Ch. 2: Kinematics of...

Ch. 8: Kinetics of Particles

8.0 Outline 415 Introduction 416 Newton’s Second Law 417 Equations of Motion 418 Rectilinear Motion 421 Curvilinear Motion 444

8.0 Outline

415


8.1 Introduction

8.1 Introduction

Kinetics is the study of the relations between the forces and the motion. Here we will not seriously concern whether the forces cause the motion or the motion generates the forces (causality). In this chapter, the focus is on the particles. That is the body whose physical dimensions are so small compared with the radius of curvature of its path. There are at least 3 approaches toe the solution of kinetic problems: (a) Newton’s second law (b) work and energy method (c) impulse and momentum method.

416


8.2 Newton’s Second Law

8.2 Newton’s Second Law

For most engineering problems on earth, the acceleration measured w.r.t. reference frame fixed to the earth’s surface may be treated as absolute. And Newton’s 2nd law of motion holds. Newton’s 2nd law breaks when the velocities of the order of the speed of light are involved theory of relativity

mm mass (resistance to rate of change of velocity) of the

acting on the particleresulting acceleration measured in a of reference

particleresultant force

nonaccelerating frame

=F = a

F =a =

417


8.3 Equation of Motion and Solution of Problems

8.3 Equation of Motion and Solution

Two problems of dynamics (1) specified kinematic conditions, find forces straightforward application of Newton’s law as algebraic equations (2) specified forces, find motion Difficulty depends on the form of force function (t, s, v, a), as the solutions are found by solving a system of differential equations. For simple functions, we can find closed form solutions of motion as in rectilinear motion (sec. 2.2).

m --- equation of motionscalar components decomposition according to a specified coordinate∑F = a

418



Unconstrained motion Motion of the particle is determined by its initial motion and the forces from external sources. It is free of constraints and so has three degrees of freedom to specify the position. Three scalar equations of motion would have to be applied and integrated to obtain the motion. Constrained motion Motion of the particle is partially or totally determined by restraining guides, other than its initial motion and the forces from external sources. Therefore, all forces, both applied and reactive, that act on the particle must be accounted for in Newton’s law. The number of d.o.f. and equations are reduced regarding to the type of constraints.

419



Free body diagram All forces acting on the particle needed to be accounted in the equations of motion. Free body diagram unveils every force that acts on the isolated particle. Only after the FBD has been completed should the equations of motion be written. The appropriate coordinate axes and directions should be indicated and consistently used throughout the problem. Treatment of the body as particle is valid when the forces may be treated as concurrent through the mass center.

420


8.4 Rectilinear Motion


x x y z

If the x-axis is the direction of the rectilinear motion,F ma F 0 F 0

If we are not free to choose a coordinate direction along the motion,the nonzero acceleration component will be sh

= = =∑ ∑ ∑

x x y y z z

A B A

own up in all equations:F ma F ma F ma

Other coordinate system such as n-t or r-

may be determined via the use of relative motionFor pure translating moving reference frame

θ

= = =∑ ∑ ∑

a

a = a + a /B

421



P. 8/1 The coefficient of static friction between the flat bed of the truck and the crate it carries is 0.30. Determine the minimum stopping distance s that the truck can have from a speed of 70 km/h with constant deceleration if the crate is not to slip forward.

422



P. 8/1

If the crate is not to slip, crate and truck must have same acceleration.If the crate is not to slip, friction static friction at impending status.Minimum stopping distance when the deceleration is th

=

( ) ( )

x x x x

22 2

o o

e max allowable value.

F ma 0.3mg ma , a 0.3g constant for minimum distance

10v v 2a s s 0 70 2 0.3g s, s 64.2 m36

= − = = −

= + − = × + − =

∑

mg

N

F < 0.3N

+x

423



P. 8/2 If the truck of Prob. 3/17 comes to stop from an initial forward speed of 70 km/h in a distance of 50 m with uniform deceleration, determine whether or not the crate strikes the wall at the forward end of the flat bed. If the crate does strike the wall, calculate its speed relative to the truck as the impact occurs. Use the friction coefficients μs = 0.3 and μk = 0.25.

424



P. 8/2

( )

( )

22 2 2

o o truck truck

o o stop

stopping distance 50 m, which is less than minimum value 64.2 mthe crate slips

10v v 2a s s 0 70 2a 50, a 3.781 m/s36

10v v a t t 0 70 3.781 t, t 5.36

=∴

= + − = × + × = − = + − = × − × =

( )

s k

truck crate

x x s

k

k crate c

14 s

Friction force: F 0.3mg 2.943m and F 0.25mg 2.45mAssume crate and truck go together a a

F ma F m 3.781 required friction 3.781m F

the crate slips and F F

F ma , a

= = = =→ =

= − = − → = > ∴ =

− =

∑

[ ] ( )

( ) ( )

2rate

2crate/truck crate truck crate/truck

2 2o o o o

2.45 m/s

a a a a 2.45 3.781 1.331 m/sthe crate slips forward but will it strike the wall?

1s s v t t a t t relative motion calculation2

13 1.3312

= −

= − = − − − =

∴

= + − + −

= ×

( )

2strike stop

o o

crate/truck

t , t 2.123 s t crate will strike the wall before the truck stops

v v a t t relative motion calculation

v 0 1.331 2.123 2.826 m/s

× = < ∴

= + − = + × =

N

F < 0.3N

+x

425



P. 8/3 If the coefficients of static and kinetic friction between the 20-kg block A and the 100-kg cart B are both essentially the same value of 0.50, determine the acceleration of each part for (a) P = 60 N and (b) P = 40 N.

426



P. 8/3

A max A

2x x A A

2B B

max

x

(a) N 20g, F 0.5N 98.1 N 120 Nblock A moves forward relative to B

F ma 120 98.1 20a , a 1.095 m/s

98.1 100a , a 0.981 m/s

(b) F 80 N block A does not move relative to B

F

= = = <∴

= − = = = =

> ∴

=

∑

∑ x

2

max

ma A & B move together

80 120a, a 0.667 m/sFind developed friction by isolated FBD at A or B80 F 20a, F 66.67 N F assumption is validF 100a, F 66.67 N

= =

− = = < ∴= =

2P 20g

100g

NB

NA NA F

F

427



P. 8/4 A simple pendulum is pivoted at O and is free to swing in the vertical plane of the plate. If the plate is given a constant acceleration a up the incline θ, write an expression for the steady angle β assumed by the pendulum after all initial start-up oscillations have ceased. Neglect the mass of the slender supporting rod.

428



P. 8/4

y

x x

1

F 0 Tcos mgcos 0

F ma Tsin mgsin ma

a gsintangcos

β θ

β θ

θβθ

−

= − = = − =

+=

∑∑

T

mg θ

β x

y

429



P. 8/5 For the friction coefficients μs = 0.25 and μk = 0.20, calculate the acceleration of each body and the tension T in the cable.

430



P. 8/5

A B A B

max s

max

max

s 2s c a 2a 0N 60gcos30, F N 127.4 NAssume motion impends at block A F F and equilibrium

F 0 60gsin30 F T 0, T 166.9 N

but cylinder B will not be in equilibrium 20g 2T 0 move

lµ

+ + = → + == = =

→ =

= − − = = − < →

∑( )

k A B

2 2B B A

upAssum block A slides down and block B moves up

F ma 60gsin30 F T 60a 120a

20g 2T 20a , T 105.35 N, a 0.725 m/s , a 1.45 m/s

= − − = = − − = = = − =

∑

T

2T

20g

60g

N F

431



P. 8/6 A bar of length l and negligible mass connects the cart of mass M and the particle of mass m. If the cart is subjected to a constant acceleration a to the right, what is the resulting steady-state angle θthat the freely pivoting bar makes with the vertical? Determine the net force P (not shown) that must be applied to the cart to cause the specified acceleration.

432



P. 8/6

y

1x x

x x

From the given statements, pendulum and cart have same accelerationAt the pendulum,

F 0 Tcos mg 0, T mg/cos

aF ma Tsin ma, tang

At the cart,

F ma P Tsin Ma, P

θ θ

θ θ

θ

−

= − = =

= = =

= − =

∑

∑

∑ ( )m M gtanθ= +

x

y

mg

Mg

T T θ

P

N

433



P. 8/7 Determine the accelerations of bodies A and B and the tension in the cable due to the application of the 250 N force. Neglect all friction and the masses of the pulleys.

434



P. 8/7

A B A B

x x A B

2 2A B

2s 3s c 2a 3a 0

F ma 2T 70a and 300 3T 35a

a 2.34 m/s , a 1.56 m/s , T 81.8 N

l+ + = → + =

= − = − = = − = =

∑

sA sB

2T 3T

70g 35g

300 N

NA NB

435



P. 8/8 The sliders A and B are connected by a light rigid bar and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position shown, the velocity of A is 0.4 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.

436



P. 8/8

( )

A B A B A B2 2 2

A B A B

A A B B A B B A

A B2 2A A A B B B A B B

Kinematics: triangle OABs s and 0.5 s cos15 s cos15, s s 0.2588 m

s s 2s s cos150diff: 0 2s v 2s v 2cos150 s v s vgiven: v 0.4 m/s v 0.4 m

diff: 0 v s a v s a cos150 s a s a

l= = + = =

= + −

= + − +

= → = −

= + + + − +( )( )

( )

A A B

A B

A B

2 2A B

2v v

0 0.04287 0.4829a 0.4829a 1Kinetics:

F ma 40 Tcos15 2a and Tcos15 3a into 1

a 7.95 m/s , a 8.04 m/s , T 25.0 N

+

= + +

= − = − = = = − =

∑

sB

sA

NB

NA T

T

40 N

437



P. 8/9 With the blocks initially at rest, the force P is increased slowly from zero to 260 N. Plot the accelerations of both masses as functions of P.

438



P. 8/9

max max

k k

A B A

A A B B

A A B B

N 35g, N N 42g 77gF 0.2N 68.67 N, F 0.15N 113.3 N

F 0.15N 51.5 N, F 0.10N 75.54 N

Three possible situations: no motion, B & A move together, and B & A move separately.Two impossibl

= = + == = = =

= = = =

( )max

max

A

A

B B

e situations: B moves alone then F will 0 A will move eventuallyand A moves alone P is applied at block B and force P is increased slowly from zero

not jump right to F

1) 0 P F : F will be

≠ →

≤ ≤

A A B

developed to cancel with the applied P,

F will stay zero, and so there is no motion a 0 & a 0→ = =

35g

42g

NA FA

FA NA

NB FB

P

439



P. 8/9

( ) ( )

max k

k

max

max

A A B B

A A B

2min B A

A A

2) assume both A and B go together in this phase F F and F F

F ma F 35a & P F F 42a

at P P F increased slowly , a 0.49 m/s and F 17.16 N jumping

at F F about to slip r

∴ ≤ =

= = − − = = = = =

=

∑

( )

( )

k

max

2

B

2B A B

elative to each other , P 226.6 N and a 1.962 m/s

P Fbetween these extremum values, a : linear function of P

77F P 226.6 : a a which varies linearly from 0.49 to 1.962 m/s

3) A slide backward

= =

−=

∴ < ≤ =

k

k k k

A A

A A A B B

2A B

relative to B increasing P makes B accelerates more and moreP 226.6 N makes A slips F F

F ma F 35a & P F F 42a

P 127.04a 1.47 m/s constant and a : linear function of P42

226.6

> → =

= = − − = −

= =

∴

∑

( )2 2A BP 260.0 : a 1.47 m/s constant and 2.37 a 3.166 m/s jumping< ≤ = < ≤

440



P. 8/9

441



P. 8/10 The system is released from rest in the position shown. Calculate the tension T in the cord and the acceleration a of the 30 kg block. The small pulley attached to the block has negligible mass and friction. (Suggestion: First establish the kinematic relationship between the accelerations of the two bodies.)

442



P. 8/10

( )( )

2 2 2

2 2

Kinematics: b x and b y

diff: bb xx and b y 0

b bb x xx and b y 0 1

at this instant: x/b 4 / 5, x 0, b 0 initially restassume cylinder moves down,

c l= + + =

= + =

+ = + + =

= = =

hence block moves to the leftKinetics: for 30 kg block

F ma T 3/5 T 30g N 0, N 30g 2T/5

F T 4/5 30xassume the block moves F 0.25Nfor 15 kg cylinder

15g T 15y 15b

r

= × − − + = = + − × =

→ =

− = = −

∑

( ) ( )

2

T/15 g 30x 4 becall 1 , , T 137.9 Nb 5 x 7.5g 0.7T

7.5g 0.7Tx 0.766 m/s30

− ×= = = =

−−

= = −

c

+b

+x +y

30g

15g

T

T

T

F N

443


8.5 Curvilinear Motion

8.5 Curvilinear Motion Choose appropriate coordinate system (x-y, n-t, or r-θ) for the given problem. Determine the motion along those axes. Then set up the Newton’s law along those axes. The positive sense of the force and acceleration must be consistent.

( ) ( )( ) ( )

x y

2 2n t

2r

x-y system: F mx F my

n-t system: F m m v / F mv

r- system: F m r r F m r 2rθ

ρβ ρ

θ θ θ θ

= =

= = =

= − = +

∑ ∑∑ ∑∑ ∑

444



P. 8/11 The member OA rotates about a horizontal axis through O with a constant counterclockwise velocity ω= 3 rad/s. As it passes the position θ= 0, a small block of mass m is placed on it at a radial distance r = 450 mm. If the block is observed to slip at θ= 50°, determine the coefficient of static friction μs between the block and the member.

445



P. 8/11

( )

( )( )

t t

2n n

s s

use n-t coordinate systemgiven: 0.45 m, 50 , 0 no slip until 50 ,

3 rad/s, 0

F ma N mgcos50 m , N mgcos50

F ma mgsin50 F m

At 50 , F F N and directs u

ρ β ρ β

β β

ρβ ρβ

ρβ

µ

= = ° = = °

= =

= − = + =

= − = ° = =

∑∑

( )

2

2

pward because gsin50which means bar OA rotates too slow than required to keepthe block stays on the bar. The friction will develop to resist

the block from sliding down or to match F with .

I

ρβ

ρβ

>

∑

( )2 2

2

f the bar rotates very very slow, friction force cannot make

F to match F cannot be reduced any more, F .

And then the block will slide down, hence decreases, to the positionwhere v / lar

ρβ ρβ

ρ

ρ

>∑ ∑ ∑

( )2s s

ge enough to match F (i.e., to satisfy Newton's law)

mgsin50 mgcos50 m , 0.549µ ρβ µ− = =

∑

mg

N F

n

t

446



P. 8/12 A 2 kg sphere S is being moved in a vertical plane by a robotic arm. When the arm angle θis 30°, its angular velocity about a horizontal axis through O is 50 deg/s CW and its angular acceleration is 200 deg/s2 CCW. In addition, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping surfaces is 0.5. Compare P to the minimum gripping force Ps required to hold the sphere in static equilibrium in the 30˚ position.

447



P. 8/12

( )( )

r r

2

2r r f f

f f

given: m 1 kg, r 1 m, r 0.5 m/s, r 0

30 , 50 0.873 rad/s, 200 3.49 rad/s180 180

F ma mgsin 2F m r r , F 4.143 N

F ma 2F mgcos m r 2r , F 12.859θ θθ θ

π πθ θ θ

θ θ

θ θ θ

= = = − =

= ° = − × = − = × =

= − + = − =

= − = + =

∑∑

r

2 2f f f s

s s s s

N

F F F 13.51 P P 27.02 N

static equilibrium: 2F 2 P mg P 19.62 Nθ

µ

µ

= + = = → =

= = → =

r θ mg

2Fr

2Fθ

mg

2Fs

448



P. 8/13 A flatbed truck going 100 km/h rounds a horizontal curve of 300 m radius inwardly banked at 10°. The coefficient of static friction between the truck bed and the 200 kg crate it carries is 0.70. Calculate the friction force F acting on the crate.

449



P. 8/13

2

n n

y

assume the crate tends to slide up the truck bedfriction directs downslope

the crate has absolute curve motion into the paper on the horizontal plane

m 10F ma Nsin10 Fcos10 100300 36

F

∴

= + = ×

=

∑

max

0 mg Ncos10 Fsin10 0

N 2021.52 N and F 165.9 Ncheck if this friction can be providedF 0.7N 1415 N F

the crate tends to slide up due to high speed curved motionbut still too far from slidi

− + − = = =

= = >∴

∑

ng up (can increase the truck speedyet the crate does not move relative to the truck bed)

mg

N

F

n

y

450



P. 8/14 The flatbed truck starts from rest on a road whose constant radius of curvature is 30 m and whose bank angle is 10°. If the constant forward acceleration of the truck is 2 m/s2, determine the time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is μs = 0.3, and the truck motion occurs in a horizontal plane.

451



P. 8/14

static motion

200g

F N N

Fsn

Fst (inward)

200g

n

y

452



P. 8/14 s s s

s

s

s

Static case: N 200gcos10 1932.2 N, F 0.3N 579.66 NF 200gsin10 340.7 N upward to prevent sliding down the incline, and F

Slipping when friction F but in what direction?F can be divided in two co

= = = == = <

=

t

n

s t

s n

s

mponents: along n- and t-axisF points in positive t (inward the paper) to match the positive a

F points down the incline to match the component of a down the incline

When the truck moves, N N to >

[ ]( )

n

n

n

2t

t t

y s

2

n n s

match the positive component of a up the truck bed

given: 30 m, 0, 0, a 2 m/sa v v a t 2t

F 0 Ncos10 200g F sin10 0 1

4tF ma F cos10 Nsin10 2003

ρ ρ ρ= = = =

= = =

= − − =

= + = ×

∑

∑

( )

( )

( )

t

n t n t

n

n

t t s

22 2 2 2 2s s s s s

2s

s

s

20

F ma F 200 2 400 N

F F F F F 0.3N

F 0.09N 160000 and substitute into 1

N 2076.47, 1919.24 N but 1919.24 N which is impossibleN 2076.47 N, F

= = × =

+ = + =

= −

= <∴ = =

∑

ts477.55 N, F 400 N, t 5.58 s= =

453



P. 8/15 The small object is placed on the inner surface of the conical dish at the radius shown. If the coefficient of static friction between the object and the conical surface is 0.30, for what range of angular velocities ωabout the vertical axis will the block remain on the dish without slipping? Assume that speed changes are made slowly so that any angular acceleration may be neglected.

454



P. 8/15

min n s n

max n s n

min y

n n

given: 0, 0.2 m, 0, 0 causes small a F upward to reduce F

causes large a F downward to increase F

: F 0 Ncos30 0.3Nsin30 mg 0

F ma Nsin30 0.3Ncos30 m 0

ω ρ ρ ρω

ω

ω

= = = =

→

→

= + − = = − =

∑∑

∑∑

( )

( )

2min min

max y

2n n max max

.2 , 3.405 rad/s

: F 0 Ncos30 0.3Nsin30 mg 0

F ma Nsin30 0.3Ncos30 m 0.2 , 7.214 rad/s

3.405 7.214 rad/s

ω ω

ω

ω ω

ω

=

= − − = = + = = ∴ < <

∑∑

mg

N Fs

ωmin

mg

N

Fs ωmax

455



P. 8/16 The 2 kg slider fits loosely in the smooth slot of the disk, which rotates about a vertical axis through point O. The slider is free to move slightly along the slot before one of the wires becomes taut. If the disk starts from rest at time t = 0 and has a constant clockwise angular acceleration of 0.5 rad/s2, plot the tensions in wires 1 and 2 and the magnitude N of the force normal to the slot as functions of time t for the interval 0<=t<=5 s.

456



P. 8/16

2 2

r r

given: 0.5 rad/s constant 0.5t, 0.25tr 0.1 m & free to move move with the disk r 0, r 0assume N and T to be in the indicated direction and use r- coordinate

F ma Nc

slightlyθ θ θ

θ

= → = == ≈ → = =

= − ∑

( )( )( )

2 2

2 2

1

os45 Tcos45 2 0.1 0.5t 0.05t

F ma Nsin45 Tsin45 2 0.1 0.5 0.1

0.05t 0.1 0.05t 0.1N T2 2

N is always positive the assumed direction is correctT will be negative for t 1.414 s

0, 0T

θ θ

− = × − × = −

= − = × × = + −

= =

∴<

∴ =

∑

2

22

t 1.414 s 0.1 0.05t , 0 t 1.414 s and T 20.05t 0.1, t 1.414 s

0, t 1.414 s2

≤ ≤ −≤ < =−

> ≥

r

θ

45°

N

T θ

457



P. 8/16

458



P. 8/17 A small rocket-propelled vehicle of mass m travels down the circular path of effective radius r under the action of its weight and a constant thrust T from its rocket motor. If the vehicle starts from rest at A, determine its speed v when it reaches B and the magnitude N of the force exerted by the guide on the wheels just prior to reaching B. Neglect any friction and any loss of mass of the rocket.

459



P. 8/17

[ ] ( )

2n n

t t t t

2 2t t

0

= /2 = /2

F ma N mgsin mv / r

T mgcosF ma T mgcos ma , am

Tvdv a ds v / 2 a rd , v 2r gsinm

N 3mgsin 2T

Tv r 2 N 3mg Tm

g

θ

θ π θ π

θ

θθ

θθ θ

θ θ

π π

= − = + = + = =

= = = +

= +

= + = +

∑

∑

∫

mg

n

t

T

N

460



P. 8/18 A hollow tube rotates about the horizontal axis through point O with constant angular velocity ωo . A particle of mass m is introduced with zero relative velocity at r = 0 when θ= 0 and slides outward through the smooth tube. Determine r as a function of θ.

461



P. 8/18

( )( )

( )( )

( )

o

o

2r r

2o o

p h

2p o o

p

given: , 0at t 0, r 0, r 0, 0 t t

F ma mgsin m r r

r r gsin t differential equation of r t

r t r r

particular solution r is a solution of r r gsin t

r t force

θ ω θθ θ ω

θ θ

ω ω

ω ω

= =

= = = = ∴ =

= = − − = ←

= +

− =

=

∑

( ) ( )

o o

2 2o o o o o 2

o2

h ost

h

2 s

d response of gsin t gsin t

1sub. into diff. eq. gsin t gsin t gsin t2

homogeneous solution r is a solution of r r 0

r t free natural response

sub. into diff. eq. s

C

C C C

Ae

A e

ω ω

ω ω ω ω ωω

ω

=

− − = → = −

− =

= =

( )

( )

( ) ( )

( )

o o

o o

t 2 sto o o

t th

t tp h o2

o

o oo

2 2 2o o o

0, s ,

r t1r t r r gsin t, which must satisfy i.c.

2gr 0 0 and r 0 0

2g g g, r sinh sin

4 4 2

A e

Ae Be

Ae Be

A B A B

A B

ω ω

ω ω

ω ω ω

ωω

ω ωω

θ θω ω ω

−

−

− = = −

∴ = +

= + = + −

= = + = = − −

→ = = − ∴ = −

r θ

N mg

462



P. 8/19 The small pendulum of mass m is suspended from a trolley that runs on a horizontal rail. The trolley and pendulum are initially at rest with θ= 0. If the trolley is given a constant acceleration a = g, determine the maximum angle θmax through which the pendulum swings. Also find the tension T in the cord in terms of θ.

463



P. 8/19

[ ]

( )( )

P C P/C

2C P/C n t

t t

use n-t coordinate to avoid unknown T in t-direction translating axes attached to the cart to observe pendulum

g

F ma mgsin m gcos

g cos sin a

l l

l

l

θ θ

θ θ θ

θ θ θ

= +

= = +

= − = − +

= −

∑

a a a

a i a e e

( ) ( )

( )( )

2 2

0

max min max

2n n

s function of

g gd d / 2 cos sin d , 2 sin cos 1

or when 0 sin cos 1 / 2

F ma T mgcos m gsin

T mg 3sin 3cos 2

l l

l

θ

θ

θ θ θ θ θ θ θ θ θ θ θ

θ θ θ θ θ θ π

θ θ θ

θ θ

= = − = + −

= → + = ∴ =

= − = + = + −

∫

∑

t

n

T

mg

2nlθ e

tlθe

giaP

464



P. 8/20 A small object is released from rest at A and slides with friction down the circular path. If the coefficient of friction is 0.2, determine the velocity of the object as it passes B. (Hint: Write the equations of motion in the n- and t- directions, eliminate N, and substitute vdv = atrdθ. The resulting equation is a linear nonhomogeneous differential equation of the form , the solution of which is well known.) ( ) ( )dy/dx f x y g x+ =

465



P. 8/20 ( )

( )( )

( ){ } ( )( ) ( ) ( )

2n n

t t

2

2 2

22 2

F ma N mgsin m 3

F ma mgcos 0.2N m 3

eliminate N: gcos 0.2 gsin 3 3

1 1d d d gcos 0.2 gsin 3 d d3 2

d 20.4 g cos 0.2sin , as a fund 3

θ θ

θ θ

θ θ θ θ

θ θ θ θ θ θ θ θ θ θ θ

θθ θ θ θ

θ

= − =

= − =

− × + =

= = − × + =

+ = −

∑∑

( ) ( )

( ) ( )

( ) ( )

2

p h

p

ction of

let u and to solve the differential equation for uu u u

2u forced response of g cos 0.2sin cos sin3

2sub. into diff. eq. sin cos 0.4 cos sin g cos 0.2sin3

match the c

A B

A B A B

θ

θ θ θ

θ θ θ θ θ

θ θ θ θ θ θ

=

= +

= − = +

− + + + = −

( )

( ) ( )

sh

s s

0.4

1.2 2 0.48oeff. of sin and cos : g g3.48 3 3.48

u solution of the homogeneous equation

s 0.4 0, s 0.41.2 2 0.48u gcos gsin with u 0 03.48 3 3.48

1.2 g 0 u3.48

A B

Ce

C e Ce

Ce

C

θ

θ θ

θ

θ θ

θ

θ θ θ −

= = −

= =

+ × = = −

∴ = + − + =

+ = → ( ) 2 0.4

2B

1.2 2 0.48 1.2gcos gsin g3.48 3 3.48 3.48

at / 2, 3.382 v r 5.52 m/sReal world where friction exists makes the phenomena difficult

e θθ θ θ θ

θ π θ θ

− = = + − −

= = → = =

n

t

F=0.2N

N

mg

466



P. 8/21 A small collar of mass m is given an initial velocity of magnitude vo on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friciton is μk, determine the distance traveled before the collar comes to rest. (Hint: Recognize that the friction force depends on the net normal force.)

467



P. 8/21

[ ]

( )o

v h

y v

2

n n h

2 2t t k v h t

2 2 2 2 4kt

2 4 2 20 s2o o

22 2 2 kv 0k

Normal force has component N and N

F 0 N mg

vF ma N mr

F ma F N N ma

vdv a ds vdv r m g m v dsmr

v v r grdv rds, s ln2 rg2 r g v

µ

µ

µµ

= =

= =

= − = − + =

= = − +

+ +− = =+

∑

∑

∑

∫ ∫

y

t

n mg

Nv

Nh

F

468



P. 8/22 The slotted arm OB rotates in a horizontal plane about point O of the fixed circular cam with constant angular velocity = 15 rad/s. The spring has a stiffness of 5 kN/m and is uncompressed when θ= 0. The smooth roller A has a mass of 0.5 kg. Determine the normal force N that the cam exerts on A and also the force R exerted on A by the sides of the slot when θ= 45°. All surfaces are smooth. Neglect the small diameter of the roller.

θ

469



P. 8/22

2 2 2

2

2

Kinematics: 0.2 0.1 r 0.2rcosdiff: 0 2rr 0.2rcos 0.2r sin 0 2r 2rr 0.2rcos 0.2r sin 0.2r sin 0.2r sin 0.2r cosgiven: / 4, 15 rad/s, 0

r 0.

θ

θ θ θ

θ θ θ

θ θ θ θ θ θ

θ π θ θ

= + +

= + −

= + + −

− − −

= = =

∴ =

( )( )

( )

2

2r r

1164 m, r 0.66 m/s, r 15.05 m/s0.2 0.1 , 20.7

sin135 sinKinetics: spring force at / 4 : F 5000 r 0.1 compressed

F ma F Ncos20.7 m r r

F ma R Nsin20.7 m r 2r

N 81.7θ θ

ββ

θ π

θ

θ θ

= =

= = °

= = × −

= − + = −

= − = + =

∑∑

N R 38.7 N=

0.2

0.1

r θ β

N

R

F

r

θ

β

470



P. 8/23 The small cart is nudged with negligible velocity from its horizontal position at A onto the parabolic path that lies in a vertical plane. Neglect friction and show that the cart maintains contact with the path for all values of k.

471



P. 8/23

( )( )

3/ 222

22

3/ 22 2

2

n n

2

If the cart maintains contact, N 0use n-t coordinate since N aligns with the n-axis

1 ' dy d y y kx k 2x tan 2k'' dx dx

1 4k x2k

vF ma N mgcos m

1 tan

y

yρ θ

ρ

θρ

θ

>

+ = = = = =

+ =

= − + =

+

∑

[ ]

2

2 2

t t t

2 2t

23/ 2 3/ 22 2 2 2 2 2

1sec cos1 4k x

F ma mgsin ma

vdv a ds vdv gsin ds gdy, v 2gy 2kgxmg 2k mgN 2mkgx 0

1 4k x 1 4k x 1 4k x

θ θ

θ

θ

= = + = =

= = = = =

∴ = − × = >+ + +

∑

dx

dy ds θ

mg

N

n

t

472


8.6 Work and Energy

8.6 Work and Energy

473

Many of the dynamic problems need to determine the velocity or the displacement quantities. This requires integrating the equations formulated from the Newton’s second law with the use of kinematical analysis for the acceleration term. However, this integration step can be performed beforehand. If we integrate the Newton’s law directly w.r.t. the displacement, i.e.

∫ ∫

−==

2

21

22 vvmmvdvFds


8.6 Work and Energy

474

The left hand side is defined to be the work done on the particle. The right hand side corresponds to the change in the kinetic energy. This integrated equation of motion is called work-energy equation. On the other hand, if we integrate the Newton’s law directly w.r.t. time, i.e. For this one, the left hand side quantity is called the impulse applied to the particle, while the right hand one is the change of the momentum of the particle. This integrated equation of motion is then called impulse-momentum equation.

( )∫ ∫ −== 12 vvmmdvFdt


8.6 Work and Energy

475

Definition of Work If the force acting on a particle at point A of which its position is described by the position vector , then

is the work done by the force during the differential displacement , of which its magnitude is equal to the force times the displacement in the direction of force. Alternatively, the magnitude will also be equal to the displacement times the force in the direction of the displacement. In other words, components of force in the direction normal to the displacement vector, , will do no work.

Fr

rF ddU ⋅=F

rd

nF


8.6 Work and Energy

476


8.6 Work and Energy

477

Calculation of Work When the force , which may not be constant, has been applied to the particle during a finite movement, the work performed by this force on the particle will be which may be further elaborated as if the x-y coordinate frame is used, or

F

∫ ⋅=2

1

rF dU

( )∫ +=2

1

dyFdxFU yx


8.6 Work and Energy

478

if the n-t coordinate frame is used. Moreover, if the function of the force is not known explicitly, the numerical integration approach for approximating the area under graph may be used.

∫=2

1

s

stdsFU

sFt −


8.6 Work and Energy

479

Work done by spring force From the figure, the spring force exerted on the particle when it is displaced from the unstretch by is . Therefore the work done by spring as the particle undergoes an arbitrary displacement from to would be

xiF kx−=

1x2x

( ) ( )22

2121 2

12

1

xxkdxkxUx

x

−=⋅−= ∫− ii


8.6 Work and Energy

480


8.6 Work and Energy

481

Work done by gravity force a) g = const.

b) g != const.

( ) ( ) ( )21

2

121 yymgdydxmgU −=+⋅−= ∫− jij

( )

−=⋅

−= ∫−

12

2

1221

11rr

mGmdrr

mGmU erre ee


8.6 Work and Energy

482


8.6 Work and Energy

483

Work done by force lead us to define the kinetic energy of the particle, which can be interpret as the amount of work given to the particle to bring it from rest to a velocity . Hence the above equation may be rewritten as

( )∫ ∫∫∫ −===⋅=⋅=−

2

1

2

1

21

22

2

1

2

121 2

1 vvmmvdvdsmadmdU trarF

F

2

21 mvT =

v

TTTU ∆=−=− 1221


8.6 Work and Energy

484


8.6 Work and Energy

485

This equation is known as work-energy equation for the particle. It says that the total work done by all forces acting on the particle as it moves from 1 2 equals to the change in kinetic energy of the particle. Alternatively, saying that the final kinetic energy equals the initial kinetic energy plus the work done on the particle during that period. The equation also applies to a system of particles of which their relative distances are kept unchanged.

2211 TUT =+ −


8.6 Work and Energy

486

However, would be the net work done on the system by external forces and would be the total kinetic energy of the system. Before applying the work-energy equation, it is required to make clear the isolated system under consideration by drawing the free-body diagram of it, showing all externally applied forces which do work (active forces).

21−UT


8.6 Work and Energy

487

Potential Energy Potential energy is here to treat the work, especially those done by gravity forces and by spring forces. Potential energy may be converted into kinetic energy and do work later. Gravitational Potential Energy is the work done against the gravitational force to elevate the particle a distance above some reference datum, where the potential energy is taken to be zero

h

mghVg =


8.6 Work and Energy

488

The change in potential energy when the particle goes from the level to the new level is which is the negative of the work done by the gravitational force on the particle . If the level change is significant compared to the radius of the earth, the gravity force is not equal to the simple , rather it will be . Accordingly, the potential energy is the work done against this gravity force which, in this case, may be calculated from

1h

( ) hmghhmgVg ∆=−=∆ 12

2h

hmg∆−

mg 222 // rmgRrGmme =

rmgRVg

2

−=


8.6 Work and Energy

489

where the reference datum for zero potential is taken to be at . The change in potential energy when the particle goes from to is which is the negative of the work done by the gravitational force on the particle .

∞→r

−=∆

21

2 11rr

mgRVg

1r 2r

( )122 /1/1 rrmgR −


8.6 Work and Energy

490


8.6 Work and Energy

491

Elastic Potential Energy is the work done against the spring force to deform the spring. Therefore, the potential energy of the spring when it is deformed by the amount of from the unstretched posture is The change in potential energy when the spring undergoes the deformation from to is which is the negative of the work done by the spring force.

x

2

0 21 kxdkV

x

e == ∫ δδ

1x 2x

( )21

222

1 xxkVe −=∆


8.6 Work and Energy

492

Work-Energy Equation Revisited The original work-energy equation may be modified to account for the potential energy explicitly as where now the scope of the system is changed to include the spring and the gravity field. is then the work done by all external forces on the new system. Alternatively

1221 TTU −=−

'21−U

( ) ( ) 12'

21 TTVVU eg −=∆−+∆−+−

22'

2111 VTUVT +=++ −


8.6 Work and Energy

493

P. 8/24 A constant horizontal force P = 700 N is applied to the linkage shown. With the 14-kg ball initially at rest on its support with , calculate the velocity of the ball as approaches zero where the ball reaches its highest position.

60=θθ

Ch. 2: Kinematics of...

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