Fig. 22-16

6 5 8 CHAPTFR 22 ViRRATioNs

*22.5 Viscous Damped Forced Vibration

The most generał case of single-degree-of-freedom vibrating motion occurs when the system incIudes the effects of forced motion and induced damping. The analysis of this particular type of vibration is of practical value when applied to systems having signifieant damping characteristics.

If a dashpot is attached to the błock and spring shown in Fig, 22-11a, the differen fial equation which describes the motion hecomes

ek = Fo sin &lot (22-35)

A similar cquation can be written for a block and spring having a periodic support displacement, Fig, 22-13a, which includes the effeets of damping, In that case, however, F0 is replaced by k.50. Since Eq. 22-35 is nonhomogeneous, thc generał solution is thc sum of a complcmentary solution, xc, and a particular solution, xp. The complementary solution is determined by sełting the right side of Eq. 22-35 equal to zero and solving the hornogeneous equation, which is equvalent to Eq. 22-27. The sołuton is therefore given by Eq, 22-30,22-31, or 22-32, depending on the valucs of hi and /li. Because ali systerns arc subjected to friction. then this solution will darnpcn out with time. Only the particular solution, which describes the steady-stare vibration of the system, will remain, Since the applicd forcing function is harmonie, the stcady-state motion will also be harmonie. Consequently, the particular solution will be of the form

Xp sin(wot - O') (22-36)

The constants X' and O' are determined by taking the first and second time derivatives and substitutiag them into Eq. 22-35, which after simptification yields

— nawk; sin(wot — da')

X 'ewo cos(eisot — (y) + r k sin(wot — da`) = Fo sin

Since this equation holds for all time, the constant coefficients can beobtained by setting wat = 0 and (.00/ - 49' = 7r f 2, which eauses theabove equation to beeorne

rewu = F o sin

—X' sre4 + X'k = F() cos rfi'

2 2

22.5 ■Ascous PAMPEn FORCE0 ViRRATION659

The amplituda is obtained by squaring these equations. adding theresults. and using the identity sin20' cas2cb' = 1, which gives

X' = (22-37)V(k, F0 in(4)2 c2o;(-)

Dividing the first equation by the second givesc f -w o

cp = taił (22-38)k - inr.o()

Since w„ = k jrn and cc = 2fflo, filen the above equatłons can also bewritten as

F 0 / k

11[1 - [2(c/c0)(wplWrk

o, s._ tan_i[2(eicA(wo/w,)] rgJ

(e/cj(w01w,.)1 — (‹Liu/w,,Y.

(22-39)

The angle cf,' represents the phase difference between the applied force and the resiffing steady-stałe vibrałion of the damped system.

The magnification factor MF has Been defined In Sec. 22.3 as the rafio of the arnpliłude (›f deflection caused by the fbreed Yibration w the defleetion caused by u static forte F0, Thus.

' MF = X (22-40)Foik (wop,,,,)212 1 [2(c/c,..)(wD/wa

The MF is plotted in Fig. 22-17 versus the frequency ratio wdw„ forvarious values of the damping factor h can be scen from this graphthat the rnagnification of the amplitude increases as the damping factor decreases. Resonance obviously occurs on]y when the damping factor is zero and the frequency rafio equals 1.

M F

5

4

3

O Fig, 22-17

6 6 0 CHAPTFR 22 ViRRATIoN5

EXAMPLE E l

The 30-kg electric motor shown in Fig. 22-18 is confined to move verticaiiy, and is supported by foar springs, each spring having a stiffness cif 200 Nim. If the rotor is unbalanced such that its effect is equivalent to a 4-kg mass locałed 60 mm from the axis of rotation, determine the amplitude of vibration when the rotor is turning at (oo = 10 rad/s. The damping factor is c/c, = 0.15.

ii•444tbiokFig. 22-18

SOLUTIONThe periodic force which causes the motor to vibrate is the centrifugal force due to the un balanced rotor. This force has a constant rnagnitude of

ma„ = mrwc2

) = 4 kg(0.06 m)(10 rad/s)2 = 24 N

The stiffness of the entire system of four springs is k = 4(200 Nim) = 800 N/m. Therefore, the natural frequency of vibration is

— 5 .164 g

Since the darnping factor is known, the steady-state arnplitude can be determineci trom the first of Eqs. 22-39, i.e„

F o / k1,1[1 (coo/oi,)22 [2(c/c,.)(wQ/w„)fz

24/8001v/1[1 — (10/5.164)212[2(0.15)(10/5164)f

= 0.0107 m = 10.7 mm Atu

\1800 N/m

in 30 k

x ' -

22.6 ELCTIRIC.6.L CIRCLI1T ANALCGS 659

*22.6 Ełectrical Circuit Analogs

The characteristics of a vibrating mechanical system can be represented by an electric circuit. Consider the circuit shown in Fig, 22-19a, which consists of an inductor L. a resistor R, and a capacitor C. When a voltage E(ł) is applied, it causes a current of magnitude i to flow through the circuit. As the current flows past thc inductor the voltage drop is L(.di/di), when it flows across the resistor the drop is Ri, and when it arrives at thc capacitor the drop is (1 x) f I di. Sincc current cannot flow past a capacitor, it is only possible to measure the charge q acting on the capacitor.11te charge can, however, be related to the current by the equation i = dg Thus, the voltage drops which occur across the ind-uctor, resistor, and capacitor bec.onie L d2q ł d.t2, R dw dł, and q/C, re_spectively. According to Kirchhoffs voltage Iaw, the applied voltage balances the sum of the voltage drops around thccircuit.Therefore,

c p q d q 1L—, + R— ± —q = E(1) (22-41)d r d r C

Consider now the model of a single-degree-of-freedom mechanical system, Fig. 22-19b, which is subjected to both a generał forcing function F(r) and clarnping.The equation of motion for this system was established in the previous section and can be written as

d 2 x d xdr + c— + kx = ni)(22-42)

By comparison. it is scen that Eqs. 22-41 and 22-42 have the same form, and henee mathematically the procedurc of analyzing an electric circuit is thc same as that of analyzing a vibrating mechanical systcni. The analogs between the twa equations are given in Table 22-I.

This analogy has important application to experimental work, for it is much easier to simulate the vibration of a complex mechanical system using an electric cireuit,which can be constructed on an analog computer, Chan to make an equivałent mechanical spring-and-dashpot model.

TABLE 22-1Electrical-Mechanical Analogs

(b)

Fig. 22-19

Electrical

Electric charge

Electric currentVoltage

Inductance

Resistance

Reciprocal of capacitance

rvlechanical 1-

",splacement

Vełocitydxl&

E(t) Applied furce F(t)

L Mass rra

R Viscous damping ccoefficient

1/C Spring stiffness k

2 2

662 CHAPTFR 22 VIRRATIONS

E PROBLEMS

22-41. If the block is subjected to the periodic forte F = F0 cos cor, show that the differeniial equation of motion is :55 + (ki m)y = (Foi rn) cos utr, where y is measured from the equilihrium position of the block_ What is the generaI sol ation of this equation?

Prob. 22-41

22-42. The block silami in Fig. 22-15 has a mass of 20 kg, and the spring bas a stiffness k -= 600 NI/rn. When the block is displaced and released. twa suecessive amplitudes are measured as xl = 150 mm and x, = 87 mrn. Determine the coefficient of viscous damping. c.

22-43. A 4-lb weight is attached to a spring having a stiffness k = 10 ibift. The weight is drawn downward distance of 4 in, and released from rest. If the support moves with a vertical displacement 8 = (0.5 sin 4t) in., where t is in seconds, determine the equation which describes the posiiion of the weight as a function of time.

22-44. A 4-kg błock is suspended from a spring that has a stiffness af k = (100 N/m. The block is drawn downward 50 mm from the equilibrium position and released from rest when r = 0. if the support rnoves with an impresseddisplacement of 8 = sin mm, where t is in seconds.determine the equation that describes the vertical motion of the block. Assume posiłive displacementis downward.

22-45. Use a block-and-spring model like that shown in Fig. 22-13a, but suspended from a vertical position and subjected to a periodic support dispłacement 8 = SSa sin ok,/, determine the equation af motion for the system, and obtain its general solution. Define the thsplacement y measured from the ststic eguitibrium posilion of the block when r = 0.

22-16.. A 5-kg block is suspended from a spring having a stiffness of 300 Nim. «If the block is acted upoi by avertical kirce F = (7 sin where t is in seconds.determine the equation which describes the motion of the błock when it is pulled down 100 mm from the equilibrium position and released from rest at r = 0. Assume Chat positive disphicement is downward.

k = 300 Nfrn

22-47. The electric motor has a mass of 50 kg and is supported by forgr spring.% each spring having a stiffness of 100 Nim. 1f the motor turns a disk D which is mounted eccentricaliy. 20 mm from the disk's center. determine the angular velocity 0.1 at which resonance occurs. Assume that the motor ordy vibrates in the vertical direction.

-20 mrn

ri...

F= 7 sin 81

Prob. 22-46

t k = LiX} N /m k = 11717 .,1"/m:

Prob. 22-47

22.6 ELECTIRICAL CIRCIJIT ANALOGS 663

*22-48. The 20-11, block is attached to a spring having alliffness of 20 Ib i ft. A furce F = cos 2t) Ib, where i is inseconds. is applied to the block. Determine the maximum speed of the block after frietional forces eause the free vibrations to dampen out.

22-49. The light elastic rod supparts a 4-kg sphere.en an 18-N verticat force is applied to the sphere, the rod deflects 14 mm. If the wali oscillates with harmonie frequency of 2 Hz and has an amplitude of 15 mm, &tenni ne the amplitude of vibration for the sphere.

Prob. 22-49

22-50. The instrument is centered uniforinly on a platform P, which in t urn is supporled by four springs,each spring having stiffness k = 130 Nim. If the fluor is subjected to a vibratinn w = 7 Hz, having a vertical displacement amplitude

= 0.17 ft. determine the vertical displacement amplitude of the platform and instrument. The instrument and the platform have a total weight of 18 lb.

22-51. The uniform rod has a mass of m. 1f it is acted upon by a periodie force of F =, F0 sio ew, determine the amplitude of the steady-stałe vibratinn.

ł

L2

L2

F= F0 sin

tor Prob. 22-51

*22_52. Using a block-and-spring model, likc Chat shown in

Fig. 22-13a, but suspended from a vertical position and subjected to a periodic support displacement of 8 = 30 cos determine the equałion of motion for the system, and obtain its generał solution. Dcfiiie the displacement y incasured from the stalic equilibrium position of the błock when t = G,

22-53. The fan bas a mass of 25 kg and is fixed to the end of a horizontal beam Chat bas a negligible mass. The fart blade is mounted cceentrically on the shaft such that it is equivalent to an unbalanced 3.5-kg mass Iocated 100 mm from the axis ol rotation. If the Stalic deflection of the beam is 50 mm as a resuIt of the weight of the fan, determine the angular velociły of the fan blade at which resonance wili occur. Hiat See the first part of Example 22.8.

22-54. In Prob. 22-53. dctmnvine the amplitude of steadystate vibration of the fan if its angular velocity is 10 radis.

22-55. What will he the amplitude of steady-state vibration of the fan in Prob. 22-53 if the angular velocity of the fan blade is 18 radis? Hint:See the first part of Example 22.8.

Prob.. 22-4

Prohs, 22-53/54/55Prob. 22-50

2 2

664 CHAPTFR 22 ViRnATioNs

*22-56. The smali block al A has a mass of 4 kg and is rnounted on the bent rod having negligible mass.If the rotor at B eauses a harmonie movernent Sp = (0.1 cos 151) where c is in seconds, determine the słeady-state amplitudo of vibration of the błock,

Prob. 22-56

*22-60. The engine is mounted on a foundation bloek which is spring supported. Describe the steady-state vibration of the system if the hlock and engine have a total weight of 1500 Ib and the engine, when running, creates an impressed fotce F = (50 sin 2r1 Jb, where is in seconds. Assume that the system vibrates only in the vertical direction, with the positive displacement rneasured downward, and that thc totał stiffness of the springs can be represented as k = 2000 Ibift.

22--61. Determine the rotational speed w of the engine in Prob. 22-60 which will cause resonance.

2 2

22-57. The elecłric muter turns an eceentric flywheel which is enuivalent to an unbalanced 0.25-lb weight Iocated 10 in. from the axis of rotation. 1f the stalic defiection of the beam is 1 in. because of the weight of the motor, determine the angular velocity of the flywheel at which resonance will occur. The motor weighs 159 lb. Negleet the masa of thc beatu.

22-58. What will be the amplitude of steady-stale vibration of the motor M Prob. 22-57 if the angular vetocity of the flywheel is 20 rad/ s?

22-59. Determine the angular velocity of the flywheel in Prob, 22-57 which will produee an amplitude of vibration of 025 in.

Probs. 22-57/58/59

Prohs. 22-60/61

22-62. The motor of mass r1fi is supported by a simply supported beam ol negligible masa. If Mack A of mass m is clipped onło the rotor. which is turning at constanł angular velocity of w, determine the amplitude of the steady-state vibration. Mfw When the beam is subjected to a concentrated lorce of P at its mid-span, it deflects

PL3,48E1 at this point. Here E is Young's modulus of elasticity, a property of ehe materia', and I is the moment of hienia of ehe beam's c-ross-seetional arca,

Prob. 22-62

22.6 ELEttRicAL CincutT ANALOGS665

22-63. A block having a mass ("f 0.8 kg is suspended from a spring having n stiffness of 120 N/m. If a dashpot provides damping force of 2.5 N when the speed of the block is 0.2 mis, deteriTi i ne, the period cif free vihration.

2764. The block, having a weight of 15 Ib, is immersed in a liquid such that the damping lorce acting on the block has magnitude of F = (0.81-vi) 113, where v is the velocity of the block in ft/s. If the błock is pulted down 0.8 ft and rcicascd from Test, determine the posdion of the block as a functinn of tinle.The spring has a stiffness of k = 40 ibift. Consider positive displacement to be downward.

Prob. 22-64

22-68. The 4-kg eircular disk is ałtached to three springs, eaeh spring having a stiffness k = 180 Nim. If the disk is immersed in a fluid and given a downward velocity of 0.3 m/s at the equilibrium position, determine Che equation which describes the mafiom. Consider positive displacement to be measured downward, and that fluid resistance acting on the disk furnishes a damping force having a rnagnitude F = (601v1) N, where v is the veloeity of the błock in m/s.

22-69. If the 12-kg rod is subjeeted to a perioclic force of F = (30 siu 6t) N. where t is in seconds, delermine the steady-stale vihration amplitude 0„,„ af the rod about the piu R. Assume N is smali.

Prob. 22-68

22-65. A 7-lb block is suspended from a spring having stiffness of k = 75 lbift. The suppori to which the spring is attached is given simplc harmonie molton which may beexpressed as S = (0.15 siuft, where t is in scconds„ If thedamping factor is c jc, = 0.8, determine the phase angie, of forced vibration.

2766. Determine the nlagnification factor af the block, spring, and dashpot combination in Prob. 22-65,

22-67. A block having a mass of 7 kg is suspended from a spring that hasa stiffness k = 600 Nim. If the block is given ap upward velocity of 0.6 mis from its equilibrium position at t = 0, determine ds position as a functian cif dme. Assume that positłve ciisplacement of the błock is downward and that morion takes place in a medium which fumishesdamping force F(501v1) N. where v is in m/ s.

Prob. 22-69

666CHAPTFR 22 ViRnATioNs

22-70. The damping factor, c/cc, may be determined experimen tatły by measuring the successive ampiitudes of vibrating motion of a system. If two of these maximumdisplacernents can be approximałed by x andasshown M Fig. 22-16. show that the ratio In (xl /x-.) =

21r(ć-icr)/V1-(cfc,)2. The quantity In (xlix-,) is called the logarithmic decr.E,ment.

22-71. If the amplituda of the 50-lb cylinclers steady-vibration is 6 in., determine the wheel's angolar velocity

Prob. 22-71

22-73. The 20-kg block is subjectcd ło the action of the harmonie furce F = (90 cna 61)N. where 1 is in seconds. Write the equation which describes the steady-state motion.

k = 400 Nim•...

F - 90 cos fir-,:---N.\\\.\--'," —10.. 20 kg

-...::-. \ \. N, \-1

k = 400 N,.'m 1115 N • s../ino0 b

Prob. 22-73

22-74. A bulle! of mass m hasa veloci ty of vo ]ust before it strikes the target of mass M. If (be bullei embeds in the target, and the vibration is to be criticalły damped, determine thc dashpoes critical damping coefficient, and the springs maximum compression. The target is free to move along the two horizontal guides that ara "nested" in the springs.

22-75. A bulleł af mass m has a veloeity just before itstrikes the target of mass M.the bulle! embeds in thełarget, and the dashpots damping coefficient is O < cc,determine the springs maximum compression.The target is free to = c along Ule two horizontal guides that ara "nested" in the springs.

k = 200 Ibift

1

r = 25 lbsin

= 200 lbift

22

22172. The IQ-kg block-spring-damper system is dampcd. If the block is displaced to = 50 mm and re]eased from rent, determine the Orne required for it to return to the position x = 2 mul.

c= 80 N -

k= 60N/m

-

Prob. 22-72Prob. 22-74/75

22.6 ELECTRICAL CIRCUIT ANALOGS667

*22-76. Determine the differential equation of motion for the damped vibratory system shown. What type of motion oectirs? Take k = 100 Nim, = 200 N • sim, m = 25 kg.

22-78. Draw the elcctrical circuit that is cquivalent to the mechanical system shown. What is the differential equation which describes the charge q in the circuit?

Prob. 22-78

Prob. 22-76 22-79. Draw the electrical circuit that is equivalcnt to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit.

22-77. Draw the electrical circuit that is equivalent to the mechanical system shown. Determine the differential equation which describes the charge q in the circuit.

Frot

2 2

c

Prob. 22-77

Prob. 22-79

668CHAPTER 22 VIBRATIONS

CHAPTER REVIEW

Equilibrium position

+ w n2 x = O 2r w == =r 2 / r

Equ pos]. lon

F o i kxr — y SIn woi

ł — (60e/wrr)

F = Fi sin mur

Undamped Free Vibration

A body lias free vibration when gravitationaI or elastic restoring forces cause the motion. This motion is undamped when frietion forces are neglected. The periodic motion af an undamped. freely vibrating body can be studied by displacing the bady from the equlibrium position and then applying the equation af motion aIong the palb.For a one-degree-of-freedom system, the resulting differentiaI equation can be written in terrns of its natural frequency cg„.

Energy M eth odsProvided thc restoring forces acting on the body are gravitational and elastic, then conservation of energy can :ulw be used ta determine its simplc harmonie motion. To do this, the body is displaccd a smali amount from its equilibriurn position, and ari expression for its kinefic and potential energy is written. The time derivative of this equation can then be rearranged in the standardfonii r.o„2 x = 0.

Undamped Fnreed Vibration

When the equation of motion is apphed to a body_ which is subjected to a periodic force, or the support hasa displacement with a frequenc■,, rva, then the solution of the differential equation consists af a compfementary solution and a particuar solułion. The complemen tary solution is caused by the free vibration and can be neglected. The particułar solution is eaused by the forced vibration.Resonance will occur if the natural frequency of vibration w, is equal to the forcing frequency wo. This should be avoided, since the morion will tend to become unbounded.

CHA PTEft RE ',/114N 669

Viscous Damped Free Vihration

A. viscous damping force is caused by fluid drag on the system as it vibrates. If the motion is siow, this drag form will be proportional to the velocity, that is,F =, Here c is the coefficient ofviscous damping. By comparing its value to the criticai damping coefficient e, = 2/nr.o„, we can specify the type of vibration that oecurs. If c 7 e„, it is anoverdarnped system: if c cc, it is acrifically damped system; if c < c,. it is an underdamped system.

Viseous Damped Foreed Vibrat'«The most generał type of vibration for a one-degree-of-freedom system occurs when the system is damped and subjected to periodic forced motion, The soTution provides insighł as to how the damping factor, cieć, and the frequency ratio, mojw„, influence the vibration.

R e s o n a n c e i s a v o l d e d p r o v i d e dand wo/ei.„ 1.

Electrical eireni1 Analogs

The vibrating motion of a complex mechanical system can be studied by modeling it asan electrical circuit. This is possible since the differential equations that govern the hehavior of each system are the same.

Eguilibrium position

lik

APPENDIX

AMathematicalExpressions

Quadratic Formule- I)

1i. ax- + bx + c = O, theii x =

Derivatives\1b2 - 4ae d —Cu " , du

) = „2rJ dx cfx

Hyperbol ic Funct ions

e' - e' et + 0,-x!iilli x , Lod-i x = , lanh x

2 2

d d v d u- ( u v ) = u - + v -

sini-L x dx dx dxcosh x

d u d v v - - u -d x d xd v d

x ( u )

Trigonometric Identities

A Csin O —, esc S= -

A

B C uos O - -c, see O =- -B

A Btan 0 = • colt O = -A

sine E + cos' E =

v2

d , du- (c.oł u) = -esc- u -dxdx

d—(see u) = tan u sec u du

dx dx

Nin(0 ± 13) = sin O cos rt,± cos O sin d du- { c s c _ - e s c u c n i u d x d x

sin 26 = 2 sin O cos

cos(0 4,) = cos O cos F sin O sin d du—

CDS 20 = COS2 (sin u) = cos 44 — sin2 O dxdxI cos O ±+ cos 20 sin \(1 cos 20

2 ' 2 (1{cos u) = —sin 44—

d usin o dx dx

d d u- d x ( t a n u )

= s e c u d x P o w e r - S e r i e s E x p a n s i o n s

d du-{s inh u} = eosh u x

s i n x = x — — x 3 + s i n h , x - = x + — x 3 - F . - .x

3! 3!2 2xcos x = 1 - - + cash x X dI + — + - - - —{cosh u} = sinh u du—

2! 2! dr dx

tan Ocos O

I + tan2 0 = sec2 0 I + cot2 0 = csc2

670

APPENDix A MATHEmATEcAL ExPREssIoNs671

Integrals

f x" dx n + 1 + C, n -1

j r a + dx

bx1- 1n{a + bx) + C

jb

f dx1ln a + x-V - ab J a + b9C22-\,5aa - x',.

jr x dx_,,+ b x 2 b In(bx - + a) + C

, , . . f x2 dxxlab tan + C, abJ a + bx2 b b a V7-2-b jr a2 dx x2 - I in [a + xl

+ C, a2 K2- 2a a - x

b+ + — ] + C , O

= l V -cb2 - 4ac

if sn x dx = -ros x + C

fcos x dx = sili. x + C

dx 2Va + bx+ C

Ni a + bxb

x dx =+ CI a2

d x j V a + bx + ex2

1 11,1 [ bxCJC2V C

+ C, ab < O

a+ bx dx = 2 Ni (a + bx)3 + C1x 3b x cos(ax) dx = cos(ax) + - sin (ar) + C

a--2(2a - 3bx)1,7(a + bx)

Via + bx dx -+ C2x15b2x- cos(ax) dx =cos(ax)

a-

fx2V a + bx dx -2(8a2 - 12abx + 15b2x2)17(a + bx)3

I05b3+ C a-x- - 2

3sn(ax) + C

;5.k-2 dx =-1[xila2 - x2 ++ C, a2a e" dx = e" + C

ax.'vE. a2Vc7c2a2)3

x2'vł' a2 - x2 dx =Ni(a2 - x2)34

+ a —(x1+,/a2 - x2 + a2sin-1 -x + C, a Oa

1 r x2 ± a2 dx = -2 [xVx2 ± a2 ± a2 Is-t(x + 1x2 ± a2)] + C

x1/a~ - x2 dx = 3-\1(a2 - x2) + C

XC/Z2 ± 2 .IX2N7 Xa dx = -4 V (x2 ± a2)3 + —

8 x-Nix2 ± a24

—a 8 1n(x + V x2 ± a2) + C A

d:Ure"' dx = —_, (ax - 1)

+ C a-

isinh x dx = cosh x + C feash x dx = sini" x + C

APPENDIX

Vector Analysis

B

The following discussion provides a brief review of vector analysis. A more detailed treatment of these topics is given in Engineering Mechanics: Statics.

Vector. A vector, A, is a quantity which has magnitude and direction, and adds according to the parallelogram law. As shown in Fig. B-1, A = B + C, where A is the resultant vector and B and C are cornponent vectors.

Unit Vector. A unit vector,uA. has a magnitude of one "dimensionless" unit and acts in the sanic direction as A. It is determined by dividing A by its magnitude A, i.e,

A1 1 A = A

( B - 1 )

= B +

B

Fig. B-1

672

APPErunix B VEcToR ANALYsi 67 3

Cartesian Vector Notation. The directions °f the positive x, y. z axes arc defined by the Cartesian unit vectors i, j, k, respeetively.

As shown in Fig. B-2, vector Ais forrnulated by the addition of its x, y, z components as

A = Ail + AJ + A,k (13-2)

The magniłcade of A is determined from

A VA1 + Ay2 + A' (B-3)

The direction of A is defined in term of its coordinate direcikm angles, a, p, y, measured from the taił of A to the pasidve x, y, z Rxes, Fig. B-3.These angles are determincd from the direction cosines which represen t the i, j, k componcnts of the unit vector ikt; i.e., from Eqs. B-1 and B-2

A „. Ay AuA = —j +

A A A

so that the diteetioo cosines arc

A, A y A.c o s r x = — c o s P — c o s =

A A A

Hence,g4 = cos ai + cos Ę3j cos yk, and using Eq. B-3, it is scen that

cos= a + cos' p + cos2 y = I (13-6)

The Cross Product. The cross product of two vectors A and B, which yields the resultant vector C, is written as

C=AXB (13-7)

and reads C equals A "cross" B. The magnitude of C is

C = AB sin (B-8)

where O is the angle made between the tails of A and B ((r --5 ł3-.5 IWY), The directivn of C is deterrnined by the right-hand rule, whereby the fingers of the right band arc curled from A to B and the thumb points -in the direction of C, Fig. 13-4. This vector is perpendicular to the piane containing vectors A and B.

(13-4)

(13-5)

y

Fig. B-2

Fig. B-3

Fig. 8-4

674 APPENDIX B VECTOR ANALYSIS

The vector cross product is not commutative, i.e., A X B B x A.Rather,

A x B = -B x A (B-9)

The distributive law is valid;

AX(B+D)=AXB-FAXD (13-10)And the cross product may be multiplied by a scalar m in any manner:i.

e., m(A x B) = (mA) xB=AX (inB)= (A X B)in (B-li)

Fig. B-5

Equation B-7 can be used to tind the cross product of any pair of Cartesian unit vectors, For example_ to ind i X j, the magnitude is (i)(j) si n 90' = (1X I )( l) = 1. and its direction +k is determincd from the right-hand rule, applied to i X j, Fig. B-2. A simple scheme shown in Fig. B-5 may be helpful in obtaining this and other results when the need arises. If the circIe is consłructed as shown. then "erossing" two of the unit vectors M a counrenlockivise fashion around the circie yields a positive third unit vector, e.g., k X i = j. Moving clockwise, a neprive unit vector is obtained, e.g., i x k = -j,

If A and B are expressed in Cartesiari comporient form, then the cross product, Eq, B-7, may be evaluated by cxpanding the, determinant

i j k

C = A X B = A, B ,

A yB,

A ,

B ,

(B-12)

which yields

C = (A .B_ A 12Y - (ARB.-. - A .8,),1 ± (A „B, - A ,B)k

Recall that the cross product is used in statics to define the moment of a furce F about point 0, in which case

Mo = r X F (B-13)

Bwhere r is a position vector directed from point O to any point on the line ()f action of F.

AppErvoix. B VECTCR ANALysis 67 5

The Dot Prod u ct. The dat product of two vectors A and B, which yields a scalar, is defined as

A-B = AB cos O (B-14)

and rcadl A "dot" B. The angic 9 is farmer] between the rails of A and B(Cr 180').

The dot product is commutative;i.e.,

A•B = B-A (B-15)

The distributiyc law is valicl;

A•(B + D) = A-B A•D (B-16)

And scalar multipiication can be performed in any manner, i.e.,

m (A • B) = (rn A) - B = A • (rnB) = (A • B)In (B-17)

Us ing Eq . B-14 , the do t p roduc t be tween any two Car tes ianvectors can be dełerminecl. For example, i • i = (1)(1) cos= 1 andi • j = (1)(1) cos= O.

H A and B are expressed in Cartesian component form, then the dot product, Eq. C-14, can be determined from

A •B = A + A + A ,I3,1 (B-18)

The dot product may be used to determine the crtzgle O formed berween rwo vecroirs, From Eq. B-14,

A • B• = . c o s - 1 ( —

A B(B-19)

676 APPFNDFX O VECTOR ANA.LYSS.

It is also possibIe to find the componenr of a vector in a given direction using the (lot product. For example. the magnitude of the component (ar projecłion) of vector A in the direction of B. Fig. B-6, is defined by A cos O, From Eq. B-14, this magnitude is

l i R

Fig. B-6 A cns9 =- A •11- = A • uli (B-20)

where ug represents a unit vector acting in the direction of B, Fig. B-6.

Differentiation and Integration of Vector Functions. The rules for differentiation and integration ()f the surns and products of scalar functions also apply to vector functions. Consider, for example, the two vector functions A(s) and B(s). Provided these functions are smooth and continuous for all s, then

d dA dB— (A -V B) -= —ds ds ds

(B-21)

f (A + B) ds = f A ds + i B ds (B-22)

For the cross product,

—d(A X B) = (—dA x B) + (A x —`33) (B-23)ds ds ds

Similarly, for the dot product,

dB—d(A • B) = —dA • B + A • —

ds ds ds. (B-24)

The Chain Rule

APPENDIX

C

The cham role of calculus can be used to determine łhe time derivative of a composite -function. For example. if y is a function of x and x is a function of t, then we can Pind the derivative of y with respect to t as fullows

dy dy dx3

d r T d r d l

In other words, to find 5: we take the ordinary derivative (dy/dx) and m uitiply it by the time derivative

lf severai variables are functions of tlnie and they are multiplied together, then the product role d(uv) = du v + u dv m ust be used along with the chain tule when taking the time derivatives. Hera are some example&

( C - 1 )

677

678 APPEND[K C THE CHAiN ROLE

If y = xex, find y.

SOLUTIONSince x and e3 are both functions of time the product and chain rules must be applied. Have bi -= x and

3; = + x[e-

The second time derivative also rcquires application of the product and chain mies. Note that the product Fuk applies to the thrcc time variables in the last term, i.e., e, e', and x..

y = (Uje' + } + [i]e.x.;c + + xe-15/

ex[:«1 + + _12(2 + x)IIf x = t2 then x = 2t, x = 2 so that in terrns in /, we have = e''

1-2(1 + t2) + 4t2(2 + /2)-1

C - 1

If y = x3 and x = t4, find the second derivative of y with respectto time.

SOLUTIONUsing thc chain rule, Eą. C-1.

= 3x25c

To obtain the second time derivative we must use the product rule since x and x. are both functions of time, and also, for 3x2 the chain rule musi be applied. Thus, with u = 3x2 and y = x, we have

= + 3x2f.;(:)

= 3x[2i2 + xx]

Since x = (1, then x = 4/3 and = 1212 so that-y" = 3(t4)[2(4.(3)2 + t4(1212)]

= 13V°

Note that ibis result can also be obtained by combining the functions, then taking thc time derivatives, Chat is,

3;= 12t11 v

= 132tI<)

EXAMPLE

EXAMPLE

C - 2

C

APPENDIX C THE CHAIN RULE 679

EXAMPLE

C-4

If r2 = 603. End r.

SOLUTIONHere Ule chin and product rules arc applied as follows.

= 663

2ri- = 1802Ó

21(r; + r(r)] = 18[(200 + 02(9)j2

r + r;r' = 9(2002 + 026)

To find "r- at a specified value of O which is a known function of time, we can first find t'9' and B. Then using these values, evaluate r from the first equation, r from the second equation and '?; using thc Last cquation.

EXAMPLE

the path in radia' coordinates is g-iven as r = 502., where is a known functio-n of time, find r.

SOLUTIONFirst, using the chain rale then thc chain and product rules where 100 and v =

Ó, we have

r =

= 11MÓ

= iW0)ri + 0(d)]

= 1002 + 1009

Fundamental ProblemsPartial Solutions And Answers

Ans.

Ans.

Ans .

Ans .

Ans.

Ans.

Ans.

Chapter 12

F12-1. = vo + ac/

14 = 35 + a,(15)a, = —1.67 m/s2 = 1.67 m/s2

F12-2. s = + -F 3,:y2o = o + 15r + (-9.8o12 ł = 3.06 s

F12-3. ds = v dt's ds = (4( — 3(2)dt

s = (2(2 — r3)ni

s = 2(42) — 43

= —32 m 32 m <—F12-4. a = d = d (0.513 — )a = ( 1.512 — 8) tn/s2

When ł = 2 s,R= 1.5(22) — 8 = —2m/s2 = 2 m/s2

F12-5. v = T, =1, (2i2 — St + 6) = (4t — S) m/s• O = - ( 4 ( — 8 ) t

= 2 5 s l , . 0 = 2 ( 0 2 ) —

8 ( 0 ) + 6 = 6 m

• = 2( 22) — 8(2) + 6 = —2 m

• = 2(32) — 8(3) + = Om( = 8 m 4- 2 m = 10 m

F12-6. ft, = I a ds

Lv ov = (10 — 0.2s)ds "

= V :20s — 0.2s2 + 25 ) m/s

At s -= 10 m. v =V20(10)

— 0.2(102) + 25

= 14.3 m/s -

F12-7. v = (412 — 2) dr

v = 2t + Ci

= (112 2t + Ci ) dts = t4 t2 + Cit + C2 = 0, s = —2, = —2 = 2, s -= —20. Ci = —9.67 = 4, s = 28.7 m

6 8 0

F12-8. a -=--= (20 — 0.05s1) ( —0.1s)

At s = 15 m.a = —13.1 m/s-2 = 13.1 tnis2 Ans.

F12-9. t.) = (0.5(3) 1.512

= = (108) = O Ans.

-v (m/s)

54

10

FI2-10. ds = v dźo' ds = i

J(-4( + 80) dt• o

=-- —2,2 -L 801a= = ( + 80) = —4 ft/s2 = 4 ft/s2

a = — 0,0,8°_ftś' — —4 ft/s2

s f t )

r (s)20

(ft/S2)

20((s)

4

6

r (s) r (s)

1125

375

s 10

FUNDAMENTAL PROBLEMS681

F12-11. a ds = v dva = v = 0.25s4 (0.25s) = 0.0625s

Is = tn = 0,0625(40 m) = 23 m/s2

o(misa)

2.5

s (m)

F12-13. O t < 5 s,I 1 ) d v = r

dv = a dt a 20 dtov = (200 m/s5 s < I s t',( j-, )= a dt dv = —10 dł

inis 15sv-100 -= (50 — 10t) m/s,

O = 150 — 101" t' = 15 sAlso,A v = G = Area under the a—t graph

= (20 m / s2)(5 s) + [—( IP m/s)(1' — 5) A t' = 15 s

FU-12. 0 s r

S s <

5 s<

s < v (

m/s)

3 0

a (m

< 5 s,

= vr =

t 10 s,

v =

< 5 s,

a =11; =t 10s,

= =t < 5 s, a =-

-s 10 s,a

d (3t2)(6/)

(30/ — 75)7s5:

m

mis

= 30 mis

— 30 mis

m/s2

m/s2

0

F12-14.

(

100

0 5- ds =

s = 5s<ts

) ds =N =

s = =

A l s ,

A s = =

= s

(nri/s)

t (s)

Ans.

Ans.

(6t) = 6

(30) =❑v/t = 6

= Av/At

=

5- S

s, v dt

15r2lio (15?) m

15

v dt;

(-7.5t2(-7.5X15)21125 m

Area under

1125 m (m)

4 (1501n/s)(15

ffds = I 30t dt

. o

s,J375 ds = (-15t + 225)d/

+ 225/ — 5623)m+ 225(15) — 562.5 m

the v-1 graphs)

/s)

t (s)10

682 PARTIAL SOLLITIONS AND ANSWERS

n2-15. = 32/ dif o x - o

= (16/2) m

dy - i 8 di

Y=-

8Substituting Eq. (2) into Eq. (1), get

y =-- 2V:tF12-16. y = 0.75(8/) = 6/

• =- = (8/) = 8 mis -->

= = (61) = 6 mis 1 The magnitude

ol the particle's velocity is

v = = V(8 m/512 + (6 m/s)2

= 10 m/s Ans.

F12-17. y = (4t2) m

v, = = JA (4/4) = (16/3) m/sv, = (4/2) = (8/) m/s tWhen = 0.5 s,=V.,

2, = V(2 m/s)2 + (4 m/s)2

= 4.47 m/s a, = vX = .<5: (16?) = (48/2) m/s2a„ (81) = 8 m/s'When t = 0.5 s.• = i + = "V(12 m/s2)2 + (8 m/s2)2 = 14.4

m/s2

F12-18. y = 0.5x=

v,. =

When t = 4 s,

v„ = 32 m/s v,. = 16 m/s• = + v = 35.8 m/s Ans.

ax = = 41

• = 2/ When= 4 s,

ax -= 16 m/s2 = 8 m/s2

a = 141,2 + a = V"162 + 82 = 17.9 m/s2 Ans.

F12-19. y = t4) mv, k (4t) mis v,. =y = (4/3) mis

When t = 2 s,

v,, = 8 m/s v = 32 mis= + v?: = 33.0 m/s Ans.

a, = = 4 m/s2

When t = 2 s,a, = 4 m / si a, = 48 in/s2

a -= a?". = 42 ÷ 482 = 48.2 m/s2 Ans

F12-211. y = 0.1k-ky, = 0.1(5)(-31 = -1.5 m/s = 1,5 m/s i Ans.

= o .1 +x51

a. = 0 .1[( -3)2 + 5(-- i .5) = 0.15 rn/s2 1 Ans.

F12-21. (./.?R) = (vA);; + 2a,.(y8 - yA)02 = (5 m/s)2 + 2(-9,81 m/s2)(h - 0)

h = 1,27 m Ans.

F12-22. Yc = Y A ± (1? AM AC + a4c= 0 + mi s)tAc + (-9.81 iri/s)dc tAe = 1,

0194 s

(vc),. (vA), + api AC

(Vr)y = 5 m/s + (-9.81 m/s2)(1.0194 s)= -5 m/s = 5 m/s vc

= -\/(vc)?" + (vr)

= x/(8.660 m/s)2 + (5 m I s)2 = 10 m/s Ans. R

= xA + (vA),tAc = O + (8.660 m/s)(1.0194 s)

= 8.83 m Ans.

F12-23. s = su + vo/

10 = O + vq cos 30°:s = so + yot + a‹.1

2

3 = 1.5 + vA sin 30°: + .4, (-9 .81 )/ 2

t = 0.9334 s, vA = 12.4 m/s Ans.

F12-24. s = so + votR(ś) = O + 20(fit

s = so + vot + (,12

-R() = O + 20()/ + 1 (-9.81)12

= 5.10 sR = 76.5 m Ans.

(

1 )

( 2 )

A n s .

Ans.

Ans.

FUNDAMENTA L PRORLEM$ 683

F12-15. Xg = xA + (vA),./u

12 ft = 0 + (0.8660

vtitAB = 13.856 (1)

YB = YA. (VA)vrAs + 4 a,1,21,s

(8 - 3) ft = O + 0.5v,,i/AB + ł (-32.2 ft/s2)r yBB

1.35ffig Eq. (1).

5 = 0,5{11.856)- 16.11:42 B

tAB = 0.3461 s

1),1 = 40,0 ft/s Ans

1-12-29. = + 2a,{se - sĄ)(15 m/s)2 = (25 m/s)2 + 2a,(300 m - 0) a,

= -0.6667 m/s2

= 2),2k 2ar(4 - 5A)vg = (25 m/s)2 + 2(-0.6667 m/s2)(2,50 m - 0) =

17.08 m/s

(a8), = p t7; = 300 m (17.08 m/a12

- 0.9722 m/s2

as = V(as ) ; 1 ( a s ) ; ,

= V"(-0.6667 m/s2)2 + (0.9722 m =

3.18 m/s2s'Y

Ans.

Ans.

F12-26. yB = yt + +-150 m = O -F (90 m/s)rAR + 4 -9,81 m/s214, /

AB = 19.89 s

xs = (vA)..,(As

R -= O + 120 mis(19.89 s) = 2386.37 m

= 2.391cm Ans.

F12-27. a, = v = = (0.0625/2) = (0.125/) m/s21,,i0,

= 1.25 m/s2

(0.0625/2)2ra, - L97.656(10 %/1 m/s21,=.10,

P 40 m

= 0,9766 m/s2

F12-30. tantv = tdv = =

O =tan I (14 X)

.r=10= tan - ] (-) = 39.81' = 39.8° 7

[1 + (dy afx)213/2 [ I -F (; x)213/2P =

d2y/d,.2 I

= 26.468 ft

1/2 (20 ft/s)2a, - - - 15.11 ft/s2

p 26,468 ft

= V(01)2+ ta,,)2 = V(6 ft/s2)2 + (15.1116.3 ft/s2 Ans.

. T = Wit

F 1 2 - 3 1 . < 4 2 B ) , = - 0 . 0 0 1 s = ( - 0 . 0 0 1 ) ( 3 0 0 m ) ( ' S

r a d ) r r i / s 2 = - - 0 . 4 7 1 2 m / s 2 d v = a s

d s

LIR 150u m

)25 mis F12-28. = 2s „lo = 20 m/s = -0.401 s ds

yz (20 m/s)2

a„ = - P 50m - 8mjs- VB = 20.07 m/sa; (20.07 m/s12

dv i (as), - = = 1.343 m/s-al = = 4s1,- io = 40 mis2

613 P 300 m

a = )42 + a,2 = \/(40 m/s2)2 + (8 m/s2)2 oB = {o B}12 (o.s.)2

= 40.8 m Ans. = )/(-0.47 In/s2)2 + (L343 m/s2)2.

= 1.42 m/s2

-\0.-f a = \7(1.25 m/s2)2 -1- (0.9766 m/V)2

= 1.59 m/s2 Ans,

Ans.

684 PAĄTIpL SOLLITIONS AND ANSWERS

F12-32. ar ds = v dvAr = v - = (0.2s)(0.2) -=- (0.04s)

in/s. ar = 0.04(50 m) -= 2mfs2 v

= 0.2(50 m) = 10 m/s

v2 (10 m/s)2

a .

Ans.

--0.2 m/s2p 500 m

a = = \r/(2m1s2)2 + (0,2 rnis2)=

= 2.01 m/s2

a, = "?' rÓ2 = 2 ft/s2 - (-ł- ft )(3 rad/s)2 = -12.14 Et/s2

= + Ó= ft)(1 rad/s2) + 2(6 ft/s)(3 rad/s)

= 37.57 rt/s2

a = a,q2

= Vi(-12.14 flis2)2 + (37.57 ft/s2)2

= 39.5 ft/s2Ans.

F12-33. 71r = i =

vy rÓ = (400(4 ksF12-36.. r = e'

= eł'i1-H t,Z•P' = + e°Ó2

55 ft/s = V02 + 1(4(10Ó) ft/s12

Ó = 0.1375 rad/s Ans. a, = - = + eW) = 6.'14(4)

= 8.77 m/ s2 Ans.FU-34. r = 0.1131„15 s = 0.3375 m = rB + 2i-4 = (e°)+ (2(A)0) + 262)

= 0.3t211.,1.5„. = 0.675 m /s = e"/4(4 + 2(2)2)0,64- i 5, = 0,900 rn/s2

= 26.3 m/s2 Ans.= 4t3/21,-1,5, = 7.348 rad

Ó = 6t1/21,_ 5» = 7,34R rad /A F12-37. r = 10.2(1 + cos 0)1 M = 0.3732 In

= 3/-1/21i. = 2.449 rad/s2 = [ -0.2 (sin 01Ó] m/A16=30.

F12-38. 30 m = r sin O✓ .= )

(30 csc P) m

r = (30 c.s.c. 0)1,= 45. = 42.426 m-30 esc cm O Óło45.(42.4266) m/svr =r = -(42.426,1) m/s

vo = = (42.426O) m/s

2) = N/4›,2 + -H2

2 = 'V4-42.426(.3)2 + (42.426Ó)2

= 0.0333 radis

F12-39. l, = 3s, + 5A

4=3%+va

= 3 % + 3 m / s y r ,

= - l m / s = 1 m / s l '

/ . = = 0.675 m/s• = rfł = (0.3375 m)(7.348 rad/s) = 2.480 m/s

ur = rO2= (0.900 rn/A2) - (03375 m)(7,348 rad/s)2-= -17.325 m/s2

• = r6 + 2i-O = (0.3375 m)(2.449 rad/s2) + 2(0.

675 m/s)(7.348 rad/s) = 10.747 m/s2

= \/(0.675 m/ -F (2.480 m/s).2

= 2.57 m/s Ans.

= + ao2

-= V(-17,325 m j s2)2 + (10.747 m /s2)2

= 20.4 rn/s2 Ans.

F12-35. r = 20

2Ó

2O

M O = rr/4 rad.✓ - -2-()=.;irl = 2(3

rad/s) = ft/A = 2(1 razi/s) = 2 ft/s2

= -0.2 sin 30'(3 rad/s)

= -0.3 m/sv, = r= -0.3tnfs

• rÓ =- (0.3732 m)(3 rad/s) = L120 m/s• = = V(-0.3 m/s)2 + (1.120 mis)2

= i.16 rn/s Ans.

Ans.

Ans.


F12-43.

F12-44.

• + 2sA + 2h =vB + 271A = O6 -F 2vA = v4 = -3 mis = 3 mis I Ans.

3sA + sB = 1 3vA + v8 = O

3vA + 13 = O vA = -0.5 m/s = 0.5 m/s Ans.

ly -= 4 sA + sp

O = 4 vA

O 4 V A 3 m i svA = -0.75 m/s = 0/5 m/s t Ani

sa + 2(sa - + 4.sA - sp} = f4sA - sp = ł + 2a 4vA - = O 4v

A - (-4) =

4vA + 4 = O 1.5 = -1 m/s = 1 m/s ł Ans.

sC = I C E D

(8A - 8C) + (SB - sc) + S

B sA + 258 2sc = łACDF usvc + VB = O

VA 22.152.ve =- O

ElimmatingvA + 4v# = OThus,4 ft/s + 4vo = Ovy = -1 ft/s = I ftp.

F12-45. v - v + v A B/A

1001 = 80j + vałA vs/A = 1001 - 80j7B/A = \1 7 ( " B / A ) . ( V , W 4 ) , .

= V(100 km/h)2 + (-80 km/h)2

= 128 krri/h Ans.

O = tan-i [-(rj3141 Lin-1( gp 38.7°(vyiA), 100

km/h Ans.

v e v A v e / A(-400i - 692.82j) = (6501) + vR/AvBłA = r- losui - 692.82j] km/hv e m = l ' A v s / A ) , ( u 8 . , 4 )

-V(1050 krn /h)2 + (692.82 km/h)2

= 1258 km/h Ans.

tan_ tan )[ (vBiA), (692.82 km/h(VB/A), 1050 km/h - 33.4° Ans

V8 = 7 A + ven.(51 + 8.660j) = (12.991 + 7.5j) + vBiA vB

/A = 1.-7.9901 + I .160j1 m/s v87A = -

14-7.990 rnis)2 + (1.160 m/s)2 = 8.074

mis

dAB = = (8.074 m/5)(4 s) = 32.3 m Ans.

ve Tam

-20 cos 451 + 20 sal 45] = 651 + vA/B

vA/B = -79.141 + 14.141

vAjs = 1.7( - 79.14)2 + (14.14)2

= 80.4 km/h

14.4 = ae ame(20)2 (20)2

0.1 cos 451 + sili 451 -+- 12001 +0.1aAIB = 16281 +

2828j

aA/B 1/(1628)2 + (2828)2

= 3.26(103) km/112'Chapter 13

s = sc ± vol -I- 4 cc, t2

6m= O + -I- 4 a(3 s")2

= 1.333 in/s:-'

= ma,,; NA - 20(9.81)N eas 30° = ONA = 169.91 N

SP, = nia,; T - 20(9.81) N sM 30°- 03(169.91 N) =- (20 kg )(1.333 m/s21

T = 176 N Ans.

F13-2. (Ft). = pi,NA = 0.3(245.25 N) = 73.575 N.Sinee F = 100 N > (Fs),,, when r = O, the cratewill start to rnove immediately after F is applied.

= - 25(9.81) N =

NA = 245.25 N

nra„;

1012 + 100 - 0.25(245.25 N) = (25 kg)ci

a = (0.4r2 -I- 1.5475) rn/s2

dv = n cit

• 4 S '

dv = (0.4t2 + 1.5475 )dt

= 14.7 m/s. -› Ans.

F12-47.

F12-48.

(1)

4 CDF

( 2 )

A n s .

Ans.

aam

A n s .

686 PARTIAL SOLUTIONS AND ANSWERS

Ans.

F13-3. = ma„;( )500 N - (500s)N = (10

kg)a a = (40 - 50s) m/s 2

vdv = ads0.5 m

j o v y d v = i ( 4 0 - 5 0 s ) d s

4:145 = (40s - 25s2 )18."v = 5.24 mis

ma„ 100(s + 1) N = (2000 kg)a

a = (0.05(s + 1)) m/s2

dv = a dsjf 10m

V = 0.05(5 + 1) dsa

J

= 2.45 m/s

F13-5. Fbr = - 10) = (200 N/m)(0.5 m - 0.3 m)

F13-11. EF, = mar; 10(9.81) N cos 45° = (10 kg)a,a, = 6.94 m/s2 Ans.

= ma„;(3 m/s)2

T- 10(9.81) N sin 45° = (10 kg) m

T = 114N Ans

F13-12. EF„ = ma,;(15 m/5)2

Fr, = (500 kg) - 562.5 N200 m

SF, = ma,;

Fr = (500 kg)(1.5 m/s2) = 750 N

F = 1,7F"; + F.;2 =1%(562.5 N)2 + (750 N)2

= 938 N Ans.

F13-13. ur = r - rÓ2 = O - (1.5 m + (8 m )sin 45°)82 = (-

7.157 Ó2) m/s2

F13-4.

F13-45.

F13-7.

F13-8,

F13-9.

F13-10.

= 40N

= tan 1(E%) = 36-860

ma„;100 N - (40 N)cos 36.86° = (25 kg)a

a = 2.72 m/s2

Blocks A and 8:ma,; 6 = f.411 a; a = 2.76 ft/s2

Check if slipping occurs between A and B.

±,SF„ = ma„; 6 - F = (2.76);

F = 4,29 Ib c 0.4(20) = 8 Ib

a,4 = = 2.76 ft/s2

ł F„ = m-5; (0.3)09.81) = 2

12 = 2.43 m/s

+ 1 ł F„ = mam; m(32.2) = m( *3) 89,7 ft/s

150

Ans.

Ans.

Ans.

Ans.

EF= nuiz;T cos 45° - m(9.81) = m(0) T = 13.87 m

= mar.;

-(13.87m) sin 45° = m( - 7.157 Ó2)

= 1.17 rad/s Ans.

F13-14. O =- irt21,=039 -= (2r/4) rad

Ó = 27rtkos, = 7r rad/s

Ó = 27r rad/s2

r = 0.6 sin = 0.4243 m

= 0.6 (cos (i)Ó1,,-„./.4r,d-= 1.3329 m/s

F = 0.6 1(cos - (sin0)(j2110-ir/4 rad = -1.5216 m/s'-ar = r - -1.5216 m/s2 - (0.4243 m)(ir rad/s)2

= -5,7089 m/52

= rH + 2i- B = 0.4243 m(2n- rad/s2)

+ 2(1.3329 m/s)(71- rad/s)

= 11.0404 m/s2

ma,.;

Fcos 45° - N cos 45° -0.2(9.81)cos 45°

= 0.2( -5.7089)

+ F"„ = mar,; 150 + N = - ((120)232.2 400

Nr = 17.7 lb SF,

= ma„;N. sin 30° + 0.2k cos 30° = m -

500▪= 0;

cos 30° - 0.2N, sin 30° - m(32.2) = O

▪= 119 ft/s

Ans .

SFs = mas;F sili 45° + N sin 45° -0.2(9.81 )sin 45°

-= 0.2(11.0404)

N = 2.37 N F = 2.72 N Ans.

FUNDAMENTA L PRoBLEms 687

F13-15. r = 50e 10= 7116 rad = [50e2("'] m = 142.48 m

= 50( 2e2" Ó) = 100e2°

= 100e2("/6)(0.05) = 14.248 m/sF = 100 ( (2e"w(;)(> + c"n(410=1,/4rmi

= 100 [ 2£26} (0.0521 + e2t'/6)(0.01)1

= 4.274 M/S2

ar = r - = 4.274 m/s2 - 142.48 m(0.05 rad/s)2

= 3.918 m/s2

ao = r + 2"ih = 142.48 m(0,01 rad/s2)

+ 2(14.248 m/s)(0.05 rad/s)

= 2.850 m/s2

ł F, = magFr = (2000 kg)(3.918 m/s') = 7836.55 N

2.Fo = man;

Fo = (2000 kg)(2.850 m/82) = 5699.31 N

F =1•1%, Fo2

= 1/(7836.55 N)2 + (5699.31 N)2

= 9689.87 N = 9.69 kN

+ 300 N(10 tn) - 0.3(169.91 N) (10 m) - 20(9.81)N (10 m) sin 30° =4 (20 kg)v2

v = 12.3 m/s Ans.

F14-3. T, + = T215 m

0 + 2[ (600 + 2se) N dsl - 100(9.81) N(15 m)o

= 3 (mo kg)v2

= 12.5 m/s Ans.

F14-4. + 51/1-2 = T2( i 800 ke.)(125 m/s)'- [5a °(1° N 2.-f) °I)".1 (400 m)]

(1800 kg)v2

= 8.33 m/s Ans.

F14 -5. r , + ł U 1 = (10 kg)(5 m/s)2 + 100 Ns' + [10(9.81) NI s' sin 30°

-4 (200 N/m)(s')2 = O

s' = 2.09m

s = 0.6 m + 2.09 m = 2.69 m Ans.

F13-16. r =- (0.6 cos 20) m10-0, = [0.6 cos 2(0°)1 rn = 0,6 m F-14_6. T4 4" UA_B = Tg

i = (-1.2 sin20() Consider difference in cord length AC - BC,= [ sin2(0°)(-3)1 m/s = O which is distance F moves.

= -1.2( sin2ć6+ 2cos20Ó2) m/s18.o.O + 10 lb( V(3 f02 + (4 1'02 - 3 ft)= -21.6m/s2=slug)vi,

Thus,va = 16.0 ft/s Ans.a, = - r-Ó2 = -21.6 m/s2 - 0.6m(-3 rad/s)2

F14-7. ±. 1F, = ma,;= -27 m/s2

30(1) = 20a a = 1.2 m/s'-au = rii + 2i- H = 0.6 m(0) + 2(0)(-3 rad/s) = O

v = v a + a u rFo =- mao; F - 0.2(9.81) N = 0.2 kg(0)

= O + 1.2(4) = 4.8 m/sF= 1.96 N t Ans.

P= F • v= F (cos (ł)v= 3 t ) ( 5 } ( 4 . 8 ) C h a p t e r 1 4

= 115 W Ans.El 4-1. Ti - 2 -

O + l ś}(500 N)(0.5 m) - i (500 N/rn)(0.5 m)2 = (10 kg)tr2

• = 5 . 2 4 m / s A n s .

J= f o 5 m 0 . 5 s d s

F14-2. IF, = mar; NA - 20(9.81) N cos 30° = ONA = 169.91 N v = rn/s

• + P = F • v = [10(5)1(3.674) = 184 W Ans.

-= ma,:lOs = 20a a = 0.5s rn/s2 -->

vdv = ads

688PARTIAL SOLLITIONS ANO ANSWERS

F14-9. ( +I 0;T1 - 100 lb = OT1 = 100 lb(+1)E, =100 Ib + 10016 - = 0 T2 = 200 IbP,„,, = TE • VE = (20016)(3 ft/s) = 1.091 hp

P„,„ 1.091 hpP = = = 1.36 hp,ne 0.8

F14-10. 2 F = ma N - 20(9.81) cos 30°N = 169.91 N

F,. = ma,.;F - 20(9.81) sin 30° - 0.2(169.91) = OF= 132.08 NP = F - v =- 132.08(5) = 660W Ans.

F14-11. + fi E F,. = trak;T - 50(9.81)=- 50(0) T= 490.5N1),„„ = T • v -= 490.5(1.5) = 735.75 WAlso, for a point on the other cableP„„ = (490.5) (1 .5)(2) = 735.75 W

490 5P„,„ 735.75

0.8 - 920 WAns.Pip ".=-

F14-12. 2sĄ sp = 1

2aA + ap = O2LiA + 6 = OuA = -3 m/s2 = 3 m/s2 t5F, = ma,.; TA - 4903 N = (50 kg)(3 m/s2)

TA = 640.5 NP,,„, = T • v =- (640.5 N/2)(12) = 3843 W

Pe , 384 .80 3

P,,= 4803.75 W = 4.80 kW Ans.

F14.13. TA ± VA = TE -F VE

O + 2(9.81)(1.5) = (2)(v8)2 + O

v8 = 5.42 m/s((5,42)2'\+ t 5F,, = nm„: T - 2(9.81) = 2 k,. 1.5 i

T -= 58.9 N Ans.F14-14. TA + VA = VB

m ,41;■ + m ghA = -.1m8v3 + m gks{-1 (2 kg)( I m/s)'] + [2 (9.81) N(4 m)1

[ .!, (2 kg)v,r1 ] + [0]v8 = 8.915 m/s = 8.92 m/s

+ t 5.F„ = ma.; NE - 2(9.811N=-kg)((8.915 m/s)2

2 m No = 99.1 N

F14-15. T, + V1 = T2 + V24 (2)(442 + 4 (30)(2 - 1)2

= 3 (2)(02 - 2(9.8I)(1) + 4 (30) evrŚ -

1)2 = 5.26 m/s

F14-17. Ti ± Vi - T2 ± V2 IMV?

+ m gy + 3ks7

--- 4 + mgy: + 4[01 + [01 + [01 = [O] +

[-75 lb(5 ft + s)] + [ 2(4 (1000 lb/f0s2)

+ 4 (1500 lb/ft)(s - 0.25

ft121 s = sA = se = 0.580 fł

Also,

s8 = 0.5803 ft - 0.25 ft = 0.330 ft Ans.

F14-18. TA ± VA = TE VB

ni + (4 k ,S,?, ± m gy A )

+ ks;-; + m,gyR)

(4 kg)(2 In/s)2 + s (400 N/rn)(0. I m - 0.2 m)2 + O= (4 kg)v.f2 + 4 (400v(0.4+ (0.3 m)2

- 0.2 m12-1- [4(9.81) N](-(0.1 m + 0.3 m))

v8 = 1.962 m/s = 1.96 m/s Ans.

Chapter 15

(±..) m(v1)_, + 5i dr = tn(v.2),-ri

(0.5 kg)(25 m/s) cos 45° - J F„..(11

= (0.5 kg)(10 m/s)cos 30°

=F, dt = 4.509 N- sr r2(+ t ) m(v,), + ŹJF,. dl = m(02},,

d

- (0.5 kg)(25 m/s)sin 450+ 1 F. di

= (0.5 kg)(10 rn/s)sin 30°

=F,, dt = 11.339 N • s

= if F dt 11(4.509 N • s)2 + (11.339 N- s)2

= 12.2 N -s Ans.

Ans.

Ans.

= 20(0)(41_± -?k, (4) (1 - 0.5)2

vs = 16.0 ft/s

F14-16. TA + VA = TE VB

Ans. O + 4 (4)(2.5 - 0.5)2 + 5(2.5)

Ans.

Ans.

Ans.


F15-2. (+ t) m(y1), +F, d1 = rn(v2),rl

O + N(4 s)(14016)(4 s)sin- (150 łb)(4 s) =

N= lIXllbr2

±.)+F, dl = rr7(v2),ra

O+ (10016)(4 s)eos 30° - 0.2(100 lb)(4 s)

= (+A shig)/2

v= 57.2 ftjsAns.

F15-3. Time to start molion,

+! £F„ = 0; N - 25(9.81)N = O N = 245.25 N2F, = 0; 2012 - 0,3(245.25 N) = U t = I.918 s

r2( i.)+F", di = m(v2).

d

O + j20g - (0,25(245.25 N))(4 s - 1.918 s)1911h.

= (25 kg)v= 10.1 m/s

r2F15-4. () m(vF, fil = ni(v2:),

rl(1500 ke)(0) + Cl, (6000 N)(2 s) + (6000 N)(6 s -

= (1500 kg) v

= 20 m/s

F15-5. SUV and trailer,

Mv] +Fx dt = rn(112),ff r,

O (9000 N)(20 s) = (1500 k2. + 2500 kg)v▪= 45.0 mis

Trailer,

m ( v j ) , r n ( v , y , ri

O + T(20 S) = (1500 kg)(45.0 mis)

T= 3375 N= 3.375 kN

F15-6. Blask B:(+I ) mvi + i F dr = mv-)

O+ 8(5) - T(5) =(1)T = 7.95 lb

Biodr A:rny, + j Frit = rnvi

o + 7.95(S) f- r4(10)(5} =(1)= 0.789

F15-7. ( J.) tnA (vA)t + mn (v8)1 -= owĄ (vA)2 + nre (vs4-

2 {20{103) kg)(3 m/s) + (15(103) kg)( -1.5 m/s)

= (20(103) kg)(v4), + (15(10) kg)(2 m/s)

(vA)2 = 0.375 m/s -›Ans.

±) on(vB).[ ±r2r dr = ,n(vB)2

([5(103) kg)( -1.5 m/s) +(0.5 s)

= (15(103) kg)(2 m/s)

F„, = 105(103)N = 105 kNAns.

) mp 1-(v,) i j, -H md(v)i b =+ nie»,5[W(fili + O -= (5 +

1.52 = 1.6 m/sAns.

F15-9. T1 + Vi =+ V)

ma+ (V =+ (V5)2(5)(5).1 + 5(9.81)(l.5) = 1 (5)(vA)

i (vA)2 = 7.378 m/s

() m,A(v.4)2ms(vn)2 = (mA + m f)zi

5(7.378) ± O = (5 + 8)Y

= 2.84 m/sAns.

F15-/111 {) PnA (V,;) + mi3(vA = mA(VA)2 ms(va3)2+ 4 = 10(/‚,4 + 15(vB)2(1)

+ V, = T2 + V2

MA (?),07"!:111(12B)łVe)r4 1P2,4 (vA + rttB(PB)i(Y,)2

O -f O + .1[5(103)](0.22)

= .(10)(VA)g ± i (15)(v )g ± O5(v.a.)7.5 (v B)? = 100(2)

SolvMg Eqs. (1) and (2).

(v0)2 = 231 m/sAns.(vA)2 = - 3.464 mis = 3.46 misAns.

F15-11. ()tni3(vB, = ("4 + n}v,O + 10(15) = (15 + 10)1,2

-= 6 mis

T , + = +(rn A + m R)21 + (V,)2 = (mA + 8)2) + d)3

1(15 + 10)(62) +O = O + [10(

s = 0.3 m = 300 mmAns.

Ans.

2s)]

Ans.

Ans.

Ans.

Ans.

Ans.


F15-12. ( .-±,, ) n + o --- m,,(2,,), - mćv, 300),)3 + so(v,), = 775.95 (1)0 = (20 kg) (),,,), - (250 kg)v, ( --4L- ) e - (vs).3 - (v.4):,

(vp).,_ = 12.5 y, (1) (1),..i2 - (v8)2(va)3 - (v.4)3

vp = v, + vpir 0.6 = -25.87 1)/s - O

(up), i + (vp),,j = -v, .i + I-(400 m/s) cos 30ri (yE)3 - (VA )3 = 15.52 (2)

+ (400 m/s) słn 3011 Salving Eqs, (1) and (2), yields

(vp),-1 + (vp)vi = (346.41 - v,Iii + 200j (vB)3 = 11.3 ft/s 4-

(15)., = 346.41 - y, (Y4)3 = -4.23 tk/s = 4.23 ks --> Ans.(vr,),. -= 200 m/s• F15-16. (+ t ) mi(vb),] = mkyk,),1,(vp), = 320.75 m/s y, = 25.66 m/sYp = Ni(vp);-, + (yA [(N)al,. = r(44)11,. = (20 m/s) sin30° = 10 Triis t

= -\4320.75 m /ł s)2 + (200 mis)2 ( -,. )= 378 /s Ans.

e - (1-1){"°}:)i-10-1 - -7vb i'') :2{ vi: h}21..,

(v8)2 0) 0.75 -4)2 (20 misjom 30°- O

(vA)1 - (1's)3 (v'b)-21., = - 12.99 m/s = 12.99 m/s ‹-(9 m/s) - (1 m/s)

( v , ) 2 = = V V ( T ( 1 v 2 b ) 9 2 9 ) ` m + / s [ ) ( 9 4 ) .- U

( 8 m / s ) - ( - 2 T M + (10 m/s)2F1.5-14. ( '›F". ) mA(vA)1 + rnn(vs) = mA(Y4), -I- m34(vB)2 = 16.4 m/s Ans.

[15(103) kg1(5 m/s) + [25(103)1( -7 m/s)

= [15(103)1(81(v4)2 + 1-25(103)»t3)2 k.[(vb)211, i \., 12.99 m/s,"

15(y4), + 25(y5)2 = -100 ( t) = 37_6° Ans.

Us'ing ihe coefficient of restination equation. F15-17. .r-n(v,), -= Xm(//,),(vE)2 - (vA)2

(VA)1 - 04)] ° + ° = 3÷2 (1) + 3.-.2 (V 13k)2( żo ) e -O )8_,-)2 - 0.1818 ft/s

(77B)2 - (v.4).2 0.6 - m(v,), = ,,n(v,..>25 m/s - (-7 mis)(N), - (v4)2 = 72 (2) T (3) - O = D + *› (vB,)>

Solving, - 0.545 ft/s(08,)2 -(up):, = 0.2 m/s-s Ans. (y8"),2 = .175

(-0.18181' + (0.545)1

(v4)2 = -7 m/s = 7 m/s <- Ans. = 0.575 ft/s Ans+7 (3) (fi - 7-5 (4)(fi

=3±, (v8)2, + 33.2 (vA )1,

+7 0.5 = :(vA)2, - 'vs)2,]/[(3)(ś) (-4)ffil

Solving,(vi )z, = 0.550 ft/s, (us) , = -1.95 ft/sDisc A,

. 2 ( 4 ) « ) = ( V A ) . ‚ y

- (34, )1(v.)3 (Q7, siog )(v(?) ( = -2.4-0 ft/s

F 1 S - 1 3 . 1 L ) e -

F15-15. T, +V1= +V,

4 m(vA)i rn,g(iin)3 = 4 m(vA); + m8(11A.}2

slug)(.5 ft/s12 + (30 18)(1.0ft)= (;;; slug )(v4) + O =

25.87 ftis •E-

( ) m4(v4)2 (118(vs.)2 = tn.4(vA)3 ms(v8)3s1ug)(25.87 ft/s) + O

F15 - 18 .

F UNDAMENTAL P ROBLEMS691

D i s c $ .

-(3) (4) = +...(vs)av,

(vp)2,. = -2,40 ft/s

(vA)2 = V(0.550)2 + (2.40)2 = 2.46 ft/s An.s

018)2 = - -V(1.95 + (2.40)2 = 3.09 ft/sAns.

F15-19. 110 = Snn2;

f l o = [2(10)(« (4) - 2(10)( i )1(3)

= 28 kg • ni2/5

F15-20. H pri?Vd;H = (2(15} sil.] 301(2) - [2(15) cos 301(

5) = -99.9 kg • m2 /s = 99.9 kg • m2/s ) F15-21. (IQ + J M. = (1-1)25(2)(1.5) 5(1,5)(3) = 5v(1.5)• = 5 m/ s Ans .

F15-22. (// z) i + i M , d r = (1 -1 , ) 2 4 . 5

O + 1 (100(1)(15)(it = 5v(1.5)

v = 12.8 m/s Ans.

FIS-23. (11,)1 + S i H. dr = (11,)2

55+ 0.9t2 di = 2v(0,6)

o• = 31.2 misAns.

H5-24. (H,)] +M, dr = (N:}2

14b+ if &rh + 2(10)(0.5)(4) = 21-10v(0.5)1

(}v = 1O.4 m/sAns.

Chapter 16F16-1. 8 = (20 rev)(22.`1) - 407r rad

•+ 2-ce, (0 - do)

(30 rad/s)2 = 02 + 2.(Ar ł(407r rad) - Oja, = 3.581 rad/s2 = 3.58 rad/s2 Ans.

rr = Lro + CYci

30 rudis = O + (3.581 rad/s2)(

ż = 8.38 sAns.

F16-2. 2<n.005o) ---- (mo)== (0.005 02)(0.010) = 50(10-603 rad/s2

When 0 = 20 rev(277 mel/ 1 rev ) = 40u rad.

a = 50( 10-9-(407r)31 rad/s2

= 99,22 racl/s2 = 99.2 rad/s2Ans.

F16-3. w = 401(2

150 rad/s = 4 O] /2

0 = 1406.25 rad

= -5/./

f do =i .1 401(2 rk) = f)'/'-' lrait = 4 -

= 4 04136.25)1P- - = 18.25s Ans.

F16-4. w = = .5 r 2 + 15 ) r ad / s

= = ( 3 0 r a d / s

ri = FI .5(32) + 15] rad/s = 28.5 rad/s

ce = 3(3) rad/s2 = 9 rad/s2

v = wr = (28.5 rad/s}(0.75 ft) = 21.4 ft/s Ans.

re = cer = (9 radfs2 )(0.75 ft) = 6.75 ft/s2Ans.

F16-5. u) dr9 = a dOr

w dw =ELS() d0.12 mdi:,ur0.250-▪ lo

---- (0.5 + 4)1'2 mdfs

When = 2 rev = 41r rad,

-= 10.5(4 D.).2 + 4] U2 rad/s = 9.108 rad/s

vp = rur = (9.108 rad /s)(0.2 tn) = 1.82 m/s Ans.

(a F), = ar = (0..58 rad/s2 )(0.2 rn)16=4,,,d

= 1.257111/52

(Op), = w2r = (9,108 rad/s)2(0.2 m) = 16.59 m/s ap

= \1(ap1,2 + (typ)N2

= V(1.257 rufs2)2 + (16.59 m/ 52)2

= 16.6 rn/s2 Ans.

692 PARTIAL SOLLITIONS AND ANSWERS

F16-6. a s

COB

W8

=

(rA) ocA -1.8

= (4.5 racl/s2)(:, = 1.5 rad/s2

(w8)0 + aBt= O + (1.5 rad/s2)(3 s) = 43 rad/s

F16-9. Vector AnalysisvB = vA + «o X rs/A(4 ft/s}i = (-2 ft/s)i + (-wk) x (3 ft)j41 = (-2 + 3w)1

= 2 rad/s Ans .

OB = (yB)0 (COB)01 47(B12

O B = O + G + 3 . ( 1 . 5 r a d / s 2 ) ( 3 s ) ' d R = 6 . 7 5 r a d

vc = wBrD = (4.5 rad/s)(0.125 m)

= 0.5625 m/s Ans.sc = OBrD = (6.75 rad)(0.125 = 0.84375 m

= 844 mm Ans.

F16-7. Vector Analysis

V B V A - f - W X r B / A

-vB j = (3i) m/s

+ (wk) >c (-L5 cas 301 + 1.5 sin 30°j)

-vB j = - wAR (1.5 sin 30°)Ił - w(1.5 cos 30°)jO = 3 - w(1.5 sin 30°)(1)

-v = O - w(1.5 cos 30°)(2)w = 4 rad/svB = 5.20m/sAns.Scalar SolutionV8 = VA ÷ V BiA

LI•VB] = [ 2>1+ [W(1.5) A30°] This

yields Eqs. (1) and (2).


VB V A + W X rBIA

(vB)x1 + (vB),.j = O + (-10k) x (-0.61 + 0.6j)(vB + (vB)sj = 61 + 6j(vR)s = 6 m/s and (vB),, = 6 m/s

vR = \/(1,0). + (vB)

= \/(6 m/s)2 + (6 m/s)

= 8.49 m/s Ans.

Scalar Solution

VB =

[ [(vB),, 1 = [0] [450 10 ( 116 )1cos 45°

ii = 0 + 10 (0.6/cos 45°) cos 45° = 6 rn/s -›

+1` (v8), = 0 + 10(0.6/cos 45°) słn 45° = 6 misi'

Scalar Solution118 = VA + vR/AL

[ 1 = [1.2)14 = -2 + Q.,(3);w = 2 radis

F16-10. Vector AnalysisV A = W 0 A X C A

= (12 rad/s)k X (0.3 tnij

= 1-3.611 m/s

vB. = VA + ca,AB X

vft j = (-3.6 m/s)i+ (wAR k) X (0.6 cos 30°1 - 0.6 sin 301) ni

vB j = /WAB (0.6 sin 30°) - 3.61i + wAB (0.6 cos 30°)jO = WAB (0.6 sin 30°) - 3.6

vB = wAB (0.6 cos 30°)

wAB = 12 rad/s vB = 6.24 m/s TScalar Solution

VB = VA VB/A[va t = [124--(0.3)] + EV'30'w(0.6)]

This yields Eqs. (1) and (2).


v c = V B ( d B C X k / 8

VC i = ( - 601) ft/s+ (-L'sck) x (-2.5 cos 301 + 2.5 sin 301) ftvej = (-60)1 + 2.165wsej + 1.25~1

O = -60 + 1.25wBc (1)

vC = 2.165 wBc (2)

wBc = 48 rad/s Ans.

vc = 104 ft/s

Scalar Solution

v c = VB Vc/B

Evct = [ + [V30° w (2.5)]


FUNDAMENTAL PRORLEMS693

F16-12. Vec tor Arialysis vB -= VA± w X rgiA-'1y8 ms 30' i 4 y), sin 30'j = (-3 m/s)j + (-

03k) X (-2 Sin 45°1 - 2 i 451) m -0.

8641)1,8 i + 0.5v8 j

= -1.4142w1 + (1.4142w - 3)j-0.8660va = -1.4142w 0.

5v8 = 1.4142w - 3 vr = 5.

02 ladis .z;B = 8.20 mis

Scal ar Sol uton

vB. + vB/A[5,30° vB1 [ 3] -1- 15:45°03(2)1


F16-13. f.1)AR = vĄ = = 2 rad/s Ansr..Vic 3

= tart - ' ( -) = 53.13'

1/(3)z- + (2.5)z - 2(3)(2.5) cos 53.13' -=- 2.5 m

vc = EdAa re/re = 2(23) = 5 m/s

= 90° - = 90" - 53.13 ' = 36.9°

F16-14. vB = WAS = 12(0.6) = 7.2 m/s

vc = O7.2wBc - = - = 6 rac1/s

r vB B,,,K. 1.2

F16-15. w = = - = 20 rad/srVoojw 0.3

rAile = \ (1.32 + 0.62 = 0.6708 m

= lary t ) 26.57°

wr,vm, = 20(0.6708) = 13.4 m/s

= 90' - q5 = 90' - 26.57' = 63.4°

F16-16. Tle Ineation of /C ean kie determined using sitni lar t riangles.0.5 - refie

3v c

ID =

Also, r0"-- = 11.3 - rcim- = 0,3 - 0.1667 = 0.1333 m.

vo = wropc. = 9(0.1333) = 1.20 m/s Ans.

Ans.

Ans.

F16-19. w = 6 - - 2 rad/sr„4/B-.. 3

Vector Analysis38 = a, + a x ra/A (.02 gm

u p i = - 5j + (ak) X {31 - 4 j ) - 2 2 (31 -

4 j ) a a i=i4a-12 )1+(3a+11) j

= 4or - 12 (1)

O = 31-.r + 11 (2)

ry = -167 rad/s2 Ans,

a8 = -26.7 m/sz Ans.

Scalar Solution

• = a A a k / A

[as' -=[ J 51+Eu (3) ;5.31 [4.2 (2)2(5)]


A n s A n s .

A n .

s

A n s .

Ans.

A n s . A n s .

- 9 rad/sre/pc 0.1667

r c / i c1.51,5

rcirc. = 0.1667 m

Ans

F16-17. 115 wr-RiA =- 6(0.2) = 1.2 m/sr 1‹qc = 0.8 tan 60° = 1.3856

m ✓<- - ej,;)śc. = 1.6

<4.8c = = - 0.8660 rad/srBiir 1.3856

= 0.866 rad/s Ans.Then,

✓c -= conc. rcy[c. = 0.8660(1.6) = 1.39 m/s Ans.

(1 )

(2 ) A n s .

F16-18. vB = wAS (.13m = 10(0.2) = 2 m/s

vc wc» t-c/n = &.'a) (0.2) -s

rme = = 0.4619 rri

reitc = 0.4 tan 30.° = 0.2309 m yx

= - = 4.330 rad/sTir"- 0.4619

= 4.33 radis

vc coBc rcpcwc,) (0.2) = 4.330(0.2309)

wca = 5 radis

6 9 4 PARTIAL SOLUTIONS AND ANSWERS

F16-20, Vector Analysis Scalar Ana1ysis

a, = a(, + a X rAło - ak2 r Ałe,JA - ac JLA/C= [0.71 + [(cr,,1 [a405)1

+ [(932(0.5)1L

1.5 -= -0.75 + (0,5 ) i

a = 4.5 radfs2

= 13.61 - 43.2j m,./s2Ans.

Scalar Analysis

[(a.41[( A), tj - [00.31[(6)(031

= 1.81 + (-6k)(0.3j) - 122 (0.3j)[ 1.

5

arL = aoaAło

4(12)2(0.3)1(aA) = 1.8 + 1.8 = 3.6 misi -> (

aA )4 = -43.2 m/s=

F16-21. Usingvo = rur, 6 = w (03)

w = 20 rad/s

ao = ar; 3 = a( 0.3)

-= 10 rad/s2 Ans.

Vector Analysis

aA = ao + rr X ryfa -

= 31 4- (-10k) X (-0.61) - 202(-0.63)

= (2431 + 6j I. rn/s2 Ans.

Scalar Analys4s

aA + aAjo

[(aA),] [(aA),1 [ 3 1 + [10(06)1 + [(20)2(0.6)]

-› j -.r j 1 j

(a,,), = 3 + 240 = 243 m/s2

+1 (aA), = 10(0.6) = Óm/s2 fi

F16-23. v8 = wrBi. A = 12(0.3) = 3.6 m/svE 3.6

ĆJję = - =s rad/ srfqic 1.2

Vector Analysis

rt,p, --= (-6k) x (0.3i) - 122(0.31)

= 1-43.21 - 1.8j1 mi,s2

ac = aR + .a.,qc X rem. - wkrci6

ac i = (-43.21 - 1.8j)

(LYBc k) X (1.21) - 32(1.21)

.ac. i = -54i + (1.2%c - 1.81j

ac = -54 inis2 = 54 rri js1 Ans.

fl = 1.2aBc - 1.8 agę = 1.5 rad/s2 A ilS

Scalar Analysis

ac = aE + acm

[6(0i3)1+ [(12)2(0.3)1 [aEc(1.2)] [(3)2(1.2)1. 1 .

ac = 43.2 + 10,8 = 54 Ints1

I u = -6(0.3) + 1.2aik-

ase = 1.5 rad/52

F16-24. v8 = w rBłĄ = 6(0.2) = 12m/s -> rBpc

= 0.8 tan 60° = 1.3856 m

vE 1.2c d ,

r87R-.3856 - 0.8660 rad/s Vcctor

Analysis

an = a X 1"BiA w2rgfA

= (-3k) x (0.2j) - 62(0,2)

= - m/s

a c a . 6 % C X r C i ncos 34°i + ac $in 31rj= (0.6i - 7.2j) + (asc k x 0.8i) - 0.86602(0.8i)

F16-22.rAiic 0.5 FA//c

rAfic = 0.3333 m3

= - 9 rad/s

ri‚A Aiic 0.3333

Vector Analysisa,, = ac + tY x - rAic

1.51 - (a4)„ j = -0.751 4- (ac)„ j

+ (-ak) X 0.5j - 92 (0,5j)

1.51 - (aA), j = - 075}1 + r(ac)„ - 40.5 b

1.5 = 0.5a - 0.75

a = 4.5 n:14/N2 Ans.

[ :c

FUNDAMENTAL PROBLEMS 695

0.8ó60ar i + 0,5ac j = (0.8%c - 7.2)j F17-4. FA A = 0.2NA FB = IL,NB = 0,2NB

0.8660ar = O (1) + k-" = m(ao.)..;

0.5ac = 0.8.01Bc - 7.2 (2) 02N.t + 0.2N8 = 100a

ac -= O %ric = 9 radis2 Ans tt2F,, = m(ac);&dar Analysis NA + NB - 100(9.81) = O

ac = k, + Et Cm C +.11.1c = 0;

[..,:3cool _ [ I 3(0.2)] + [(6.20).2)] + [,,,,f<0..8)1_,.[ I J 0.2N4(0.75) -F NA(0.9) + 0.2NB(0.75)

- NB(0.6) = O+ [(0.866012(0.8)] Soilving Eqs. (1), (2), and (3),

This yiclds Eqs. (1) and (2), NA = 294.3 N = 294 N686.7 N = 687 N

= 1.96 m/s2 Ans.

(2 )

(3 )

Chapter 17F17-1. ±) X F, = Mack; 100() -= 100a

a = 0.8 m/s2 Ans

+15F, = Mac.),;

NA + NB - 100(3) - 100(9.81) = (1)

c + 2,mQ = 0;

NA(0.6) + I00(fi(0.7)

- NB(0.4) - 100(Ś)(0.7] = 0(2)NA = 430.4 N = 430 N AnsNB = 610.6 N = 611 N Ans

F17-2. = m(ac)z; 80(9.81) sin 15° = 80a

a = 2.54 m/s/ Ans.• = m(aa›,..;

NA + Na - 813(9,81) cos 15° = O (1)+2Ma = 0;

N4(0.5) - NB(0.5) = (2)

NA = NB = 379 N Ans.

F17-3. +2.MA = 2(M.óA; 10M(7) = 3 03-5)

• = 19.3 ft/s2 An.s

= m(ac),; A, + 10( -fl = : ..2(19.32)

A, = 6 lb Ans.

+ ł SF„ = ni(aG),; A, - 20 + )0(5) =

Ay = 12 lb Ans

%nee Na is positivc, the labie will indccd słidc before it tips,

F17-5. (ac), = ar -= a<1.5 m)

(ac), = cu2r = (5 rad/s)2(1.5 m) = 37.5 m/s2

= m(ac), 100 N = 50 kgra(1.5 m))

= 1.33 rad/s2 Ans.EF, = m(aG)A, TAB + TcD - 50(9.811 N

= 50 kg(37.5 m/52)

7.4s Tc,r) - 2365.5

C -F ZMG = O; Tcp ( 1 m) - TAB ( I m) = O

TAB = Tcp = 1182.75 N = 1.18 kN Ans

F17-6. C +łAfc = O;alo -= =ayD,.(0.6) - 450 = O U,. = 750 N Ans

(ac), = cu2r = 62(0.6) = 21.6 m/s2 (ac), = ar = a(0.

61

+ = m(uc)h

750 - 50(9.81) = 50fa(0.6)1cr = 8.65 rad/s2 Ans.

21r;. = intac,),";

FAS + D, = 50(21.6) (1)

C -FZNIG = 0;

D,(0.4) 750(0.1) - FAB(0.4) = O (2)

D, = 446.25 N =-- 446 N Ans.

FAB = 633.75 N = 634 N Ans,

696 PARTiAL SOLLITIONS AND ANSWERS

F17-7. /0 = mk = 100(0.52) = 25 kg- m2 F17-11. 10 = fimi= = (15 kg)(0.9 rn)= = 1.0125 kg - m=

C +2.M0 = fou; -100(0.6) = -25a (a0), cogra = O

Ci = 2.4 rud/s2 (a£;), = ot(0.15 m)

= wty a,,t = /G + tnd-L->-)G

w = O + 2A(3) = 7.2 rad/s Ans. = 1,0125 kg + 15 kg(0,15 m)2

= 1.35 kg - m=

1F1/-8. ,ro = mr2 = (50) (0.32) = 125 kg • M2 C +5M0 = forY;

+ X.Lid„ = 10a; [15(9.81) N1(0.15 = (1.35 kg- m2-;n a = 16.35 raelfs2 Ans.

-91 = -2.25o a = (4t) rad/s2

+ = m(aG), ; - Oś + 15(9.81)Nd w - d t = (15 kg)E16.35 rad/s-2(0.15 rn)1r r i c r ) = f r4r dr

o O, = 110.36 N = 110N Ans.= infao),,; = O Ans.

w = (2/2) tsclis

w = 2(42) = 32 rad/s Ans. F17-12. (aG), = arG = ot(0.45)

(au,)„ = w=r-G = 62(0,45) = 16.2 mis=

F17-9. ((to), = urG = (0.15) ir0 = 4 m(' = 4 00)(0.42) = 8.1 kg • m2

(aG), = o2rG = 62(0,15) = 5,4 m/s2 C +5M0 = luu;16, = 1G + mdi = (30)(092) + 30(0.152)

= 2.7 kg • m2

C +2.A .10 lua; 60 - 30(9.81)(0.15) = 2.7a

a = 5,872 rad/s2 5.87 rad/6-2 AI1.5.

= rn(aG),; O, = 30(5.4) = 162 N Ans.

300(1)(1E1.6) - 30(9,81)(0.45) 88.1a

a = 1.428 rad/s2 = 1.43 rad/s2 Ans.2F„ = ro(oo)„; O, A- 30Q(i) = 30(16.2)

0,, = 306 N Ans.+ "SF, = rn(aG),-, O, + 300(i ) - 30(9.81)

+ S F, = m(aa)r; = 30[1.428(0.45)1

O, - 30(9.81) = 30[5.872(0.15)1 O, = 73.58 N = 73.6 N Ans.

O, = 320.725 N = 321 N Ans.F17-13. iG = +,ndi = 1,(60)(32) = 45 kg - m2

F17-10. (aG), = arG = 4(0.3)

(ao), = w2ro = 102(0.3) = 30 m/s2

= fn(aG)v:.80 - 20 = 6,0aG aG = 1m/s-1

= /G + mc/2 = (30)(032) + 30(0.32) C +2Mc = .1G0; 80(1) + 20(0/5) = 45a

= 4.05 kg• m2 oś = 2.11 rad/s2 Ans.

C -FIX0 -= ou;F17-14. = (Mk)A.:

50(fi(0.3) 4- 50(1)(0.3) = 4.05a-200(0.3) = -100aG((13) - 4,5a

= 5.185 rad/s2 = 5.19 rad/s2 Ans.30aG + 4.5a -= 60 (1)

M.aG)a; aG = ar = 4(0.3) (2)O, 50( fi - 30(9.81) = 30(30) a = 4,44 radfs.2 aG = 1.33 m/s= Ans.O. = 1164.3 N = 1.16kN Ans.

£f"/ = m(aa)r; F17-15. r,. = tv(aG"),.;+ 50(1) = 30[5.185(0.3)) N - 20(9.81) -=- O N = 196.2 N

O, = 6.67 N Ans. .±› = m(aG),; 0.5(196.2) = 20a0

FUNDAMENTA L PRORLEM$ 697

ac = 4.905 tnis2-,-

+5M0 łow,0.5(196.2)(0.4) - 100 = -1.8a

•= 33.8 radjs'l

Ans Chapter 18F18-1. fo == 80{0.42} = 12.8 kg - rn2

T, = O

Ans T, = 410f42( 2.8)w2 = 6.4W2

F17-16. Sphere tG = ś [241{0.15)2 = 0.18 kg •rn2s =-= 20(277)(0.6) = 24-frC..±5.41"ic = ("),c;T1 -1- W] =

20(9.81)sin.30°(0,15) = 0.18a + (20ao)(0.15)O + 50(244r) T 6.4u)2

0,18a + 3ao = 14.715rr = 24.3 nidisao= a(0.15)•= 23.36 rad /s2 = 23,4 radis? Ans.

F18-2. T1 = O•3.504 m/s'- = 3,50m/s2 Ans.

= 4 m(vc)?, + 41/0.;F17-17. + 2E3. = orr(ac).;

fV" --200(9.81) = O N - 1962 N= 4(+% slug )(2.5w2.)2SF.1. =

T.› =+6,.317[ 107.204,7 sjug)(5 n):1w3

T - 0.2(1962) = 200ao Or,+231,4 = (M.04; 450 - 0.2(1962)(1)

/0 = mi2 = (ffi sług)(5 fl)2= 18a + 200ao(0.4) =- 12.9400 slug • ft2

aG = a,4 + tr X reriA - (.02rGjaaGi = + ak X (-0.4j) - w2-(-0.4j)

0.4ai + (0.4w2 - a.k)j• = 0 . 4 a

Solving BO. (1), (2), and (3),

a = 1.15 rad/s-2 ac = 0.461 mfs2 T =

485 NF17-18.£F,= m(ac)„; O = 12(uG)., (ac), = O=

-12(9.81)(0.3) = 12(aG),(0.3) -I 2)(0,6)2a

0.36a - 3.6(ao),.. = 35.316

w -= 0

at; = RA + dx X rom -

(a.G),.J = aAl + (-ak) X (0.31) - O(aG), = (aA)i - 0.3j

a,,h0Ans,

(no), = -0.3a(2)

Solvk•ig .(łs., (1) ariel (2)

rr = 24.5 rad/s=

(ac), = -7.36 infs.:' = 7.36 m/s2,1 Ans

(g4)t a,4(cW,,

F18-3.(1)

Ans_

A n s .

So that

= 4 -4 (12.9400 slug • ft)w;

6.470NA

T1+ U1-2 =T2

T, -f [-Wyte + MOT = T2O -I - [-(501h)-(2.5 ft) + (100

Ib • 0)(1;) = 6.4700w-22 w2

= 2.23 rad/s(1){;)2 = W2rGi1C = (•,'2(2 .5)

= [71!2" (50)(52) = 104.17 kg -m2

T,=4

• = 4 +

= 4 (50) [ W2(2.5) ] 2 + 1(104.17)3 = 208.33cA

Up = Pap = 600(3) = 1800 ł

Uw = -W6 = -50(9.811(2.5 - 2) = -245.25 J

T, + X1.11_2 =O + 1800 + (-245.25) = 208.33wZ

= 2.732 rad/s u 2.73 radis Ans.

698 PARTiAL SOLUTIONS AND ANSWERs

F18--4. T = 4 nY72,, -H 4 low2 T2 = 4 ,n(vG).-' + 4- (50 kg)(0.4w)2 + 4 [50 kg(0.3 rn)2]w2 = 4 (30)[,2(0.3)]2 + 3 (1.35). =

= 6.25o? J rrVg) = Wyj = O

Or. Vi2 = - Wy2 = -30(9.81){0.3} = -88.291T =- 411c

2

+ V i -= T, + V,= 3 50 kg(0.3 m)-.2 + 50 kg(0.4 m)2](02

O + O = 2.025W3 + (-88.29)= 6.25 w13

sU 6.603 radis 6.60 rad/s= Or = 10(2/r rad)(0.4 = Sir m w2 =

T, + = T. F/8-8. y0 trirojżew(0.2)+ P cos 30° su -= T,

lo = m1420 = 50( 0.32) = 4.5 kg • m2

+ (50 N.Icos 30°(87r = 6.25(923= O

w = 13.2 radis Ans.T, = 4 m(2),A + 4 /Gw;

FI8-5. = rni2 = (30)(32) =- 22.5 kg • m2(50) [.,(o.2)] ± (4.5)(;i

T, = O= 3.25w3

T2 = mt.,"- -i- i,co2

=- (30)kii(0.5)12 + 4 (22.5).2 = 15,2 (v,), = wy, =(Vg)2 WY2 = -50(9-81» sin 306)

= -1471.51to = /G + nu:12 = lil(30)(32) + 30(0.52)

= 30 kg m2 + v, - T2 + V,T, = i iow

2 = 2 (30)w2 = 15w2 O -I- O = 3.25w?5. + (-14715)

s, = r i = 877(0.5) = m -= 21.28 rad/s 21.3 rad/s

s, = = 87r(1.5) 12Tr rnUpi = Pis, -= 30(4/r) = 1207r J F18 9. v0 = wro = w11.5)

/G =- + =- 2(60)(32)45 kg • m2= P252 = 20(127r) = 24077-J= MB = 20[4(2/r)] = 1607r J T = O

T, + = T, T2 = rn(vo)j /‹.p.Ao + 1207r + 2407r 1607r = 156)2 = 4 (60)[w,(1.5)12 + 4 (45)w3:

= 10444 radis = 10.4 radis Ans. =

F18-6. Up •.= rar =- w(0_4)

T2 = ioett.3 = .41 [ 45 + (3.0 ( 1.52 ) == = 20(0.32) = 1.8 kg-m2

T i = 0 (V g)IW- Y ł°T, =+(14)2= - Wv2 = -60(9.81)(1.5 sin 4.5n)

-624.30 J= (20[id(0.4)] 2 (1.8)(.4/2

.910.2 (14)1 = =UM = MH = M = 50 (k) = 25410) (V,), = ks;-' = -1

2(150)(3 sin45°).2 =- 337.5 J

+T2 T, + Vi = T2 + V2

O + 2500 = 2.5w2 O + O = [ -624.30 -F 337.5]

w = 31.62 radis = 31.6 radjs Ans. = 1.783 radis -= I.79 rad/s

F18-7. v =tyr = ‹.0(0.3) F18-10. V G- = wrG (4(0.75)

mr2 = 1(30}1;0.3'} = 1.35 kg • m2 fG = (30)(1.52) = 5.625 kg • m=

T, = O T, O

Ans.

Ans.

Ans.

-= i m(%),2 + 3/Gr;

4 (301.0.75)12 + 4 (5.62.5)wZ = I I .25w.Or,T2 =1oG.#3 =[5.1525 + 30(13.752)1N2

=.25w"-';'

() i = Wy i flVz)2 = -Wy2 = -30(9.81)(0_75)

= -220.725 J

(V,)i =ks =0= 1601


Cha pter 1 919-1. +/owi + i 10(1)2

F Mo dt =

0 + i 312 = [60(0.3)2]w2

• o= 11.$5 rAdis = 11.9 rad/

F19-2. C +(HA), + ł f2MAdt = (HA)2

O + 300(6) = 300 (0.42)«h + 3001«,(0.6)](0.6) arą = 11.54 radis = 11.5 radis 12

m(v1), + J F,dt = m(v2),

699

Ans.

Ans.cv,4, = {$0)(1/22 + 1,52 - 0.5)22

Tł += T, +

+ O -= 11.25J, -1- ( -220.725 + 160)

= 2.323 rad/s = 2.32 rad/sAns

FIS-H. = «krom, = «?!(0.75).FG = (30)(1.52) = 5.625 kg - rn2

T, = O

Ti = 4 in(9)6);, +

{30)rw,(0.75)12 + 3(5.625)w?11.25w?-,( V g) = Wyj = 30(9.81)(0.75 sin 45°) = 156.08 J(-Wy,2 = O

( V, ) = 4,./s.ss-1 = O

(V,), = 1 ksZ = 1(300)(1.5 - 1.5 cos 45°)2

= 28.95 Jr, -E- = T2 ł V,

+ (156.{78 + G) = I 1.254 -I- (0 + 28.95)

= 3.362 rad/s = 3.36 rad/s Ans.

F18-12. (l/s)i = Wyj = -[20(9.81m) = -196.2J

(Vg)2 '= O

=-- 4(100 Nin-4)( V(3 m)2 + (2 m)2 - 0.

5 m ics'f

= 482.22 J

(V,)2 = lks2 = (100 N/n-0(1 m - 0.5 n-02= 12.

5J T1 = O

T2. = 1lpw2 = 2 [1 (20 kg)(2 m)21(..1213.3333w2

7'1 +=+ 1/2

O + 1.-196.2 J + 482.22 J1

13.3333,rd:)2 + j0 + 123

= 4.53 radisAns

+ Ff(6) = 300111.54(0.6)1

Ff = 346 N Ans.

&J.Ar.ijrc i-d'A(19,15.1C+ZAl„ 0; 9 - 40.45) = O A, = 20N +(f-fc.), +

j SLfr df = (11C)2

O + [20(5)1(0.15) 10[(

trA(0.15)}(0.15} + :

10(0.12NA

wA = 46.2 radjs Ans.

F19-4. 1A = łpdc2A = 10(0.081 = 0.064 kg • m2

/B == 50(0.152) = 1.125 kg • rn2

«) .4 = M«.,8 = (4,3 = 2.«"B0.10_2

C-i- 1.4(04) + MAdr = 1,1(wA):

Equating Eqs, (1) and (2).500 - I .28(ws),5,625(w,),

{um}, = 72.41 radis =- 72.4 radisAns.

F19-3.

.5sO -I- 10{5} -F(0.1)dr = 0.06412(wE)21

L Fdt = 500 - 1.28(wi4)2SSo

-1-4Q.013)1 łZr MB dt = ift(W,02

5 NO + j F(0.2)c1r = I .125(wB)2

a3s

Fdt = 5.625(438)2

( 1 )

( 2 )


o + [(150 N)(0.2 + 0.3) m)(3 s)

= 1(50 kg)(0.175 m).2 + (50 kg)(0.3 m).11w,

£02 = 37.3 radis Ans.

F19-6. (-El )m[(vG)i]y. SIF„ch m[(vG)21).

O + NA(3 s) — (1501b)(3 s) = O

Nh -= 150 b

C + (firc)i (Hic)2

O + (25 lb • ft)(3 s) — 10.15(150 Ib)(3 s)1(0.5 ft)= siug(1.25 rt)2],-2+ (A sług) [4.ft2(1 ft)](1 ft)

w2 = 3.46 rad/s Ans.

P/9-5. nr:(/a0),11 + = m[(v(r),1,

a+ (130 N1(3 s) + FA(3

(50 kg)(0.3w2)

C +16.cd, + SiModt = .1,0w2

O + (150 N)(0_2 m)(3 s) — FĄ(03 m)(3 s)1(50 kg)(0.I75 m)21

= 37.3 rad/sAns.

F A.-= 3633 N

A l s o .Z

/ / cal f[ + Mwdt = [ww,

Answers to Selected Problems

Cha pter 1212-1. v2 = 59.

5 ftist= 1.29s

12-2. v = 3.93 m/ss = 9.98 m

12-3. The car Toast be dropped from the 9th floor.12-5. t = 25 s

s = 312.5 m12-6. 0,s = 48.3 ft12-7. = 1.74 m/s2

= 4.80 s12-9. v = 29.4 m/s

h = 44.1 m12-10. sir=6, = -27.0 ft

si„,„ = 69.0 ft12-11. s = 20 ft12-13, Normal: d = 517 ft

Drunk: d = 616 ft12-14. t = 21.9 s12-15. s = 28.4 km12-17. v = 32 m/s

s = 67 md = 66 m

12-18. s = 123 fta = 2.99 ft/s2

12-19. s = 7.87 m12-21. (a) Whenr = 5s,

= 45,5 m/s(b) v„„,„, = 100 m/s

12-22. At t = 6 s, s = -18 ft.sr = 46 ft

U-23. dA = 41.0 ftda = 200 ft0sAa = 152 ft

12-25. s = 54.0 m12-26. - 2vAva - v2A 2aA

12-27. S I r=2 6r,7 = 3.56 ni 12-29. t = 2.47s

sit=2, = 18 m 12-30. v = 14 m/s

= 250s12-31. (a) s = -30.5 m (

h) sTor = 56.0m(c) It = 10 m/s

12-33. r = 10.3 sh = 4.11 km

12-34. vi = 3.68 tnis I t2= 1.98s v2= 15.8 m/s I

1 2 - 3 5 . t = . 5 4 9 (

g

12-37. v = 11.2 km/s

12-38. v = -R i 2gt)(Y4) - Y)

N (R + y)(R + yo1

v = 3.02 km/s ł12-39. t' = 27.3 s

When t = 27.3 s, v = 13.7 ft/s.12-41. v = 16.7 m/s

v = vm,, for 2 min < t < 4 min.12-42. a 1 r=ti = -4 m/52

a r=-2, = 0al ,=4 „ 4 mi s2

v 1-01 = 3 mis v1,2, = -1 m/s

s = 3m/s12-43. s = 2 sin (El.)5 + 4

v=2

5 cos (Irt)1

a=- 5t12-45. (It)

2 5 12-45. t = 7,48 s

When .t = 2.14 s,v = v.„ = 10.7 ft/s, h = 11.4 ft.

12-46. s = 600 mFor O 5 r < 40 s, a -= 0.For 40 s < 80 s, a = -0,250 m/s2.

12-47. Ats = 50 m, a = 0.32 m/s'. Ats = 150 m, a = -0.32 tn/s2.

Ats = 100 m, a changes from a = 0.64 m/s'to a mm = -0.64 m/s'.

12-49. When t = 0.1 s, s = 0.5 m and a changes from 100 m/s' to -100 m/s2.

When = 0.2 s, s = 1m.

701

702 ANSWERS TO SELECTED PROBLEMS

12-50. For O s i „ 5 s, a= 4 m/s2.For 20 s < t 30 s, a = -2 rn/s2.M t= 5 50 rn.At /= 20 s,s= 350 m.M t = 30 s. s= 450 m.

12-51. = 133 sWhen t = 30 s, v = 45 mis and s = 450 m.When t = 75 s, v = v„,„„ = 112.5 m/s and

s = 4500 m.When t= 133 s. v = O and s = 8857 m.

12-53. When r = 15 s:= 270 m/ss = 2.025 kmWhen r = M s:= 395 m/ss = 3.69 km

12-54. v., = 400 ft/st' = 33.3 sWhen ł = 5 s, v = v„ia„ 400 ft/s and s = 1000 ft.When = 33.3 s, s = 8542 fł.

12-55. When / = 6 s, v = 12.0 mis.When t = 10 s, v= 36.0 m/s and s = 114 m.

12-57. ( v, =- 37.8 ft/sWhen t = 10 s,s = 133 ft.When i = 3U s.s = sT = 1.13(103) ft.

12-58. When = 25 s, a = ain.„ = 9 m isols2 and s =469 m.When t= 50 s,s= 1406 m.

12-59. t' = 33.3 s• = 550 ft• = 15000 s

I r_20 s = 1800 ft

s I 3 s = 2067 ft12-61. When t = 30 s, s = 90 m.

Whenr = 48 s, s = 144 m.12-62. Whenr = 5 s, s = 83.3 ft and ci = 20 ft/s2.

When t = 10 s, s = 583 ft.12-63. sT = 980 m12-65. When t 5 s, sB = 62.5 m.

When t= 10 s, A = (v = 40 m/s andsA 200 m.

When t= 15 s, sA = 400 nn and sB = 312.5 rn.As = sA - str =- 87.5 m

12-66. When t = 30 s. v = 90 m/s.When t = 60 s, v =- 540 m/s.

12-67. When t = 30 s, s -= 675 m.When t = 60 s, = 10.125 m.

12-69. ab=0 = 1,00 flisz

a I ,,..5{jo = 6.00 ft/s2

t = 17.9s12-70. When s = 100 m,t = 10 s.

When s = 400 m, r = 16.9 s. a1,- ton m = 4 mis2

16m/s2

12-71. At s = 100 s, a changes from a,„„„ = 1.5 ft/s2

to ami,, -0.6 ft/s2.12-73. y = 9.68 m/s

a = 16.8 m/s2

12-74. Dr = 6i + 4j} m12-75. (4 ft. 2 ft. 6 ft)12-77. Ar 3.61 km

d -= 5 km12-78. s = 9 km

Ar = 3.61 kmv„„1 = 2.61 m/s( v.„)„.j; = 6.52 mis

12-79. aAB = {0.4x41 + 7.07j} m/s2

aAc -= {2.50i} m/s2

12-81. v = 8.55 ft/sa = 5.82 m/s2

12-82. v = 1003 m/sa = 103 tri/s2

12-83. d = 4.00 ft= 37.8 ft/s2

12-85. v+W -={ 101 - 10j} m/s/s12-86. v = 201 m/s

a = 405 mis2

12-87. v = 10.4 m/sa = 38.5 m/s2

12-89. HA = 30.5°vA = 23.2 m/s

12-90. O -= 58.3°(1-10).w = 9-76 ni/s

12-9L vA = 27.3 ft/s12-93. O = 15.0°, t = 1.45 s or

O = 75.0°. t = 5.40 s12-94. vo = 67.7 ft/s

O = 58.9°

12-95. d = - ( s n 21 - 2 tan O cos 24))g cas O

12-97. h = 11.1 ft 12-98. O = 6.41' 12-99. vA = 36.7 ft/s

h = 11.5 ft12-101. vA = 19.4 m/s

vB = 40.4 m/s12-102. d = 166 ft 12-103. x = 32.3 ft

y = 6.17 ftv = 71.8 ft/s

12-105. /JA = 39.7 ft/ss =- 6.11 ft

12-106. x = 1.95 fty = -0.153 ft

12-107. v A = 16.5 ft/svB = 29.2 ft/s

ANSVVERS TO SELECTED PROBLEMS703

12-109. 9r - g (COS 02 + cos 01 )

12-110. 1 2 = 0.189 mR „„,„ = 1.19 m

12-111. 0, = 24.9°02 = 85.2°

12-113. 0 = 38.4°

h = 14.8 ft12-114. a = 9.50 m/s2

12-115. u = 38.7 m/s 12-117. a = 0.488 m/s2

12-118. v = 4.58 m/sa = 0.653 m/s2

12-119. a = 42.6 ft/s2

12-121. a = 8.43 m/s2

O 38.2°12-122. a = 6.03 m/s2

12-123. a = 1.05 m /s2

12-125. a = 2.75 m/s2

U-126. 1.68 m/s2

12-127. a = 5.02 m/s2

12-129. v = 51.1 ft/s= 10.3 ft/s2

12-130. vs = 62.9 ft/sa = 0.0133 ft/s2

12-131. a, = 3.62 m/s2

p = 29.6 m12-133. v = 3.19 m/s

a = 4.22 tn/s2

12-134. a = 7.42 ft/s2

12-135. a = 2.36 m/s2

12-137. a = 3.05 m/ s2

12-138. a = 0.525 m/s2

12-139. a = 0.763 m/s2

12-141. y = -0.0766.1?= 8.37 m/s

a, = 9.38 m/s2

a, = 2.88 m/s2

12-142. a = 3.05 m/s2

12-143. x = 0.y = -4 rn(a) max = 50 mi s2

12-145. a = 7.48 ft/s2

12-146. i = 1.21 s12-147. t -= 2.63s12-149. sA = 1.40 m

ss = 3 mrĄ = {L381+0,195j}mrB = { -2 .821 + 0.873j} m

❑r = 4.26 m 12-150. 1 = 14.3 s as = 0.45 m/s2

12-15L a = 32.2 m/s2

12-153. a = 35.0 m/s2

s -= 67.1 ft12-154. y = {0.839x - 0.131x2}

a, = -3.94 m/s2

= -8.98 m/s2

12-155. s -= 10.1 s= 47.6 m/s

a= 11.8 m/s2

12-157. 'un = Ov, = 7.21 m/sa,, = 0.555 m/s2

a, = 2.77 rn/s2

12-158. a , a ,b-

12-159, a = 14.3 in ./s2

12-161. v, = -1.66 m/svs = -2.07 m/s= 4.20 m/s2

as = 2.97 m/s2

12-162. v = 464 ft/sa = 43.2(103) ft/s2

12-163. v =- 120 flisa = 76.8 ft/s2

12461 v, -2 sin ivs = cos t

a, = Ź t cos

un = -2 sin12-166. v, = -2.33 m/s

vs = 7.91 m/s

as = -18.6m/s2

12-167. v, = Ovs = 10 ft/sar -0.25 ft/s2

uy = -3.20 ft/s2

12-169. A = ( - 3.;N2` 3ri)-1)+ + r0. - r()3)u„ + (i)u,

12-170. a = 48.3 in./s2

12-171. v, = 1.20 m/svs = 1.26 m/sa , -3 .77 m/s 2

uM = 7.20 m/s2

12-173. v, = 1.20 m/svy = 1.50 m/sa„ = -4.50 m/ s2

ao = 7.20 m/52

12-174. v, = 16.0 ft/s= 1.94 ft/s

a, = 7.76 ft/s2

as = 1.94 ft/s2

2vo sin (O, - 02)


2 0v, -

NITT- 02

200v o -

v„ = -14,1 ft/svo = 14.1 ft/s

12-177. B = 0.378 rad/s12-178. V = { -116u. - 163u_} mm/s

a = -5.81u, - 8.14u,} mm/s2

12-179. v = 12.6 m/sa -= 83.2 m/s2

12-181. v = 16.8 ft/sa = 199 ft/s2

12-182. Ó = 0.75 rad/s12-183. v,. = at-t

vo = a6()ar = -a002

a, = 2aH2

12-185. v = 10.7 ft/sa= 24.6 ft/s2

12-186. y = 10.7 ft/sa = 40.6 ft/s2

12-187. v,. = 25.9 mm/sa, -= -195 mm/s-

12-189. a = 7.26 m/s2

12-190. v = 4.16 m/sa = 33.1 m/s2

12-191. v,. =- -2.80 m/svy = 19.8 m/s

12-193. v, = 32.0 ft/svo = 50.3 ft/sar -= -201 ft/s2

a, = 256 ft/s2

12-194. v, = 32.0 ft/svo -= 50.3 ft/sar = -161 ft/s2

ag = 319 11/s2

12-195. y = 5.95 ft/sa = 3A4 ft/s'

12-197. y = 8.21 mm/sa -= 665 mm/s2

12-198. v = 8.21 mm/s= 659 mm/s2

12-199. y8 0.5 m/s12-201. v, = 12 ft/s12-202. r = 160 s12-203. AsB = 1.33 ft -12-205. vA = 1,67 mis12-206. vA = 4 ft/s12-207. vo = I ft/412-209. As8 = 2 ft,

12-210. vA -= 24 ft/s 12-211. vA -= 1.33 m/s 12-213. vo = 0.671 m/s 12-214. v, = 1.41 misi 12-215. vB -= 2.40 ft/s ->

aB = 3.85 ft/s2

12-217. ve = 1.2 m/4ac = 0.512 m/s2t

12-218. vb = 5.56 m/sO = 84.4°

12-219. Asr = 2 ft12-221. = 0.809 ft/s

12-222. VBIA = 875 km/h O = 41.5°

12-223. vo/A = 28.5 mi/h= 44.5°aBp, = 3418 mi/ h2 O = 80.6° .d

12-225. vg1B = 49.1 km/h O = 67.2° 7'

12-226. v«. =- 58.3 km/h O = 59.0°

12-227. VAIR = 21.7 ft/s= 18.0°

r = 36.9 s12-229. vAio = 98.4 ft/s

= 67.6°aA/B = 19.8 ft/s2

O = 57.4° 12-230. vAio = vV2 (1 sin O)12-231. vo/A = 36.6 mi/h

0 = 46.9°aB/A = 3737 mi/h2

efi = 12.9°

12-233. v, = 34.6 km/h12-234. v,„ = 4.87 ft/s

t = 10.3s12-235. vNys = 19.9 m/s

= 74.0° 7'12-237. voic = 18.6 mis

O, -= 66.2°

a,/c = 0.959 m/s2

0„ = 8.57° ."%12-238. vo = 5.75 m/s

vcio = 17.8 m/sO = 76.2°

acta = 9.81 m/s21.

12-239. vA,B V‚v.2,1 + t H - 2vAvacos

IBĄ - VBC0s0)O = tan-1

vo sin O

12-175.

Chapter 1311. = 97.4 ft13-2. - 46.2 ft/s

- 66.2 ft13-3. v = 3.

36 rti/ss = 5.04m

13-5. v = 77.9 ft/s13-6. v = 3.29 m/s 13-7. P = 392 N13-9. a = 1.66 m/s2

13-10. a = 1.75 m/s 13-11. as = 2.30 ft/s2i

= )(1TI M2F-4510

V71'M

s = ( F ( 1 4 ) ( I - s i t t ( 1zm /T to

13-14. T= 11.25 kNF = 33.75 kN

13-15. aF = 0.75 mis=1`T= 1.32 kN

13-17. a = 3.69 ft/s2

13-18. R = 5.30 ftTotal dme = 1.82 s

1349. R = 5.08 ftTotal LiTI1C = 1.48 s

13-21. s = 543 m13-22. mA = 13.7 kg13-23. T = 4.92 kN13-25. T = 49.2 N13-26. v = 30 m/s13-27. A, = 35.6 Ib

A,. = 236 Ib= 678 lb • ft

13-29. u = 2.01 ftp;13-30. y = 0.301 m/s13-31. T = 1.63 kN13-33. as = 0

ar = 4.11 m/s.1 -›aty = 0.162 m/s' -›

13-34. (a) = - 3ri.g

P(b) £1.4 = -2m - 2gy

13-35. t = 2.04 sa R + 8,)4\-3.1d.2

13-37. F -

13-13. c o s

4y13-38. v$.= 4,52 m/s.

13-39.

13-41.

13-42.

13-43.

ANSWERS TO SELECTED PROBLEMS

705

v = V1.09F ? + 2F4:,trnvo + or2vZ

= Y

0.3(a) = 6.94 m/s2

(b) = 6.94 m/52

(c) ac = 7.08 m/s' = 56.5°7 ac = 7.49 m/s2

= 22.8° 7i = 5.66 s

13-45. v = 32.2 ft/s 13-46. P 2.91 g um 9

sin H -I- ii, cos 9)P = 2m g( - Oj

13-47. 13-49. v$fA.e = an (sili e) r s

= 2a(,(sin 0) ,f2

m Vei= " ( k i r ' ' )

13-50.

k -)( 1 -gim)t)

cos 00(1 - emg k irn-(v.0 tan 61.0 +k

mvacosBo

k

13-51.(r + ms)g d -

k13-53. r = 1.36 m 13-54. v = 10.5 m/s 13-55. P -= 17.3 N

= 120 N1 13-57. T = 1.82 N

NB = 0.844 N 13-58. t3 = 0.969 m/s 13-59. v = 1.48 mis 13-61. v= 9.29 ft/s

T = 38.0 ib 13-62. hr = 5.8816 ar

= 23.0 ft/s2

13-63. N = 277 lb F= 13.41b

13-65. -v = 6.30 m/s = 283 N F, = O Fb = 490 N

13-66. i = 7.39 s 13-67. O = 26.7° 13-69. = 111 kt4

N=6.73kN 13-70. vc = 19.9 fił,s

Nc = 7.91 lb v$= 21.0 ft/s


13-71. T = 414 N= 37.2°

13-71 v =N = 2mg

13-74. v = 80.2 ft/s13-75. 9 = cos-1(17--)

tnA

Vy = \ig(1 - h1(m24 ni-1)

13-77. N = 1.02 kN13-78. 8 = 31.3°

/ = 2.585 ft13-79. Np = 2.65 kN

p -= 68.3 m13-81. L = 50.8 kN

r = 3.60 km13-82. N = 7.69 kN 13--85. FA = 4.46 lb 13-86. F = 210 N 13-87. F = 1.60 lb 13-89. N = 2.77 Ib 13-90. F, = -29.4 N

Fp = 0= 392 N

13-91. F = 0.143 Ib 13-93. F()Ą = 12.016 13-94. k. = 54.4 N

F = 54.4 N13-95. B = 4.00 raci/s

T = 8 N1397. /V = 6.37 N

F= 2.93 N13-98. N = 24.8 N

F = 24.8 N13-99. F, = -20.0 N

ro = O= 2.45 kN

4r13-101. 9 =)

13-102. N = 0.883 NF = 3.92 N 13-

103. T = 509 Ib 13-105. F = -0.0155 lb 13-106. F = -0.0108 lb 13-107. F = 0.163 lb 13-1119. F = 7,67 N 13-110. F = 7.82 N 13-111. r = 0.198 m 13-113. v0 = 30.8 krnis

- = 0,502(10-12) (x-s O + 6.11(10'2) r

13-114. h = 35.9 Mmv, = 3.07 km/s

13-115. vo = 7.45 km/s13-118. y8 = 7.71 km/s

vA = 4.63 km/s13-119. u„ = 2.59 km/s

T = 9.33 hr13-121. yA = 7.47 km/s13-122. vp = 466 m/s

Av8 = 2.27 km/s13-123. ro = 10.8(109) km13-125. (a) r = 194(103) mi

(b) r = 392(103) mi(c)194(103) mi ° r < 392(103) mi(d) r a 392(103) mi

13-126. O = 77 cos (k2n1

13-127. cw = 530 mis 13-129. = 2.57 km/s

13-130. vA = 3.08 km/s T = 15.1 hr13-131. v =

r c ( V P 2 - N f r F e )

Chapter 1414-1. v = 10.7 m/s14-2. x„,,„, = 3.24 ft14-3. s = 1.3514-5. v2 2,12 km/s14-6. vq = 10.5 m/s14-7. Observer A: v2 = 6.08 m/s

Observer B: yi -= 4.08 m/s14-9. P = 207 N 14-10. s = 178 m 14-11. gk = 0.255 14-13. v8 = 31.5 ft/s.

d= 22.6 ft vc = 54.1 ft/s

14-14. 88 = 3.34 m/s 14-15. s = 9.29 ft 14-17. v8 = 27.8 ftis 14-18. vA = 3.52 ft/s 14-19. Ny = 20.51b 14-2L kp = 111 kN/m 14-22. sT,, t = 3.88 ft 14-23. s s = 0.628 m

s, = 0.377 m14-25. v$ = 30.0 m/s

s = 130 m14-26. vA = 28.3 m/s 14-27. s = 2.48 ft

TĄ = 36.0 lb

ANSWERS TO SELECTED PROBLEMS 707

14-29. s = 0.730 m 14-30. s = 3.33 ft 14-31. R = 2.83 in

v(-= 7.67 m/s 14-33. UD = ł 7.7 m/s

= 33.0m 14-34. s = 1.90 t] 14-35. ?Je = 2.36 m/s 14-37. h, = 22.5 m

itc = 12.5 m 14-38. 1t8 = 26.1 ft/s

Ny = 135 lb14-39. v, = 1.98 mis

14-41. = 2.77 ft14-42. O = 47.2°14-43. power input = 4.20 bp14-45. P = 8.32 (103) bp14-46. t = 46.2 min14-47. P = 12.6 kW14-49. Ais = 113 kW

Pin )u = 56.5 kW14-50. P = 1.12 kW14-51. power input = 1.60 kW14-53. P = 0,229 hp14-54. P = 0.0364 hp14-55. P = 22.2 kW

14-57. = 622 kW14-58. P = 58.1 kW 14-59. P = 5.33t MW 14-61. P = 1400( I 03)fr W 14-62. P = (160 r - 533i2) kW

U = 1.69 kł14-63. P = 10.

7 kW14-65. P = 0237 hp 14-66. F = 227 N 14-67. kj5= 287 Nim 14-69. u2 = 11.7 ftis 14-70. d = 8.53 m

vp = ldmfs 14-71. y = 80.2 flis

Nu = 952 lb 14-73. tiA = 3.46 m/s 14-74. V = 32.3 ft/S 14-75. k = 8.57 lb/ft 14-77. y = 213 mm 14-78. vc = 19.4 ftis 14-79. vB = 25.4 ft/s 14-81. h = 416 mm

14-83. = Gitijnr, r2

14-85. 1,5 -= 34.8 Mm/h

14-86. t = 94 ft

= 219 fta = 483 ft/s2

14-87. t;,H, = 29.5 ftis14-89. N -= 22.3°

s = 0.587 m

14-90. va = 'vi,g(5p - 26)14-91. yi = 7.43 m/s

N = 86.7 N14-93. ( o,)2 = 1.68 m/s14-94. a, = 1.29 ft14-95. x- 17.7 ft

y = 8.83 ft14-97. d = 1.34 m

Chapter 1515-1. t = 0.280 s

v$= 15.6 ft/s15-2. v = 29.4 ftis15-3, t = 0.439 s15-5. f Frit = 0306 N • s .40°

15-6. F = 19.4 kNT = 12.5 kN

15-7. F,1& = 16.7 lb= 13.4 ftis

15-9. ( = 16.6 kN15-10. P --- 205 N 15-11. I = 63.4 N• s 15-13. juk = 0.340

2Ct'15-14. V2 =

TriTICr''

=rrnr

15-15. = 89.8 ft/s

15-17. Fdt = 0.311 lb -s

15-18. 1.J2 = 12.0 rn/s -15--19. v, = 27.6 ft/st

v8 = 55.2 ft/4

15-21. T =- 14.9 kNF =- 24.8 kN

15-22. ti = 8.80 flis

15-23. I Fdt = 63.4 N -s

15-25. F„0. = 847 lb15-26. Observer A: v = 7.40 m/s

Observer B: y = 5.40 m/s 15-27. 252 = 21.8 m/s


15-29. v = 14.3 m/s

FD = 15.7 kN 15-30. vB = 2.10 m/s1

vA = 8.41 m/s 15-31. ( vA )2 = 10.5 ft/s -4 15-33. v2 = 7.65 m/s 15-34. v2 = 1.92 m/s 15-35. v = 3.5 ft/s 15-37. v2= 5.21 m/s <-

9E = 32.6 kJ 15-38. v2 = 0.5 m/s

AT = 16,9 kJ 15-39. u2 = 0.600 ft/s 15-41. vi, = 0.379 m/s 15-42. vbi„ = 0.632 m/s -) 15-43. vA8c = 2.18 mis «-

15-45. v2 = Vfy.12 + 2gh- 1

= sin ( n )\iv; + 2gh

15-46. sb = 13.2 ft15-47. s = 481 mmm°x15-49. (Y8)2 = 3.29 ft/s 15-50. W8 = 75 lb 15-51. v = 4.29 m/s 15-53. vb = 2.73 ft/s

vs = 2.27 ft/ssc = 1.09 ft

15-54. v,. = 0.276 ft/s <--15-55. x = 0.221 m 15-57. d = 6.87 mm 15-58. e = 0.901 15-59. e = 0.75

AE =-- 9.65 kJ15-6L (vA)2 = 0.400 m/s

(v8)2 = 2.40 m/s d = 0.951 m

15-63. v8 = vevA = ve v" = O

15-65. t = 0.226 s 15-66. d = 8.98 ft 15-67. ( vA ) 2 = 1.04 ft/s

88.)3 = 0.964 ft/s(vc); = 11.9 ft/s

15-69. v'B = 22.2 m/s= 13.0°

e(1 + e)15-70. (vB), - vo15-71. vA = 29.3 ftis

v82 -= 33.1 ft/s= 27.7'd

15-73. d = 13.8 ft15-74. e = 0.502

d = 723 ft15-75. d = 3.84 m15-77. (Y8)3 = 3.24 m/s

O = 43.9°

15-78. v'B = 31.8 ft/s15-79.

(1 + e\- 7

15-81. (a) (vB)I = 8.81 m/sO = 10.5°

(b) ( vb )2 = 4.62 m/s(f) = 20 .3°

(c)s = 3.96 m15-82. s = 0.456 ft15-83. d = 1.15 ft

= 0.770 ft15-85. 1. = 0.2515-86. ( )2 = 4.06 ft/s

( vB )2 = 6.24 ft/s15-87. vA = 1.35 m/s

= 5.89 m/sO -= 32.9°

15-89. v8 = 3.50 m/svA = 6.47 m/s

15-90. (vB), = "Vr3 -(1 + e)v

- 408 = 30°

3my2 ,Atik --= k i - e2)

1615-91. v ' A = 11.0 ft/s

0A == 11.0 ft/s

Og = 58.3° -":C15-93. v = 8.19m/s

v#8 = 9.38 m/s15-94. Ho = (42i + 21k } kg m2/ s15-95. Ho = (-16.81 + 14.9j - 23.6k slug • ft2/s15-97. Ho = (12.5k} kg • m2/s15-98. (1/A)0 = 72.0 kg • m2/s)

(118)0 = 59.5 kg - m2/s15-99. (HA )1, = 66.0 kg • m2/s.)

(118)p = -73.9 kg • m2/s15-101. t = 3.41 s15-102. t = 0.910 s

y -= 17.8 ft/s15-103. v2 = 9.22 ft/s

Ul = 3.04 ft • Ib15-105. v2 = 19.3 ft/s 15-107. v2 = 13.8 ft/s

ANSWERS TO SELECTE0 PROBLEMS 709

15-109. v2 = 45.1 ft/sUF = 2641 ft •

lb 15-110. v2 = 4.60 h/s 15-111. u = 20.2 ft/s

h = 6.36 ft15-113. ve = 10.2 km/s

rg = 13.8 Mm 15-114. T = 40.1 kN 15-115. F, = 55.3 Ib

F,. = 25.8 lb15-117. ć, = 4.97 kND„ = 2.23 kN D, = 7.20 kN 15.-118: P = 42.8 hp 15-119. F, = 9.87 Ib

F. = 4.931615-Ul. F, = 19.5 Ib

• = 1.96 Ib

15-122. F, = 6.28 kip

F, = 2.28 kip 15-123. P =- 22.4 lb 15-125. T = 82.8 N N = 396 N15-126. A, = 3.98 kN

A, = 3.81 kN M= 1.99 kN • rn

15-127. m' = -m

s15-129. C, = 4.26 kN

• = 2,12 kN Me = 5.16 kN • m

15-130. T = 1.12 kN15-131. A, = 4.18 kN

Br = 65.0 NB, -= 3.72 kN 't

15-133. JI;;ID = 10.7 kip ftDr = 5.82 kip

= 2.54 kip15-134. v .„ = 625 m/s 15-135. P = 452 Pa

P 2

m o 3

15-138. a, = 133 ft/s2

= 200 ft/s2

15-139. v„,„, = 580 ft/s15-141. F = 3.55 kN15-142. Fa = 11.5 kN15-143. a = 37.5 ft/s2

15-145. F = m' v215-146. t max = 2068 ft/s15-147. F = (7.85r + 0.320)N

15-149. . \114(1 .3)

33:2

R eview 1R1-1v = 77.5 ft/sR1-2. (a) aA = ar{ = 2.76 ft/s2 (

b) aA = 70.8 ft/s2 aa = 3.86 ft/s2

R1-3. v8 = 47.8 ft/s R1-5. (vR)2 = 0.840m/s R1-6. v2 = 21.5 ft/s R1-7. F = 85.7 NR1-9. F,-68 N

N = 153 N R1-10. 1= 1.02 s R1-11. v, = 31.7 mis R1-13. P = 946 kW R1-14, s = 640 m R1-15. s t

O O2.01 0.253.83 0.505.49 0.757.03 1.008.43 1.259.87 1.5011.2 1.7512.5 2.00

R1-17. T = 158NR1-18. v... = 0.838 m/s

vE,„„, = 1.76 misR1-19. s = 0.933 m

R1-21. h = 0,390 mR1-22. r = 23.8 sR1-23. y = 0.0208x2 + 0.333xR1-25. w = 2.68 ft/sR1-26. vA = 26.8 ft/s,.R1-27. (v8)2 = 1.34 m/s 4-

(vA )2 -= 1.30 mis(0,4), = 8.47° A

R1-29. v8 = 10.4 m/sR1-30. uA = 0.755 M/52

= 1.51 m/s2

TA = 90.6 NTĄ = 45.3 N

R1-31. v = 0.686 m/sTg = 206 NTc = 103 N

R1-33. N = 14.6°d> = 14.0°

R1-34. a = 0.603 ft/s2

R1-35. ei = 13.8 ft/s

15-137. v =


v = 25.4 ft/s s = 1.84 m

= -24 m/s ,9s = -880 m sT = 912 m v8= 0.5 m/s▪= 5.38 ft/s▪= 532 ft/s 4=4

= 1.06 m/s2

• = 312 - 6: + 2

= - 6 : - 6 P r= 1 . 8 0 h p e = 0 . 9 0 1 v 8 = 3 . 3 3 f t / s v 8 i c= 1 3 . 3 f t / s '

Chapter 1616-1. vA = 2.60 m/saA = 9.35 m/s'16-2. ar = 0.562 ft/s2

• = 3600 ft/s2

s = 3000 ft16-3. v = 22 ft/s

• = 12.0 ft/s2

a„ = 242 ft/s'16-5. vc = 8.

81 in./5

a c = 32.6 in./s2 16-6. O = 3.32 mv

= 1.67 s16-7. i = 6.98 s

OD = 34.9 rev16-9. v„/ = 70.9 ft/s

v8 = 35.4 ft/s (a)A = 252 ft/s2 (a)8 = 126 ft/s2

16-14. vA = 40.0 ft/sv8 = 20.0 ft/s (a)A = 80.6 ft/s2 (0)8 = 40.3 ft/s2

16-11, vA = v8 = 2.4 ft/s a A = 0.4 ft/s2 c c = 17.3 ft/s2

16-13. (OB = 12 rad/s 16-14. w = 42.7 rad/s

= 42.7 rad16-15. a, = 2.83 m/s' a„

= 35,6 m/s2 16-17. v8= 0.394 m/s 16-18. v = 1.34 m/s 16-19. co8= 156 rad/s 16-21. y8= 1.37 m/s a B = 0.472 m/s2

16-22. coD = 22.4 rad/s 16-23. wp = 11.8 rad/s 16-25. vp = 2.42 ft/s

ap = 34.4 ft/s2

16-26. wc = 1.68 rad/s08 = 1.68 rad

16-27. co8 = 528 rad/sHR = 288 rad (wa)2 = 16-29. 10.6

rad/s

16-30. a =)s ,17

16-31. V F = 0.318 mmis 16-33. ros = 312 rad/s alt = 176 rad/s2

16-34. a= 2w77

2( ri )d

16-35. = -4.81 - 3.6j - 1.2k } m/sac 138.4i - 64.8j 4- 40.8k 1 m/s2

16-37. vc = 2.50 m/sa = 13.1 m/s2

vc = 21.2 Ris= I 06 ft/s!

16-39. vA8 = art cosC I AB = -w2/sin

16-41. w = 8.70 rad/sa = -50.5 rad/s2

v tan 1,1 6 - 4 2 . 3

eas20 + + 2r,r2

16-46. w = 0.0808 rad/s 16-47. vA = ecu sin O

aA ec02- cos O16-49. ve .---- GwT

a c = 0.577 Lw2

t 16-50. w = - sia l i ' O

= ( v °

a

16-51. w = 1,08 radis v8= 4.39 ft/s

v sinLcas (4, - 0)

R1-37. R1-38. R1-39.

R1-41. R1-42. R1-43. R1-45. R1-46.

R1-47. R1-49. R1-50.

v-

Y - 9L cos3

16-43. vp = 18.5 ft/s <-riw sin 20

1 6 - 4 5 . -

16-53. H -

+ rew sin 0)

ANSWERS TO SELECTED PROBLEMS711

, r l ( r + d c o s H ) 16-54. x

(d + r cos 0)-

Jr sin U (2r' - d2 + rd cos 0) 2= (d r cos )3

16-55. wA= 2( Vr2 cos U-- 1 )

3 - 2VcosB16-57. #=U

2v- cos O h

16-58. WAB = 2.00 rad/s 16-59. vB = 2.83 ft/s

wRc = 2.83 rad/stoay = 2.83 md/s

16-61. vc = i.64 m/s 16-62. wBc = 0.693 rad/s 16-63. wcB = 2.45 rad/s `5

vc = 2.20 ft/s16-65. w = 20 rad/s

vA = 2 ft/s ->16-66. w = 3.1 I rad/s

vo = 0.667 ft/s -› 16-67. vA = 5.16 ft/s

O = 39.8°16-69. vA = 2,45 mis I 16-70. wsc = 2.69 rad/s

wAc = 4.39 rad/s16-71. vc = 0.776 m/s gt

vo = 1.06 m/s',..16-73. vA - (-R r )v -3 ,

16-74. vc = 1.80 m/s wac = 6.24 rad/s

16-75. vs = 3.00 m/s --> 16-77. wp = 5 rad/s wA = 1.67 rad/s

16-78. wc = 50 rad/s

16-79. ve = (R - r) T16-82. WAB =-1.24 rad/s16-83. vA = 0

v5 = 1.2 m/s J,v, = 0.849 m/s " 45'

16-85. wAB = 6 rad/s 5trE = 4,76 m/s

= 40.90

16-86. v6 = 6.00 m/s 4--16-87. = 1.

04 mis16-89. vH = 18.0 ft/s

16-90. = 2.50 ft/svo = 9.43 ft/s O = 55,8°

16-91. = 2.50 ft/svE -= 7.91 ft/s

O = 18.4° A16-93. WAB = 13.1 rad/s

16-94. w = 1.12 rad/s vc = 3.38 ft/s

16-95. vA = 60.0 ft/s -•vc -= 220 ft/s <-vB = 161 ft/s 6°-3'5:

2.. 16-97. vE = 2 ft/s <15-98. w = 5.33 rad/s

va -= 2 ft/s (-16-99. wc = 26.7 rad/s 5

w8 = 28.75 rad/s )= 14.0 rad/s 5 16-

101. wBc = 1.98 rad/s ) 16-102. wcH = 0.0510 rad/s 16-103. a = 0.332 rad/s2 )

aR = 7.88 ft/s2

16-105. a = 0.0962 radfs2

(JA, = 0.385 ft/s- -› 16-106. ac = 3.82 m/s2/

aRe = 9.60 rad/s2 16-107. aA/B = 25.4 rad/s2

aR = 5.21 m/s21. 16-109. amt = 6.59 rad/s2

aBC = 5.16 rad/s2

16-110. aA = 12.5 rn/s24- 16-111. ac = 10.3 m/s24- 16-113. a5= 2.25 m/s2

O = 32.6°16-114. ap = 10.0 m/s2

6 =-- 2.02° 716-115. vB = 4v

vA = 2V5_77.45.2v2 ,

a B =2v2

r

aR = 63.55 ft/s2

= 873°ac = 10.7 rad/s'Sap = 14.1 in ./s2

as = 1.58 - 1.77w2awAt- = Gwp = 10,7 rad/s aAc

= 28.7 rad/s2 ) aB -= 1.43 rad/s2 (tE = 0.0714 na/521

= 4.73 rad/s 5 a = 131 rad/s2 )

a A =

16-117.

16-118.

16-119.16-121.

16-122.16-123.16-125.

71 2 ANSWERS TO SELECTED PROBLEMS

16-126. w = 0.25 rad/s ) vg = 5.00 ft/s i= 0.875 rad/s2 ) a

g =- 1.51 ft/s2 Ofi = 852°

16--127. GA8 = 3.70 rad/s2 )16-129. vc -8.

121 - 8.12j} ft/sac = {61.91 - 61.0j} ft/s2 16-130. i‚g = 1.30 ft/s a8 = 0.6204 ft/s2

16-131. v„, = {7.51 - 5j} ft/sa,,, = 51 + 3.75j } ft/s2

16-133. vc = -31 + 3j } m/sac = {-36.751 - 17.5j} m/s2

16-134. aA -5.601 - 16j} m/s2

16-135. (a) a8 = -11} m/s2

(b) a8 = I-1.691} m/s2

16-137. wa, = 9.00 rad/s )(rep = 249 rad/s2

16-138. v8 = 7.7 mis f-- a8= 201 m/s2

16-139. cdu, = 0.750 rad/s )ac!, = 1.95 rad/s2

16-141. rato = 0.866 rad/s )atm = 3.23 rad/s2

16-142. vc = {-0.9441 + 2.02j } m/sac = { -11.21 - 4.15j} m/s2

16-143. (diw = 5 rad/salAa= 2.5 rad/52

16-145. yo = -31.5i - 13.0j} ft/sap = (13.01 - 79.5j } ft/s2

16-146. cuci) = 13.8 radis16-147. (yrd), = {29j} m/s

(arei)xyz (4.31 - 0.2j) m/s2

16-149. w,.,8 = 3 rad/soiA8= 32.7 rad/s2 )

16-150. copc = 2.96 rad/s16-151. wgc = 0.720 rad/s

aBC = 2.02 rad/s2

17-6. ł, = 10mre2

17-7.k, = t.56 in.

17-9. t, = m(R2 + 4 -3a2)

17-10. /0 = 5.27 kg • m2

17-11. IA = 2.17 sług • ft2 17-13. /A = 7.67 kg - m2 17-14. 1, = 222 slug • ft2

17-15. /0 = 6.23 kg - m2

17-17. L = 639 m/o = 53.2 kg • m2 17-

18. /0 = 6.99 kg • m2

17-19. 5; = 0.888 m/a -= 5.61 kg • m2 17-

21. v = 1.78 mift; = 4.45 kg • m2

17-22. = 3.25 g • rn2

17-23. fx, =- 7.19 g- ru2

17-25. F = 5.96 lbNfi -= 99.0 lbNA = 101 lb

17-26. MA = 5.08(103) Ib•ft17-27. a G = 1.07 ft/s2

NB = 86.7 lbNA = 21316

17-29. a = 10.3 f[/s2`17-30. T = 15.7 kN

= 8.92 kNC„ = 16.3 kN

17-31. aG = 2.33 m/s2

17-33. NB = 237 NNA = 744 N

17-M. a = 4 misa -0N8= 1.14 kNNA 327 N

17-35. u = 13.2 ft/s2

NA = 6.10 lbNB = 91016

17-37. NA = 2.35 kipNB = 2.15 kip

a = 13.4 ft/s2

17-38. NA = 2.40 kipNfi = 2.10 kip max

= 12.0 ft/s2

17-39. O = tan - 1 I )

g

17-41. a -= 3.96 m/s2 17-42. a = 2.01 m/s2 The

crate slips,

Chapter 17

17-1. 1,. = m f-

17-2, m = rrh R2(k + aR2)

7 7 1 1 R 4 H 2aR 2 13

17-3. = m R2

17-5. /, = 2 ,

-

ANSWERS TO SELECTED PROBLEMS 7 13

17-43. a = 2.68 ft/s2

NA = 26.9 lb NB = 1231b

17-45. NR = 10,7 kNFA = 2.25 kN

NA = 8.89 kN 17-46. t = 3.94 s 17-47. h,. = 3.16 ft FA

= 2481b-= 40016

17-49. FAR = i 12 NC, = 26.2 NC,, = 49.8 N

17-50. P = 765 N 17-51. T = 1.52 kN

• = 18.6°

17-53. NB = 402 NNA = 391 N

17-54. a = 5.95 rad/s2 17-55. a = 8.72 rad/s2

FAR = 402 Ib17-57. w = 56.2 rad/s

A, = OA, = 98.1 N

17-58. A, = OA, = 262 N

17-59. FA = - mg

17-61. ł -= 6.71 s1 7 - 6 2 . w = 1 0 . 9 r a d / s 1 7 - 6 3 . w = 9 . 4 5 r a d / s

1 7 - 6 5 . a = 2 4 . 5 r a d / s 2 = 9 0 N

A, = 29.4 N17-66. (ai,),roa + irGa = (aG),(roc; + r6p)

17-67. rp = 2.67 ftA , r = O

17-69. w = 2.71 rad/s 17-70. k = 27.2 N • m/rad 17-71. Fo = 299 N

mg17-73 . N A = 4 c o s

5mgF =s

2 m= tan-1(

1017-74. a = 14.2 rad/s2 17-75. Ax = 89,2 N

A. = 66.9 Ni= 1.25 s

17-77. w, = 38.3 rad/s iós = 57.5 rad/s

17-78. a =(2 - .1.4)g

17-79. a = 2.97 m/s2

1741. A, = OA,. = 289 N

a= 23,1 rad/s'17-82. NA = 177 kN

VA = 5.86 kNMA = 50.7 kN • m

1743. M = 03 gm!17-85. N = wx[°2-(L - -;x) e• os Ol

gS = wx sin

1 ,M = 2wx- sin

17-86. w = 4.83 rad/s ft = 92.2°

17-87. y = 4.57 m/s.1, 17-89. w = 800 rad/s 17-91. a = 5.62 rad/s2

T = 196N17-92. a = 7.86 rad/s2

aG = 34.1 ft/s2 17-93. a = 0.274 rad/s2

ap -= 49.0 ft/s2

O = 80.3°17-94. a = 4.32 rad/s2

17-95. U = 46.9°17-97. a = 0.500 rad/s2 17-98. a = 15.6 md/s2 17-99. w = 5.21 rad/s 17-1111. a A = 167 ft/s2 17-102. T = 5.61 N

a = 28.0 rad/s2 17-103. a = 0.109 rad/s2

17-105. a = 18.9 rad/s2

P = 76.4 lb17-106. a = 1.25 rad/s

T = 2.32 kN17-107. T = 3.13 kN

er = 1.68 rad/sac = 1.35 m/s2

17-109. a = 3 rad/s2 17-110. s = 1.40 in. 17-111. aq = 43.6 rad/s2

ap = 43.6 rad/s2

T= 15.7N

7 1 4 17-113.

17-114.1=

3w a wor 3

/4 .g17-115. s = 4.06 ft

17-117. T = 19.6 N

17-118. a -= 23.4 rad/s2

B, = 9.6216

17-119. a = 6.67 rad/s'-

Chapter 1818-2. w = 14.0 rad/s18-3. w = 14.1 rad/s18-5. s -= 0.661 m

18-6. s = 0.859 m18-7. T= 283 ft • lb18-9. w = 7.57 rad/s

1 71-PW18-10 . w = -

kG m 18-11. T = 0.0188 ft • lb 18-13. yr = 16.9 ft/s t 18-14. yr = 11.8 ft/st 18-15, v c = 1.97 m/s

i18-17. w = 4.2.):( + ; 4 . s sin O

r-

18-18. Yr = 6.11 ft/s18-19. w = 4.90 rad/s18-21. s = 0.304 ft18-22. yr = 19.6 ft/s18-23. wAB = wa, = 2.24 rad/s1825. w --- 3.48 rad/s18-26. w, = 5.37 radis18-27. t‚B = 5.05 ft/s18-29. YA = 26.7 ft/s18-30. w = 2.50 rad/s

3g ,18 -31 . w = - t s tn O

18-33. O = 0.891 rev, regardlessoforientatIon1 i , 10

w,18-34. , = r - v z; - -7 gR- 18 -35. v G = 3 - 3 R

7'18-37. 9 = 25.4° 18-38. s = 0.301 m

T = 163 N

ANSWERS TO SELECTED PROBLEMS

a c = gkg2,48n = -

r

18-39. 18-41. 18-42. 18-43. 18-45. 18-46. 18-47. 18-49. 18-50. 18-51.

18-53.

18-54 . 18-55 . 18-57 . 18-58 . 18-59 . 18-61 . 18-62 . 18-63 . 18-65 . 18-66 . 18-67 .

vA = 1.29 m/s= 13.3 ft/s

?te = 3.07 ftis ue = 31.9 m/s

= 1.73 rad/s w = 2.57 rad/s d = 3.38 m

= 39.3 rad/s vA = 1.40 m/s = 3.09 rad/s

r gcam = a ac =

ub = 15.5 ft./s I.(wABc)2 = 7.24 rad/sw = 12.8 rad/sk = 42.8 Nirnk = 206 N/mH=48.2°

= 1.80 rad/sw = 85.1 rad/sw = 1.82 rad/sk = 1001bfftk = 814 N • m/rad

Chapter 1919-5. J M dr = 0.833 kg -1112/s

19-6. (2,)7 = 0.386 rad/s19-7. w = 0.0178 rad/s19-9. (to), = 6 m/s

= 26.7 rad/s19-10. t = 0.6125 s 19-11. w = 16 rad/s

v = 4 m/s

19-13. v = -23 /'

19-14. wo -= 2.5 (

19-15. co, = 9.40 rad/s 19-17. wg = 127 rad/s 19-18. w = 6,04 rad/s 19-19. vA = 24.1 m/s 19-21. yB = 34.0 ftis 19-22. rp = 1,39 ft 19-23. t = 9.74 s 19-25. vA = 1.66 m/s

vB = 1.30 m/s 19-26. w = 9 rad/s

V2Y = - 3a19-27.

ANSWERS TO SELECTED PROBLEMS 715

19-29. (a) rom = OR2-11. tIG = 3.22 ftis21 NB = 14,81b

(b) com = - wL NA = 65.316

2/ R2-13. au = 39.2 m/s2(C) &im = -to

i,. It cannot be done. sincemAk2A-coA + m8k21,cos 39.2 m/s2 > a Ri, = 5.89 m/s2.219-30. w =R2-14. FB, = 4.50 NmAi(,4' + 07181(7,NA. = 1.78 kN

19-31. w = 0.244 rad/s Ng. = 5.58 kN19-33. w, = 2.55 rev/s R2-15. wp = 784 rev/min

19-34. wl, = 10.9 rad/s R2-17. vo = 2.58 m/s19-35. w = 0.0906 rad/s P = 141 N

19-37. w, = 6.94 rad/s R2-18. s = 9/5 m19-38. w = 3,70 rad/s R2-19. h = 0.2 m: aG = 1.41 m/s2

3 3g h = 0: a0 = 1.38 m/s2

19-39. (03 - ._.72V L R2-21. aB = 40.2 ft/s21 O = 53.3°7'

19-41. w2 = 3w,R2-22. '4./8 = 0.860 m/s19-42. co2 = 57 rad/s R2-23. T = 218 NUF = 367 Ja = 21.0 rad/s2

19-43. w, = 3.47 rad/s R2-25. Fc = 12.4 lb19-45. y = 5.96 ft/sN8. -- 3.09 kip19-46. O = 22.4°NA. = 1.31 kip19-47. Ol ..-= 39.8°FA, = 68.3 lb19-49. ( 'DD ) 3 = 1.54 m/sR2-26. a = 73.3 rad/s2

co3 = 0.934 rad/st = 0.296 s19-50. co, = 7.73 rad/s

R2-27. a = 12.1 rad/s2F = 30,0 lb19-51. 0 = tan l( „ffe)R2-29. w = 100 rad/s

5 R2-30. v = 4.78 m/s19-53. w3 = 2.73 rad/s 122-31. vc = 9.38 in./s Ź

i-ac = 54.7 in./s2 Ź19-54. w, = 1.02., '3- R2-33. (ą) ar = 1.38 m/s2T

(b) aG = 1.56 m/s219-55. ( v6.),.2 = e (1.,c; )y, łR2-34. ( vc, )2 = 9.42 m/s,5 -,

( vG,,, = -(, vG hi -coir)«- I.30g- 75R2-35. a = i O, = 0.325 mg Review 2 O,. = 0.438 mg

R2-1. R = 33.316 R2-37. w., = 1.70 rad/s

R2-2. N' ,A = 383 N <os = 5.10 rad/s

N'8 = 620 N R2-38. w = 28.6 rad/s

R2-3. a = 16.1 ft/s2 O --= 24.1 rad

122-5. w = 39.5 rad/s At y = 0,5 m:

R2-6. a,.4 = 282 ft/s2 -* vp = 7,16 m/s

R2-7. ac = 50.6 m/s2 ap = 205 rn/s2

R2-10. as = -3(1-71 -I- ri)

k

2rrrs- JE = 7.291b • ft

O = 9.09° .« R2-39. y= 1.5 cot O

as = 8.00 m/s2 łR2-41..v, = 3.46m/s122-9. w = 0.275 rad/s 122-42. h = 1.80 m

a = 0.0922 rad/s2 R2-43. E, = 9.87 lb

71 6 ANSWERS TO SELECTED PROBLEMS

Es(k20 ÷ '2)1L k2b

v -r w sin a = -r w" cos O

R2-49. B, = 180N= 252N

A. = 139NR2-50. B, = I43N

200NA., = 34.3N

R2-51. vw = 14.2 ft/s aw= 0.25 ft/s2

Chapter 2020-1. y, - { -7.61i - 1.18j + 2.54k } m/s

aA = {10.41- 51.6j - 0.463k } mis' 20-2. w = { -8.0j + 4.0k} rad/s a = {321} rad/s2

20-3. vA = { -5.201 - 12j + 20.8k} ft/saA = {-24.11 - I3.3j - 720k } ft/s2

2 0 - 5 . w = 4 1 .2

r a d / sv p = 4 . 0 0 m / s a = 4 0 0

r a d / s ' a p = l 0 0 m / s ' '20-6. vp = { -1.601} m/s

ap = { -0.6401 - 12.0j - 8.00k} mis"20-7. vp = {-7.

791 - 2.

25j + 3.90k } ft/sap = {8.301- 35.2j + 7.02k) ft/s2

20-9. vB = { -7.U61- 7.52k} ft/saB = {77.31 - 28.3j - 0.657k } ft/s"

20-10. y8 = {-90,1 - 15j + 6k } ft/saB = {2431 - 1353j + 1.5k } ft/s2

20-1L vB = {4101 - 15j + 6k} ft/saB = {2931 - 1353j + 1.5k} ft/s2

20-13. vA = {51}ft/saA = { -1.51 - 6j - 3,21t } ft/s2

20-14 vc = { 1.8j - 1.5k} m/sac = {-36.61 + 0.45j - 0.9k} m/s"

20-15. vA = - 8.661 + 8.00j - 13.9k} ft/saA = (-24.81 + 8.29j - 30.9k } ft/s2

20-17. vA -= [100j - 40k] ft/saA = {-5801+60j - 30k } ks'

20-18. wp = { - -10j } rad/saB = { -64001} rad/s"

20-19. w = {4.351 + 12.7j} rad/sa = { -26.1k } rad/s2

20-21. w = {30j - 5k }rad/sa = {1501 }rad/s"

20-22. vA = { 101 + 14.7j - 19.6k} ft/saA = (-6.12i+ 3j - 2k } ft/s'-

20-23. wA = 47.8 rad/swg = 7.78 rad/s

20-25. wBc = {0.2041 - 0.612j + 1.36k } rad/svB = { -0.333j} m/s

20-26. w = {0.9801 - 1,06j - 1,47k } rad/svB = 12.0 ft/s

20-27. a5 = {-96.51} ft/s2

20-28. vB = [ -I.33i + lj] m/s20-29. aB = {4.991 - 3.74j} m/s2

20-30. wBc = { -2k} rad/s20-31. ww, = { -2.001} rad/s20-33. vB = 4.71 ft/s

wAB = {1.171 + I .27j - 0.779k} rad/s20-34. aB = 17.6 ft/s2

= { -2.781 - 0.628j - 2.91k } rad/s2

20-35. 0.12c = {0.7691 - 2.31j + 0.513k} rad/svB = { -0.333j } m/s

20-37. vc = -1.001 + 5,00j + 0,800k} m/sac = { -28.81 - 5.45j + 32,3k} m/s'

20-38. vc {-li + 5j + 0.8k} m/sac = { -28.21 - 5.45j + 32.3k } m/s2

20-39. vB = {-2.751 - 2.50j + 3.17k} m/saB = {2301 - 2.24j - 0.00389k} ft/s2

20-41. vA = {5.201 - 5.20j - 3.00k } ft/sa4 = {251 - 26.8j + 8.78k } ft/s2

20-42. vB = { -17.81 - 3j + 5.20k } m/saB = {9i - 29.4j - 1.5k) mis'

20-43, vB = { -17.81 - 3j + 5.20k} m/saB = {3.051- 30.9j + 1.10k} m/s'

20-45. vc = { -6.751 - 6.25j } m/sac = {28.751 - 26.25j - 4k } m/s2

20-46. vB = (-17.31 + 18.8j + 10.0k) ft/saB = { -19.1i + 24.0j - 8.66k} ft/s"

20-47. vB -17.31 + 18.8j + 10.0k) ft/saB = { -45.01 + 24.0j - 6.34k}

ft/s' 20-49, vc = -2.71 - 6k m/sac = {-721 - 13.5j + 7.8k} mis"

20-50. vp = -25.51 - 13.4j + 20.5k } ft/sap = {1611 - 249j - 39.6k } ft/s2

20-51. vp = { -25.51 - I3.4j + 20.5k} ft/sap = 1611 - 243j - 33.9k) 20-

53. vi, = { -10.21 - 28j + 52.0k} m/saB = { -33.01 - 159j - 90k } m/s"

20-54. vc {1.81 + 0.9j + 2.25k) m/sac = {0.91 - 9.075j - 12.75k) m/s2

Chapter 2121-2 = (h' + 4<J2) 80

= -20(2h- + 3a-)

R2-45.R2-46 .

R2-47 .6 =

vc = 1.34 m/s

tan

ANSWE.R5 TO SELECTED PROBLEMS 7 1 7

1 ,21-3. IY = -3mr

11. = 176 (r2 + 3a1)

m2 1 - 5 . l , - 6= -ah-

21-6. I" = 2412

21-7. /, = 636p2 , 1 ,- . m r - m n -

4(

I , 2 ,21-9. -ma- -ma-

4 3I , I ,4m 4

- a " - - m a -

21-10. I. --- f'., (a2 + /12)

21-11. I,„, = - (3a2 + 4h2)12

21-13. /5., = 021-14. 4,. = 0.32 kg - m2

l',.,= 0.08 kg • tri2/„ = O

21-15. 4. = 0.0966 slug - ft2/,. = 0.0483 slug • ft2

I„ = 0.0372 slug • ft2 21-17. 5, = 0.5 fi

= -0.667 ft/„., = 0.0272 slug • ft2I,., = 0.0155 sług • f1.2

1 ,\ - rrbrI,-- ma-42 , -3m.fr /

21-43.

21-45.

21-46.

21-47.

21-49.

21-50.

21-51

21-53.

21-54.

M , = O

21-37. VG = { -10.2k } ft/s21-38. H = 0.625 kg • m2/s21-39. w,. = 28.7 rad/s21-41.

£Mx = (111:ox Iz,,w,. - 4:wz) - OM" - /N:-(02 lyx£0,r)+- - /„.,.w,)B. = 4 lbA, = -2.00 lb A,. = 0.627 IbB, = 2.00 IbB. = -1.37 lb A z = 1.46 lb Ax = Bx = O Bz = 13.5 lb FA = 277N FB = 166N FA= 213N FB = 128N A, = B, = OA. = B, = 24.5 ND,, = -12.9 N

= 200 rad/s2

Dx = -37.5 NC, = -37.5 N

= -11.1NC. = 36.8 N(;),,. = -102 rad/s2

A, =- B, = OA, = OA, = 297 NB. = -143 N

/,..XM., = .,:trG

P i, ,- G

,A,/Y = P2rt..

itr,. = 25.9 rad/s2

/3 - (I,., + 4-5, + 1:.:)/2 + (1-riyy + Iw 1 zz + 1=4._ 1:'[.'s - ' 1. - 1k )1 B, = -0.0791 N

B. = 12.3 N- (4.41,7 - 243,/yzi‚v - /exk: - /.1,3,Ps- -- /7.A) = O A, = -1.17 N

A. -= 12.3 N21-25. 9/rth2[ 1 , 21-55. Ay = OT "' 20 L I + 6h'i m- Az = Bz = 98.1 N

21-26. ar, = 61,7 rad/s Ax = -3.75 kN21-27. w, =- 87.2 rad/s Bx = 3.75 kN21-29. wol, = 15,1 rad/s 43 21-30. h = 2.24 in. 21-57. M, = --ml-w.l ue cos O21-31. vA = 18.2 ft/s I .,21-33. wp -= 4.82 rad/s M,. = -ml- w 2 sin 2021-34. HA = { -2000i - 55 000j + 22500k) kg • m-/s 3 p

21-35. T = 37.0 M./ M. = O

I. -= 0.0427 sług - ft2 21-18. I.,_ = 4.50 kg • m-

1,. = 4.38 kg • m2I = 0.125 kg • m2

21-19. 1„ = 1.13 slug - ft2

21-21. J. = 0.429 kg • m2 21-22.

718ANSWERS TO SELECTED PROBLEMS

21-58. A, = -1.09 kN8, = 1.38 kN

21-59. SM, -= OSM,. = (-0.036 sin tł) N • mM. = (0.003 sin 20)N • m

21-61. a = 69.3°p = 128°y = 45°No. the orientation will not be the same for anyorder. Fin te rotation s are not vectors.

21-62. (a) A, = 1.49 kN= 2.43 kN

(b) A„ = -1.24 kN= -5.17 kN

(e) A,. = 1.49 kN= 2.43 kN

21-63. M, = -56.25 N • m21-65. w.y = 105 rad/s21-66. AF = 53.4 N

21-67."gcot0 y/2reos0.11

21-69. wp = 1.19 rad/s 21-70. w< = 3.63(103) rad/s21-71. H = 68.1°21-73. 4 = 23.3 rad/s 21-75. ¢ = 12.8 rad/s 21-77. Ir.f) = 17.8 rad/s

df = 22.3 radis/3 = 19.8°

21-78. Ho = 0.352 kg • m,-/s = 35.1 radis

21-79. 11,; = 17.2 Mg • m2is

Chapter 2222-1.= -0.05 cos( 12.20 m22-2.f = 4.98 Hz

r = 0201 s22-3.y = 0.

107 sin (7.00ł) + 0.100 cos (7.000cf) = 43.0°

22-5.= 49.5 rad/sT = 0.127 s

22-6.x = -0.126 sin (3.16!) - 0.09 cos(3.16t) mC = 0.155 m

22-7.w„ = 19.7 rad/sC= lin.y = (0.0833 cos 19.71) ft

22-9.x -= -0.05 cos (8.160C = 50 mm

1 li;

22-11. T = 1.18 s1kU+ dl

22-13.= 21rgri

22-14. d = 1-l6 mmkG = 0.627 m

22-15. T = 2rg

22-17. Let ki be the larger value.ki = 2067 N/mk, = 302 Nim

22-18. r = 0.167 s22-19. (on = 9.47 rad/s22-21. y = 1.10 sin (5.481) m

3g ( 4R2 - f2).?-

22-22. wp =

= 2.67 s+ k2

co„ -T = ‚ i rkojE

C

22-29.

22-30. :ii + 333x = O22-31. r = 1.52 s22-33. T = 0.339 s22-34. T = IT -k-

i l ;

22-35. T = 0.487 s 22-37. ło, = 11.3 rad/s 22-38. w,, = 14.1 rad/s

F22-39. T = 4.25 \.

g22-41. y = A sin w,it + B cos wnt coswt

22-42. c = 18.9 N- s/rn 22-43.y = (-0.0232 sin 8.971 + 0.333 cos 8.97: + 0.0520 sin 4t) ft

8022-45. y = A sin cd,/ + B cos w„t + , sin

_ )(1)1,

22-46. y = (361 sin 7.75: + 100 cos 7.75:- 350 sin 81) mm

22-10. mr

6 R 2 1 2

22-23.

22-25.

22-26.

22-27.k a = \ I ( 7 2 2

M r ,= - Ti

47T2k -

T i -

= rr' 2 7 r L

Fo

(k - mw2)

ANSWERS TO SELECTEn PROBLEMS71 9

22-66. MF = 0.99722-67. v = E-0.0702 e 3.STrsin (8$4« m22-69. 0,„,0.106 rad22-71. w -= 21.1 radi22-73. x = 0.119 cos (6/ - 83.9°) ] rn22-74, c, = \.78/161e'

[M I

rIt12.%\ 1 2 k ( M + M ) 1 v i j

2m .0022-75.ri., - „ ,-wg.1(2-Nie,k(m+m}-c2)

V8k(m + M) - c2

22-77. 1,q + Rq + ()q = E0 cos. 4.40t

22-78. Lii + R4 +(Cł. q = E0 cos wt1

2 2 - 7 9 . L e j - q = 0 C

22-47,= 2.83 cadfs22-49.(xp),„„., = 29.5 nim22-50. (xp)„. = 1.89

3F0

22-51. C = :41(2 mg + Lit) - m Lw=

22-53. w = 14.0 radis22-54. (...x,) max = 14,6 mm22-55. (xp)„,,„, = 35.5 mm22-57. w = 19.7 rad/s22-58. C = 0,490 in.22-59. w -= 19.0 radis22-61. w -= 6,55 rad/ s

22-62. Y =no2 L3 48E1 - MaJ2L-3

22-63. rd = 0,666 s22-65. 43" = 9.89°

Indexa-s (acceleration-position) graphs, 21

12-t (acceleration-łime) graphs, 19-20

Absolute accc 'era tion, 393

Absolute (dependent) motion arta lysis.81-86,105,329-332.392

particles. 81-86, 105

procedures for. 82, 329

rigid bodies, 329-332,

392

rotation and transtatión for,329-332, 392

Absolute velocity, 338

Acceleration (a), 7-8, 19-21, 33.35, 54-56, 69.88, 102, 106-167, 313, 315, 317-318, 363-370, 380-381.393-453,534. 551-552,568

absolute, 393Angular (a), 69, 315, 534, 551-552

average. 7, 33

circular motion and, 363-365

con sta n t, 8. 315. 534continuom motion and, 7-8

Coriolis, 38I

curvi1ine.ar motion and, 33, 35. 54-56, 69

cylindrical components and, 69. 144-148, 167

es.tuations of motion ion for, 114-120,131-136, 144-148.167. 409-418,425-431, 440-445, 453

erratie motion and. 19-21fixed-axis rotation and. 425-

431. 453.534Exed-point rotalion and. 551-552

force (F) and, 106-167, 394-453

general piane motion and, £140-145

graphs of variables, 19-21, 102

hodograp hs and. 33 instantaneous. 7, 33

kinernatics of particie s and, 7-8.33.35.54-56, 69. 88. 102

kinetics of p artieles, 106-167

magnitude and, 35. 363.380

mornents of inertia (I) and,395-403, 440-441, 453

norma' (n) compone n ts of, 54-56, 131-13b,167

norm al force (N) and, 144-148

planar kinetics of rigid bodies,394-453.534

planar kinem atics ot rigid bodiesand, 313, 315.317-318

rectangular componems and. 35, 1141-1211, 167

rectilinear kinematics and. 7-8.19-21.102. 298

relative-motian analysis and. 88.363-370,393.568

rotatiug axel, 380-381, 393. 568rotation and, 315, 317-318,410-411.

425-431.453

sign convention for, 7tangertitial (t) components. 54-56.

131-136,167

tangential force and, 14,1-148

three-dimensionaI rigid-body motion, 551-552, 568

translating axel, 88, 568

traostatioo and, 313, 363-370.380-381, 393, 4119, 412-418, 453

variable, 534velocity (y) and, 7-8

Amplitude trf vibra Gon. 633-634

Angular aceeleration (a), 69, 315, 534, 551-552

Angular displaeement (d9), 314

Angular momentom (H},262-271.297.496-5{)2, 532-533.589-592, ó17.628

angolar impuise and, 266-271,

297 arbitrary point A for, 590

center of mass (G) and. 590

conservati on of, 268, 517-520,

533 cDnstaat, 298

eccent ric irnpact and. 521-524.533

fixed-axis rotation and. 498.532

fixed-point O for. 590

free-body diagrams for.262-263.268

general piane modem and, 499.

532 gyroscopic motion and, 617

kinetics of a particie, 262-271 ,

297 magnitude of. 262

moment of a fotce relations with. 263-265

moment of momentum, 262, 297principle of, 266.-271, 297,501-508,

592, 628

procedures for analysis of. 268. 503,518

rectangular com pon en ts of momentom. 590-591

rigid-body planar morion (H).49b-502.532-533

scalar f ormul at ion. 262, 267

system of p article s. 264

three-dimens i ona I rigid bodies.589-592.617.628

tran slation and, 498,

532 variable. 298

vector formulat i o n, 262.267A n2ular motion, 314-315, 534

Angular position (0), 314 Angular

velocity (w). 68. 314.551

721

722INDEX

Apogee, earth orbit, 160

Areal veloo ty, 155

Average accele rałion , 7.33

Average speed, 6

Average ve lochy. 6. 32

Axes. 87-91. 105. 299. 312. 314-321.3374344. 377-385, 392-393.

425-43 i 453. 457. 491, 498,532, 582-583, 602-604

angular motion and, 314-315coordinating fixed and translating

reference fram es, 337-344.393uation s of m ofion for, 425-43 I , 453, 602-604

fixed reference frame, 87-91, 312,314-321, 392, 602-603

inertia (I), principle axel of. 582-583

kinematics of a particie. 87-91,

105 kinetie energy and. 457. 491

momentum and, 498, 532

moments of inertia (T) about, 583

pinned-end memhers, 337-344, 363-370

planar kinemałics of rigid bodies.312. 314-321, 337-344, 377-385, 392-393

relative-motion analysis using.87-91. 105, 377-385, 393

rigid-body planar motion, 425-431,453.457.491, 498, 532

rotati n g, 377-385. 393rołation about, 312, 314-321, 392,

425-431, 453, 457, 491, 498, 532symmetrical spinning axes, 603-604three-ffimensional rigid-body

motion, 602-604

translatin g. 87-91, 105. 299.337-344, 393

Cartesian vector notation, 673

Center of mass (G). ł 13,593

Center of percussion, 508

Central-forte motion, 155-162, 167

areal velocity, 155

circular orbit. 159

directrix, 157

ecce n łricity (e). 157

elliptical orbit, 160-161

fotos (F), 157

grayitational (G) attraetion, 156-157

Kepler's laws. 16I path uf

motion, 155-156

trajectories, 156-162. 167

Central impast. 248-251.297

Centripetal force. 131-136 Ch

ai n Tule. 677-679

Circuiar motion, 338-339, 351-357,363-365, 393

acce lera tku] (a},363-365instantaneous center (IC) of zero

velociły, 351-357, 393

planar rigid-body motion. 316-318,338-339, 363-365

procedura for analysis of, 353relative-motion analysis of,

338-339. 363-365rotation and, 316-318

veloci ty (y). 338-339. 351-357, 393

Circular orbit, 159

Closed volume. 277Coeffic ten t of restitution. 249-250,

297 Coefficient of viscous da rn ping,

655

Composite bodies, moments of inertia for, 401

Conservation of energy, 205-209, 219.477-482. 493, 645-648, 668

conservative forces and. 205-209.219, X477-478.645

kinetics of a particie. 205-209, 2 ł 9potential energy (V) and. 205-209.

219.477.45x3

procedures for analysis using, 206. 479, 646

rigid-body planar motion.477-482, 493

system of pani cl es, 206

vibrat łon and, 645-648, 668

weigh t (W), displacement of, 205work (W) and. 205-209. 219,

477-482.493

Conservation of momentum, 236-242,296. 517-518. 533

angular. 2.68. 517520.533impulse forces and, 236-237 linear, 236-242.296, 517-520. 533 particie systems, 236-247. 268. 296

procedure s for analysis of, 237, 268, 518

rigid-body planar motion.517-520,533

Conservative furce, 201-209. 219,477-482, 493, 645

con servati on of energy. 205-209.219.477--452, 4y3.645

potentiai energy (V) and, 201-204,219, 477-482 493

spring force as, 2.01-202. 219

vibration and. 645

weigh I (W). displacemen of,241-204.219

work (U) and. 201-204,477-482. 493

Constan i acceIeration. 8. 298. 315, 534

INDEx723

Consłant forte, work of, 171, 201, 218, 458.492

Constant velocity, 7Continuous motion, 5-14

accelerat i on (a), 7-.8

displacernent (A), 5

par licie rectiline ar k inc małics and,5-14

position (s). 5, 8procedure for analysis of, 9

velocity (v), 6-8

Control voIumes, 277-286.297

gaim °f mass (m). 283-284

impulse and mornenturn and,277-256, 297

kinernatics of a particie.277-286., 297

loss of mass (m).282-2.83

mass flow, 278-279,282-?84,297

procedure for analysis of. 279

propulsion and. 282-286.297

steady flow, 277-281, 297 thr

ust, 282-283

vol umetrie flow (discharge), 279

Coordinates, 5, 34-36, 53-59, 67-74, 87-94. 103-105, 114-120. 131-136, 144-148, 167, 298-299, 556-558, 390-591, 600-609, 629

acceleration (a) and, 35, 54-56.69. 114-120, 131-136.144-148, 167

angular momentum (H) and, 590-591

c urvilinear motion, 34-36.53-59.67-74, 103-104, 298-299

cylindrical (r. 6. z), 67-74, 104,144-148, 167. 299

dependent motion analysis and, 81-86, 105

equations of motion and,114-120, 1M-136. 144-148,167, 600-609, 629

fixed origin (0), 5

kinernatics of a particie. 5.34-36.53-59, 67-74, 103-105

kinctics of a particie, 114-120,131-136. 144-148.167

norrn al (n), 53-59, 104.131-136. 167

plan ar moli on, 53-

55 polar, 67-70

position-coordinate equations, 81-86.105

position ve dor (r), 34, 68

procedures for analysis using. 36.56.70.114-115. 132, 145

radł al (r). 67-68rectangular (x, y, x), 34-36.

103. 114-120, 167, 590-591, 600-609, 629

relative-motion analysis and,87-91.105.299

tangenttal (t), 53-59, 104,131-1.36,167

rec-dimen sional motion,55. 556-558. 590-591,600-609, 629

translating axes and, 87-91. 105

lranslating systems, 556-558 Ira

nsverse (0). 67-70

veloc1ty (v) and, 34-35,53, 68

Coriolis accelcration, 381

Couple moment (M). work (W) cd a,460-461.492

Critically damped vibration systems,

565 Cross produci. 673-674

Curvilincar m otton.. 32-39. 53-59, 67-74, 103-104

acceleration (a). 33,35.54-56, 69

coordinates for, 34-36,53-59,67-74,103-104

cylindrieal (r. 0,coordhlates.ó7-74.104

displacement (A), 32

general, 32-39

normal (n) axes. 53-59. I 04

kinematics of a particie, 32-39,53-59, 67-74. 103-104

planur molton, 53-55polar coordinates, 67-70. 104

position (s), 32, 34. 68

procedures for analysis of. 36, 56, 70

radius of curvature (p), 53

rectanguIar (x, y, z) coordinates, 34-36.103

tangential (t) axes, 53-59, 104

finie derivatives of, 70, 104

threc-dirnensional morion. 55

velocity (y). 32-35. 53, 68

Curyilinear translation, 312,392, 413

Cylindrical (r. O, z) coordinates„ 67-74.104:144-148, 167, 299

acceierałton (a) and, 69,144-148, 167

curvilinear molion, 67-74, IN, 299directional angle Ok), 144-145

cylindrical (r. fl. z) coordinates.144-148, 167

equations of motion and,144-148. /67

norma] force (N) and. 44-148

polar coordinates for, 67-70.104

procedures for analysis using, 70,145

724 INDEX

Cylindrical (r, 6, z) coordinates (cont'

d) tangential force and, 144-148

lim e derivatives of. 70.104

ve1ocity (v) and, 68

D'Alembert prin cipie. 110

Damped vibrations. 631,

655-660, 699

eritically, 565

motion of, 631viscous forced, 658-660, 669

viscous free, 655-657, 669

Damping force. 655

Das hpot. 655

Deceleration,7

Dc form ation, 174-175, 177. 248-251,297, 521-523, 533

coefficient of restitution (e).249-250, 297, 521-523. 533

Impact and, 248-251, 521-523, 533

maximum. 248

period of, 248

principles of work and energy and, 174-175, 177

restitution and, 241-240.521

separation of contact points, 523

Diagrams for impułse and momentum, 223-224

Dire efi on al angle (4/), 144-145

Directrix, 157

Displacement (d). 5, 32, 314, 316..337,459, 632-634, 653

angolar (cki), 314kinentatics of a particie 6,), 5, 32

periodic supporl and, 653

plan ar kinem atics of rigid bodies and. 337

mtation and, 314,316,337

transl ati on and, 337

vertical, work o f a weigla and, 459

vibrations and, 632-634. 653

Dot notation, 34-35Dot produc L. 170, 675-676

Dynamie equiIibritun, 114

Dynam ies. 3-4

principl es of, 3-4

procedure for problem soiving, 4

Ecce trie impact, 521-524,533

Eceentriciily (e). 157

Efficiency (e), 192-195, 219energy (E) and. 192-195.219

mechanica1, 192-193

power {P} and. 192-195, 219

procedura for analysis of. 193

Elastic impact, 250

Elastic poien bal energy. 202. 219, 477, 493

Ele ctrical circuit analogs, vibrations and, 661, 669

Eiliptical orbit, 160-161

Energy (E), 168-219, 300, 454-493,536, 592-595, 628-629

conservatio n of, 205-209. 219.477-482, 493. 536

efficiency (e) and, 192-195, 219

iniernal. 177

kinetic, 174-175.218, 455-458,491, 592-595. 628-629

kinetics of a particie.168-219, 300

potential (V), 201-204. 219, 477,493

powcr (P) and, 192-165.219

principłe of work and, 174-182,218-219. 462-468,493

procedure s of analysis for. 175.193, 206, 463,479

rigid-body planar motion and,454-493.536

wark (U) and, 168-219,454-493.536

Equations of morion. 108. 110-120. 131-136, 144-148, 155-162. 167. 300,409-418.425-431, 440-445, 453.535-536. 600-609, 629

central-force motion, 155-162, 167

centripetal force, 131-136

cylindrical (r, 0. z) coordinates.144-148,167

equilibrium and. IW

external force, 112-120

fixed-axis rotation, 425-431.453, 536, 602-603

free-body diagrarns for.110-111, 167, 410-412

friction (F) furce. 115. 144 gen eral

pIane muflon, 440-445.453

gravitabon al attraction, 108-109

inertial reference frame for,110-111.167internal lorce, 112-113

kinetic diagram for. 110

kinetics of a padicle, 108, 110-120,131-136, 144-148,167, 300

moments of inertia (1) and, 440-441

Newton 's second law, 108, [67

normal (n) coordinates,131-136, 167

normal (N) force, 144-148, 167

procedure s for analysis using.114-115, 132, 145, 414. 427, 441, 604

INDEX725

rectangular (x, y, z) coordinałes,114-120, 167. 600-609, 629

rigid-body planar muflon.409-4 I 8. 425-43 h 440-445. 453, 535-536

rotaticrnal, 410-41 I, 425-431, 453, 6011.-60 I

slipping and, 440 145

spring force. 115

syrnmetricat spinning axes, 603-604

systems of partieles, 112-113

tangential (t) coordin tes,131-136, 167

tangential force, 14-4- / 48, 167

three-dimensional rigid bodies.600-609. 629

trajectories, 155-162translationa l, 409, 412-418,

453.600

Equilibri um, equation of motion and. 110

EyuilIhrium position, vibrations. 632,634

Erratic morion, 19-24

a-s (acceleration-position) graph for, 21

a-1 (acceleration-time) graph for, 19-20

particie rectilinear kinematies for, 19-24

s-t (posidon-tirne) graph for, 19-20v-s (veIocity-position) graph

for. 2

v-t (veloeitv-time) graph for. 19-20

Euler angles, 614

Euler's equations. 602-603

Euler's theorem, 550

External force, 112-120, 410-411

Fini te rotation, 550

Fixed origin (0). 5, 426, 590.593

Fixed-axis rotation, 312.314-321,392,425-431, 453, 457, 491.498, 532, 534, 602-603

acceleration (a) and, 425-431, 453

angular acceleration (a) 315, 534

angular displacement (d0). 314

angular pos 0 lon (0), 314

angular veloeity M.3 14

equations of motion for, 425-431. 453,602-603

Euler's equations for. 602-603irnpulse and momentom for.

498,532kinetic energy and, 457, 491

magnvtude of. 425-426

moment equation about point 0. 426

procedure for analysis of, 319,

427 reference frame, 87-91

three-drnensi on a I rigid bodies, 602-6133

Fixed-point rotation. 549-556. 577. 590, 593

acceleration (a) and, 552angular acce I crat i n n (a) of,

551-552angular momentom of,

590 angular velocity (0))

of, 551 Euler's theorern for,

550 finite rotation, 550

infinitesimal rotation.

551 kinetic energy and,

593

tirae derivatives for, 552-556, 577 velocity (v) and. 552

Fluid stream, steady flow of,277-281, 297

Focus (F). 157

Force (F), 106-167, 169-173, 201-209,218-219, 263-265, 394-453,458-459. 492. 631-632. 645. 651-654, Sty. also

Central -furce motionacceleration (a) and. 106-167.

394-453

angular momenlom relatioase willi, 263-265

centripetal, 13 I -136

conservałion of energy and,205-209, 219.477-478, 645

conservative. 2171--209.219.

477#78 constan t, 171, 201. 218.

458,492

equations of motion for, 108, 110,112-120, 131-136, 144-148,409-418. 425-43 I , 440-445, 453

exbernal, 112-120, 410-.411

fixed-axis rotation and.425-431.453

free-body diagramy for. 110-11 I, 167.410-412

friction (F), 115, I44. 177-178

genem] piane motion and,110

145 inertia vector. 110

internaI. 112-113.410

kinetic of a particie. 106-167,169-173,201-204.218-219

moments of a, 263-265

moments of inertia (1) and,395-403, 440-441, 453

External imputse force s, 228,216spheres. 549-556, 577

External work. I 77t hree-dimensional rigid bodies.549-556, 577, 590,593

726INDEX

Furce (eoni 'd)Newton 's laws and,107-I 09. 167normal (N). 144-148, 167

periodic. 651-654

potential energy (y) and,201-204, 219

resul tani, 177

rigid-body planar motion and,394-453, 458-459, 492

rohtion and, 410-411, 425-431,453

slipping (no work) and, 459.492

spring. 115, 172-173. 202, 218-219, 459, 492. 632

system of particles, 112-1 ł 3 tangential. 144-148, 167 trajectorles, 155-162, 167 translation and, 409,412-418. 453 unbalanced, 107-108 units of, 170

variable, 170. 458vibrations and, 631-632, 645, 651-

654

weight (W), 109, 171, 201-205.218-219, 459, 492

work (U) of, 169-173, 176-182,218-219,458-459,492

Forced vibratqons, 631,651-654. 658-660. 668-669

motion of. 631un dam ped, 651-654.668 vlscollS damped, 658-660, 669

Free vibrations. 631-639. 655.-657, 668-669

motku] of, 631un dam ped. 63ł-639.668 viscous clamped, 655-657.66y

Free-body diagrams. 110-1 I 1. 167,262-263.2b8, 410-412

Frequency (f), vi bration and, 632Friction force (F), 115, 144, 177-178

equations of motion for, 115, 144 work of caused by sliding, 177-178

General piane modo'', 311 329-393, 440-145, 453. 457, 491, 499, 532, 535. See aha Honor motion

absolute motion analysis for,329-332, 392

accelerati on (a). 363-370, 380-381, 393,440-445

equations of motion for.44[1-445.453

force (F) and, 440-445

impuise and momentom for. 499, 532

in stantaneo us center (IC) of zero velocity, 351-357. 393

kinetic energy and.. 457. 49 I

moment eq uation about the instantaneons center (IC), 440-441

proce dure for analysis of, 329.365, 382, 441

relative-motion analysis for,337-344, 363-370, 393, 535

velocity (v), 337-344, 393

General three-t:limensional motion.557-558,577

Graphs, rectilinear kinematie solutions using, 19-24, 102

Gravitational aceeleration (g). I09

Gravitat ion a I attraction (G), 108-109, 156-157

central-force motion and. 156-157

Gravitational potential e nc rgy, 201-202, 219.477,493

Gyroscopic motion. 614-619, 629

H odographs, particie acceleration and. 33

Horizontal projectile motion, 39-43

Horsepower (hp), unit of. 192

Impact, 248-254, 296-297,521-524,533

central, 248-250, 251.297

coefficien t of restitu1lon (e).249-250, 297, 521-523, 533

deformaticrn and. 248-251, 521-

523 eccent ric, 521-524,533

elastic, 250

kineties of a particie. 248-254, 296-297

line of. 248, 296, 521

oblig ue, 248, 251.297

plasti c, 750

procedure s for analysis of, 251

res lit u Gon from. 248-250,521

rigid-body planar motion.521-524.533

separation of contact points due to,523

1mpulse, 220-297.301,494-533, 536-537

angolar, 266-271.297,501-502

conservation of momentum and, 236-237,537

control vol umes, 277-286,

297 di ogram s. 223-224 ternal

[orce% 228, 236

impact and. 248-254. 296-297,521-524.533

intern al forces, 236-237Free-Right trajectory, 157 Newton's law of, 108-109

INDEX 727

kineties of a particie, 220-297, 301

linear, 221-228.296, 501-502

momentum and, 2z1-297.301,494-533, 536-537

principlc of momentum and.221-228. 266-271, 277-28 I .296-297. 501-508, 532

procedures for analysis cif, 224,237, 268.279.503

propulsion and, 282-286, 297

rigid-body plan ar m otion, 494-533, 536-537

ste ady flow and, 277-281, 297

Tnertia (ł), I I 0-111. 167. 395-403,4413-441, 453, 579-584. 628

accelcration (a) and, 395-403.440-441,4U

angular acceleration (a) and, 395

arhdrary axis, moment of aboul,583

ccm Eros ile bodic s, 401equations of matron and. 449-441

fotce vector, 110

kinetics of a particie, 110-111, 167

mass moments af, 395-403

moments (M) 0095-403.443 +11.453,580-583.628

theorem, 400-401, 473,581-552

parallel-plane theorem, 582

principle axel of, 582-583

procedure for analysis of, 397

produci of, 584-581, 628

radius of gyration, 401

reference framc, 11 0-111. 167

reference franie, 110-111. 167

rigid-body planar motion and,395-403, 440-441. 453

tensor, 582-583

tfircc-dimensi on al rłgid-body motion, 579-584, 628

voIume cle nients for integration of, 396-397

Infinitesimal rotation. 551

Instan tan eous acecleration, 7.33

Instan tan eous center (IC), 351-357.393, 440-441

moment equation about, 440-441 zero velocity, 351-357.393

Instantan QOUS velocity, 6, 32, 34

Inłernal energy, 177

Interna] force, 112-113. 410

Internal i rnpulse forces, 236-237

Kepler's Iaws, 161

Ki n cm at ics, 2-105, 298-393, 534-535,548-577, See u1so Pl a n ar motion

absolute dependent motion analysis,.81-86, 105

con lin uous motion, 5-14

coordinates for. 34-36. 53-56.67-70. 103-104. 556-558

curvilinear =dan. 32-39, 53-.59,67-74. I 03-104, 298-299

cylindricał (r. 9, z) coordinates, 67-74, 194

erraiic motion, 19-24fłxed-axis rotation. 312, 314-321 ,

392, 534graphs for solution of, 19-24. 102 normal (n) axes, 53-59, 104 particies and, 2-105, 298-309 planar, 310-393.534-535 p-dar coord i nates, 67-70 principle. of, 3

proce dures for analysis of, 9. 36.40, 56. 319, 329, 340. 353. 365, 382.569

projectilc motion, 39-43, 103rectanguł ar (x, y. z) coordinates,

34-36. 103

rectilinear. 5-15, 19-24, 102, 298rei ative-motio n a n ałysi s, 87-91,

105. 299, 337-344, 363-370.377-385, 393, 535, 566-573. 577

rigid bodies, 310-393, 548-577rotating axes, 377-385, 393.

566-573,577rotation and, 3 2. 314-321 .

337-344. 392. 534sign conventions for. 5-7tang.= lial (t) axes, 53-59,104 tbree-

di men sional m otion, 548-577

translating axes, 87-91, 105,552-556.566-577

translalion and. 312-313, 337-344, 392-393, 534

Kinetic diagram. 110

Kinetic energy, 174-175, 218.455-458,491.592-595, 628-629

cen ler of mass (G) for, 593

fixed-point for, 593

general piane ~jon and, 457.491

p articles, 174-175, 218

principle of work and energy,174-175, 218.593, 628-629

rigid bodies. 455-458, 49 I

rotation ahout a fixcd axis and, 457, 491

system of bod e s, 458

th ree-dimensi on al rigid-body motion, 592-595, 628-629

translation and, 457,491

728INDEX

Ki netics, 3, 10(r-219, 220-297, 299-301, 394-533, 535-537, 578-629. See afro Manat- rnotionn Space mech anics

acceleration (a) and. 104-1.67, 394-453

angolar momentum (H).262 271, 297, 496-502,532 533, 589-592, 628

central-furce motion, 155-162, 167

conservation momentum,236-247, 268, 296, 517-52G. 533, 537

conservative forces and. 2U1-2U9.219

control volumes, 277-286. 297

cylindricał (r, 9. z) coordinates,144-148,167

effi eney (e) and. 192-195. 219

e nergy (E) and, 168-219, 300,454-493, 536, 592-595, 628-629

equations of morion, 110-120.131-136,144-148. 300, 409-418,425-431,+10 145, 453, 535-536,

600-609,629 forte (F) and. 106-167, 169-173.

201-209. 218-219, 394-453free-body diagrams for. 110-1

11, 167. 410-412

gyroseopic ~ton, 614-619. 629

i m pact and, 248-254, 296-297.521-524,533

impułse and momentum, 220-297,301, 494-533. 536-537. 592, 628

Media (/). 110-111, 167.395-403.440-441, 453, 579-584. 628

Newton 's ławr and, 107-109, 167

norma] (n) coordinates.131-136. 167

particles. 106-297, 299-301

planar motk,. 394-533, 535-

537

powor (P), 192-195, 219 principle of, 3

procedures for analysis, 114-115,

427,441,6D4

propulslon. 252-286.297rect angolar (x, y, z) coordinates,

I 14-120. 167

Figi d-body planar motion. 394-533, 535-537

rotation and, 410-411, 4Z5--•331.453

steady tłow and, 277-281, 297

tangential (t) coordinates,131-136.167

three-dimension al rigid bodies. 578-629

torque-free motion. 620-623, 629

trajectories, 156-162. 167

lransiation and, 409. 412-418. 453. 534

work (U) and. 168-219. 300,454--493.536, 593, 628-629

Line of action, 352, 411

Line of impact. 248. 296, 521

Lincar momentami (L), 221-228,236-.242,296_ 495. 498-502,517-518. 532-533

conservation of momentu-m,236-242,517-518,533

agram s for. 223extern al furce and. 228, 236

fixed-axis rotation and, 498, 532 (

orce (F.) and. 221-228

generał piane motion and. 499.532

kinetics of a parficle. 221-228.236-242, 296

pr in cipie of, 221-228,501-508,532procedures for analysis of, 224,

237, 503

systems of partide s, 228.236-242. 296

translation and, 498, 532

Magnification futor (MF).652-653.659

Magnitude. 6, 32,34-35, 68, 262, 314,363. 378. 380.. 460. 492

acceleration (a) and. 35.363,

380 angular momentem, 262

angular motion and, 314

constant. 460, 492

couple moment (M). work of and, 460, 492

fixed-axis rotation and, 425-

426 speed as, 6,32, 34, 68

position vector (r) and. 34

rotaling axel, changes in. 378,

380

velocity (v). 6,32.34, 68, 378

Mass (m), 107-109, 112-113. 277-279,282-284, 297.393-403, 593

center (G) of, 113,

593 condnuity of, 279

control volumes and. 277-279.282-284, 297

floty, 278-279

gala of. 283-

284

gravitałion al at tracłton and, 108-109

lass of. 282-283moments (M) of inertia (ł), 395-

403

par i kle body. 107-109

132, 145. 175, 193.2U6, 224, rigid-body planar moiion, 495.237, 251, 268, 279, 397, 414, 498-500

INDEx 729

swady flow of fluid systems and, 277-278,247

system of particles and, 112-113

Mathernatteal expressions, 670-671

Maximum defomation, 248

diagrams, 223-224

impact and. 248-254, 296-297,521-524. 533

impulse and, 220-297.341 kinetics

of a particie, 220-297, 301

curvilinear motion and, 53-59

equations of motion and.131-135.167

Normal (N) furce, 144-1 48. 167

Nutation, 614-615

linear (L). 221-228. 236-242, 296,Mech ani cal energy, 205-209. See aiso 495, 498-502, 533

Conservalion of energyprinciples of impulse and, 221-228,

Moments (M), 395-403.140 441. 453, 266-271, 277-281501-508, 532. 592

.296-297,

460-461. 492, 580-583,628

scceIeratinn (a) and, 395-403,440-441.453

arbiłrary axis. about. 583ca m pos i te bodies. 401couple, 460-461, 492equations cif motion and. 441144/inerti a (I), 395-403, 440-441, 453,

460-461 , 492, 580-583, 628

mass, 395-403

parallel-axis theorem for, 400-101, 453. 581-582

parallel-plane theorem for. 582procedure for analysis of. 397

radius of gyrał i on for„ 401

rigid-body planar motion and.395-403. 440-441. 453

sli ppi n g and, 440-441three-dimensional rigid-body

modom, 580-583, 628volume elements for integration of,

396-397

'work (W) cif. 460-461, 492

Momentum, 220-297, 301. 496-502,532-533,589-592,62$

ang ułar (H), 262-271, 297.496-502,532-533. 589-592.628

con servati on of, 236-242. 268. -)96, 517-520,533

control vo I u rn es, 277-286, 297

procedures for analysis of, 224,237, 268. 279, 503, 518

propnista', and, 282-286.2y7

rigid-body planar modom496-5[12, 532-533

steady now and. 277-281, 297

systems of partieles, 228. 216-242. 264,296

chree-dimensional rigid bodies, 589-592, 628

Natura] frequency (w,), 632. 634,645-648, 668

N ew ton 's l aws, I 07-110, 167

body mass and weight from, 109equation of motion, 108, 167

first law °f motion, 1 I 0

gravitati on a I attraction. 108-109

kinetics cif particles and,107-106, 167

second Iaw of motion. 107-108,

167 static equilibrium and. 110

N onconservative furce, 201

Nonimpulsive forces, 236

Non rigid bodies, prin cipie of wark and energy for. 176

Normal (n) coordinate axes. 53-59.131-136,167

Oblique impact, 248, 251.297

Orbit, trajectory and, 159-161

Osculadn g piane. 53

Qvcrdamped vibrati on systems, 656

Parallei-axis theorem. 400-40.1. 453, 581-582

Parallel-plane theorem, 582

Particles. 2-309

absolute dependent motion analysis, 81-86, 105

acce I e ra tion (a), 7-8, 33, 35.54-56.69.88. 106-167, 298-299

central-forte motion of,155-162, 167

conservativc forces and.201-209, 219

ccm d n uous modom of, 5-14

cocrrdinates for, 34-36, 53-56,67-70,103-104.114-120,

131-136.144-148,298-299

eunihnear motion ot 32-39.53-59. 67-74,103-104. 298-299

displacement (A), 5, 32

energy (Ej and, 168-219, 300

equations of motion, 108.11U-120,ł31-13ó ,144-148,155-162. 167,300

erratic motion oL 19-24forte (F) and, 106-167. 169-173,

201-204, 219

730INDEX

Particles (cont'd)

free-body diagrams. I10-1 I L J 67

grayitational attracrion (G),108-109,156-157

hpdographs, 33

impast, 248-254, 296-297

impułse and momentum of,220-297,3x1

i n ertial reference frame,110-111, 167

kinematics of. 2-105, 298-299

ki netics of, 106-297, 299-301

mass (m).107--109

Newton's second law of motion, 107-109. 167

plan ar molion of, 53-55position (5), 5, 8.32. 34. 68.87

power (P) and, 192-165. 219

principle of work and energy for,174-182. 219, 300

projectile morion of, 39-43. 103

rectilinear kinematics of, 5-14.19-24. 102, 298

re Ia Live motion ust ng 11-ansi a Ling axes. 87-91, 105. 299

system of. 112-113, 176-182, 206, 228,236-242

t h rce-dim ensional motion of, 55

velocity (v), 6-8, 32-35, 53, 68, 87

work (U) and. 168-219, 300

Path of motion. 155-156

Perigee, carl]] orbit, 160

Period of deformation, 248

Period ic support displacement, 653

Phase angle (0), 634

Pinned-end m em bers, relati ve-mott on analysis of. 337-344, 363-370

Plan ar malim. 55-53,310-547absolute (dependent) motion

analysis, 329-332, 392

ac,celeration (a) and. 313.3 15,317-318, 363-370, 380-381, 393-453, 534

angular Biatlon and, 314-315, 393conservatio n of energy,

4 7 7 - 4 8 2 , 4 9 3

conservation of momentum,517-520, 533, 537

curvilin e ar. 53-55displacement, 314, 316, 337

energy (E) and. 454-493,

536

equations of molton for_ 409-418,425-431, 440-445. 453. 535-536

fi xed-axl s rota tion • 312, 314-321,

453.535

impast (eccentric), 521-524, 533impulse and momentum, 494-

533. 536-537

in stantaneous center of zerovelodty. 351-357, 393

kinernatics, 310-393.534-535

kinefic energy and, 455-458, 49I

kin eties, 394-533, 535-537

moments of inertia (I) and,395-403, 440-441, 453

position (r) and, 313-314.316, 337,377

princIples elf impulse and momentum, 501-508,532

prindple of work and energy,462-468, 493

procedures for analysis of. 319.329, 340, 353, 365.3A2, 397,414, 427,441, 463.479,503,518

relative-motion analysis, 337-344,363-370, 377-385, 393, 535

rigid bt dies, 310-453rotation and.410-411, 425-431,

453 rotating axes, 377-385, 393

translation, 312-313, 337-344.363-370,392,409,412-418, 453,534

velocity {v) and, 313-314.316,337-344, 351-357. 378-379, 393

work (U) and, 454-493,

536 Plastic impact. 250

Polar coordmates. 67-70. 104

Position (s), 5, 8, 21 32, 34, 68, 81-91,105.313,314.316,337.377.567

absolute dependent motion and. 8]-86

angular (0), 314continuous motion and. 5, 8

curviTinear motion and, 32, 34, 68

dependent-motion analysis and. 81-86. 105

erra t ic motion and, 21-

20 graphs of variables. 21

kiriematics of p articie s and, 5, 8, 32, 68, 87

magnitude and, 34planar kinematics of rigid bodiesand. 313. 314, 316, 337.377

posi tion-coordi n ate equations, 81-86. I05

rectangular componeMs, 34

relative-motion analysis and, 87-9],105, 337, 377, 567

procedures for analysis of. 9. 36. 392, 425-431.453. 498.532, 534

40.56. 82.88, 114-115, 132, forte ( F) and. 394-453,145, 175, I 93, 206, 224, 237, 458-461,492251,268,279 gener al, 312.329-393, 440-445,

In 731

rota ling a xes, 377.567

rotation and, 314, 316, 337

three-dimension a I rigid-body motki n. 567

time (t), as a function of. 8

tran slating axel, 87, 567

translation and, 313.337vectors (r), 32.34, 68, 87, 313,

337. 377

veiocity (y) as a function of, 8

Position coordinate (0).5. 426, 590.593

Potential energy (V), 201-204.219,477-482.493

conservation of energy and.205-209. 219. 477-4-82, 493

eon servati ve forces and, 201-204,219.477-482, 493

el asti c, 202, 219. 477. 493gravita tion al, 2.01-202. 219, 477.

493 particles, 201-204. 219

potenti a I function for, 203-204

procedure for analysis of, 206.479

rigid-body planar mnticsn,477-482, 493

spring fotce and, 201-202, 219.477

weight (W), displacement of,201-204, 219.477

work (U) and, 201, 203-204,477-482, 493

Power (P). 192-195. 21.9efficiency (e) and, 192-195. 219

energy (E) and, l 92-195, 219

procedure for analysis of, 193

units of, 192

Power-flight trajectory, 158

Precessi on, 614-616

Principle axes of inertia (1). 582-583

Principle of work and energy. 174-182.218-219.462-468, 493.593. 628-629

deformation and, 174-175, t 77

equation for, 174, 218

kinetic energy and. 174-175, 218, 593. 628-629

kinetics of partides. 174-182. 218-219

proccdures for analysis using. 175,463

rigid-body planar mot ion.462-468. 493

three-dimensional rigid bodies.593. 628-629

systems of pa rtides, 176-182

units of, 174

work of friction caused by siiding. 177-178

Principles of iinpulse and momentum, 221-228.

266-271, 277-281,296-297, 501-508, 332, 592, 628

angular momentum (H) and, 592, 628

diagram for, 223-224angular. 266-271, 297,501-508, 532

impulsc forces and, 228

k inetics of particies,221-228,266-27 I , 296-297

linear, 221-228, 296, 541-5U8.532

procedures for analysis uslng, 224, 268.279.503

steady flow and, 277-281, 297

systems of particies, 228

three-di mcnsi on a I rigid-body motian, 592, 628

Problem solvin g procedure, 4

Projectile motion. particie kinernatics and, 39-43, 103

Propulsion, 282-286, 297. See aiso Control volum e

Radial (r) coordinale, 67-68

Radius of curvature (p). 53

Radius of gyration, 401

Rectanguiar (x, y, z) coordinatcs,34-36, 103, 114-120, 167,590-591, 600-609. 629

angular momentum components, 590-591

curvil ine ar muflon. 34-36

dat notation for, 34-35

equations of motion and, 114-120,167,600-609.629

kin cm aties of a particie. 34-36,103

procedures for analysis using. 36. 114-115

threc-dimensional rigid-plano mation and, 590-591, 529.

600-609. 629

Rectilinear kinernatics. 5-15. 19-24, 102,298

acceierati on (a), 7-8. 19-21. 102.298con t inuous mo t i on . 5 -15

d i sp l ace rnen l ( a ) , 5 e r r a t i c

rno t ion , 19 -24 g raphs fo r

so lu t ion o f . 19 -24 , 102

particies and. 5-15, 19-24, 102.298

position (.9).3. 8, 19-21

procedure for analvsis of, 9

linie (ś) and, 8. 19-20. 102

velocity (v), 6-8, 19-21. 102

Rectilinear translation. 312. 392, 412-413

Reference Franies, 87-91,1 I 0-I 11. 167,312,314-321, 337-344, 392-393, 552-553

coordinating fixed and translaling axel, 337-344, 393

732 INDEX

Reference frarnes (eont'd)equations of motion and.

11Q-111,167fixed, 87-91.312, 314-321, 392, 553 inertial. 110-111.167 moving, 553

i hree-dimens kmal rigid-body motion, 552-553

Reladve-motion anałysis, 87-91, 105,299, 337-344.363-370, 377-385.393, 566-573, 577

aceeleration (a) and, 88. 363-370,380-381. 393, 568

cireular motion. 338-339.351-357. 363-365, 393

clisplacem c nt and. 337kinem a tics of a particie. 87-91,

105, 299

pinned-end members. 337-344, 363-370

position vectors (r) and. 87, 337. 377, 557

procedures for analysis using,340. 365, 382, 569

rigichbody planar motion. 337-344, 363-376,393

rotation and, 337-344, 377-384.393

rotatin g axes. 377-385, 393,566x573.577

three-dimensional rigicl-body motion, 566-573,577

translating axes. 87-91. 105, 299,337-344. 363-370. 393,566-573,577

iransl ati on for, 87-91, 105, 299.337-.344, 363-370.377-384, 393

velocity (v) and. 87, 337-344,378-379, 393, 567

R estit Mian, deforni ati on from im pac t and, 248-250

Resub ant furce, 177 Right-band rule, 262

Rigid bodies. 176. 310-493, 534-628

absol ule (dependent} motion analysis, 329-332, 392

acceleration (a) and. 313, 315,317-318, 363-370. 380-381, 393-453, 534

eireular motion. 338-339.351-357, 363365,393

conservat lon of energy,477-482, 493

conservation of rnomen [um.517-520. 533, 537

energy (E) and.454-493. 536equations of motion for. 409-418.

425-431,440-445.453, 535-536, 600-609, 629

fixed-axis rot ation, 312, 314-321.392,425-431.453,535

fixeo.1-point rotation. 549-556. 577, 590, 593

forte (F) and, 394-453,458-461, 492

free-body diagrams for, 410-412genc rat piane motion, 312, 329-39.

440445.453,491,532,535

gyToscopi c madom 614-619, 629

impulse and momentum, 494-533,536-537, 592, 628

inertia and. 579-584, 628

instantaneous center (IC) of zero velocty. 351-357, 393

kin em atics of. 310-393, 534-535. 548-577

kinetic energy and, 455-458. 491,592-595, 628-629

kinetics of, 394-533. 535-537. 578-629

moments of inertia (1) and,395-403, 440-441, 453

planar motion, 310-453prindple of work and energy,

176. 4-62-468. 493

procedllreS for analysis ol. 319.329, 340. 353.365. 382. 414,427,441, 463,479, 569. 604

relative-motion anaTysis,337-34-4. 363-370, 377-385,393. 565-573. 577

rotating axes, 377-385, 393,566-573, 577

rołation. 312. 314-321, 337-344,392, 410-41 I , 425-431, 453,457, 491, 498. 532, 534

systems of, 176, 458 ihree-

dimension al. 548-629 fime

derivatives for. 552-556, 577

iorque-fre e motion. 620-623, 629

translating axes. 566-573,577

translation, 312-313, 337-344,392-393, 409, 412-418, 453, 534

velocity (v) and, 313-314.316,337-344, 35/-357. 378-379. 393

wark (U) and. 454-493, 536

zero velociły, 351-357.393

velocity (y) of. 567

Rotating axes, 377-385.393, 552-556,5b6-573,577

acceleration (o) of, 380-381, 568

rnagnitude changes and. 378.380

posł ti(m vectors (r) for, 377.567

procedure for analysis of. 382, 569

relati ve--rn otion analysis using.377-385, 393, 566-573, 577

threc-diiWcnsinna1 motion and,552-556, 566-573. 577

time derivatives for systems, 552-556-

velocity (v) of, 378379, 567

Rotation, 312, 314-321, 329-332, 337-344. 351-357, 377-384, 392-393, 410-411, 425-43 I „ 453, 457, 491, 498,532, 534, 549-558, 577. 600-604, 629

synumctrical spinning axes, 603-604

three-dimensional rigid bodies.549-558, 577, 600-604, 629

linie derivatives for, 552-556

tran siatki n and, 377-384

veloeity (v) and. 314.316.338-339. 378-379

INDEX733

t h rust, 282-283torque-free motion, 620-623, 629 t

rajectories. 156-162. 167

Speed (rn agni ude), 6.32

Spheres. fixed-poi nt rotation and.549-556, 577

Spin, 614-616absolute (dependent) motion

analysis. 329-332.392

acceleration (a) and, 315, 317, 380-381

angular motion and, 314-315

circular motion and, 316-318.351357,393

displacement and, 314, 316.337

equations of motion for, 410-411,425-431, 453, 600-604, 629

finitc, 550

fixed-axis, 312, 314-321, 392,425-431.453, 457. 491. 602-603

fixed-poin t. 549-556. 577

ge eral hree-dimen skin al motion, 557-558, 602

impulse and momentum and, 498, 532

infinitesimal, 551

instantaneous axis of, 551-552

instantaneons center of zera veloei fy, 351-357.393

kinetic energy and, 457.491

position and, 314, 316.337

procedures for analysis of, 319,329, 340, 353, 382, 604

relati ve-m o lian analysis, 337-344, 377-384, 393

rigid-body planar motion and, 312, 314-321, 337-344, 392, 410-411, 425-131, 453, 457, 491, 498,532, 534

(position-time) graph s, 19-20

Scalarformufation, angular momentum. 262, 267

Separation of coni ad points after impaet, 523

Simple harmonie morion, vibrations, 632

Slidłng, work of friction by, 177-178

Slipping. 440-445, 459.492equations of motion and, 440-445 forces that do no work. 459, 492 zero velocity and. 459

Space rn echan ics, 155-162.167, 282-286.297.579-584. 620-623. 629

central-foree motku' and,155-162,167

circular orbit, 159control vol um e of particles.

282-2$6.297elliptical orbit, 160-161 free-flight trajectory, 157 inertia (1) and. 579-584

kineties of partielcs and,155-162, 167

parabolic path. 159 power-flight trajectory, ! 58 propulsion, 282-286.297

t hree-dim en sio n al rigid-body motion and, 579-584,620-623. 629

Spinning axcs, 603-604

Spring tarce, 115. 172-173, 201-202,218-219, 459. 477, 492-493, 632

conscrvation of energy and, 477, 493

canservative furce of 201elastic potenlial energy and, 202,

219, 477, 493

equation of motion for. 115p artieles, 115, 172-173, 201-202,

218-219

rigid-body planar m o [łon, 459,477.412-493

vibrations and, 632work of, 172-173.201.21$, 459,

477, 492-493

Static equilihrium, 110Steady flow, 277-281, 297

closed valume, 277 control valume, 277-281, 297 mass flow, 278-279

prineiples of impulse and =inent= for, 277-281. 297

procedure for analysis of, 279 vol um etric flow (discharge), 279

Steady-stafe vibrati on , 658

Syrnmetrieal spinning axel, 603-604

Systems, 112-113, 176-182, 206, 228, 264.458, 552-556, 577. S2e algo Vihratiahs

734INDEX

Systems (cont'd)angolar momentum of. 264

conservation of energy, 206

conse ryz tive forees and.

206 equation of motion for,

112-113 kinematics of, 1 r2-

113 kin et ic energy and. 458

kineties of, 176-182.206,228, 264

nonrig1d bodies. 176

particles. 112-1 ł 3, 176-182. 206, 228,264

principle of impulse and momentum for, 228

principle of work and energy for, 176-182

rigid bodies, 176, 552-556,

577 sliding and, 177-178

time derivatives for. 552-556, 577

translatingfrotating, 552-556, 577

work of fricti on and, 177-178

Tangential (,t) coordinate axes, 53-59.131-136, 167

curvilinear morion and, 53-59

equations of motion and,131-136,167

Tangential Tarce, 144-148, 167 Three-

dim ension al morion, 55, 548-629

angolar momentom of, 589-592. 617. 628

curvilinear. 55

equations of motion for,600-609, 629

fixed-point rotation, 549-556.577

frames of reference for, 552-553

generał muflon of, 557-558, 577

gyrostopic motiOn, 614-619, 629

impułse and momentom, 592.628

in ertia and, 579-584, 628

kinematics of, 548-577

kinetic energy of, 597-595.

628-629 kinctics of, 578-629

particles. 55

principle of wosk and energy of, 593, 628-629

proecdores for analysis of, 569,604

rectangular (x, y, z) coordinates,590-591, b00-609.629

reIative-rnotion analysis of,566-573,577

rotating axes, 552-556.566-573, 577

time derivatives for, 552-556, 577

lorque-free motion. 620-623, 629

translating axes, 552-556. 566-573

transIatingirotating systems,

continuous motion and, 8

erratic ~łon and, 19-20graphs of variables. 19-20, 102

position (s), as a function of. 8

reciiIinear kinematics and, 8,

19-20, 102

velodty (v) as a function of, 8Time derivatives, 74, 144, 552-556,

577 curvihnear motion, 70, 104

ihree-dimensional motion,

552-556, 577Trajeetories, 156-162,

167 circular orbit,

I59

orbit, 160-

161 free-flight. 157

paraholic path, 159

power-flight. 158

Translatin g axes. 87-91. 105, 552-556. 566-577

accele rat jon (a). 88,568

coordinates for. 87-91.105

fixed reference frame. 87-

91 particie 87-88.

105

position vet tors (r) for, 87,567

procedure for analysis of. 88. 569

relative-motion analysis of, 87-91,105, 566-573, 577

three-dimensional rigid bodies,552-556, 566-573.577

Inne derivatives for systems, 552-556

velocity (y) of, 87, 567

Translating/rotating coordinate systems, 556-558. 577

absolule (dependent) motion analysis, 329-332,392

acceleration (u) and. 313.363-370, 393

coordinating fixed and translating reference frames (axes), 337-344, 393

curvifinear, 312, 392.413

displacement and. 337

equations of motion for, 409,412-418. 453, 600. 629

impulse and moment urn for, 498,532

kinetic energy and, 457,

491 part kies, 299

position vectors (r). 313, 337

procedure for analysis using, 240,329,340,365,414

552-556, 577 Translalion. 299. 312-313. 329-332, 337-344. 363-370, 377-384.

Thrust. 282-283 392. 409, 412-418, 453, 457.

Time (1), 8. 19-20, 102 491, 534, 600, 629

INDEx 735

recfilinear, 312.392, 412-413

relative-motłon analysis, 299.337-344, 363-370,377-384, 393

rigid-body planar motion, 312-313, 329-332, 337-344, 363-370. 392-393,409,412-418,453.457, 491, 498, 532.534

rotation and. 377-348three-dimensional rigid-body

motion. 600. 629wiochy (y) and, 313, 337-344, 393

Transverse (0) coordinate, 67-70

Un balan ced forte, 107-108

Undamped vibrations. 631-639.651-654. 668

forced, 651-654, 668

frce. 631-639.668

Underdamped vibration systems. 657

Unit vectors, 672

1)-s (velocity-position) grap hs, 21(velocity-time) graphs. 19-20

Vari able acceleration. 298, 534

Vari able forte, work of. 174.458

Vector analysis, 672-676

Vector forrnul a ti on, angular momentum. 262, 267

Vector function s, di f feren tiation and i n tegration of, 676

Vector (paniny, particle displacemen l as. 5

Vełocity (v). 6-8.32-35, 53, 68. 87, 102. 155. 313. 314. 316, 337-344. 351-357,378-37y,393.459. 551-552,567

abso lute, 338acceleration (a) and, 7-8

angular (w), 68, 314.551

arcal, 155

average. 6. 32circuiar motion and, 339

consta nt, 7

ton lintious muflon and, 6-8

curvilinear motion and. 32-35, 53, 68

cylintirical components and, 68

erratic motion and, 19-21 fixecl-

point rotation and, 551-552

graphs of variables, 19-21., 102

ins tan tan e ous center (IC) of zero. 351-357. 393

ins tan taneous, 6,32

kinematics of particles and, 6-8,32-35, 53, 68, 87,102

magniłude (specd), 6, 32,34. 68, 378 position (s). as a function of. 8

radia] component (vr), 68

rectangul ar components and. 34-35

rectilinear kinematics and, 6-8. 19-21,1U2

relative, 338

relative-motion analysis and. 87.337-344, 378-379,393, 567

rigid-body planar ~lon.313-314.316, 337-344,351-357.378-379,393

rot a ting axis, 378-379, 393.567rotation and, 314, 316, 337-344,

378-379sign convention for. 6

slipping and. 459

three-dimensi ona I rigid-body motion, 551-552, 567

lime (ł), as a function of. 8

translating axis, 87, 567

translation and, 313, 337-344,

393 transverse cornponent (v0),

68 zero, 351-357, 393.459

Vertical projectile rnotio n. 39-43

Vi brations, 630-669

amp li łude cif, 633-634critically damped systems. 565 damped, 631. 655-660, 699 displacement and, 632-634

electlical circuit analogs and, 661.669

energy methods for conservation of, 645-648, 668

equi i brium posiłion, 632, 634forced, 631. 651-654, 658-660.

668-669Irce, 631-639, 655-657, 668-669 frcquency (j), 632

magnification factor (MF) for,652-653, 659

natura' frequency (w„). 632. 634, b45-648.668

overclaraped systems, 656 perl odic furce and, 651-654

periodic supporł displacement of. 653

phase armie (43), 634procedures for analysis

of, 635, 646

struple harmonie muflon of. 632undainped forced, 651-654.668un dam ped free, 631-639, 668 u n derdamped systems, 657

viscous damped forced,658-660. 669

viscous damped free. 655-657.669

736 INDEX

Vi scous vibration, 655-660, 669

coefficient of damping, 655

critically darnped systems, 565

dampe d f orce d. 655-660. 669

dumping „orce, 655

furce d, 658-660, 669

free. 655-657.669overdarnped systems, 636

steady-state, 658

underdamped systems, 657

Volume cieln ents, integration ofmoments of inertia using, 396-397

Volumetris flow (clischarge}, 279

Watt (W). unit of, [92

Weight (W), 109,171. 201-205.218-219, 459, 477.492

conservation of energy and.201-205, 219, 477

conhervative forces and

displacement of. 201-205. 219

constant, 201

gxavitati on al attraction and, 109

gravitational potcutial energy (V) from, 201-204,477

ver lic al displacernent (4) of. 459

work (U) of a, 171, 203-2114.218, 459,492

Work (U) and. 168-219, 300,454-493, 536

conservation of energy and.205-209. 219, 477-4$2.493

conservative forces and.201-209, 219

constant furce, 171, 218, 458, 492

couple moment (M), of a,460-461,492

deformation and, 174-175.177

energy (E) and, 168-219, 300.454-493, 536

external, 177

lorce (F). of a. 169-173.176-182.218-219.458-459,492

friction caused by sliding, 177-178

ki n clic energy and. 455-458, 491

kinetcs of a particie, 168-219, 300

polential energy (Y) and, 201.203-204, 477-482, 493

potentiai function and, 203-204

pr inciple of energy and, 174-182.218-219, 462-468, 493

procedure for analysis of. 175. 463, 479

rig1d body planar ~11on,454-493, 536

s lipping and, 459

spring force. 172-I 73. 201, 218, 459, 492

system of parti cl es, 176-182

units of. 170

variable forte, of. 170, 458

weight (W), of a, 171, 203-204.218, 459, 492

nero ve loci ty and (no work}, 459

Zero velocity, 351-357, 393, 459

instan tan eous center (1C) of,351-357,393

slipping (no work) and, 459

Fundamental Equations °f DynamicsKINEMATICS

Particie Rectilinear MotionVarlabie a Consurm a =

Equations of »AmonPamete 2F = ma

d v=

d rd s

= , d i

a ds = v dv

s — sfi vot jiat2

V2 = V8 2a,(s — so)

Rigid Body = In(aG),(Piane Motion? in(tre,)_,

IM,c7-=_T0a. ar M p= ł (Ak) p

Principle of Work and EnerR,

T1 + 2.U[_2 = Tx

Kinezie Energy Particie

Particie Curvilinear ~łonx, y, z Coordinares r, e, z Coordinates

)

= y

v , =

V r =

lig =a = a y

= y

a = z

a,. —r — ri)2

oo = rij +

a ,

n, r, b Coordinałes• = dv

a, = u = v 7,

v2 [1 + (dyidx)2]3/2

ulip —P IdzYidx21

T — IGw2

UF = F cos O ds

U F = o m O ) ❑

s U w = W A y

— a ks21}

U m = M

Rig id Body (P iane ~on)

WorkVariabie furce

Consranr furce Weight

Spring

Coupie monłensPower and Efticiency

Relative li.fofionPB = VA VB.,A = a A a w A

Rigid Body Motion About a Fixed Axis

Varinbie a Constara a = tv,dw

a - = - d r

dU PonttiontP= — = -v e =—dinConservation of Enery Theorem +

— T2 ł V2

P(Olentiał EnergyV = V g + 14, where V g = ±Wy,V,, = 2ks2

Principle of Linear Impulsy and Momentumw = w f t f a g

Particie mus ± ł F dr --- invzdO

=dr

w dło = udHm(vc)iRigid Body

O = Oo łł "IC4I 2

w2 = wf1 + 2a,(0 Oo)

+ 2/F dr = m(uc)z

Fur Point. PConservation of Linear Moment=ł(syst. mv) X(syst. mv)2

s = a r v = w r u t = a r o , = w 2 r

Relative General Piane Motion—Translating Axes

11,8 = VA ł VEIA{ pul) aE = aA + aB,A(r,,,)Relative General Piane Motion — Trans. and Rot. Axis

Coetlicient of Restitution(vAz — (v A)2

e — (vA)i. (vs)i

Principle of Angular Impulse and Mornentum

Particie (110) ± j Mo dt = (110)zV g = V A + fl x aB

= RA + .1'1 x rsm,where T/0 = (c1)(my)

(v8./.4)xyz

+ SI .x (n. x rB/A) +

211 x (vB/A),, + (aDIA)„,.,

KINETICSRigid Body (Płone Inotion)

Mass Moment of Incrtia i = j r2 dm

Paraliel-Axis Theorern I = Ta -i- md2

(HG)1 + 2 j MGdt = (HG)2 where =

i(HA + Mo = (110)2

Ractins of Gyration k --

where Ho = irowConservation of Angular Momentum (

syst.1-1)] = :(syst. 11)2

S1 Prefixes

Multiple Exprmentiuf Farne Prefix Symbol

1 000 000 000 109 giga G 1 000 000 10') mega M 1 000 103 kilo k

Submultiple

0.001 10 3 milli 0.000 001 10-6 mitro 0.000 (XX.) 0101 10-9 nano n

Conversion Factors (FPS) to (SI) Unie of Unit of

QuaneiłyMeasuremene (FPS)EqualsAleasureniene (SI)

Furce lb 4.448 NMass slug 14.59 kgLength ft 0.3048 m

Conversion Factors (FPS)I ft = 12 in. (inches)

1 mi. (mile) = 5280 fl1 kip (kilopound) = 1000 lb

I ton = 2000 lb

Hasiak ksiazka

Documents

Transcript of Hasiak ksiazka