Chapter 12

Kinematics of a Particle

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

acceleration.

To study particle motion along a straight line and represent thismotion graphically.

To investigate particle motion along a curved path using differentcoordinate systems.

To present an analysis of dependent motion of two particles. To examine the principles of relative motion of two particles using

translating axes.

12

12.1 IntroductionMechanics is a branch of the physical sciences that is concerned with thestate of rest or motion of bodies subjected to the action of forces.Engineering mechanics is divided into two areas of study, namely, staticsand dynamics. Statics is concerned with the equilibrium of a body that iseither at rest or moves with constant velocity. Here we will considerdynamics, which deals with the accelerated motion of a body.The subjectof dynamics will be presented in two parts: kinematics, which treats onlythe geometric aspects of the motion, and kinetics, which is the analysis ofthe forces causing the motion. To develop these principles, the dynamicsof a particle will be discussed first, followed by topics in rigid-bodydynamics in two and then three dimensions.

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 3

3HDUVRQ(GXFDWLRQ,QF8SSHU6DGGOH5LYHU1-$OO5LJKWV5HVHUYHG

4 CHAPTER 12 KINEMAT ICS OF A PART ICLE

12Historically, the principles of dynamics developed when it was possible

to make an accurate measurement of time. Galileo Galilei (15641642)was one of the first major contributors to this field. His work consisted ofexperiments using pendulums and falling bodies. The most significantcontributions in dynamics, however, were made by Isaac Newton(16421727), who is noted for his formulation of the three fundamentallaws of motion and the law of universal gravitational attraction. Shortlyafter these laws were postulated, important techniques for theirapplication were developed by Euler, DAlembert, Lagrange, and others.

There are many problems in engineering whose solutions requireapplication of the principles of dynamics. Typically the structural designof any vehicle, such as an automobile or airplane, requires considerationof the motion to which it is subjected. This is also true for manymechanical devices, such as motors, pumps, movable tools, industrialmanipulators, and machinery. Furthermore, predictions of the motions ofartificial satellites, projectiles, and spacecraft are based on the theory ofdynamics. With further advances in technology, there will be an evengreater need for knowing how to apply the principles of this subject.

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

1. Read the problem carefully and try to correlate the actual physicalsituation 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 theaccuracy of the given data.

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

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

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

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 4


12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 5

1212.2 Rectilinear Kinematics: ContinuousMotion

We will begin our study of dynamics by discussing the kinematics of aparticle that moves along a rectilinear or straight line path. Recall that aparticle has a mass but negligible size and shape.Therefore we must limitapplication to those objects that have dimensions that are of noconsequence in the analysis of the motion. In most problems, we will beinterested in bodies of finite size, such as rockets, projectiles, or vehicles.Each of these objects can be considered as a particle, as long as the motionis characterized by the motion of its mass center and any rotation of thebody is neglected.

Rectilinear Kinematics. The kinematics of a particle ischaracterized by specifying, at any given instant, the particles position,velocity, and acceleration.

Position. The straight-line path of a particle will be defined using asingle coordinate axis s, Fig. 121a. The origin O on the path is a fixedpoint, and from this point the position coordinate s is used to specify thelocation of the particle at any given instant. The magnitude of s is thedistance from O to the particle, 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 thecoordinate axis is positive to the right of the origin. Likewise, it isnegative if the particle is located to the left of O. Realize that position isa vector quantity since it has both magnitude and direction. Here,however, it is being represented by the algebraic scalar s since thedirection always remains along the coordinate axis.

Displacement. The displacement of the particle is defined as thechange in its position. For example, if the particle moves from one pointto another, Fig. 121b, the displacement is

In this case is positive since the particles final position is to the rightof its initial position, i.e., Likewise, if the final position were to theleft of its initial position, would be negative.

The displacement of a particle is also a vector quantity, and it should bedistinguished from the distance the particle travels. Specifically, thedistance traveled is a positive scalar that represents the total length ofpath over which the particle travels.

ss 7 s.

s

s = s - s

s

s

Position

(a)

O

s

s

Displacement

(b)

s

O!s

Fig. 121

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 5



12Velocity. If the particle moves through a displacement during thetime interval the average velocity of the particle during this timeinterval is

If we take smaller and smaller values of the magnitude of becomes smaller and smaller. Consequently, the instantaneous velocity isa vector defined as or

(121)

Since or dt is always positive, the sign used to define the sense of thevelocity is the same as that of or ds. For example, if the particle ismoving to the right, Fig. 121c, the velocity is positive; whereas if it ismoving to the left, the velocity is negative. (This is emphasized here bythe arrow written at the left of Eq. 121.) The magnitude of the velocityis known as the speed, and it is generally expressed in units of

Occasionally, the term average speed is used. The average speed isalways a positive scalar and is defined as the total distance traveled by aparticle, divided by the elapsed time i.e.,

For example, the particle in Fig. 121d travels along the path of length in time so its average speed is but its averagevelocity is vavg = -s>t. 1vsp2avg = sT>t,t, sT

1vsp2avg = sTtt;sT ,

m>s or ft>s.s

t

v = dsdt

1:+ 2 v = lim

t:01s>t2,st,

vavg =st

t,s

s

Velocity

(c)

O!s

v

!s

sP

sTAverage velocity and

Average speed

O

P

(d)

Fig. 121 (cont.)

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 6



12Acceleration. Provided the velocity of the particle is known at twopoints, the average acceleration of the particle during the time interval is defined as

Here represents the difference in the velocity during the timeinterval i.e., Fig. 121e.

The instantaneous acceleration at time t is a vector that is found bytaking smaller and smaller values of and corresponding smaller andsmaller values of so that , or

(122)

Substituting Eq. 121 into this result, we can also write

Both the average and instantaneous acceleration can be either positiveor negative. In particular, when the particle is slowing down, or its speedis decreasing, the particle is said to be decelerating. In this case, inFig. 121f is less than and so will be negative.Consequently, a will also be negative, and therefore it will act to the left,in the opposite sense to Also, note that when the velocity is constant,the acceleration is zero since Units commonly used toexpress the magnitude of acceleration are or

Finally, an important differential relation involving the displacement,velocity, and acceleration along the path may be obtained by eliminatingthe time differential dt between Eqs. 121 and 122, which gives

(123)

Although we have now produced three important kinematicequations, realize that the above equation is not independent ofEqs. 121 and 122.

a ds = v dv1:+ 2

ft>s2.m>s2v = v - v = 0.v.v = v - vv,

v

a = d2s

dt21:+ 2

a = dvdt

1:+ 2a = lim

t:01v>t2v, tv = v - v,t,

v

aavg =vt

t

s

Acceleration

(e)

O

a

v v

sP

Deceleration

(f)

O

P

v v

a

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 7



12Constant Acceleration, When the acceleration isconstant, each of the three kinematic equations and can be integrated to obtain formulas that relate s,and t.

Velocity as a Function of Time. Integrate assumingthat initially when

(124)

Position as a Function of Time. Integrate assuming that initially when

(125)

Velocity as a Function of Position. Either solve for t in Eq. 124and substitute into Eq. 125, or integrate assuming thatinitially at

(126)

The algebraic signs of and used in the above three equations,are determined from the positive direction of the s axis as indicated bythe arrow written at the left of each equation. Remember that theseequations are useful only when the acceleration is constant and when

A typical example of constant accelerated motionoccurs when a body falls freely toward the earth. If air resistance isneglected and the distance of fall is short, then the downwardacceleration of the body when it is close to the earth is constant andapproximately or The proof of this is given inExample 13.2.

32.2 ft>s2.9.81 m>s2v = v0 .s = s0 ,t = 0,

ac ,v0 ,s0 ,

v2 = v02 + 2ac1s - s02Constant Acceleration

1:+ 2L

v

v0

v dv = Ls

s0

ac ds

s = s0 .v = v0v dv = ac ds,

s = s0 + v0t + 12 act2

Constant Acceleration1:+ 2

Ls

s0

ds = Lt

01v0 + act2 dt

t = 0.s = s0v = ds>dt = v0 + act,

v = v0 + actConstant Acceleration

1:+ 2L

v

v0

dv = Lt

0ac dt

t = 0.v = v0ac = dv>dt,

v,ac ,ac ds = v dvv = ds>dt,ac = dv>dt,a = ac.

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 8



12

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, and a is then defined by theiralgebraic signs.

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

Kinematic Equations. If a relation is known between any two of the four variables a, s and t, then a third variable can be

obtained by using one of the kinematic equations, or since eachequation relates all three variables.*

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

Remember that Eqs. 124 through 126 have only limited use. These equations apply only when theacceleration is constant and the initial conditions are s = s0 and v = v0 when t = 0.

a ds = v dv,v = ds>dta = dv>dt, v,

v,

During the time this rocket undergoes rectilinearmotion, its altitude as a function of time can bemeasured and expressed as Its velocitycan then be found using and itsacceleration can be determined from a = dv>dt.v = ds>dt,s = s1t2.

s

Important 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 thedisplacement divided by the time.

A particle that is slowing down is decelerating. A particle can have an acceleration and yet have zero velocity. The relationship is derived from and

by eliminating dt.v = ds>dt, a = dv>dta ds = v dv

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

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 9



12

The car in Fig. 122 moves in a straight line such that for a short timeits velocity is defined by where t is in seconds.Determine its position and acceleration when When s = 0.

t = 0,t = 3 s.v = 13t2 + 2t2 ft>s,

EXAMPLE 12.1

SOLUTIONCoordinate System. The position coordinate extends from thefixed origin O to the car, positive to the right.

Position. Since the cars position can be determined fromsince this equation relates s, and t. Noting that

when we have*

When Ans.

Acceleration. Since the acceleration is determined fromsince this equation relates a, and t.

When Ans.

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

a = 6132 + 2 = 20 ft>s2:t = 3 s, = 6t + 2

a = dvdt

= ddt

13t2 + 2t2 1:+ 2 v,a = dv>dt,v = f1t2,s = 132

3 + 1322 = 36 ftt = 3 s, s = t3 + t2

s `0

s

= t3 + t2 `0

t

Ls

0ds = L

t

013t2 + 2t2dt

v = dsdt

= 13t2 + 2t2 1:+ 2t = 0,s = 0v,v = ds>dt, v = f1t2,

s

O

a, v

Fig. 122

*The same result can be obtained by evaluating a constant of integration C ratherthan using definite limits on the integral. For example, integrating yields Using the condition that at then C = 0.s = 0,t = 0,s = t3 + t2 + C.

ds = 13t2 + 2t2dt

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 10



12EXAMPLE 12.2

A small projectile is fired vertically downward into a fluid medium withan initial velocity of Due to the drag resistance of the fluid theprojectile experiences a deceleration of where is in

Determine the projectiles velocity and position 4 s after it is fired.

SOLUTIONCoordinate System. Since the motion is downward, the positioncoordinate is positive downward, with origin located at O, Fig. 123.Velocity. Here and so we must determine the velocity as afunction of time using since this equation relates a, and t.(Why not use ) Separating the variables and integrating,with when yields

Here the positive root is taken, since the projectile will continue tomove downward. When

Ans.

Position. Knowing we can obtain the projectiles positionfrom since this equation relates s, and t. Using the initialcondition when we have

When Ans.s = 4.43 m

t = 4 s,

s = 10.4

e c 116022 + 0.8t d1>2 - 160 f ms = 2

0.8 c 116022 + 0.8t d1>2 ` 0tL

s

0ds = L

t

0c 116022 + 0.8t d-1>2dt

v = dsdt

= c 116022 + 0.8t d-1>21+ T2t = 0,s = 0,

v,v = ds>dt, v = f1t2,v = 0.559 m>sTt = 4 s,

v = e c 116022 + 0.8t d-1>2 f m>s1

0.8 c 1v2

- 116022 d = t1

-0.4 a 1-2 b 1v2 ` 60v = t - 0L

v

60 m>s dv-0.4v3 = Lt

0dt

a = dvdt

= -0.4v31+ T2 t = 0,v0 = 60 m>sv = v0 + act?

v,a = dv>dt,a = f1v2m>s. va = 1-0.4v32 m>s2,60 m>s.

s

O

Fig. 123

77919_01_ch12_p002-105.qxd 3/27/09 10:23 AM Page 11



12

During a test a rocket travels upward at and when it is 40 mfrom the ground its engine fails. Determine the maximum height reached by the rocket and its speed just before it hits the ground.While in motion the rocket is subjected to a constant downwardacceleration of due to gravity. Neglect the effect of airresistance.

SOLUTIONCoordinate System. The origin O for the position coordinate s istaken at ground level with positive upward, Fig. 124.

Maximum Height. Since the rocket is traveling upward,when At the maximum height the velocity

For the entire motion, the acceleration is (negative since it acts in the opposite sense to positive velocity orpositive displacement). Since is constant the rockets position maybe related to its velocity at the two points A and B on the path by usingEq. 126, namely,

Ans.

Velocity. To obtain the velocity of the rocket just before it hits theground, we can apply Eq. 126 between points B and C, Fig. 124.

Ans.

The negative root was chosen since the rocket is moving downward.Similarly, Eq. 126 may also be applied between points A and C, i.e.,

Ans.

NOTE: It should be realized that the rocket is subjected to adeceleration from A to B of and then from B to C it isaccelerated at this rate. Furthermore, even though the rocketmomentarily comes to rest at the acceleration at B is still

downward!9.81 m>s2 B 1vB = 029.81 m>s2,

vC = -80.1 m>s = 80.1 m>s T = 175 m>s22 + 21-9.81 m>s2210 - 40 m2 vC2 = vA2 + 2ac1sC - sA2 1+ c2

vC = -80.1 m>s = 80.1 m>s T = 0 + 21-9.81 m>s2210 - 327 m2 vC2 = vB2 + 2ac1sC - sB2 1+ c2

sB = 327 m

0 = 175 m>s22 + 21-9.81 m>s221sB - 40 m2 vB2 = vA2 + 2ac1sB - sA2 1+ c2ac

ac = -9.81 m>s2vB = 0. s = sBt = 0.vA = +75m>s

9.81 m>s2sB

75 m>s,EXAMPLE 12.3

A

O

vA ! 75 m/s

vB ! 0

sA ! 40 m

s

sB

B

C

Fig. 124

77919_01_ch12_p002-105.qxd 3/27/09 10:24 AM Page 12



12EXAMPLE 12.4

A metallic particle is subjected to the influence of a magnetic field as ittravels downward through a fluid that extends from plate A to plate B,Fig. 125. If the particle is released from rest at the midpoint C,

and the acceleration is where s is inmeters, determine the velocity of the particle when it reaches plate B,

and the time it takes to travel from C to B.

SOLUTIONCoordinate System. As shown in Fig. 125, s is positive downward,measured from plate A.Velocity. Since the velocity as a function of position canbe obtained by using Realizing that at we have

(1)

At Ans.

The positive root is chosen since the particle is traveling downward,i.e., in the direction.Time. The time for the particle to travel from C to B can be obtainedusing and Eq. 1, where when FromAppendix A,

At

Ans.

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

t =ln A410.222 - 0.01 + 0.2 B + 2.303

2= 0.658 s

s = 0.2 m,

ln A4s2 - 0.01 + s B + 2.303 = 2t ln A4s2 - 0.01 + s B ` 0.1s

= 2t `0

t

Ls

0.1

ds1s2 - 0.0121>2 = L t0 2 dt = 21s2 - 0.0121>2dtds = v dt 1+ T2

t = 0.s = 0.1 mv = ds>dt+svB = 0.346 m>s = 346 mm>s Ts = 200 mm = 0.2 m,

v = 21s2 - 0.0121>2 m>s 12

v2 `0

v

= 42

s2 `0.1 m

s

Lv

0v dv = L

s

0.1 m4s ds

v dv = a ds 1+ T2s = 0.1 m,v = 0v dv = a ds.

a = f1s2,s = 200 mm,

a = 14s2 m>s2,s = 100 mm,

A

200 mm

100 mm

B

sC

Fig. 125

77919_01_ch12_p002-105.qxd 3/27/09 10:24 AM Page 13



12

A particle moves along a horizontal path with a velocity ofwhere t is the time in seconds. If it is initially

located at the origin O, determine the distance traveled in 3.5 s, and theparticles average velocity and average speed during the time interval.

SOLUTIONCoordinate System. Here positive motion is to the right, measuredfrom the origin O, Fig. 126a.

Distance Traveled. Since the position as a function oftime may be found by integrating with

(1)

In order to determine the distance traveled in 3.5 s, it is necessary toinvestigate the path of motion. If we consider a graph of the velocityfunction, Fig. 126b, then it reveals that for the velocity isnegative, which means the particle is traveling to the left, and for the velocity is positive, and hence the particle is traveling to the right.Also, note that at The particles position when

and can now be determined from Eq. 1.This yields

The path is shown in Fig. 126a. Hence, the distance traveled in 3.5 s is

Ans.

Velocity. The displacement from to is

and so the average velocity is

Ans.

The average speed is defined in terms of the distance traveled Thispositive scalar is

Ans.

Note: In this problem, the acceleration is which is not constant.

a = dv>dt = 16t - 62 m>s2,1vsp2avg =sTt =

14.125 m3.5 s - 0 = 4.04 m>s

sT .

vavg =st =

6.125 m3.5 s - 0 = 1.75 m>s:

s = s t=3.5 s - s t=0 = 6.125 m - 0 = 6.125 m

t = 3.5 st = 0

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

s t=0 = 0 s t=2 s = -4.0 m s t=3.5 s = 6.125 mt = 3.5 st = 2 s,t = 0,t = 2 s.v = 0

t 7 2 s0 6 t 6 2 s

s = 1t3 - 3t22m L

s

0ds = L

t

0(3t2 - 6t) dt

= 13t2 - 6t2dt ds = v dt 1:+ 2s = 0.t = 0,v = ds>dtv = f1t2,

v = 13t2 - 6t2 m>s,EXAMPLE 12.5

O

s ! "4.0 m s ! 6.125 m

t ! 2 s t ! 0 s t ! 3.5 s

(a)

(0, 0)

v (m/s)

v ! 3t2 " 6t

(2 s, 0)t (s)

(1 s, "3 m/s)

(b)

Fig. 126

77919_01_ch12_p002-105.qxd 3/27/09 10:24 AM Page 14



12FUNDAMENTAL PROBLEMS

F121

s

F122

s

F123

s

F124

s

F125

s

s

F126

s

F127

s

F128

F125. The position of the particle is given bywhere is in seconds. Determine the

time when the velocity of the particle is zero, and the totaldistance traveled by the particle when t = 3 s.

ts = (2t2 - 8t + 6) m,

F123. A particle travels along a straight line with a velocityof where is in seconds. Determine theposition of the particle when .s = 0 when t = 0t = 4 s.

tv = (4t - 3t2) m>s,

F121. Initially, the car travels along a straight road with aspeed of If the brakes are applied and the speed ofthe car is reduced to determine the constantdeceleration of the car.

10 m>s in 15 s,35 m>s.

F127. A particle moves along a straight line such that itsacceleration is where is in seconds.When the particle is located to the left of theorigin, and when it is to the left of the origin.Determine the position of the particle when t = 4 s.

20 mt = 2 s,2 mt = 0,

ta = (4t2 - 2) m>s2,

F126. A particle travels along a straight line with anacceleration of where s is measured inmeters. Determine the velocity of the particle when if at s = 0.v = 5 m>s s = 10 ma = (10 - 0.2s) m>s

2,

F124. A particle travels along a straight line with a speedwhere is in seconds. Determine the

acceleration of the particle when t = 2 s.tv = (0.5t3 - 8t) m>s,

F122. A ball is thrown vertically upward with a speed ofDetermine the time of flight when it returns to its

original position.15 m>s.

F128. A particle travels along a straight line with avelocity of where is in meters.Determine the acceleration of the particle at s = 15 m.

sv = (20 - 0.05s2) m>s,

77919_01_ch12_p002-105.qxd 3/27/09 10:24 AM Page 15


Chapter 12

Documents

Transcript of Chapter 12