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CHAPTER 17CHAPTER 17

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

PROBLEM 17.1

It is known that 1500 revolutions are required for the 3000 kg flywheel to coast to rest from an angular velocity

of 300 rpm. Knowing that the radius of gyration of the flywheel is 1 m determine the average magnitude of the

couple due to kinetic friction in the bearings.

SOLUTION

Angular velocity: w p

pw

0 3002

60

10

0

= ÊËÁ

ˆ¯̃

==

rpm

rad

2

Moment of inertia: I mk= =

= ◊

2 23000 1

3000

( ) ( )kg m

kg m2

Kinetic energy: T I

T

1 02

2

6

2

1

2

1

23000 10

1 48044 10

0

=

=

= ¥ ◊=

w

p( )( )

. N m

Work: U MM

M

1 2

1500 2

9424 8

Æ = -= -= -

qp( )( )

.

rev rad/rev

Principle of work and energy: T U T

M1 1 2 2

61 48044 10 9424 8 0

+ =

¥ - =Æ

. .

Average friction couple: M = ◊157 08. N m M = ◊157 1. N m �

4


PROBLEM 17.2

The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off.

The 50-kg rotor, which has a centroidal radius of gyration of 180 mm, then coasts to rest. Knowing that the

kinetic friction of the rotor produces a couple of magnitude 3 5. ,N m◊ determine the number of revolutions that

the rotor executes before coming to rest.

SOLUTION

Angular velocity: w p

pw

0 36002

60

120

0

= ÊËÁ

ˆ¯̃

==

rpm

rad/s

2

Moment of inertia: I mk=

=

= ◊

2

250 0 180

1 620

( )( . )

.

kg m

kg m2

Kinetic energy: T I

T

1 02

2

2

1

2

1

21 620 120

115 12

0

=

=

==

w

p( . )( )

. ,kJ

Work: U M1 2

3 5

Æ = -= - ◊

qq( . )N m

Principle of work and energy:

T U T1 1 2 2 115 12 3 5 0+ = - ◊ =Æ : . ( . )kJ N m q

Rotation angle: q = ¥32 891 103. rad q = 5230 rev �

5


PROBLEM 17.3

Two disks of the same material are attached to a shaft as shown. Disk A is of radius r

and has a thickness b, while disk B is of radius nr and thickness 3b. A couple M of

constant magnitude is applied when the system is at rest and is removed after the

system has executed 2 revolutions. Determine the value of n which results in the

largest final speed for a point on the rim of disk B.

SOLUTION

For any disk: m r t

I mr

tr

=

=

=

r p

pr

( )2

2

4

1

2

1

2

Moment of inertia.

Disk A: I b rA = 1

2

4p r

Disk B: I b nr

n br

n II I I

n

B

A

A B

=

= ÈÎÍ

˘˚̇

== +

= +

1

23

31

2

3

1 3

4

4 4

4

4

p r

pr

( )( )

(

total

))I A

Work-energy. T U M M

T I

1 1 2

2 22

0 4

1

2

= = =

=

Æ q p

w

( )rad

total

T U T M n I A1 1 2 24

220 4

1

21 3+ = + = +Æ : ( ) ( )p w

w p22

4

8

1 3=

+Mn I A( )

For Point D on rim of disk B

v nrD = ( )w2 or v n r MrI

nnD

A

2 2 222

2 2

4

8

1 3= = ◊

+w p

6


PROBLEM 17.3 (Continued)

Value of n for maximum final speed.

For maximum v ddn

nnD :

2

41 30

+

Ê

ËÁˆ

¯̃=

1

1 312 1 3 2 0

12 2 6 0

2 3 1 0

4 2

2 3 4

5 5

4

( )[ ( ) ( )( )]

( )

+- + =

- - =

- =

nn n n n

n n n

n n

n = 0 and n = ÊËÁ

ˆ¯̃

=1

30 7598

0 25.

. n = 0 760. �

7


PROBLEM 17.4

Two disks of the same material are attached to a shaft as shown. Disk A has a mass

of 15 kg and a radius r = 125 mm. Disk B is three times as thick as disk A. Knowing

that a couple M of magnitude 20 N ◊ m is to be applied to disk A when the system is

at rest, determine the radius nr of disk B if the angular velocity of the system is to be

600 rpm after 4 revolutions.

SOLUTION

For any disk: m r t

I mr

tr

=

=

=

r p

pr

( )2

2

4

1

2

1

2

Moment of inertia.

Disk A: I brA = 1

2

4pr

Disk B: I b nr

n br

n I

B

A

=

= ÈÎÍ

˘˚̇

=

1

23

31

2

3

4

4 4

4

pr

pr

( )( )

I I I n IA B Atotal = + = +( )1 3 4 (1)

Angular velocity: ww

p

1

2

0

600

20

===

rpm

rad/s

Rotation: q p= =4 8rev rad

Kinetic energy: T

T I

1

2 22

0

1

2

=

= total w

Work: U M1 2

20 8

502 65

Æ == ◊=

qp( )( )

.

N m rad

J

8




I

I

1 1 2 2

20 502 651

220

0 25465

+ =

+ =

= ◊

Æ

. ( )

.

total

total2kg m

p

But, I m rA A A=

=

= ◊

1

2

1

215 0 125

0 117188

2

2( )( . )

.

kg m

kg m2

From (1) 0 25465 1 3 0 117188

0 390998

0 79076

4

4

. ( )( . )

.

.

= +

==

n

nn

Radius of disk B: r nrB A= = ( . )( )0 79076 125 mm rB = 98 8. mm �

9


PROBLEM 17.5

The flywheel of a punching machine has a mass of 300 kg and a radius of gyration of 600 mm. Each punching

operation requires 2500 J of work. (a) Knowing that the speed of the flywheel is 300 rpm just before a punching,

determine the speed immediately after the punching. (b) If a constant 25-N ◊ m couple is applied to the shaft of

the flywheel, determine the number of revolutions executed before the speed is again 300 rpm.

SOLUTION

Moment of inertia. I mk=

=

= ◊

2

2300 0 6

108

( )( . )kg m

kg m2

Kinetic energy. Position 1. wp

w

p

1

1 12

2

3

300

10

1

2

1

2108 10

53 296 10

==

=

=

= ¥

rpm

rad/s

J

T I

( )( )

.

Position 2. T I2 22

221

254= =w w

Work. U1 2 2500Æ = - J

Principle of work and energy for punching.

T U T1 1 2 23

2253 296 10 2500 54+ = ¥ - =Æ : . w

(a) w22 940 66= .

w2 30 67= . rad/s w2 293= rpm �

Principle of work and energy for speed recovery.

T U TU

MU M

2 2 1 1

2 1

2 1

2500

25

2500 25 100

+ === ◊= = =

Æ

Æ

Æ

J

N m

radq q q

(b) q = 15 92. rev �

10


PROBLEM 17.6

The flywheel of a small punching machine rotates at 360 rpm. Each punching operation requires 2250 N ◊ m

of work and it is desired that the speed of the flywheel after each punching be not less than 95 percent of the

original speed. (a) Determine the required moment of inertia of the flywheel. (b) If a constant 27 N ◊ m couple

is applied to the shaft of the flywheel, determine the number of revolutions that must occur between two

successive punchings, knowing that the initial velocity is to be 360 rpm at the start of each punching.

SOLUTION

Angular speeds. w pw ww p

1

2 1

2

360 12

0 95

11 4

= ===

rpm rad/s

rad/s

.

.

Work. U1 2 2250Æ = ◊N m

Principle of work and energy for punching

T U T

I U I

1 1 2 2

12

1 2 221

2

1

2

+ =

+ =

Æ

Æw w

(a) Solving for I IU

=-

-Æ2 1 2

22

12w w

= - --

= ◊( )( )

( ) ( . ).

2 2250

12 11 432 475

2 2

2

p pkg m I = ◊32 5. kg m2 �

Principle of work and energy for speed recovery.

T U T2 2 3 3+ =Æ

But, T TU T T

T TU

3 1

2 3 3 2

1 2

1 2

2250

== -= -= -= ◊

Æ

Æ

N m

Work, U M2 3Æ = q

(b) q =

=

ÆUM2 3

2250

27

= 83 33. rad q = 13 26. rev �

11


PROBLEM 17.7

Disk A is of constant thickness and is at rest when it is placed in contact

with belt BC, which moves with a constant velocity v. Denoting by mk

the coefficient of kinetic friction between the disk and the belt, derive an

expression for the number of revolutions executed by the disk before it attains

a constant angular velocity.

SOLUTION

Work of external friction force on disk A.

Only force doing work is F. Since its moment about A is M rF= , we have

U MrFr mgk

1 2Æ ===

qqm q( )

Kinetic energy of disk A.

Angular velocity becomes constant when w

w

2

1

2 22

22

2

0

1

2

1

2

1

2

4

=

=

=

= ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

vr

T

T I

mr vr

mv

Principle of work and energy for disk A.

T U T r mg mvk1 1 2 2

2

04

+ = + =Æ : m q

Angle change. qm

= vr gk

2

4rad q

p m= v

r gk

2

8rev �

12


PROBLEM 17.8

Disk A, of weight 5 kg and radius r = 150 mm, is at rest when it is placed

in contact with belt BC, which moves to the right with a constant speed

v = 12 m/s. Knowing that mk = 0 20. between the disk and the belt, determine

the number of revolutions executed by the disk before it attains a constant

angular velocity.

SOLUTION

Work of external friction force on disk A.

Only force doing work is F. Since its moment about A is M rF= , we have

U MrFr mgk

1 2Æ ===

qqm q( )

Kinetic energy of disk A.

Angular velocity becomes constant when w

w

2

1

2 22

22

2

0

1

2

1

2

1

2

4

=

=

=

= ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

vr

T

T I

mr vr

mv

Principle of work and energy for disk A.

T U T r mg mvk1 1 2 2

2

04

+ = + =- : m q

Angle change qm

= vr gk

2

4rad q

p m= v

r gk

2

8rev

Data: r

vk

===

0 15

0 20

12

.

.

m

m/s

m

qp

=( )

( . )( . )( . )

12

8 0 15 0 20 9 81

2m/s

m m/s2 q = 19 47. rev �

13


PROBLEM 17.9

Each of the gears A and B has a mass of 2.4 kg and a radius of

gyration of 60 mm, while gear C has a mass of 12 kg and a radius

of gyration of 150 mm. A couple M of constant magnitude 10 N ◊ m

is applied to gear C. Determine (a) the number of revolutions of

gear C required for its angular velocity to increase from 100 to

450 rpm, (b) the corresponding tangential force acting on gear A.

SOLUTION

Moments of inertia.

Gears A and B: I I mkA B= = = = ¥ ◊-2 2 32 4 0 06 8 64 10( . )( . ) . kg m2

Gear C: IC = = ¥ ◊-( )( . )12 0 15 270 102 3 kg m2

Kinematics. r r rA A B B C C

A B C C

A B C

w w w

w w w w

q q q

= =

= = =

= =

200

802 5

2 5

.

.

Kinetic energy. T I= 1

2

2w :

Position 1. w p

w w p

C

A B

= =

= = =

10010

3

25025

3

rpm rad/s

rpm rad/s

Gear A: ( ) ( . ) .T A13

21

28 64 10

25

32 9609= ¥ Ê

ËÁˆ¯̃

=- pJ

Gear B: ( ) ( . ) .T B13

21

28 64 10

25

32 9609= ¥ Ê

ËÁˆ¯̃

=- pJ

Gear C: ( ) ( ) .T C13

21

2270 10

10

314 8044= ¥ Ê

ËÁˆ¯̃

=- pJ

System: T T T TA B C1 1 1 1 20 726= + + =( ) ( ) ( ) . J

Position 2. w pw w p

C

A B

= == =

450 15

37 5

rpm rad/s

rad/s.

Gear A: ( ) ( . )( . ) .T A23 21

28 64 10 37 5 59 957= ¥ =- p J

14



Gear B: ( ) ( . )( . ) .T B23 21

28 64 10 37 5 59 957= ¥ =- p J

Gear C: ( ) ( )( ) .T C23 21

2270 10 15 299 789= ¥ =- p J

System: T T T TA B C2 2 2 2 419 7= + + =( ) ( ) ( ) . J

Work of couple. U M C C1 2 10Æ = =q q

Principle of work and energy for system.

T U T C1 1 2 2 20 726 10 419 7+ = + =Æ : . .q

qC = 39 898. radians

(a) Rotation of gear C. qC = 6 35. rev �

Rotation of gear A. qA ==

( . )( . )

.

2 5 39 898

99 744 radians

Principle of work and energy for gear A.

( ) ( ) : . ( . ) .T M T MA A A A A1 2 2 9609 99 744 59 957+ = + =q

M A = ◊0 57142. N m

(b) Tangential force on gear A. F Mrt

A

A= = 0 57142

0 08

.

. Ft = 7 14. N �

15


SOLUTION

Moments of inertia.

Gears A and B: I I mkA B= = = = ¥ ◊-2 2 32 4 0 06 8 64 10( . )( . ) . kg m2

Gear C: IC = = ¥ ◊-( )( . )12 0 15 270 102 3 kg m2

Kinematics. r r rA A B B C C

A B C C

A B C

w w w

w w w w

q q q

= =

= = =

= =

200

802 5

2 5

.

.

Kinetic energy. T I= 1

2

2w :

Position 1. w p

w w p

C

A B

= =

= = =

10010

3

25025

3

rpm rad/s

rpm rad/s

Gear A: ( ) ( . ) .T A13

21

28 64 10

25

32 9609= ¥ Ê

ËÁˆ¯̃

=- pJ

Gear B: ( ) ( . ) .T B13

21

28 64 10

25

32 9609= ¥ Ê

ËÁˆ¯̃

=- pJ

Gear C: ( ) ( ) .T C13

21

2270 10

10

314 8044= ¥ Ê

ËÁˆ¯̃

=- pJ

System: T T T TA B C1 1 1 1 20 726= + + =( ) ( ) ( ) . J

PROBLEM 17.10

Solve Problem 17.9, assuming that the 10-N ◊ m couple is applied

to gear B.

PROBLEM 17.9 Each of the gears A and B has a mass of 2.4 kg and

a radius of gyration of 60 mm, while gear C has a mass of 12 kg and

a radius of gyration of 150 mm. A couple M of constant magnitude

10 N ◊ m is applied to gear C. Determine (a) the number of revolutions

of gear C required for its angular velocity to increase from 100 to

450 rpm, (b) the corresponding tangential force acting on gear A.

16



Position 2. w pw w p

C

A B

= == =

450 15

37 5

rpm rad/s

rad/s.

Gear A: ( ) ( . )( . ) .T A23 21

28 64 10 37 5 59 957= ¥ =- p J

Gear B: ( ) ( . )( . ) .T B23 21

28 64 10 37 5 59 957= ¥ =- p J

Gear C: ( ) ( )( ) .T C23 21

2270 10 15 299 789= ¥ =- p J

System: T T T TA B C2 2 2 2 419 7= + + =( ) ( ) ( ) . J

Work of couple. U M B B1 2 10Æ = =q q

Principle of work and energy for system.

T U T B1 1 2 2 20 726 10 419 7+ = + =Æ : . .q

qB = 39 898. radians

(a) Rotation of gear C. qC = =39 898

2 515 959

.

.. radians qC = 2 54. rev �

Rotation of gear A. q qA B= = 39 898. radians

Principle of work and energy for gear A.

( ) ( ) : . ( . ) .T M T MA A A A A1 2 2 9609 39 898 59 957+ = + =q

M A = ◊1 4285. N m

(b) Tangential force on gear A. F Mrt

A

A= = 1 4285

0 08

.

. Ft = 17 86. N �

17


PROBLEM 17.11

The double pulley shown weighs 15 kg and has a centroidal

radius of gyration of 160 mm. Cylinder A and block B are

attached to cords that are wrapped on the pulleys as shown.

The coefficient of kinetic friction between block B and the

surface is 0.25. Knowing that the system is released from rest

in the position shown, determine (a) the velocity of cylinder A

as it strikes the ground, (b) the total distance that block B moves

before coming to rest.

SOLUTION

Let vA = speed of block A,vB = speed of block B, w = angular speed of pulley.

Kinematics. r rv rv r

A B

A A

B B

= == == =

0 25 0 15

0 25

0 15

. , .

.

.

m m

w ww w

s rs r

A A

B B

= == =

q qq q

0 25

0 15

.

.

(a) Cylinder A falls to ground. s

s

A

B

=

= ÊËÁ

ˆ¯̃

=

0 9

0 9

0 250 54

.

.

..

m

(0.15) m

Work of weight A: U m g sA A1 2 12 5 9 81 0 9Æ = =( ) ( . )( . )( . )

Normal contact force acting on block B: N m gB= = =( )( . ) .10 9 81 98 1 N

Friction force on block B: F Nf k= = =m ( . )( . ) .0 25 98 1 24 525 N

Work of friction force: U F sf B1 2 24 525 0 54 13 2435Æ = - = - = - ◊( . )( . ) . N m

Total work: U1 2 110 3625 13 2435 97 119Æ = - = ◊. . . N m

Kinetic energy: T T m v I m vA A B B1 22 2 20

1

2

1

2

1

2= = + +; w

T m r m k m rA A C B B22 2 2 2 2 2

2

1

2

1

2

1

2

1

212 5 0 25 15 0 16

= + +

= +

w w w

( . ) ( . ) ( ) ( . )22 2 2

2

10 0 15

0 695125

+ÈÎ ˘̊

=

( ) ( . )

.

w

w

18



Principle of work and energy.

T U T1 1 220 97 119 0 695125+ = + =Æ : . . w

w = 11 820. rad/s2

Velocity of cylinder A: vA = ( . )( . )0 25 11 820 vA = 2 96. m/s Ø �

(b) Block B comes to rest.

For block B and pulley C. T I m v TB B32 2

4

1

2

1

20= + =w ; :

T m k m rC B B32 2 2 2

2 2

1

2

1

2

1

215 0 16 10 0 15 11 820

= +

= +ÈÎ ˘̊

w w

( )( . ) ( )( . ) ( . ))

.

2

42 5424= ◊ N m

Work of friction force: U F s sf B B3 4 24 525Æ = - ¢ = - ¢.


T U T sB3 3 4 4 42 5424 24 525 0+ = - ¢ =Æ : . .

¢ =sB 1 7347. m

Total distance for block B. d s s dB B= + ¢ = + =: . . .0 54 1 7347 2 2747 m d = 2 27. m �

19


PROBLEM 17.12

The 160 mm-radius brake drum is attached to a larger flywheel that is

not shown. The total mass moment of inertia of the flywheel and drum

is 20 2kg m◊ and the coefficient of kinetic friction between the drum and

the brake shoe is 0.35. Knowing that the initial angular velocity of the

flywheel is 360 rpm counterclockwise, determine the vertical force P that

must be applied to the pedal C if the system is to stop in 100 revolutions.

SOLUTION

Kinetic energy. w pw

w

p

1

2

1 12

2

360 12

0

1

2

1

220 12

14 212 23

= ==

=

=

= ◊

rpm rad/s

N

T I

( )( )

, . mm

T2 0=

Work. q p

q

= == =

= - = -Æ

( )( ) .

( . )

( . )(

100 2 628 32

0 16

0 161 2

rad

mM F r FU M F

D f f

D f 6628 32

100 531

. )

.= - Ff


T U T Ff1 1 2 2 14 212 23 100 531 0+ = - =Æ : , . .

Ff = 141 371. N

Kinetic friction force. F N

NF

f k

f

k

=

= = =

m

m141 371

0 35403 918

.

.. N

Statics. + SM P F NA f= + - =0 180 40 200 0: ( ) ( ) ( )mm mm mm

180 40 141 371 200 403 918 0

417 38

PP

+ - ==

( )( . ) ( )( . )

. N P = 417 N Ø �

20


PROBLEM 17.13

Solve Problem 17.12, assuming that the initial angular velocity of the

flywheel is 360 rpm clockwise.

PROBLEM 17.12 The 160-mm-radius brake drum is attached to a

larger flywheel that is not shown. The total mass moment of inertia of

the flywheel and drum is 20 2kg m◊ and the coefficient of kinetic friction

between the drum and the brake shoe is 0.35. Knowing that the initial

angular velocity of the flywheel is 360 rpm counterclockwise, determine

the vertical force P that must be applied to the pedal C if the system is to

stop in 100 revolutions.

SOLUTION

Kinetic energy. w pw

w

p

1

2

1 12

2

360 12

0

1

2

1

220 12

14 212 23

= ==

=

=

=

rpm rad/s

N

T I

( )( )

, . ◊◊=

m

T2 0

Work. q p

q

= == =

= - = -Æ

( )( ) .

( . )

( . )(

100 2 628 32

0 16

0 16 61 2

rad

M F r FU M F

D f f

D f 228 32

100 531

. )

.= - Ff


T U T Ff1 1 2 2 14 212 23 100 531 0+ = - =Æ : , . .

Ff = 141 371. N

Kinetic friction. F N NF

f kf

k= = = =m

m:

.

..

141 371

0 35403 918 N

Statics. + + = - - =SM P F NA f0 180 40 200 0: ( ) ( ) ( )mm mm mm

180 40 141 371 200 403 918 0

480 21

PP

- - ==

( )( . ) ( )( . )

. N P = 480 N Ø �

21


PROBLEM 17.14

The gear train shown consists of four gears of the same thickness

and of the same material; two gears are of radius r, and the other

two are of radius nr. The system is at rest when the couple M0

is applied to shaft C. Denoting by I0 the moment of inertia of a

gear of radius r, determine the angular velocity of shaft A if the

couple M0 is applied for one revolution of shaft C.

SOLUTION

Mass and moment of inertia:

For a disk of radius r and thickness t: m r t tr

I mr r r tr

= =

= = + =

r p rp

rp rp

( )

( )

2 2

02 2 2 41

2

1

2

1

2

For a disk of radius nr and thickness t, I t nr I n I= =1

2

4 40rp ( )

Kinematics: If for shaft A we have w A

Then, for shaft B we have w wB A n= /

And, for shaft C we have w wC A n= / 2

Principle of work-energy:

Couple M0 applied to shaft C for one revolution. q p= 2 radians, T1 0= ,

U M M M

T I w IA A B B

1 2 0 0 0

22

2 2

1

2

1

2

- = = =

= +

q p p

w

( )

( ) ( )

radians

shaft shaft22 2

02

04

0

24

0

1

2

1

2

1

2

1

2

+

= + + ÊËÁ

ˆ¯̃

+

( )

( ) ( )

I

I w I n In

n I

C C

AA A

shaft w

w wnn

I nn

I nn

A

A

2

2

02 2

2

02

2

1

22

1

1

2

1

ÊËÁ

ˆ¯̃

= + +ÊËÁ

ˆ¯̃

= +ÊËÁ

ˆ¯̃

w

w

T U T M I nnA1 1 2 2 0 0

22

0 21

2

1+ = + = +ÊËÁ

ˆ¯̃- : p w

Angular velocity. wp

A

n

MI n

2 0

0 12

4 1=+( )

wp

An

nMI

=+

2

12

0

0

�

22


PROBLEM 17.15

The three friction disks shown are made of the same

material and have the same thickness. It is known that disk A

weighs 6 kg and that the radii of the disks are rA = 200 mm,

rB = 150 mm, and rC = 100 mm. The system is at rest when a

couple M0 of constant magnitude 7 5. N m◊ is applied to disk A.

Assuming that no slipping occurs between disks, determine the

number of revolutions required for disk A to reach an angular

velocity of 150 rpm.

SOLUTION

Kinematics.

Denote velocity of perimeter by v.

Mass and moment of inertia of disks.

Denote mass density of material by r and thickness of disks by t.

Then mass of a disk is m r t= =( ) ( )volume r p r2

and I mr t r= =1

2 2

2 4pr

Kinetic energy: T I

T t r r r

t

A A B B C C

=

= ÊËÁ

ˆ¯̃

+ +ÈÎ ˘̊

= ÊËÁ

ˆ¯

S 1

2

1

2 2

1

2 2

2

4 2 4 2 4 2

w

pr w w w

pr˜̃ +

ÊËÁ

ˆ¯̃

+ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

r r rr

r rrA A B

A

BA C

A

CA

4 2 4

2

2 4

2

2w w w

T t r r r rAA A B C=

Ê

ËÁˆ

¯̃+ +ÈÎ ˘̊1

2 2

22 2 2 2pr w

(1)

23



Recall that m r tA A= p r2 and write

Eq. (1) as: T r t r r r

T m r rr

rr

A A B C

A AB

A

C

A

= ( ) + +( )

= +ÊËÁ

ˆ¯̃

+ÊËÁ

ˆ

1

4

1

41

2 2 2 2

2

2

p r

( )¯̃̄

È

ÎÍÍ

˘

˚˙˙

2

2w A

Data: w p pA

A A B Cm r r r

= ÊËÁ

ˆ¯̃

=

= = = =

1502

605

6 0 2 0 15 0 1

rpm rad/s

kg m, m, m, . . .

MM = ◊7 5. N m

Work: U M1 2 7 5- = = ◊q q( . )N m

Principle of work and energy for system:

T U T1 1 2 2

22 2

0 7 51

46 0 2 1

0 15

0 2

0 1

0 2

+ =

+ = + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

Æ

. ( )( . ).

.

.

.q

ÈÈ

ÎÍÍ

˘

˚˙˙

= + +ÈÎÍ

˘˚̇

=

( )

. . ( )

. .

5

7 5 0 06 19

16

1

425

7 5 26 833

2

2

p

q p

q

qp

=ÊËÁ

ˆ¯̃

=

3 57772

0 5694

.

.

radrev

rad

rev q = 0 569. rev �

24


PROBLEM 17.16

A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring

of constant k = 400 N/m and of unstretched length 150 mm is attached to

the rod as shown. Knowing that the rod is released from rest in the position

shown, determine its angular velocity after it has rotated through 90°.

SOLUTION

Position 1:

Spring: x CD

Ve

1 150 370 150 220 0 22= - = - = =( ) .mm mm m

Unstretched Length674 84

== = =1

2

1

2400 0 22 9 681

2 2kx ( )( . ) .N/m m J

Gravity: V Wh mghV V V

g

e g

= = = =

= + = +

( )( . )( . ) .

. .

4 9 81 0 18 7 063

9 68 7 06

2

1

kg m/s m J

J 33 16 743J J= .

Kinetic energy: T1 0=

Position 2:

Spring: x

V kxe

2

22 2

230 150 80 0 08

1

2

1

2400 0 08 1 28

= - = =

= = =

mm mm mm m

N/m m

.

( )( . ) . JJ

Gravity: V WhV V V

g

e g

= =

= +

=

0

1 28

2

. J

25



Kinetic energy: v r

I mL

T mv

2 2 2

2 2 2

2 2

0 18

1

12

1

124 0 6 0 12

1

2

= =

= = = ◊

=

w w( . )

( )( . ) .

m

kg m kg m

2222

22

22

2 22

1

2

1

24 0 18

1

20 12

0 1248

+

= +

=

I

T

w

w w

w

( )( . ) ( . )

.

kg

Conservation of energy:

T V T V1 1 2 2

22

22

2

0 16 743 0 1248 1 28

123 9

11 131

+ = +

+ = +

==

. . .

.

.

J J

rad/

w

ww ss w2 11 13= . rad/s �

26


SOLUTION

Position 1:

Spring: x CD

Ve

1 150 370 150 220 0 22= - = - = =( ) .mm mm m

Unstretched Length674 84

== = =1

2

1

2400 0 22 9 681

2 2kx ( )( . ) .N/m m J

Gravity: V Wh mghV V V

g

e g

= = = - = -

= + = -

( )( . )( . ) .

. .

4 9 81 0 22 7 063

9 68 7

2

1

kg m/s m J

J 0063 2 617J J= .


Position 2:

Spring: x

V kxe

2

22 2

230 150 80 0 08

1

2

1

2400 0 08 1 28

= - = =

= = =

mm mm mm m

N/m m

.

( )( . ) . JJ

Gravity: V WhV V V

g

e g

= =

= +

=

0

1 28

2

. J

PROBLEM 17.17

A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring

of constant k = 400 N/m and of unstretched length 150 mm is attached to

the rod as shown. Knowing that the rod is released from rest in the position

shown, determine its angular velocity after it has rotated through 90°.

27



Kinetic energy: v r

I mL

T mv

2 2 2

2 2 2

2 2

0 18

1

12

1

124 0 6 0 12

1

2

= =

= = = ◊

=

w w( . )

( )( . ) .

m

kg m kg m

2222

22

22

2 22

1

2

1

24 0 18

1

20 12

0 1248

+

= +

=

I

T

w

w w

w

( )( . ) ( . )

.

kg

Conservation of energy:

T V T V1 1 2 2

22

22

2

0 2 617 0 1248 1 28

10 713

3 273

+ = +

+ = +

==

. . .

.

.

J J

rad/s

w

ww w2 3 27= . rad/s �

28


PROBLEM 17.18

A slender rod of length l and weight W is pivoted at one end as shown. It is

released from rest in a horizontal position and swings freely. (a) Determine

the angular velocity of the rod as it passes through a vertical position and

determine the corresponding reaction at the pivot, (b) Solve Part a for

W = 10 N and l = 1 m.

SOLUTION

Position 1: v

T

v l

1

1

1

2 2

0

0

0

2

===

=

w

w

Position 2: T mr I

m l ml

T ml

2 22

22

2

22

22

22

1

2

1

2

1

2 2

1

2

1

12

1

6

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

w

w w

w222

Work: U mg l1 2

2Æ =


mg l ml

1 1 2 2

2220

2

1

6

+ =

+ =

Æ

w

(a) Expressions for angular velocity and reactions.

w

w

22

22

3

2 2

3 3

2

=

= = ◊ =

gl

a l l gl

g

w2

3= gl

�

+ S SF F A W ma= - =( )eff :

A mg m g

A mg

- =

=

3

2

5

2 A = ≠5

2W �

29



(b) Application of data:

W lgl

g= =

= = =

10 1

3 3

129 432

2

N m

rad /s2 2

,

.w w2 5 42= . rad/s �

A W= = =5

2

5

210 25( )N N A = 25 0. N ≠ �

30


PROBLEM 17.19

A slender rod of length l is pivoted about a Point C located at a distance b from

its center G. It is released from rest in a horizontal position and swings freely.

Determine (a) the distance b for which the angular velocity of the rod as it

passes through a vertical position is maximum, (b) the corresponding values

of its angular velocity and of the reaction at C.

SOLUTION

Position 1. v T= = =0 0 01, wElevation: h V mgh= = =0 0 1

Position 2. v b

I ml

T mv I

m b l

2 2

2

2 22

22

2 222

1

12

1

2

1

2

1

2

1

12

=

=

= +

= +ÊËÁ

ˆ¯̃

w

w

w

Elevation: h b V mgb= - = - 2

Principle of conservation of energy.

T V T V m b l mgb1 1 2 22 2

220 0

1

2

1

12+ = + + = +Ê

ËÁˆ¯̃

-: w

w22

2 112

2

2=+gb

b l

(a) Value of b for maximumw2.

ddb

bb l

b l b b

b lb

2 112

2

2 112

2

2 112

22

22

0+

Ê

ËÁ

ˆ

¯˜ =

+( ) - ( )+( )

= == 1

12

2l b l=12

�

(b) Angular velocity. w22 12

12 12

2

12

2 2=

+

=

g

gl

l

l l

w21 412= / g

l w2 1 861= .

gl

�

31



Reaction at C. a bl g

lg

n =

=

=

w22

1212

+ SF ma C mg mgy n y= - =:

C mgy = 2

+ SM mba IC t= + a : 0 2= +( )mb I a

a = =0 0, at

+ SF max t= : C max t= - = 0 C = 2mg ≠�

32


SOLUTION

Position 1. (Directly above the bar).

Elevation: h1 1= m

Potential energy: V mgh1 1 80 9 81 1 784 8= = = ◊( )( . )( ) .kg m/s m N m2

Speeds: w1 10 0= =, v


(a) Position 2. (Body at level of bar after rotating 90∞).

Elevation: h2 0= .

Potential energy: V2 0=

Speeds: v2 21= ( ) .m w

Kinetic energy: T mv mk

T

2 22 2

22

2 22 2

22

22

1

2

1

2

1

280

1

280 0 4

46 4

= +

= +

=

w

w w

w

( ) ( )( . )

.


T V T V1 1 2 2 220 784 8 46 4+ = + + =: . . w

w22 16 9138= .

w2 4 11264= . rad/s w2 4 11= . rad/s �

PROBLEM 17.20

An 80-kg gymnast is executing a series of full-circle swings

on the horizontal bar. In the position shown he has a small

and negligible clockwise angular velocity and will maintain his

body straight and rigid as he swings downward. Assuming that

during the swing the centroidal radius of gyration of his body

is 0.4 m determine his angular velocity and the force exerted on

his hands after he has rotated through (a) 90°, (b) 180°.

33



Kinematics: at = ( )( )1 a

an = = ¨( )( ) .1 16 913822w m/s2

+ S SM M0 0

280 9 81 1 80 1 1 80 0 4= ( ) = +eff

: ( )( . )( ) ( )( )( )( ) ( )( . ) ( )a a

a = =8 4569 8 4569. . rad/s m/s2 2at

+ SF max n= : Rx = =( )( . ) .80 16 9138 1353 1 N

+SF may t= - : Ry - = - ≠( )( . ) ( )( . )80 9 81 80 8 4569

Ry = 108 248. N ≠ R = 1357 N 4.57∞ �

(b) Position 3. (Directly below bar after rotating 180∞).

Elevation: h3 1= - m.

Potential energy: V Wh3 3 80 9 81 1 784 8= = - = - ◊( )( . )( ) . N m

Speeds: v3 31= ( ) .w

Kinetic energy: T3 3246 4= . w


T V T V1 1 3 3 320 784 8 46 4 784 8+ = + + = -: . . .w

w32 = 33.828 rad /s2 2 w3 5 82= . rad/s �

Kinematics: an = = ≠( )( . ) .1 33 828 33 828 m/s2

From S S SM M Fx0 0 0= =( )eff and ,

a = = =0 0 0, a Rt x

+SF ma Ry n y= - =: ( )( . ) ( )( . )80 9 81 80 33 828

Ry = 3491 04. N R = 3490 N ≠ �

34


PROBLEM 17.21

Two identical slender rods AB and BC are welded together to

form an L-shaped assembly. The assembly is pressed against a

spring at D and released from the position shown. Knowing that

the maximum angle of rotation of the assembly in its subsequent

motion is 90° counterclockwise, determine the magnitude of the

angular velocity of the assembly as it passes through the position

where rod AB forms an angle of 30° with the horizontal.

SOLUTION

Moment of inertia about B. I m l m lB AB BC= +1

3

1

3

2 2

Position 2. q = ∞= +

= ∞ + - ∞ÊËÁ

ˆ¯̃

30

230

230

2 2 2V W h W h

W l W lAB AB BC BC

AB BC

( ) ( )

sin cos

TT I m m lB AB BC2 22 2

221

2

1

6= = +w w( )

Position 3. q = ∞

= =

90

203 3V W l TAB

Conservation of energy.

T V T V2 2 3 3+ = + :

1

6 230

230 0

2

222( ) sin cosm m l W l W l W l

AB BC AB BC AB+ + ∞ - ∞ = +w

w22 3 1 30 30

3

21 30 30

= ◊- ∞ + ∞

+

= - ∞ + ∞

lW W

m mgl

AB BC

AB BC

( sin ) cos

[ sin cos ]]

. ..

..= = =2 049 2 049

9 81

0 450 25

gl

w2 7 09= . rad/s �

35


PROBLEM 17.22

A collar with a mass of 1 kg is rigidly attached at a distance

d = 300 mm from the end of a uniform slender rod AB. The rod has

a mass of 3 kg and is of length L = 600 mm. Knowing that the rod

is released from rest in the position shown, determine the angular

velocity of the rod after it has rotated through 90°.

SOLUTION

Kinematics.

Rod v LR =

2w

Collar v dC = w

Position 1. w == =

0

0 01 1T V

Position 2. T m v I m v

m L m L

R R R C C

R R

22 2 2

22 2

1

2

1

2

1

2

1

2 2

1

2

1

12

= + +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

w

w w ++

= +

= - -

1

2

1

6

1

2

2

2 2

2 2 2 2

2

m d

m L m d

V W d W L

C

R C

C R

w

w w

T V T V m L m d W d W LR C C R1 1 2 2

2 2 201

6

1

2 2+ = + + = +Ê

ËÁˆ¯̃

- -: 0 w

w2

2 2 2 2

3 2

3

3 2

3=

++

=+

+( ) ( )W d W Lm d m L

g m d m Lm d m L

C C

C R

C R

C R (1)

Data: m d m LC R= = = =1 3 0 6 kg, 0.3 m, kg, m.

From Eq. (1), w2

2 23 9 81

2 1 0 3 3 0 6

3 1 0 3 3 0 6= +

+È

ÎÍ

˘

˚˙( . )

( )( )( . ) ( . )

( )( . ) ( . )

= 52 32. w = 7 23. rad/s �

36


PROBLEM 17.23

A collar with a mass of 1 kg is rigidly attached to a slender rod AB

of mass 3 kg and length L = 600 mm. The rod is released from

rest in the position shown. Determine the distance d for which the

angular velocity of the rod is maximum after it has rotated 90°.

SOLUTION

Kinematics.

Rod v LR =

2w

Collar v dC = w

Position 1. w == =

0

0 01 1T V

Position 2. T m v I m v

m L m L

R R R C C

R R

22 2 2

22 2

1

2

1

2

1

2

1

2 2

1

2

1

12

= + +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

w

w w ++

= +

= - -

+ = + + =

1

2

1

6

1

2

2

01

6

2 2

2 2 2 2

2

1 1 2 2

m d

m L m d

V W d W L

T V T V

C

R C

C R

w

w w

: 0 mm L m d W d W LR C C R

2 2 21

2 2+Ê

ËÁˆ¯̃

- -w

w2

2 2 2 2

3 2

3

3 2

3=

++

=+

+( ) ( )W d W Lm d m L

g m d m Lm d m L

C C

C R

C R

C R (1)

Let x dL

= .

w2

2

32

3

= ◊+

+

gL

x mm

x mm

R

C

R

C

Data: m m

Lg

xx

C R= =

= ++

1 3

3

2 3

3 3

2

2

kg, kg

w

37



L gw2 3/ is maximum. Set its derivative with respect to x equal to zero.

ddx

Lg

x x xx

x x

w2 2

2 2

2

3

3 3 2 2 3 6

3 30

6 18 6 0

Ê

ËÁˆ

¯̃= + - +

+=

- - + =

( )( ) ( )( )

( )

Solving the quadratic equation

x x= - =3 30 0 30278. .and

d L===

0 30278

0 30278 0 6

0 1817

.

( . )( . )

. m d = 181 7. mm �

38


PROBLEM 17.24

A 20-kg uniform cylindrical roller, initially at rest, is acted upon by a 90-N

force as shown. Knowing that the body rolls without slipping, determine

(a) the velocity of its center G after it has moved 1.5 m, (b) the friction force

required to prevent slipping.

SOLUTION

Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center.

Kinematics: v r= wPosition 1. At rest. T1 0=

Position 2. s v vvr

T mv I

mv mrvr

GG

GG

= = =

= +

= + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ

1 5

1

2

1

2

1

2

1

2

1

2

22 2

2 2

. m w

w

¯̃̄

= = =

2

2 2 23

4

3

420 15mv v vG G G( )

Work: U Ps Ff1 2 90 1 5 135Æ = = =( )( . ) J. does no work.

(a) Principle of work and energy.

T U T vG1 1 2 220 135 15+ = + =Æ :

vG2 9= vG = Æ3 00. m/s

(b) Since the forces are constant, a a

av

s

G

GG

= =

=

=

=

constant

m/s2

2

2

9

2 1 5

3

( )( . )

+ SF ma P F max f= - =:

F P maf = -

= -90 20 3( )( ) Ff = ¨30 0. N �

39


PROBLEM 17.25

A rope is wrapped around a cylinder of radius r and mass m as shown. Knowing that

the cylinder is released from rest, determine the velocity of the center of the cylinder

after it has moved downward a distance s.

SOLUTION

Point C is the instantaneous center.

v r vr

= =w w

Position 1. At rest. T1 0=

Position 2. Cylinder has fallen through distance s.

T mv I

mv mr vr

mv

22 2

2 22

2

1

2

1

2

1

2

1

2

1

2

3

4

= +

= + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

w

Work. U mgs1 2Æ =


T U T mgs mv1 1 2 220

3

4+ = + =Æ :

v gs2 4

3= v = 4

3

gs Ø �

40


PROBLEM 17.26

Solve Problem 17.25, assuming that the cylinder is replaced by a thin-walled pipe of

radius r and mass m.

PROBLEM 17.25 A rope is wrapped around a cylinder of radius r and mass m as

shown. Knowing that the cylinder is released from rest, determine the velocity of the

center of the cylinder after it has moved downward a distance s.

SOLUTION

Point C is the instantaneous center.

v r vr

= =w w

Position 1. At rest. T1 0=

Position 2. Cylinder has fallen through distance s.

T mv I

mv mr vr

mv

22 2

2 22

2

1

2

1

2

1

2

1

2

= +

= + ÊËÁ

ˆ¯̃

=

w

( )

Work. U mgs1 2Æ =


T U T mgs mv1 1 2 220+ = + =Æ :

v gs2 = v = Øgs �

41


PROBLEM 17.27

The mass center G of a 3-kg wheel of radius R = 180 mm is located at a distance

r = 60 mm from its geometric center C. The centroidal radius of gyration of the wheel

is k = 90 mm. As the wheel rolls without sliding, its angular velocity is observed

to vary. Knowing that w = 8 rad/s in the position shown, determine (a) the angular

velocity of the wheel when the mass center G is directly above the geometric center C,

(b) the reaction at the horizontal surface at the same instant.

SOLUTION

v BG

vmk

1 1

2 2

2 2

0 18 0 06 8

8 0 036

0 24

3

0

=

= +

===

=

( )

( . ) ( . ) ( )

.

.

.

w

w m/s

kg

009 m

Position 1. V

T mv I

1

1 12

12

2 2 2

0

1

2

1

2

1

23 8 0 036

1

23 0 09 8

4 233

=

= +

= +

=

w

( )( . ) ( )( . ) ( )

. 66 J

Position 2. V Whmgh

T mv I

2

2 22

22

3 9 81 0 06

1 7658

1

2

1

2

1

23 0

====

= +

=

( )( . )( . )

.

( )( .

J

w

2241

23 0 09

0 09855

22 2

22

22

w w

w

) ( )( . )

.

+

=

42



(a) Conservation of energy.

T V T V1 1 2 2

22

22

2

4 2336 0 0 09855 1 7658

25 041

5 004

+ = +

+ = +

==

. . .

.

.

J J

w

ww rrad/s w2 5 00= . rad/s �

(b) Reaction at B.

+

ma m CG

F ma N m

n

y y

=

=

= Ø= -

( )

(

.

w22

3

4 5

kg)(0.06 m)(5.00 rad/s)

N

:

2

S gg man= -

N - = -( )( . ) .3 9 81 4 5 N = ≠24.9 N �

43


PROBLEM 17.28

A collar B, of mass m and of negligible dimension, is attached to the rim of a

hoop of the same mass m and of radius r that rolls without sliding on a horizontal

surface. Determine the angular velocity w1 of the hoop in terms of g and r when B is directly above the center A, knowing that the angular velocity of the hoop is 3w1

when B is directly below A.

SOLUTION

The point of contact with ground is the instantaneous center.

Position 1. Point B is at the top.

w w w w

w

w w

= = =

= + +

= +

1 1 1

12 2 2

12

1

2

1

2

1

2

1

2

1

22

1

2

v r v r

T mv mv I

m r m r

B A

B A

( ) ( )) ( )2 212

212

1

2

3

+

=

mr

mr

w

w

Position 2. Point B is at the bottom.

w w w

w

w

= = =

= + +

= + +

2 2

22 2

22

22 2

0

1

2

1

2

1

2

01

2

1

2

v v r

T mv mv I

m r mr

B A

B A

( ) ( ))

( )

w

w

w

w

22

222

21

2

212

3

9

=

=

=

mr

mr

mr

Work. U mg h mgr1 2 2Æ = =( )D


T U T mr mgr m1 1 2 22

12

123 2 9+ = + =Æ : w w

w12

3= g

r w1 0 577= .

gr

�

44


PROBLEM 17.29

A half section of pipe of mass m and radius r is released from rest in the position

shown. Knowing that the pipe rolls without sliding, determine (a) its angular

velocity after it has rolled through 90°, (b) the reaction at the horizontal surface at

the same instant. [Hint: Note that GO r= 2 /p and that, by the parallel-axis theorem,

I mr m GO= -2 2( ) .]

SOLUTION

Position 1. w1 1 10 0 0= = =v T

Position 2. Kinematics: v AG r2 2 212= = -Ê

ËÁˆ¯̃

( )wp

w

Moment of inertia: I mr m mr m r mr= - = - ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

2 2 22

2

20 6

21

4( . )

p p

Kinetic energy: T mv I

m r mr

2 22

22

22

22 2

2 22

1

2

1

2

1

21

2 1

21

4

1

= +

= -ÊËÁ

ˆ¯̃

+ -ÊËÁ

ˆ¯̃

=

w

pw

pw

221

4 41

4

1

22

4

2

2 2

2

mr

mr

- +ÊËÁ

ˆ¯̃

+ -ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚˙

= -ÊËÁ

ˆ¯̃

p p p

p

Work: U W OG mg r mgr1 2

2 2Æ = = =( )

p pPrinciple of work and energy: T U T

mg r mr

gr

1 1 2 2

222

22

2

02 1

22

4

2

11 7519

+ =

+ = -ÊËÁ

ˆ¯̃

=-( ) ◊ =

Æ

p pw

wp p

.ggr

45



(a) Angular velocity. w2 1 324= .gr

�

(b) Reaction at A.

Kinematics: Since O moves horizontally, ( )a y0 0=

ar g

rg

n =

= ÊËÁ

ˆ¯̃

= ≠

( . )

.

.

0 6

21 7519

1 1153

22w

p

Kinetics:

+ S SF F A mg mgy y= - =( ) .eff : 1 1153 A = ≠2 12. mg �

46


PROBLEM 17.30

Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm are

connected by a belt as shown. Knowing that the initial angular velocity of

cylinder B is 30 rad/s counterclockwise, determine (a) the distance through

which cylinder A will rise before the angular velocity of cylinder B is reduced

to 5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION

Kinematics. v rAB B= w

Point C is the instantaneous center of cylinder A.

w w

w w

AAB

B

A A B

vr

v r r

= =

= =

2

1

2

1

2

Moment of inertia. I Wg

r= 1

2

2

Kinetic energy.

Cyl B: 1

2

1

2

1

2

1

4

2 2 2 2 2I Wg

r Wg

rB B Bw w w=ÊËÁ

ˆ¯̃

=

Cyl A: 1

2

1

2

1

2

1

2

1

2

1

2

1

2

3

2 22

22

mv I Wg

r Wg

rA A B B+ = ÊËÁ

ˆ¯̃

+ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

w w w

116

2 2Wg

r Bw

Total: T Wg

r B= 7

16

2 2w

(a) Distance h that cylinder A will rise.

Conservation of energy for system.

T V T V Wg

r Wg

r Wh

h rg

B B

B

1 1 2 22

12 2

22

2

1

7

160

7

16

7

16

+ = + + = +

=

: ( ) ( )

[( )

w w

w 222

2

22 27

16

0 1

9 8130 5

0 3902

-

= ÊËÁ

ˆ¯̃

-

=

( ) ]

( . )

( . )( )

.

wB

m h = 0 390. m �

47



(b) Tension in belt between the cylinders.

When cylinder A moves up a distance h, the belt moves up a distance 2h.

Work: U P h Wh1 2 2Æ = -( )

Principle of work and energy for cylinder A.

T U T Wg

r Ph Wh Wg

r

P W Wrg

B B1 1 2 22

12 2

22

2

3

162

3

16

1

2

3

32

+ = + - =

= -

Æ : ( ) ( )w w

hh

W W

W

B B( ) ( )

( )( . )

w w12

22

1

2

3

14

2

7

2

77 9 81

-ÈÎ ˘̊

= -

= = P = 19 62. N �

48


PROBLEM 17.31

Two uniform cylinders, each of mass m = 7 kg and radius r = 100 mm

are connected by a belt as shown. If the system is released from rest,

determine (a) the velocity of the center of cylinder A after it has moved

through 1 m, (b) the tension in the portion of belt connecting the two

cylinders.

SOLUTION



w w

w w

AAB

B

A A B

vr

v r r

= =

= =

2

1

2

1

2


r= 1

2

2

Kinetic energy.

Cyl B: 1

2

1

2

1

2

1

4

2 2 2 2 2I Wg

r Wg

rB B Bw w w=ÊËÁ

ˆ¯̃

=

Cyl A: 1

2

1

2

1

2

1

2

1

2

1

2

1

2

3

2 22

22

mv I Wg

r Wg

rA A B B+ = ÊËÁ

ˆ¯̃

+ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

w w w

116

2 2Wg

r Bw

Total: T Wg

r B= 7

16

2 2w

Position 1. Rest T V1 10 0= =

Position 2. Cylinder has fallen through. d = 1 m.

T Wg

r

V Wd

B22 2

2

7

16=

= -

w

49



Principle of conservation of energy: T V T VWg

r WdB

1 1 2 2

2 20 07

16

+ = +

+ = -w

where rgdrB

B

=

=

= ◊

==

0 1

16

7

16

7

9 81 1

0 1

2242 29

47 3528

2

2

2

.

( . )( )

( . )

.

.

m

rad

w

w //s

(a) Velocity of the center of cylinder A.

v rA B=

=

1

2

1

20 1 47 3528

w

( . )( . ) vA = Ø2 37. m/s �

(b) Tension in belt between the cylinders.

When cylinder A moves down a distance d, the belt moves down a distance 2d.

Work: U P d Wd1 22

2Æ = +( )

Principle of work and energy for cylinder A:

T U T P d Wd Wg

r

P W Wrg d

P

B

B

1 1 2 22 2

2 2

0 23

16

1

2

3

32

1

27 9

+ = - + =

= -

=

Æ : ( )

( )(

w

w

.. )( )( . )( . ) ( . )

( . )( )

. .

8132

7 9 81 0 1 2242 29

9 81 1

34 335 14 715 1

2

- 3

= - = 99 620.

P = 19 62. N �

50


PROBLEM 17.32

The 5-kg rod BC is attached by pins to two uniform disks as

shown. The mass of the 150-mm-radius disk is 6 kg and that

of the 75-mm-radius disk is 1.5 kg. Knowing that the system is

released from rest in the position shown, determine the velocity

of the rod after disk A has rotated through 90°.

SOLUTION

Position 1. T1 0=

Position 2.

Kinematics.

v v vBE

v v v v

v vvCF

v

B AB AB AB

A B AB

C AB CC AB

= = = = =

= = =

w

w w

0 0752 2

0 075

.

.

m

mAAB = 0

Kinetic energy. T m v I m v m v IA A A A AB AB B B B B22 2 2 21

2

1

2

1

2

1

2

1

2= + + + +w w

= + ÊËÁ

ˆ¯̃

+È1

22 2

1

26 0 15

0 07552 2

22( )( ) ( )( . )

.( )m/s kg m kgv v vAB

ABAB

ÎÎÍÍ

+ + ÊËÁ

ˆ¯̃

˘

˚˙˙

=

( . )( ) ( . )( . ).

1 51

21 5 0 075

0 075

1

2

22

kg kg mv vAB

AB

224 12 5 1 5 0 75

21 625

2

22

+ + + +[ ]=

. .

.

v

T v

AB

AB

51



Work: U W

U

AB1 2

1 2

0 1125 0 075

5 9 81 0 0375

1 8394

Æ

Æ

= -==

( . . )

( )( . )( . )

.

m m

kg m

J


v

vv

AB

AB

AB

1 1 2 2

2

2

0 1 8394 21 625

0 08506

0 2916

+ =

+ =

==

Æ

. .

.

.

J

m

Velocity of the rod. vAB = Æ292 mm/s �

52


PROBLEM 17.33

The 9-kg cradle is supported as shown by two uniform disks that roll

without sliding at all surfaces of contact. The mass of each disk is

m = 6 kg and the radius of each disk is r = 80 mm. Knowing that the

system is initially at rest, determine the velocity of the cradle after it

has moved 250 mm.

SOLUTION

Moments of inertia. I I WrgA B= =

2

2

Kinematics. w wA BC

A B Cvr

v v v= = = =

Kinetic energy. T1 0=

T m v I m v I m v

m m m m m

A A A A B B B B C C22 2 2 2 21

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2

= + + + +

= + + + +

w w

CC C

C C

C

C

v

m m v

v

v

ÈÎÍ

˘˚̇

= +

= +

=

2

2

2

2

1

23

1

23 6 9

13 5

( )

[( )( ) ]

.

Work. U Fs1 2 30 0 25 7 5Æ = = =( )( . ) .N m J

Principle of work and energy. T U T vC1 1 2 220 7 5 13 5+ = + =Æ : . . vC = Æ0 745. m/s �

53


PROBLEM 17.34





has moved 250 mm.

SOLUTION


2

2

Kinematics. w wA BC

A B

vr

v v

= =

= = 0


T m v I m v I m v

m m m

A A A A B B B B C C22 2 2 2 21

2

1

2

1

2

1

2

1

2

1

20

1

20

1

2

= + + + +

= + + + +

w w

CC C

C C

C

C

v

m m v

v

v

ÈÎÍ

˘˚̇

= +

= +

=

2

2

2

2

1

2

1

26 9

7 5

( )

( )

.

Work. U Fs1 2 30 0 25 7 5Æ = = =( )( . ) .N m J


54


PROBLEM 17.35





has moved 250 mm.

SOLUTION


2

2

Kinematics. w wA BC

A B Cv

rv v v= = = =

2

1

2


T m v I m v I m v

m m m

A A A A B B B B C C22 2 2 2 21

2

1

2

1

2

1

2

1

2

1

2

1

4

1

8

1

4

1

= + + + +

= + + +

w w

88

1

20 75

1

20 75 6 9

6 75

2

2

2

2

m m v

m m v

v

v

C C

C C

C

C

+ÈÎÍ

˘˚̇

= +

= +

=

( . )

[( . )( ) ]

.

Work. U Fs1 2 30 0 25 7 5Æ = = =( )( . ) .N m J


55


PROBLEM 17.36

The motion of the slender 10-kg rod AB is guided by collars of negligible

mass that slide freely on the vertical and horizontal rods shown. Knowing

that the bar is released from rest when q = 30°, determine the velocity of

collars A and B when q = 60°.

SOLUTION

Position 1: TV W

1

1

0

0 3

10 9 8 0 3

29 43

=== -= -

( . )

( )( . )( . )

.

m

kg

J

Position 2: V W

T mv I

2

2 22

0 5196

10 9 81 0 5196

50 974

1

2

1

2

= -= -= -

= +

( . )

( )( . )( . )

.

m

kg

J

ww

w w

w

22

22 2

22

22

1

210 0 6

1

2

1

210 1 2

2 4

= + ÊËÁ

ˆ¯̃

=

( )( . ) ( )( . )

.

kg kg m

Principle of conservation of energy. T V T V1 1 2 2+ = +

0 29 43 2 4 50 974

8 9768

22

22

- = -

=

. . .

.

J w

w w2 2 996= . rad/s

Velocity of collars when q = ∞60

v ACA = = ¥( ) ( . )( . )w2 2 0 5196 2 996m rad/s vA = Æ3 11. m/s �

v BCB = = ¥( ) ( . )( . )w2 2 0 3 2 996m rad/s vB = Ø1 798. m/s �

56


PROBLEM 17.37

The motion of the slender 10-kg rod AB is guided by collars of negligible

mass that slide freely on the vertical and horizontal rods shown. Knowing

that the bar is released from rest when q = 20°, determine the velocity of

collars A and B when q = 90°.

SOLUTION

Position 1: TV W

mg

1

1

0

0 2052

0 2052

== -= -

( . )

( . )

m

Position 2: V W mg

T m I

m m

2

2 22

22

22

0 6 0 6

1

2

1

2

1

20 6

1

2

1

121

= - = -

= +

= +

( . ) ( . )

( . ) ( .

m

v w

w 22

0 24

222

2 22

)

.

ÊËÁ

ˆ¯̃

=

w

wT m

Principle of conservation of energy. T V T V

mg m mg

g

1 1 2 2

22

22

0 0 2052 0 24 0 6

1 645 1 645 9 81

+ = +

- = -

= =

. . .

. . ( . )

w

w ===

16 137

4 0172

.

.w rad/s

Velocity of collars when q = ∞90

v ABA = =( ) ( . )( . )w2 1 2 4 017m rad/s vA = Æ4 82. m/s �

vB = 0 vB = 0 �

57


PROBLEM 17.38

The ends of a 4.5 kg rod AB are constrained to move along slots cut in a

vertical plane as shown. A spring of constant k = 600 N/m is attached to

end A in such a way that its tension is zero when q = 0 . If the rod is

released from rest when q = 0, determine the angular velocity of the rod

and the velocity of end B when q = ∞30 .

SOLUTION

Moment of inertia. Rod. I mL= 1

12

2

Position 1. q w1 1 1

1

0 0 0= = ==

elevation above slot.

vh

elongation of spring.

he e

1

1 1

0

0

== =

T mv I

V ke Wh

1 12

12

1 12

1

1

2

1

20

1

20

= + =

= + =

w

Position 2. q = ∞30

e L Le L

h L L

V ke Wh

2

2

2

2 22

2

30

1 30

230

1

4

1

2

1

+ ∞ == - ∞

= - ∞ = -

= + =

cos

( cos )

sin

221 30

1

4

2 2k L WL( cos )- ∞ -

Kinematics. Velocities at A and B are directed as shown. Point C is the instantaneous center of rotation. From

geometry, b L=2

.

v b L

v L

T mv I

m L mL

B

= =

= ∞

= +

= ÊËÁ

ˆ¯̃

+

w w

w

w

w

2

30

1

2

1

2

1

2 2

1

2

1

12

22 2

22

( cos )

ÊÊËÁ

ˆ¯̃

= 1

6

2 2Wg

L w

58




T V T V Wg

L kL WL

gL W

1 1 2 22 2 2 2

2

0 01

6

1

21 30

1

4

3 3

+ = + + = + - ∞ -

= -

: ( cos )w

w kg(( cos )1 30 2- ∞

Data: W

gLk

=

===

( . )( . )

.

.

4 5 9 81

9 81

0 6

600

N

m/s

m

N/m

2

w2 23 9 81

0 6

3 600 9 81

4 5 9 811 30

41 87

= - - ∞

=

( )( . )

( . )

( )( . )

( . )( . )( cos )

. 003 w = 6 47. rad/s �

vB = ∞( . )(cos )( . )0 6 30 6 4707 vB = 3 36. m/s �

59


PROBLEM 17.39

The ends of a 4.5 kg rod AB are constrained to move along slots cut in a

vertical plate as shown. A spring of constant k = 600 N/m is attached to

end A in such a way that its tension is zero when q = 0. If the rod is

released from rest when q = ∞50 , determine the angular velocity of the

rod and the velocity of end B when q = 0.

SOLUTION

v L

v Lx L LB

2 2

2

1

2

50

0 6 1 50

0 21433

=

== - ∞= - ∞=

w

wcos

. ( cos )

.

m

m

Position 1. V W L kx

V

1 12

1

250

1

2

4 5 9 810 6

250

1

2600

= - ∞ +

= - ÊËÁ

ˆ¯̃

∞ +

sin

( . )( . ).

sin ( ))( . )

. .

.

0 21433

10 1451 13 7812

3 6361

0

2

1

= - += ◊=

N m

T

Position 2. V V V

T mv I

m L mL

g e2 2 2

2 22

22

2

22

0

1

2

1

2

1

2 2

1

2

1

12

= + =

= +

= ÊËÁ

ˆ¯̃

+ ÊË

( ) ( )

w

w ÁÁˆ¯̃

= = =

w

w w w

22

222 2

22

221

6

1

64 5 0 6 0 27mL ( . )( . ) .kg m

60



Conservation of energy: T V T V1 1 2 2

220 3 6361 0 27

+ = +

+ ◊ =. .N m w

ww

22

2

13 467

3 6697

==

.

.

rad /s

rad/s

2 2

w2 3 67= . rad/s �

Velocity of B: v LB = =w2 0 6 3 6697( . )( . )m rad/s

= 2 2018. m/s vB = ≠2 20. m/s �

61


PROBLEM 17.40

The motion of the uniform rod AB is guided by small wheels of negligible

mass that roll on the surface shown. If the rod is released from rest when

q = 0, determine the velocities of A and B when q = ∞30 .

SOLUTION

Position 1. q

w

== ====

0

0

0

0

0

1

1

v v

TV

A B


Kinematics. Locate the instantaneous center C. Triangle ABC is equilateral.

v v Lv L

A B

G

= == ∞

ww cos30

Moment of inertia. I ml= 1

12

2

Kinetic energy. T mv I T m L mL

ml

G22 2

22 2 2

2

1

2

1

2

1

230

1

2

1

12

5

12

= + = ∞ + ÊËÁ

ˆ¯̃

=

w w w

w

: ( cos )

22

Potential energy. V mg L mgL22

301

4= - ∞ = -sin


T V T V mL mgL1 1 2 22 20 0

5

12

1

4+ = + + = -: w

w

w

2 0 6

0 775

0 775

0 775

=

=

=

=

.

.

.

.

gL

gL

v gL

v gLA

B vA gL= ¨0 775. �

vB gL= 0 775. 60∞ �

62


PROBLEM 17.41

The motion of a slender rod of length R is guided by pins at A and B which

slide freely in slots cut in a vertical plate as shown. If end B is moved

slightly to the left and then released, determine the angular velocity of the

rod and the velocity of its mass center (a) at the instant when the velocity

of end B is zero, (b) as end B passes through Point D.

SOLUTION

The rod AB moves from Position 1, where it is nearly vertical, to Position 2, where vB = 0.

In Position 2, vA is perpendicular to both CA and AB, so CAB is a straight line of length 2L and slope angle 30∞.

In Position 3 the end B passes through Point D.

Position 1: T V vh

mg R1 1

1

02

= = =

Position 2: Since instantaneous center is at B,

v R

T mv I

m R mR

2 2

2 22

22

2

22

22

1

2

1

2

1

2

1

2

1

2

1

2

1

12

=

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

w

w

w w

==

= =

1

6

4

222

2 2

mR

V Wh mg R

w

Position 3: V3 0=

Since both vA and vB are horizontal, w3 0= (1)

T mv3 221

2=

63



(a) From 1 to 2: Conservation of energy

T V T V mgR mR mgR

gR

gR

1 1 2 22

22

22

2

01

2

1

6

1

4

3

2

3

2

+ = + + = +

=

=

: w

w

w w2 1 225= .gR

�

v R gR gR2 2

1

2

1

2

3

2

3

8= = =w vR gR= 0 612. 60∞ �

(b) From 1 to 3: Conservation of energy

From Eq. (1) we have w3 0= �

T V T V mgR mv

v gR

1 1 3 3 32

32

01

2

1

2+ = + + =

=

:

v3 = ÆgR �

64


PROBLEM 17.42

Two uniform rods, each of mass m and length L, are connected to form

the linkage shown. End D of rod BD can slide freely in the horizontal

slot, while end A of rod AB is supported by a pin and bracket. If end

D is moved slightly to the left and then released, determine its velocity

(a) when it is directly below A, (b) when rod AB is vertical.

SOLUTION

Moments of inertia. I I mL

I mL

AB BD

A

= =

=

1

12

1

3

2

2

Position 1. At rest as shown. T1 0=

V mgh mgh

mg L

mgL

AB BC1

02

1

2

= +

= + -ÊËÁ

ˆ¯̃

= -

(a) Position 2. In Position 2, Point A is the instantaneous center of both AB and BD. Since Point B is

common to both bars,

w wwww

AB BD

D

G

v lv l

==== ∞

2

2

30cos

Kinetic energy.

T I mv I T mL m lA G BD2 22 2

22

22

22

2

1

2

1

2

1

2

1

2

1

3

1

23= + + = Ê

ËÁˆ¯̃

+w w w w: ( cos 001

2

1

12

7

12

2 222

222

∞ + ÊËÁ

ˆ¯̃

=

) mL

mL

w

w

Potential energy. V mgh mghAB BD2 = +

V mg L mg L

mgL

22

303

230= - ∞Ê

ËÁˆ¯̃

+ - ∞ÊËÁ

ˆ¯̃

= -

sin sin

65




T V T V mgL mL mgL

gL

gL

v LD

1 1 2 22

22

22

2

2

01

2

7

12

6

7

0 926

+ = + - = -

=

=

= =

: w

w

w

w

.

00 926. gL vD gL= ¨0 926. �

(b) Position 3. In Position 3, bar BD is in translation.

v v L LD B AB= = =w w3

Kinetic energy.

T I mv T mL m L mLA B3 32 2

32

32

32 2

321

2

1

2

1

2

1

3

1

2

2

3= + = Ê

ËÁˆ¯̃

+ =w w w w: ( )

Potential energy. V mg L mg L

mgL

32

3

2

= -ÊËÁ

ˆ¯̃

+ -

= -

( )


T V T V mgL mL mgL

gL

gL

v LD

1 1 3 32

32

32

3

3

01

2

2

3

3

2

3

2

1 225

+ = + - = -

=

=

=

: w

w

w

w

.

== 1 225. gL vD gL= 1 225. ¨ �

66


PROBLEM 17.43

The uniform rods AB and BC have a mass of 1.2 kg and 2 kg,

respectively, and the small wheel at C is of negligible weight. If the

wheel is moved slightly to the right and then released, determine the

velocity of pin B after rod AB has rotated through 90∞.

SOLUTION

Moments of inertia. mm

AB

BC

==

1 2

2

. kg

kg

Bar AB: I m LA AB AB= = = ◊1

3

1

31 2 0 45 0 0812 2 2( . )( . ) . kg m

Bar BC: I m LBC BC= = = ◊1

12

1

122 0 75 0 093752 2 2( )( . ) . kg m

Position 1. As shown with bar AB vertical. Point G is the midpoint of BC.

V W h W hAB AB BC BC1

1 2 9 810 45

22 9 81

0 45

2

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ( . )( . )

.( )( . )

.

¯̃̄= ◊7 0632. N m

Bar BC is at rest. w

w

BC

G B C

ABB

AB

v v v vvL

T

== = = =

= =

=

0

0

0

01

Position 2. Bar AB is horizontal.

V2 0=

67



Kinematics. w ABB

AB

BvL

v= =

0 45.

w

w w

BCB

BC

B

B

A AB BC BC

vL

v

v v

T I m v I

= =

=

= + +

=

0 75

1

2

1

2

1

2

1

2

1

20 081

22 2 2

.

( . )).

( ) ( . ).

v v vBB

B

0 45

1

22

1

2

1

20 09375

0 75

2 2 2ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= 00 53333 2. vB

Conservation of energy. T V T V vB1 1 2 220 7 0632 0 53333+ = + + =: . .

vB = 3 6392. m/s vB = 3 64. m/s Ø �

68


PROBLEM 17.44

The uniform rods AB and BC have a mass 1.2 kg and 2 kg,

respectively, and the small wheel at C is of negligible weight.

Knowing that in the position shown the velocity of wheel C is 2 m/s

to the right, determine the velocity of pin B after rod AB has rotated

through 90°.

SOLUTION

Moments of inertia.

Bar AB: I m LA AB AB= = = ◊1

3

1

31 2 0 45 0 0812 2 2( . )( . ) . kg m

Bar BC: I m LBC BC= = = ◊1

12

1

122 0 75 0 093752

2( )( . ) . kg m2

Position 1. As shown with bar AB vertical. Point G is the midpoint of BC.

V W h W hAB AB BC BC1

1 2 9 810 45

22 9 81

0 45

2

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ( . )( . )

.( )( . )

.

¯̃̄= ◊7 0632. N m

Kinematics, Bar BC is in translation. wBC = 0

v v v vvL

T I m

G B C

ABB

AB

A AB BC

= = = =

= = =

= +

2

2

0 45

40

9

1

2

1

21

2

m/s

rad/sw

w

.

vv I BC2 2

22

1

2

1

20 081

40

9

1

22 2 0

4 8

+

= ÊËÁ

ˆ¯̃

+ +

= ◊

w

( . ) ( )( )

. N m

Position 2. Bar AB is horizontal.

V2 0=

69



Kinematics. w ABB

AB

BvL

v= =

0 45.

w

w w

BCB

BC

B

B

A AB BC BC

vL

v

v v

T I m v I

= =

=

= + +

=

0 75

1

2

1

2

1

2

1

2

1

20 081

22 2 2

.

( . )).

( ) ( . ).

v v vBB

B

0 45

1

22

1

2

1

20 09375

0 75

2 2 2ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= 00 53333 2. vB

Conservation of energy. T V T V vB1 1 2 227 0632 4 8 0 53333 0+ = + + = +: . . .

vB = 4 716. 3 m/s vB = 4 72. m/s Ø �

70


PROBLEM 17.45

The 4-kg rod AB is attached to a collar of negligible mass at A and to a flywheel

at B. The flywheel has a mass of 16 kg and a radius of gyration of 180 mm. Knowing

that in the position shown the angular velocity of the flywheel is 60 rpm clockwise,

determine the velocity of the flywheel when Point B is directly below C.

SOLUTION

Moments of inertia.

Rod AB: I m LAB AB AB=

=

= ◊

1

12

1

124 0 72

0 1728

2

2

2

( )( . )

.

kg m

kg m

Flywheel: I mkC =

=

= ◊

2

2

2

16 0 18

0 5184

( )( . )

.

kg m

kg m

Position 1. As shown. w = w1

sin.

..

( . )cos .

(

b b

b

= = ∞

= =

==

0 12

0 7219 471

1

20 72 0 33941

4

1

1 1

h

V W hAB

m

))( . )( . )

.

9 81 0 33941

13 3185= J

Kinematics. v rB = =w w1 10 24.

Bar AB is in translation. w AB Bv v= =0,

T m v I IAB AB AB C12 2

12

12

1

1

2

1

2

1

2

1

24 0 24 0

1

20 5184

= + +

= + +

w w

w w( )( . ) ( . ) 22

120 3744= . w

71



Position 2. Point B is directly below C.

h L r

V W h

AB

AB

2

2 2

1

2

1

20 72 0 24

0 12

4 9 81 0 12

4

= -

= -

====

( . ) .

.

( )( . )( . )

.

m

77088 J


w w

w

w

ABB

B

AB AB AB C

v

v v

T m v I I

= =

= =

= + +

0 720 33333

1

20 12

1

2

1

2

1

2

2

2

22 2

..

.

ww

w w w

22

22

22

221

24 0 12

1

20 1728 0 33333

1

20 5184

0

= + +

=

( )( . ) ( . )( . ) ( . )

..2976 22w

Conservation of energy. T V T V1 1 2 2 12

220 3744 13 3185 0 2976 4 7088+ = + + = +: . . . .w w (1)

Angular speed data: w p1 60 2= =rpm rad/s

Solving Equation (1) for w2 , w2 8 8655= . rad/s w2 84 7= . rpm �

72


PROBLEM 17.46

If in Problem 17.45 the angular velocity of the flywheel is to be the same in the

position shown and when Point B is directly above C, determine the required value

of its angular velocity in the position shown.

SOLUTION

Moments of inertia.

Rod AB: I m LAB AB AB=

=

= ◊

2

21

124 0 72

0 1728

kg m

kg m2

( . )

.

Flywheel: I mkC =

=

= ◊

2

216 0 18

0 5184

kg m

kg m2

( . )

.

Position 1. As shown. w = w1

sin.

.

.

. cos .

( )(

b

b

b

=

= ∞

= =

= =

0 12

0 72

19 471

1

20 72 0 33941

4 9

1

1 1

h

V W hAB

m

.. )( . )

.

81 0 33941

13 3185=


73



Bar AB is in translation. w AB Bv v= =0,

T m v I IAB AB AB C12 2

12

12

1

2

1

2

1

2

1

24 0 24 0

1

20 55901

= + +

= + +

w w

w w( )( . ) ( . ) 112

120 3744= . w

Position 2. Point B is directly above C.

h L r

V W h

AB

AB

2

2 2

1

2

1

20 72 0 24

0 6

4 9 81 0 6

23 54

= +

= +

====

( . ) .

.

( )( . )( . )

.

m

44 J


w w

w

w

ABB

B

ABAB AB

v

v v

T Wg

v I I

= =

= =

= + +

0 720 33333

1

20 12

1

2

1

2

1

2

2

2

22 2

..

.

CCw

w w

22

22

221

2

8

32 20 500

1

20 186335 0 33333

1

20 5590

= +

+

.( . ) ( . )( . )

( . 11

0 320913

22

22

)

.

w

w=

Conservation of energy. T V T V1 1 2 2 12

220 3744 13 3135 0 2976 23 544+ = + + = +: . . . .w w (1)

Angular speed data: w w2 1=

Then, 0 0760 0 410512. .w = +

w1 11 602= . rad/s w1 110 8= . rpm �

74


PROBLEM 17.47

The 80-mm-radius gear shown has a mass of 5 kg and a centroidal radius

of gyration of 60 mm. The 4-kg rod AB is attached to the center of the gear

and to a pin at B that slides freely in a vertical slot. Knowing that the

system is released from rest when q = ∞60 , determine the velocity of the

center of the gear when q = ∞20 .

SOLUTION

Kinematics. vA Av= ¨

vB Bv= Ø

Point D is the instantaneous center of rod AB.

wqq w q

wq

ABA

B AB A

G ABA

vL

v L v

v L v

=

= =

= =

cos

( sin ) tan

cos2 2

Gear A effectively rolls with slipping, with Point C being the contact point.

vC = 0

Angular velocity of gear w AAvr

= .

Potential energy: Use the level of the center of gear A as the datum.

V W L m gLAB AB= - ÊËÁ

ˆ¯̃

= -2

1

2cos cosq q

Kinetic energy: T m v I m v IA A A A AB AB AB= + + +1

2

1

2

1

2

1

2

2 2 2 2w wq

Masses and moments of inertia: m mA AB= =5 4 kg kg,

I m k

I m L

A A

AB AB

= = = ◊

= =

2 2 2

2 2

5 0 060 0 018

1

12

1

124 0 320

( )( . ) .

( )( . )

kg m

== ◊0 03413 2. kg m

75



Conservation of energy: T V T V1 1 2 2+ = +

Position 1: q = ∞= =

= - ∞

= -

60

0 0

1

24 9 81 0 320 60

3 1392

1

1

v T

V

A

( )( . )( . )cos

. J

Position 2: q = ∞ =20 vA ?

T v v vA

A A2

22 2

1

25

1

20 018

0 080

1

24

2 20= + Ê

ËÁˆ¯̃

+∞

ÊËÁ

ˆ¯̃

+

( ) ( . ).

( )cos

11

20 03413

0 320 20

2 5 1 40625 0 56624 0 18

2

( . ). cos

( . . . .

vA

∞ÊËÁ

ˆ¯̃

= + + + 8875

4 66124

1

24 9 81 0 320 20

5 8998

2

2

2

)

.

( )( . )( . )cos

.

v

v

V

A

A=

= ∞

= - J


0 3 1392 4 66124 5 8998

0 59225

0 770

2

2

- = -

==

. . .

.

.

v

vv

A

A

A

m /s

m/s

2 2

vA = ¨0 770. m/s �

76


PROBLEM 17.48

The motor shown rotates at a frequency of 22.5 Hz and runs

a machine attached to the shaft at B. Knowing that the motor

develops 3 kW, determine the magnitude of the couple exerted

(a) by the motor on pulley A, (b) by the shaft on pulley B.

SOLUTION

wp

pw w p

A

A A B Br r

=ÊËÁ

ˆ¯̃

==

22 5

45

0 03 45

.

: ( . )(

Hz2 rad

cycle

rad/s

m rrad/s m

rad/s

) ( . )

.

==

0 180

7 5

ww p

B

B

(a) Pulley A: Power = M A Aw

3000 45

21 2

W rad/s

N m

== ◊

MM

A

A

( )

.

p �

(b) Pulley B: Power = MB Bw

3000 7 5

127 3

W rad/s

N m

== ◊

MM

B

B

( . )

.

p �

77


PROBLEM 17.49

Knowing that the maximum allowable couple that can be applied to a shaft is 2000 N · m determine the

maximum power (m kW) that can be transmitted by the shaft at (a) 180 rpm, (b) 480 rpm.

SOLUTION

M = ◊2000 N m

(a) w p

pw

p

= ÊËÁ

ˆ¯̃

=== ◊ =

180

6

2000 6

rpm2

60

rad/s

Power

N m rad/s

M( )( ) 337699 1

37 699

.

.

W

kW=

37.7 kW �

(b) w p

pw

p

= ÊËÁ

ˆ¯̃

=== ◊

480

16

2000 16

rpm2

60

rad/s

Power

N m rad/s

M( )( ))

.

==

100531

100 531

W

kW

100.5 kW �

78


PROBLEM 17.50

Three shafts and four gears are used to form a gear train which

will transmit 7.5 kW from the motor at A to a machine tool at F.

(Bearings for the shafts are omitted in the sketch.) Knowing that

the frequency of the motor is 30 Hz, determine the magnitude of

the couple which is applied to shaft (a) AB, (b) CD, (c) EF.

SOLUTION

Kinematics. wp

p

AB ===

30

30 2

60

Hz

rad/s

rad/s

( )

Gears B and C. rrB

C

==

75

180

mm

mm

r rB AB C CD CDw w p w= =: ( )( ) (75 60 180 mm rad/s mm)( )

Gears D and E. w p

w w p

CD

D

E

D CD E EF

rr

r r

===

=

25

75

180

75 25

rad/s

mm

mm

mm rad/: ( )( ss mm

rad/s

Power kW

) ( )( )

.

.

===

180

10 4167

7 5

ww p

EF

EF

(a) Shaft AB. Power W rad/s= =M MAB AB ABw p: ( )7500 60 M AB = ◊39 8. N m �

(b) Shaft CD. Power W rad/s= =M MCD CD CDw p: ( )7500 25 MCD = ◊95 5. N m �

(c) Shaft EF. Power W rad/s= =M MEF EF EFw p: ( . )7500 10 4167 MEF = ◊229 N m �

79


PROBLEM 17.51

The shaft-disk-belt arrangement shown is used to transmit 2.4 kW from Point A

to Point D. Knowing that the maximum allowable couples that can be applied

to shafts AB and CD are 25 N m◊ and 80 N m,◊ respectively, determine the

required minimum speed of shaft AB.

SOLUTION

Power. 2 4 2400

25

. kW W

N m

=◊M AB �

P MPM

MP M

AB AB

ABAB

CD

CD CD

=

= = =

◊=

w

w

w

minmax

mi

2400

2596

80

rad/s

N m�

nnmax

wCDCD

PM

= = =2400

8030 rad/s

Kinematics. r rrr

A AB C CD

ABC

ACD

w w

w w

=

=

= ÊËÁ

ˆ¯̃

=

min (min )

( )120

3030

120 rad/s

Choose the larger value for min .w AB minw AB = 120 rad/s minw AB = 1146 rpm �

80


PROBLEM 17.52

The rotor of an electric motor has a mass of 25 kg and a radius of gyration of 180 mm. It is observed that

4.2 min is required for the rotor to coast to rest from an angular velocity of 3600 rpm. Determine the average

magnitude of the couple due to kinetic friction in the bearings of the motor.

SOLUTION

Time. t = =4 2 252. min s

Moment of inertia. I mkG =

=

= ◊

2

225 0 18

0 81

( )( . )

.

m

kg m2

Angular velocities. w

w

1 3600

376 99

0

===

rev/min

rad/s

2

.

Principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp.1Æ2 = Syst. Momenta2

Moments about A. I MtGw1 0- =

Average magnitude of couple. MI

tG= =

◊w1 0 81 376 99

252

( . )( . )kg m rad/s

s

2

M = ◊1 212. N m �

81


PROBLEM 17.53

A 2000-kg flywheel with a radius of gyration of 700 mm is allowed to coast to rest from an angular velocity of

450 rpm. Knowing that kinetic friction produces a couple of magnitude 16 N · m determine the time required

for the flywheel to coast to rest.

SOLUTION

Moment of inertia. I mk=

=

= ◊

2

200 0

980

( )( .0 kg 7 m)

kg m

2

2

Angular velocities. w p1 450

2

60

47 125

16

= ÊËÁ

ˆ¯̃

== ◊

rpm

rad/s

N m

.

M



Required time.

+ Moments about A: I Mtw1 0- =

t IM

t

=

=◊

◊=

=

w1

980 47 125

16

2886 4

2886 460

( )( . )

.

.min

kg m rad/s

N m

s

ss

2

ÊÊËÁ

ˆ¯̃

t = 48 6min sec �

82


PROBLEM 17.54

Two disks of the same thickness and same material are attached to a shaft

as shown. The 4-kg disk A has a radius rA = 100 mm and disk B has a

radius rB = 150 mm. Knowing that a couple M of magnitude 2.5 N · m is

applied to disk A when the system is at rest, determine the time required

for the angular velocity of the system to reach 960 rpm.

SOLUTION

Mass of disk B. m rr

mBB

AA=

ÊËÁ

ˆ¯̃

=ÊËÁ

ˆ¯̃

=

2

20 15

4

9

.( )

m

0.1 mkg

kg

Moment of inertia. I I IA B= +

= +

= ◊

1

24 0 1

1

29 0 15

0 12125

2 2( )( . ) ( )( . )

.

kg m kg m

kg m2

Angular velocity. w p2 960

2

60100 53= Ê

ËÁˆ¯̃

=rpm rad/s.

Moment. M = ◊2 5. N m



+ Moments about C: 0 2+ =Mt I w

Required time. t IM

t

=

=◊

◊=

w2

0 12125 100 53

2 5

4 8757

( . )( . )

.

.

kg m rad/s

N m

s

2

t = 4 88. s �

83


PROBLEM 17.55

Two disks of the same thickness and same material are attached to a shaft

as shown. The 3-kg disk A has a radius rA = 100 mm, and disk B has a

radius rB = 125 mm. Knowing that the angular velocity of the system is to

be increased from 200 rpm to 800 rpm during a 3-s interval, determine the

magnitude of the couple M that must be applied to disk A.

SOLUTION

Mass of disk B. m rr

mBB

AA=

ÊËÁ

ˆ¯̃

=ÊËÁ

ˆ¯̃

=

2

2125

1003

4 6875

mm

mmkg

kg.

Moment of inertia. I I IA B= +

= +

= ◊

1

23 0 1

1

24 6875 0 125

0 05162

2 2( )( . ) ( . )( . )

.

kg m kg m

kg m2

Angular velocities. w p

w p

1 2002

6020 944

8002

6083 776

= ÊËÁ

ˆ¯̃

=

= ÊËÁ

ˆ¯̃

=

rpm rad/s

rpm r2

.

. aad/s



+ Moments about B: I Mt Iw w1 2+ =

Couple M. M It

= -( )w w2 1

=◊

-0 05162

383 776 20 944

.( . . )

kg m

srad/s rad/s

2

M = ◊1 081. N m �

84


PROBLEM 17.56

A cylinder of radius r and weight W with an initial counterclockwise angular

velocity w0 is placed in the corner formed by the floor and a vertical wall.

Denoting by mk the coefficient of kinetic friction between the cylinder and the

wall and the floor derive an expression for the time required for the cylinder to

come to rest.

SOLUTION

Since the cylinder’s centre does not move, the center is the instantaneous center. Hence at both the contacts, the

contact points move w.r.t. ground and wall and hence friction force at each contact = mkN.

For the cylinder I mr W mg= =1

2

2 ,



Linear momentum + : 0 0+ - ==

N t F tN F

B A

B A

Linear momentum +: 0 0+ + - =N t F t WtA B

N F N NN F N N W

N W

F NW

N

A B A k B

A A A k A

Ak

A k Ak

k

B

+ = += + + + =

=+

= =+

mm m

m

mm

m

2

2

2

1

1

==+

=+

mm

mm

k

k

Bk

k

W

FW

1

1

2

2

2

+ Moments about G: I F rt F rtA Bw0 0- - =

tI

F F rI

WrA B

k

k k=

+=

++

w m wm m

02

01

1( )

( )

( ) t

rg

k

k k=

++

1

2 1

20m

m mw

( ) �

85


PROBLEM 17.57

A 3-kg cylinder of radius r = 125 mm with an initial counterclockwise angular

velocity w0 90= rad/s is placed in the corner formed by the floor and a vertical

wall. Knowing that the coefficient of kinetic friction is 0.10 between the cylinder

and the wall and the floor, determine the time required for the cylinder to come

to rest.

SOLUTION

Since the cylinder’s centre does not move, the center is the instantaneous center. Hence at both the contacts, the

contact points move w.r.t. ground and wall and hence friction force at each contact = mkN.

For the cylinder I mr W mg= =1

2

2 ,



Linear momentum + : 0 0+ - ==

N t F tN F

B A

B A

Linear momentum +: 0 0+ + - =N t F t WtA B

N F N NN F N N W

N W

F NW

N

A B A k B

A A A k A

Ak

A k Ak

k

B

+ = += + + + =

=+

= =+

mm m

m

mm

m

2

2

2

1

1

==+

=+

mm

mm

k

k

Bk

k

W

FW

1

1

2

2

2

+ Moments about G: I F rt F rtA Bw0 0- - =

tr

gk

k k=

++

= ++

1

2 1

1 0 1

2 0 1 1 0 1

0 125 90

9 81

20

2

mm m

w( )

( . )

( . )( . )

( . )( )

. t = 5 26. s �

86


PROBLEM 17.58

A disk of constant thickness, initially at rest, is placed in contact with a belt that

moves with a constant velocity v. Denoting by mk the coefficient of kinetic

friction between the disk and the belt, derive an expression for the time required

for the disk to reach a constant angular velocity.

SOLUTION

Moment of inertia. I mr= 1

2

2

Final state of constant angular velocity. w2 = vr


+ y components: 0 0+ - = =Nt mgt N mg

+ Moments about A: 0 2+ =m wk Ntr I

t Imgr

mrmgr

vgk

vr

k k= = =

wm m m

212

2

2 t v

gk=

2m �

87


PROBLEM 17.59

Disk A, of weight 2.5 kg and radius r = 100 mm is at rest when it is placed in

contact with a belt which moves at a constant speed v = 15 m/s. Knowing that

mk = 0 20. between the disk and the belt, determine the time required for the

disk to reach a constant angular velocity.

SOLUTION


2

2

Final state of constant angular velocity. w2 = vr


+ y components: 0 0+ - = =Nt mgt N mg

+ Moments about A: 0

2

2

2

212

2

+ =

= = =

=

m w

wm m m

m

k

k

vr

k k

k

Ntr I

t Imgr

mrmgr

vg

t vg

Data: v

k

==

15

0 20

m/s

m .

t = 15

2 0 20 9 81( )( . )( . ) t = 3 82. s �

88


PROBLEM 17.60

The 350-kg flywheel of a small hoisting engine has a radius of gyration of

600 mm. If the power is cut off when the angular velocity of the flywheel

is 100 rpm clockwise, determine the time required for the system to come

to rest.

SOLUTION

Kinematics. v rB = w


+ Moments about A: I m v r W tr I m v r

m k m r m gtr m kA B B B A B B

A B B A

w w

w1 1 2 2

2 21

2

+ - = +

+ - = +

( ) ( )

( ) ( mm r

t m k m rm rg

B

A B

B

22

2 21 2

)

( )( )

w

w w=

+ -

Data: mkA =

= =350

600 0 6

kg

mm m.

mrB =

= == ==

120

225 0 225

100 10 472

0

1

2

kg

mm m

rpm rad/s

.

.ww

t = + -[( )( . ) ( )( . ) ]( . )

( )( . )( . )

350 0 6 120 0 225 10 472 0

120 0 225 9 81

2 2

t = 5 22. s �

89


SOLUTION

Kinematics. v rB = w


+ Moments about A: I m v r W tr I m v r

m k m r m gtr m kA B B B A B B

A B B A

w w

w1 1 2 2

2 21

2

+ - = +

+ - = +

( ) ( )

( ) ( mm r

t m k m rm rg

B

A B

B

22

2 21 2

)

( )( )

w

w w=

+ -

Data: mk

mr

A

B

== === =

350

600 0 6

120

225 0 225

kg

mm m

kg

mm m

.

.

ww

1

2

100 10 472

40 4 189

= == =

rpm rad/s

rpm rad/s

.

.

t =+ -[( )( . ) ( )( . ) ]( . . )

( )( . )( .

350 0 6 120 0 225 10 472 4 189

120 0 225 9

2 2

881) t = 3 13. s �

PROBLEM 17.61

In Problem 17.60, determine the time required for the angular velocity of the

flywheel to be reduced to 40 rpm clockwise.

PROBLEM 17.60 The 350-kg flywheel of a small hoisting engine has

a radius of gyration of 600 mm. If the power is cut off when the angular

velocity of the flywheel is 100 rpm clockwise, determine the time required

for the system to come to rest.

90


SOLUTION

Kinematics. Drums A and B rotate about fixed axes. Let v be the tape velocity in m/s.

v rA A A= =w w0 025. w A v= 40

v rB B B= =w w0 04. wB v= 25

Moments of inertia. I m kA A A= = = ¥ ◊-2 2 40 6 0 0 2 4 10( . )( . ) .kg 2 m kg m2

I m kB B B= = = ¥ ◊-2 2 41 75 0 03 15 75 10( . )( . ) .kg m kg m2

State 1. t = 0 v = 0 w wA B= = 0

State 2. t = 0 24. ,s v = 3 m/s

w A = =( )( )40 3 120 rad/s

wB = =( )( )25 3 75 rad/s

Drum A.

Syst. Momenta1 + Syst. Ext. Imp.1 2Æ = Syst. Momenta 2

PROBLEM 17.62

A tape moves over the two drums shown. Drum A weighs 0.6 kg and has

a radius of gyration of 20 mm, while drum B weighs 1.75 kg and has a

radius of gyration of 30 mm. In the lower portion of the tape the tension

is constant and equal to TA = 4 N. Knowing that the tape is initially at

rest, determine (a) the required constant tension TB if the velocity of the

tape is to be v = 3 m/s after 0.24 s, (b) the corresponding tension in the

portion of tape between the drums.

91



+ Moments about A: 0 + - =r T t r T t IA AB A A A Aw

0 0 025 0 025 4 0 24 2 4 10 120

2 112

4+ - = ¥=

-( . )( ) ( . )( )( . ) ( . )( )

.

T tT t

AB

AB NN s◊

Drum B.

Syst. Momenta1 + Syst. Ext. Imp. 1 2Æ = Syst. Momenta 2

+ Moments about B: 0 + - =r T t r T t IB B B AB B Bw

0 0 04 0 04 2 112 15 75 10 75

5 065125

4+ - = ¥= ◊

-( . )( ) ( . )( . ) ( . )( )

.

T tT t

B

B N ss

(a) T T ttBB= = 5 065125

0 24

.

. TB = 21 1. N �

(b) T T ttAB

AB= = 2 112

0 24

.

. TAB = 8 80. N �

92


SOLUTION

Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks.

Let w A and wB be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at

C common to both disks.

Kinematics: No slipping v r rC A A B B= =w w

Moments of inertia. Assume that both disks are uniform cylinders.

I m r I m rA A A B B B= =1

2

1

2

2 2


Disk A

Disk B

Syst. Momenta 1 + Syst. Ext. Imp. 1 2Æ = Syst. Momenta2

Disk A: Moments about A: 0 + =r Ft IA A Aw

PROBLEM 17.63

Disk B has an initial angular velocity w 0 when it is brought into contact

with disk A which is at rest. Show that the final angular velocity of disk B

depends only on w 0 and the ratio of the masses mA and mB of the two disks.

93



Ft Ir

m r vr

m v

m r

A A

A

A A C

A

A C

A B B

= =

=

=

w

w

1

2

1

2

1

2

2

2

Disk B: Moments about B: I r Ft IB B B Bw w0 - =

1

2

1

2

1

2

20

2m r r m r mrB B B A B B B Bw w w- ÊËÁ

ˆ¯̃

= ww

B mm

A

B

=+

0

1 �

94


PROBLEM 17.64

The 4 kg disk A has a radius rA = 150 mm and is initially at rest. The 5 kg

disk B has a radius rB = 200 mm and an angular velocity w 0 of 900 rpm

when it is brought into contact with disk A. Neglecting friction in the

bearings, determine (a) the final angular velocity of each disk, (b) the

total impulse of the friction force exerted on disk A.

SOLUTION

Let Points A and B be the centers of the two disks and Point C be the contact point between the two disks.

Let w A and wB be the final angular velocities of disks A and B, respectively, and let vC be the final velocity at

C common to both disks.

Kinematics: No slipping v r rC A A B B= =w w

Moments of inertia. Assume that both disks are uniform cylinders.

I m r I m rA A A B B B= =1

2

1

2

2 2


Disk A

Disk B

Syst. Momenta1 + Syst. Ext. Imp. 1 2Æ = Syst. Momenta2

95



Disk A: Moments about A: 0

1

2

1

2

1

2

2

2

+ =

=

= =

=

r Ft I

Ft Ir

m r vr

m v

m r

A A A

A A

A

A A C

AA C

A B B

ww

w

Disk B: Moments about B: I r Ft IB B B Bw w0 - =

1

2

1

2

1

2

1

20

2

0

m r r m r mrB B B A B B B B

B mm

A

B

w w w

ww

- ÊËÁ

ˆ¯̃

=

=+

Data: mmm

A

A

B

=

= =

4

4

50 8

kg

.

rrr

B

B

A

=

= =

= =

0 2

200

150

4

3

900 300

. m

mm

mm

rpm rad/sw p

(a) ww

p

B =+

=

=

0

1 0 8

30

1 8

52 3599

.

.

. rad/s

w wAB

AB

rr

=

= ÊËÁ

ˆ¯̃

=

4

352 3599

69 8132

( . )

. rad/s w A = 667 rpm �

w B = 500 rpm �

(b) Ft = ÊËÁ

ˆ¯̃

1

24 0 2 52 3599( )( . )( . ) Ft = ◊20 9. N s ≠�

96


PROBLEM 17.65

Show that the system of momenta for a rigid slab in plane motion reduces to a single vector, and express the

distance from the mass center G to the line of action of this vector in terms of the centroidal radius of gyration k

of the slab, the magnitude v of the velocity of G, and the angular velocity w.

SOLUTION

Syst. Momenta = Single Vector

Components parallel to mv : mv X= X v= m �

Moments about G: I mv dw = ( )

d Imv

mkmv

= =w w2

d kv

=2w

�

97


PROBLEM 17.66

Show that, when a rigid slab rotates about a fixed axis through O perpendicular

to the slab, the system of the momenta of its particles is equivalent to a single

vector of magnitude mrw , perpendicular to the line OG, and applied to a Point P

on this line, called the center of percussion, at a distance GP k r= 2/ from the

mass center of the slab.

SOLUTION

Kinematics. Point O is fixed. v r= w

System momenta.

Components parallel to mv : X mv mr= = w X mr= w �

Moments about G: ( )GP X I= w

( )GP mr mkw w= 2 ( )GP kr

=2

�

98


PROBLEM 17.67

Show that the sum HA of the moments about a Point A of the momenta of the particles of a rigid slab in plane

motion is equal to IA w , where w is the angular velocity of the slab at the instant considered and IA the moment

of inertia of the slab about A, if and only if one of the following conditions is satisfied: (a) A is the mass center

of the slab, (b) A is the instantaneous center of rotation, (c) the velocity of A is directed along a line joining

Point A and the mass center G.

SOLUTION

Kinematics.

Let w = wk

and rG A G Ar/ /= q

Then, v rG A G A G Ar/ / /( )= ¥ =w w b

Where b q= + ∞90

Also v v v= +A G A/

Define h r v

h k k

= ¥

= = =

G A G A

G A G A G A G Ar v r r/ /

/ / / /( )( ) ( ) ( )2 2w w

System momenta. Moments about A:

H v

r v

r v r v

A G A

G A A G A

G A A G A G A

G

m I

m I

m m I

= ¥ +

= ¥ + +

= ¥ + ¥ +

=

r

v

r

/

/ /

/ / /

( )

w

w

w

//

/ /

/ /( )

A A

G A A G A

G A A G A

m m I

m mr I

m mr I

¥ + +

= ¥ + +

= ¥ + +

v h

r v

r v

w

w w

w

2

2

The first term on the right hand side is equal to zero if

(a) rG A A/ = 0 ( is the mass center)

or (b) vA A= 0 ( is the instantaneous center of rotation)

or (c) r vG A A/ . is perpendicular to

In the second term, mr I IG A A/2 + =

by the parallel axis theorem. Thus, HA AI= w

when one or more of the conditions (a), (b) or (c) is satisfied.

99


SOLUTION

(a) Locate the instantaneous center C corresponding to center of percussion P. Let d GPP = .


Components parallel to FDt : 0 + =F t mvD

Moments about G: 0 + =d F t IP ( )D w

Eliminate F tD to obtain v I

md

kd

P

P

w=

=2

Kinematics. Locate Point C. GC d v kdC

P= = =

w

2

GC kGP

=2

�

(b) Place the center of percussion at ¢ =P C. Locate the corresponding instantaneous center ¢C . Let

d GP GC dP C¢ = ¢ = = .


PROBLEM 17.68

Consider a rigid slab initially at rest and subjected to an impulsive force F contained in the

plane of the slab. We define the center of percussion P as the point of intersection of the

line of action of F with the perpendicular drawn from G. (a) Show that the instantaneous

center of rotation C of the slab is located on line GP at a distance GC k GP= 2/ on

the opposite side of G. (b) Show that if the center of percussion were located at C the

instantaneous center of rotation would be located at P.

100



Components parallel to FDt : 0 + = ¢F t mvD

Moments about G: 0 + = ¢¢d F t IP ( )D w

Eliminate F tD to obtain ¢¢

= =¢ ¢

v Imd

kdP Pw

2

Kinematics. Locate Point ¢C . GC d v kd

kdC

P C¢ = = ¢

¢= =¢

¢w

2 2

Using d d kdC P

P= =¢

2

gives d d GC GPC P¢ = ¢ = or �

Thus Point ¢C coincides with Point P.

101


SOLUTION

Kinematics. Rolling motion. Instantaneous center at C.

v v rvr

G= =

=

w

w

Moment of inertia. I mk= 2

Kinetics.


moments about C: 0

2

+ = +

= +

( ) sin

( sin )

mgt r mvr I

mgr t mrv mk vr

b w

b

(a) Velocity of Point G. v =+

r gtr k

2

2 2

sin b b �

+ components parallel to incline:

0

2

2 2

2

2 2

+ - =

= -+

=+

mgt Ft mv

Ft mgt mr gtr k

k mgtr k

sin

sinsin

sin

b

b b

b

PROBLEM 17.69

A wheel of radius r and centroidal radius of gyration k is released from rest on

the incline shown at time t = 0. Assuming that the wheel rolls without sliding,

determine (a) the velocity of its center at time t, (b) the coefficient of static

friction required to prevent slipping.

102



+ components normal to incline:

0 0+ - ==

Nt mgtNt mgt

cos

cos

bb

(b) Required coefficient of static friction.

m

bb

sFNFtNt

k mgtr k mgt

³

=

=+

2

2 2

sin

( ) cos m b

skr k

³2

2 2

tan

+ �

103


PROBLEM 17.70

A flywheel is rigidly attached to a 40 mm-radius shaft that rolls without sliding

along parallel rails. Knowing that after being released from rest the system

attains a speed of 150 mm/s in 30 s, determine the centroidal radius of gyration

of the system.

SOLUTION

Kinematics. Rolling motion. Instantaneous center at C.

v v rG= = w

Moment of inertia. I mk= 2

Kinetics.


Moments about C: 0

2

+ = +

= +Ê

ËÁˆ

¯̃

( ) sin

sin

mgt r mvr I

mgtr m r kr

v

b w

b

Solving for k 2, k r gtv

2 2 1= -ÊËÁ

ˆ¯̃

sin b

Data: r = =40 0 04mm m.

gtv

k

=== =

= ∞

9 81

30

150 0 15

0 049 81 30 152 2

.

.

( . )( . )( )sin

m/s

s

mm/s m/s

2

00 151

0 81088

.

.

-ÈÎÍ

˘˚̇

= m2

k 2 0 90049= . m k = 900 mm �

104


PROBLEM 17.71

The double pulley shown has a mass of 3 kg and a radius of gyration of

100 mm. Knowing that when the pulley is at rest, a force P of magnitude 24 N

is applied to cord B, determine (a) the velocity of the center of the pulley after

1.5 s, (b) the tension in cord C.

SOLUTION

For the double pulley, rrk

C

B

===

0 150

0 080

0 100

.

.

.

m

m

m



Kinematics. Point C is the instantaneous center. v rC= w

Moments about C: 0

2

+ + - = +

= +

Pt r r mgtr I mvr

mk m r rC B C C

C C

( )

( )

w

w w

w =+ -

+( )=

-

Pt r r mgtrm k rC B C

C

( )

( )( . )( . ) ( )( . )( . )

2 2

24 1 5 0 230 3 9 81 1 5 (( . )

( . . )

.

0 150

3 0 100 0 150

17 0077

2 2+= rad/s

105



(a) v = =( . )( . ) .0 150 17 0077 2 55115 m/s v = 2 55. m/s ≠ �

+ Linear components: 0 + - + =Pt mgt Qt mv

Q mvt

mg P= + -

= + -( )( . )

.( )( . )

3 2 55115

1 53 9 81 24

(b) Tension in cord C. Q = 10 53. N �

106


PROBLEM 17.72


are connected by a belt as shown. If the system is released from rest

when t = 0, determine (a) the velocity of the center of cylinder B at

t = 3s, (b) the tension in the portion of belt connecting the two cylinders.

SOLUTION



w w

w w

AAB

B

A A B

vr

v r r

= =

= =

2

1

2

1

2


2

2

(a) Velocity of the center of A.

Cyl. B:


+

Moments about B: 0 + =Ptr I Bw (1)

Cyl. A:


107



Moments about C: 0 2

0 21

2

1

2

5

- + = +

- - = ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

Ptr mgtr mv r I

I mgtr m r r I

A A

B B B

w

w w w

22

1

2

5

2 2

1

2

2

22

I mr mgrt

mr mr mgrt

B

B

+ÊËÁ

ˆ¯̃

=

+Ê

ËÁˆ

¯̃=

w

w

7

4

4

7

r gt

gtr

B

B

w

w

=

= (2)

v r gtA B= = =

=

1

2

2

7

2

79 81 3

8 408

w ( . )( )

. 6 m/s

vA = 8 41. m/s Ø �

(b) Tension in the belt.

From (1) and (2) Ptr I gtr

= ÊËÁ

ˆ¯̃

4

7

P Igr

mg= = = =4

7

4

7

4

77 9 81 39 24

2( )( . ) . N P = 39 2. N �

108


SOLUTION



w w

w w

AAB

B

A A B

vr

v r r

= =

= =

2

1

2

1

2


r= 1

2

2

(a) Required time.

Cyl. B:


Moments about B: I Ptr IB B( ) ( )w w1 2- =

Ptr I

mr

B B

B B

= -

= -

[( ) ( ) ]

[( ) ( ) ]

w w

w w

1 2

21 2

1

2 (1)

Cyl. A:


PROBLEM 17.73


are connected by a belt as shown. Knowing that at the instant shown the

angular velocity of cylinder A is 30 rad/s counterclockwise, determine

(a) the time required for the angular velocity of cylinder A to be reduced to

5 rad/s, (b) the tension in the portion of belt connecting the two cylinders.

109



Moments about C: I m v r Ptr mgtr I m v rA A A A( ) ( ) ( ) ( )w w1 1 2 22+ + - = +

1

22 0

3

2

1

2

21 2 1 2

2

mr mr r Ptr mgtr

mr

A A A A

B

[( ) ( ) [( ) ( ) ]w w w w

w

- + - + - =

ÊËÁ

ˆ̂¯̃

-È

ÎÍ

˘

˚˙ + -Ï

ÌÓ

¸˝˛

- =1

22

1 2

2

1

22

1

20

7

4

( ) [( ) ( ) ]

[

w w wB B Bmr mgtr

mr (( ) ( ) ]w wB B mgtr1 2 0- - =

t rg

B B=-7

4

1 2[( ) ( ) ]w w (2)

Data: mr

==

7

0 1

kg

m.

From Equation (2), t = - =( )( . )( )

( )( . ).

7 0 1 30 5

4 9 810 56612 t = 0 446. s �

(b) Tension in belt between cylinders.

From Equation (1) and (2) P g= ◊ ◊

= =

1

2

4

7

1

27

4 9 81

719 62

m

( )( )( . )

. N

P = 19 62. N �

110


SOLUTION

Moment of inertia. I m rA=

=

= ◊

1

2

1

28

0 2304

2

2

(

.

kg)(0.24 m)

kg m

2

Cylinder alone:


Moments about C: 0 0+ = -I m v rA Aw

or 0 0 2304 8 0 24= -. ( )( . )w vA (1)

Cylinder and carriage:


+ Horizontal components: 0 + = +Pt m v m vA A B B

or 0 10 1 2 8 3+ = +( )( . ) v vA B (2)

PROBLEM 17.74

A 240-mm-radius cylinder of mass 8 kg rests on a 3-kg carriage. The

system is at rest when a force P of magnitude 10 N is applied as shown

for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage

and neglecting the mass of the wheels of the carriage, determine the

resulting velocity of (a) the carriage, (b) the center of the cylinder.

111



Kinematics. v v rA B= - w

v vA B= - 0 24. w (3)

Solving Equations (1), (2) and (3) simultaneously gives w = 5 68. rad/s

(a) Velocity of the carriage. vB = 2 12. m/s Æ �

(b) Velocity of the center of the cylinder. vA = 0 706. m/s Æ �

112


PROBLEM 17.75

A 240-mm-radius cylinder of mass 8 kg rests on a 3-kg carriage. The

system is at rest when a force P of magnitude 10 N is applied as shown for

1.2 s. Knowing that the cylinder rolls without sliding on the carriage and

neglecting the mass of the wheels of the carriage, determine the resulting

velocity of (a) the carriage, (b) the center of the cylinder.

SOLUTION

Moment of inertia. I m rA=

=

= ◊

1

2

1

28

0 2304

2

(

.

kg)(0.24 m)

kg m

2

2

Cylinder alone:


Moments about C: 0 + = +Ptr I m v rA Aw

or 0 10 1 2 0 24 0 2304 8 0 24+ = +( )( . )( . ) . ( )( . )w vA (1)

Cylinder and carriage:


+ Horizontal components: 0 + = +Pt m v m vA A B B

or 0 10 1 2 8 3+ = +( )( . ) v vA B (2)

113



Kinematics. v v rA B= + w

v vA B= + 0 24. w (3)

Solving Equations (1), (2) and (3) simultaneously gives w = 2 21. rad/s

(a) Velocity of the carriage. vB = 0 706. m/s Æ �

(b) Velocity of the center of the cylinder. vA = 0 235. m/s Æ �

114


PROBLEM 17.76

In the gear arrangement shown, gears A and C are attached to rod ABC,

which is free to rotate about B, while the inner gear B is fixed. Knowing

that the system is at rest, determine the magnitude of the couple M which

must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is

to be 240 rpm clockwise. Gears A and C have a mass of 1.25 kg each and

may be considered as disks of radius 50 mm; rod ABC weighs 2 kg.

SOLUTION

Kinematics of motion

Let w wABC = v v BC rA C= = =( )w w2

Since gears A and C roll on the fixed gear B,

w w w wA CCvr

rr

= = = =22


Syst. Momenta1 + Syst. Ext. Imp. 1 2Æ = Syst. Momenta2

+ Moments about D: 0 + = +( )Qt r m v r I wC C C C

( ) ( ) ( )Qt r m r r m r

Qt m rw

C C

C

= +

=

21

22

3

2w w

(1)


115



+ Moments about B: Mt Qt r I ABC- =( )4 w

Mt Qt r m r

Mt Qt r m r

ABC

ABC

- =

- =

41

124

44

3

2

2

( ) ( )

( )

w

w (2)

Substitute for (Qt) from (1) into (2):

Mt m r r m r

Mt r m m

C ABC

ABC C

- =

= +

4 34

3

4

39

2

2

( )

( )

w w

w (3)

Couple M.

Data: t = 2 5. s

rm

mABC

C

= =====

50 0 05

2

1 25

240

8

mm m

kg

kg

rpm

rad/s

.

.

wp

Eq. (3): M ( . ( . ) ( ( . )2 5 0 05 8 2 9 1 252s)4

3)= +( )p

2 5 1 1100

0 444

. .

.

M

N m

== ◊M M = ◊0 444. N m �

116


PROBLEM 17.77

A sphere of radius r and mass m is placed on a horizontal floor with no linear velocity but

with a clockwise angular velocity w 0 . Denoting by mk the coefficient of kinetic friction

between the sphere and the floor, determine (a) the time t1 at which the sphere will start

rolling without sliding, (b) the linear and angular velocities of the sphere at time t1.

SOLUTION

Moment of inertia. Solid sphere. I mr= 2

5

2


+ y components: Nt Wt N W mg1 1 0- = = = (1)

+ x components: Ft mv1 2= (2)

+ Moments about G: I Ft r Iw w0 1 2- = (3)

Since F N mgk k= =m m , Equation (2) gives

mk mgt mv1 2=

or v gtk2 1= m (4)

Using the value for I in Equation (3),

2

5

2

5

20 1

22mr mgt r mrkw m w- =

or w wm

2 015

2= - k gt

r (5)

(a) Time t1 at which sliding stops.

From kinematics, v r2 = w

m w mk kgt r gt1 0 1

5

2= - t

rgk

102

7=

wm

�

117



(b) Linear and angular velocities.

From Equation (4), v gr

gkk

202

7= m

wm

v2 0

2

7= rw Æ �

From Equation (6), w w22

0

2

7= =

vr

w2 0

2

7= w �

118


PROBLEM 17.78

A sphere of radius r and mass m is projected along a rough horizontal surface

with the initial velocities shown. If the final velocity of the sphere is to be zero,

express (a) the required magnitude of w 0 in terms of v0 and r, (b) the time

required for the sphere to come to rest in terms of v0 and coefficient of kinetic

friction mk .

SOLUTION


5

2


+ y components: Nt Wt N W mg- = = =0 (1)

+ x components: mv Ft Ft mv0 00- = = (2)

+ Moments about G: I Ftr

mr mv r

w

w

0

20 0

0

2

50

- =

- =

(3)

(a) Solving for w0, w005

2=

vr

�

(b) Time to come to rest.

From Equation (2), tmvF

mvmgk

= =0 0

m t

vgk

= 0

m �

119


PROBLEM 17.79

A 1.25-kg disk of radius 100 mm is attached to the yoke BCD by means

of short shafts fitted in bearings at B and D. The 0.75 kg yoke has a

radius of gyration of 75 mm about the x axis. Initially the assembly is

rotating at 120 rpm with the disk in the plane of the yoke (q = 0). If

the disk is slightly disturbed and rotates with respect to the yoke until

q = ∞90 , where it is stopped by a small bar at D, determine the final

angular velocity of the assembly.

SOLUTION

Moment of inertia of yoke: I mkC C= = = ¥ ◊-2 2 30 75 0 075 4 21875 10( . )( . ) .kg m kg m2

Moment of inertia of disk about x axis: q = =01

4

2: I mrA

= = ¥ ◊

= ∞ =

=

-1

41 25 0 1 3 125 10

901

2

1

21 25 0 1

2 3

2

( . )( . ) .

:

( . )( .

kg m2

q I mrA

)) .2 36 25 10= ¥ ◊- kg m2

Total moment of inertia about the x axis:

q

q

= = +

= ¥ ◊= ∞ = +

=

-

0

7 34375 10

90

10 4687

1

3 2

2

: ( )

.

: ( )

.

I I I

I I I

x C A

x C A

kg m

55 10 3 2¥ ◊- kg m

Angular momentum about the x axis:

q w

wq w

= =

= ¥= ∞ =

= ¥

-

-

0

7 34375 10

90

10 46875 10

1 1 1

31

2 2 2

: ( )

.

: ( )

.

H I

H I

x

x33

2w

Conservation of angular momentum.

H H1 23

13

27 34375 10 10 46875 10= ¥ = ¥- -: . .w w

w w2 10 7015 0 7015 120= =. ( . ) ( rpm) w2 84 2= . rpm �

120


PROBLEM 17.80

Two panels A and B are attached with hinges to a rectangular plate

and held by a wire as shown. The plate and the panels are made of

the same material and have the same thickness. The entire assembly

is rotating with an angular velocity w0 when the wire breaks.

Determine the angular velocity of the assembly after the panels

have come to rest against the plate.

SOLUTION

Geometry and kinematics:

Panels in up position Panels in down position

v b0 0= w v b2 0

3

2= w

Let r = =mass density, thicknesst

Plate: m t b b tb

I tb b b

plate

plate

= =

= +

=

r r

r

( )( )

( )[( ) ( ) ]

2 4 8

1

128 2 4

160

1

2

2 2 2

22

40

3

4

4

r

r

tb

tb=

Each panel: m t b b tbpanel = =r r( )( )2 2 2

Panel in up position ( ) ( )( )I t b b

t b t b

panel 02 2

4 4

1

122 2

8

12

2

3

=

= =

r

r r

121



Panel in down position ( ) ( )[ ( ) ]I t b b b

t b

t b

panel 12 2 2

4

4

1

122 2

10

12

5

6

= +

=

=

r

r

r

Conservation of angular momentum about the vertical spindle.

Initial momenta Final momenta

+ Moments about C:

I I m v b I Iplate panel panel plate panelw w w w0 0 0 0 1 1 12 2+ + = + +[( ) ( )] ( ) mm v b

tb tb tb b b

panel 1

40

40

20

3

2

40

32

2

32

ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

+ +Èr w r w r w( )( )ÎÎÍ

˘˚̇

= + + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

ÈÎÍ

˘˚̇

40

32

5

62

3

2

3

2

40

41

20r w r w r wtb tb tb b b

440

3

4

34

40

3

10

69

56

324

40

41

0 1

+ +ÈÎÍ

˘˚̇

= + +ÈÎÍ

˘˚̇

=

r w r w

w w

t b tb

w1

56

3 24=

( )( ) w w1 2

7

9= �

122


SOLUTION

Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB.

Masses and moments of inertia about vertical axes.

m

I

I

AB

AB

DCE

=

= ◊

= ◊

1 6

0 0025

0 30

.

.

.

kg

kg m

kg m

2

2

State 1. ( ) ( )

.

rG A/

1

mm

rad/s

1

1

2125

62 5

5

=

==w

State 2. ( ) .

.

rG A/

2

mm

2 500 62 5

437 5

= -==w w

Kinematics. ( )v v rG G Cq q w= = /


PROBLEM 17.81

A 1.6-kg tube AB can slide freely on rod DE which in turn can rotate freely in

a horizontal plane. Initially the assembly is rotating with an angular velocity

w = 5 rad/s and the tube is held in position by a cord. The moment of inertia of the

rod and bracket about the vertical axis of rotation is 0 30. kg m2◊ and the centroidal

moment of inertia of the tube about a vertical axis is 0 0025 2. . kg m◊ If the cord

suddenly breaks, determine (a) the angular velocity of the assembly after the tube

has moved to end E, (b) the energy lost during the plastic impact at E.

123



Moments about C:

I I m v r I I m v rAB DCE AB G C AB DCE AB G Cw w w wq q1 1 1 1 2 2 20+ + + = + +( ) ( ) ( ) ( )/ / 22

12

1 22

2

0

I I m r I I m rAB DCE AB G G AB DCE AB G C+ +ÈÎ ˘̊ = + +ÈÎ ˘̊( ) ( )

[ .

/ /w w

00025 0 30 1 6 0 0625 5 0 0025 0 30 1 6 0 43752 2+ + = + +. ( . )( . ) ]( ) [ . . ( . )( . ) ]www

w

2

2

2

0 30875 5 0 60875

2 5359

( . )( ) .

.

== rad/s

(a) Angular velocity after the plastic impact. 2 54. rad/s �

Kinetic energy. T I I m v

T

AB DCE AB= + +

= + +

1

2

1

2

1

2

1

20 0025 5

1

20 30 5

1

2 2 2

12 2

w w

( . )( ) ( . )( )22

1 6 0 0625 5

3 859375

1

20 0025 2 5359

1

20

2 2

22

( . )( . ) ( )

.

( . )( . ) ( .

=

= +

J

T 330 2 53591

21 6 0 4375 2 5359

1 9573

2 2 2)( . ) ( . )( . ) ( . )

.

+

= J

(b) Energy lost. T T1 2 1 902- = . J �

124


PROBLEM 17.82

Two 0.4-kg balls are to be put successively into the center C of the slender

2-kg tube AB. Knowing that when the first ball is put into the tube the initial

angular velocity of the tube is 8 rad/s and neglecting the effect of friction,

determine the angular velocity of the tube just after (a) the first ball has left

the tube, (b) the second ball has left the tube.

SOLUTION

Conservation of angular momentum about C.

I I mrvI mr r

II mrC

w ww w

w w

w

q1 2

2 2

2 2 1

1

= += +

=+

=

( )

(1)

Data: I m L= 1

12

2tube

=

= ◊

== =

+

1

122 1

1

6

0 4

0 5

2

2

( )( )

( .

.

kg m

one ball) kg

500 mm m

2

mr

I mr == +

= ◊

=( )( ) =

1

60 4 0 5

4

15

0 625

2

16

415

( . )( . )

.

kg m2

C

125



(a) First ball moves through the tube.

w1 8= rad/s

By Equation (1), w2 0 625 8= ( . )( ) w2 5 00= . rad/s �

(b) Second ball moves through the tube.

w1 5 00= . rad/s

By Equation (1), w2 0 625 5 00= ( . )( . ) w2 3 13= . rad/s �

126


PROBLEM 17.83

A 3-kg rod of length 800 mm can slide freely in the 240-mm cylinder DE, which

in turn can rotate freely in a horizontal plane. In the position shown the assembly

is rotating with an angular velocity of magnitude w = 40 rad/s and end B of the

rod is moving toward the cylinder at a speed of 75 mm/s relative to the cylinder.

Knowing that the centroidal mass moment of inertia of the cylinder about a

vertical axis is 0 025. kg m2◊ and neglecting the effect of friction, determine the

angular velocity of the assembly as end B of the rod strikes end E of the cylinder.

SOLUTION

Kinematics and geometry.

vv

1 1

1

0 04 0 4

1 6

= ==

( . ( .

.

m) m)(40 rad/s)

m/s

w v2 0 28= ( . m) 2w

Initial position Final position


+ Moments about C: I AB = = ◊1

123 0 8 0 16( )( . . kg m) kg m2 2

I mv I I mv IAB DE AB DEw w w w1 1 1 2 2 20 04 0 028+ + = + +( . ( . m) m)

( . )( ( ( . ( .0 16 40 8 0 04 0 025 kg m rad/s) kg)(1.6 m/s) m) kg2◊ + + ◊ mm )(40 rad/s)2

= ◊ + + ◊+

( . ) ( )( . ) ( . )

( .

0 16 3 0 28 0 025

6 4 0

2 2 kg m kg)(0.28 kg m22

2w w w.. . ) ( . . . )

. . ; .

192 1 00 0 16 0 2352 0 025

7 592 0 4202 18 068

2

2 2

+ = + += =

ww w rrad/s

Angular velocity. w2 18 07= . rad/s �

127


PROBLEM 17.84

In the helicopter shown, a vertical tail propeller is used to prevent

rotation of the cab as the speed of the main blades is changed.

Assuming that the tail propeller is not operating, determine the final

angular velocity of the cab after the speed of the main blades has

been changed from 180 to 240 rpm. (The speed of the main blades is

measured relative to the cab, and the cab has a centroidal moment of

inertia of 1000 kg m2◊ . Each of the four main blades is assumed to

be a slender 4.2-m rod of mass 25 kg.)

SOLUTION

Let W be the angular velocity of the cab and w be the angular velocity of the blades relative to the cab. The

absolute angular velocity of the blades is W + w .

w pw p

1

2

180 6

240 8

= == =

rpm rad/s

rpm rad/s

Moments of inertia.

Cab: IC = ◊1000 kg m2

Blades: I mLB = ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= ◊

41

3

41

325 4 2

588

2

2( ) ( )( . )

kg m2

Assume W1 0= .

Conservation of angular momentum about shaft.

I I I II

I I

B C B C

B

C B

( ) ( )

( )

( )(

w ww w

p

1 1 1 2 2 2

22 1

588 8

+ + = + +

= --

+

= - -

W W W W

W

66

588 1000

2 3265

p )

( )

.

+= - rad/s W2 22 2= - . rpm �

128


PROBLEM 17.85

Assuming that the tail propeller in Problem 17.84 is operating and that the

angular velocity of the cab remains zero, determine the final horizontal

velocity of the cab when the speed of the main blades is changed from

180 to 240 rpm. The cab weighs 625 kg and is initially at rest. Also determine

the force exerted by the tail propeller if the change in speed takes place

uniformly in 12 s.

SOLUTION

Let W be the angular velocity of the cab and w be the angular velocity of the blades relative to the cab. The

absolute angular velocity of the blades is W + w .

w pw p

1

2

180 6

240 8

= == =

rpm rad/s

rpm rad/s

Moments of inertia.

Cab: IC = ◊1000 kg m2

Blades: I mLB = ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

= ◊

41

34

1

325 4 2

588

2 2( ) ( )( . )

kg m2

The cab does not rotate. W W1 2 0= =

Syst Momenta Syst Ext Imp Syst Momenta. . . . . 1 1 2 2+ =Æ

Moments about shaft: I I Frt I IFrt I

B C B C

B

( ) ( )

( )

( )(

w ww w

p p

1 1 1 2 2 2

2 1

588 8 6

+ + + = + += -= -

W W W W

))

.

..

= ◊ ◊

= = = ◊

3694 51

3694 51

5738 903

N m s

N sFt Frtr

Linear components: mv Ft mv

v v Ftm

1 2

2 1

738 903

625 4 25

1 0192

+ =

- = =+

=

.

( )( )

. m/s

(a) Assume v1 0= . v2 1 019= . m/s �

(b) Force. F Ftt

= = 738 903

12

. F = 61 6. N �

129


PROBLEM 17.86

The 4-kg disk B is attached to the shaft of a motor mounted on plate A, which can

rotate freely about the vertical shaft C. The motor-plate-shaft unit has a moment

of inertia of 0 20 2. kg m◊ with respect to the axis of the shaft. If the motor is

started when the system is at rest, determine the angular velocities of the disk and

of the plate after the motor has attained its normal operating speed of 360 rpm.

SOLUTION

Moments of inertia. motor-plate-shaft:

IC = ◊0 20. kg m2

Disk: I m rB B B=

=

= ◊

1

2

1

24

0 0162

2

(

.

kg)(0.090 m)

kg m

2

2

Kinematics. v rB B C C C= =/ .w w0 090

Kinetics.


Moments about C: 0 0+ = + -I m v IC C B B B Bw w

I m r r I

I m r IC C B B C C B C B B

C B B C C B B

w w w

w w

+ - =

+ =

+

( )

( )

[ . ( )( .

/ /

/

0

0 20 4 0 0

2

990 0 01622) ] .w wC B= wC = 0 069707.

130



Angular velocity of motor. w w ww

M B C

B

= += 1 06971.

w

ww

M

BM

=

=

=

=

360

1 06971

360

336 54

rpm

rpm

1.06971

rpm

.

. wB = 337 rpm �

wC = ( . )( .0 069707 336 54 rpm) wC = 32 5. rpm �

131


PROBLEM 17.87

The circular platform A is fitted with a rim of 200-mm inner radius and can

rotate freely about the vertical shaft. It is known that the platform-rim unit has

a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft. At

a time when the platform is rotating with an angular velocity of 50 rpm, a 3-kg

disk B of radius 80 mm is placed on the platform with no velocity. Knowing

that disk B then slides until it comes to rest relative to the platform against the

rim, determine the final angular velocity of the platform.

SOLUTION

Moments of inertia. I m k

I m r

A A

B B B

=

=

= ◊

=

=

2

2

5

0 153125

1

2

1

23

(

.

(

kg)(0.175 m)

kg m

kg)

2

2

((0.08 m)

kg m

2

2= ¥ ◊-9 6 10 3.

State 1 Disk B is at rest.

State 2 Disk B moves with platform A.

Kinematics. In State 2, vB = ( .0 12 m) 2w

Principle of conservation of angular momentum.

+ Moments about D: I I I m vA A B B Bw w w1 2 2 0 12= + + ( . m)

( . ) ( . )

( . ) (

0 153125 0 153125

9 6 10 3

1 2

32

kg m kg m

kg m

2 2

2

◊ = ◊

+ ¥ ◊ +-

w w

w kkg m)

r

2)( .

. .

.

. (

0 12

0 153125 0 20593

0 7436

0 7436 50

2

2 1

2 1

ww ww w

=== ppm)

Final angular velocity w2 37 2= . rpm �

Syst. Momenta,

Syst. Momenta,

132


PROBLEM 17.88

A small 2-kg collar C can slide freely on a thin ring of mass 3 kg and radius

250 mm. The ring is welded to a short vertical shaft, which can rotate freely in a

fixed bearing. Initially the ring has an angular velocity of 35 rad/s and the collar

is at the top of the ring ( )q = 0 when it is given a slight nudge. Neglecting the

effect of friction, determine (a) the angular velocity of the ring as the collar passes

through the position q = ∞90 , (b) the corresponding velocity of the collar relative

to the ring.

SOLUTION

Moment of inertia of ring. I m RR R= 1

2

2

Position 1 Position 2

Position 1. q ==

0

0

.

vC

Position 2. qw

= ∞= =

90

2( )v v RC y y

Conservation of angular momentum about y axis for system.

I I m v R

m R m R m R

m R m m R

R R C y

R R C

R R C

w w

w w w

w w

1 2

21

22

22

21

22

1

2

1

2

2

= +

= +

= +( )

ww w2 12

=+m

m mR

R C (1)

133



Potential energy. Datum is the center of the ring.

V m gR VC1 2 0= =

Kinetic energy: T I m R

m R

T I m v v

R R

R

R C x

1 12 2

12

212

2 22 2

1

2

1

2

1

2

1

4

1

2

1

2

= = ÊËÁ

ˆ¯̃

=

= + +

w w

w

w yy

R C C ym R m R v m v

2

222 2

22 21

4

1

2

1

2

( )= + +w

Principle of conservation of energy:

T V T V

m R m gR m m R m vR C R C C y

1 1 2 2

212 2

22 21

4

1

4

1

2

1

2

+ = +

+ = +ÊËÁ

ˆ¯̃

+w w (2)

Data: mm

R

C

R

====

2

3

0 25

35

kg

kg

m

rad/s1

.

w

(a) Angular velocity.

From Eq. (1), w2

3

2 235=

+ kg

3 kg kg) rad/s)

(( w2 15 00= . rad/s �

(b) Velocity of collar relative to ring.

From Eq. (2), 1

43 35 2 0 25( ( ( )( . kg)(0.25 m) rad/s) kg)(9.81 m/s m)2 2 2+

= +ÈÎÍ

˘˚̇

+1

43

1

22 15

1

22

57

2( ( ( (

.

kg) kg) (0.25 m) rad/s) kg)2 2 vy

4422 4 905 24 609

37 716

2

2

+ = +

=

. .

.

v

vy

y vy = 6 14. m/s �

134


PROBLEM 17.89

Collar C has a mass of 8 kg and can slide freely on rod AB, which in

turn can rotate freely in a horizontal plane. The assembly is rotating

with an angular velocity w of 1.5 rad/s when a spring located between

A and C is released, projecting the collar along the rod with an initial

relative speed vr = 1 5. m/s. Knowing that the combined mass moment

of inertia about B of the rod and spring is 1 2. , kg m2◊ determine (a) the

minimum distance between the collar and Point B in the ensuing

motion, (b) the corresponding angular velocity of the assembly.

SOLUTION

Kinematics. v rq w=

Moments of inertia. I I m r

rB C= +

= +

2

21 2 8.


( ) ( ) : ( ) ( )

( ) (

H H I m v r I m v r

I m r IB B B C B C

B C B

1 2 1 1 1 2 2 2

12

1

= + = +

+ = +

w w

wq q

mm r I IC 22

2 1 1 2 2)w w wor =

ww

21 1

2

=II

(1)

Potential energy. V V1 20 0= =

Kinetic energy. T m v m v I

m v I

C r C B

C r

= + +

= +

1

2

1

2

1

2

1

2

1

2

2 2 2

2 2

q w

w

Position 1. Just after spring is released. r rv v

I I

r r

====

1

1

1

1

( )

w w

Position 2. Distance r is minimum. r rv v

I I

r r

== ===

2

2

2

0( ) ( )

w w

135




T V T V m v I I

m v I II

C r

C r

1 1 2 2 12

1 12

2 22

12

21 1

1

2

1

20 0

1

2+ = + + + = +

=

: ( )

( )

w w

w

22

2

1 12

11

2121

ÊËÁ

ˆ¯̃

-

= -ÊËÁ

ˆ¯̃

I

I II

w

w (2)

Data. r

v

Ir

1

1

12

600 0 6

15

1 5

1 2 8 0 6

4 08

= ===

= +

=

mm m

rad/s

( m

1

.

) .

. ( )( . )

.

w

kg m2◊

Equation (2): ( )( . ) . ( . )8 1 5 4 08 1 1 52 1

2

2= -ÊËÁ

ˆ¯̃

II

II

I r

1

2

2 22

2 9608

4 08

2 96081 378 1 2 8

=

= = = +

.

.

.. .

(a) r2 0 14917= . m r2 149 2= . mm �

(b) Equation (1): w2

4 08

1 3781 5= .

.( . ) w2 4 44= . rad/s �

136


PROBLEM 17.90

In Problem 17.89, determine the required magnitude of the initial

relative speed vr if during the ensuing motion the minimum distance

between collar C and Point B is to be 300 mm.

SOLUTION

Kinematics. v rq w=

Moments of inertia. I I m r

rB C= +

= +

2

21 2 8.


( ) ( ) : ( ) ( )

( ) (

H H I m v r I m v r

I m r IB B B C B C

B C B

1 2 1 1 1 2 2 2

12

1

= + = +

+ = +

w w

wq q

mm r I III

C 22

2 1 1 2 2

21 1

2

)w w w

ww

or =

= (1)

Potential energy. V V1 20 0= =

Kinetic energy. T m v m v I

m v I

C r C B

C r

= + +

= +

1

2

1

2

1

2

1

2

1

2

2 2 2

2 2

q w

w

Position 1. Just after spring is released. r rv v

I I

r r

====

1

1

1

1

( )

w w

Position 2. Distance r is minimum. r rv v

I I

r r

== ===

2

2

2

0( ) ( )

w w

137




T V T V m v I I

m v I II

C r

C r

1 1 2 2 12

1 12

2 22

12

21 1

1

2

1

20 0

1

2+ = + + + = +

=

: ( )

( )

w w

w

22

2

1 12

11

2121

ÊËÁ

ˆ¯̃

-

= -ÊËÁ

ˆ¯̃

I

I II

w

w (2)

Data. r

r

I

1

2

12

600

1 5

300 0 3

1 2 8 0 6

= === =

= +

mm 0.6 m

rad/s

mm m

1w .

.

. ( )( . ) == ◊

= + = ◊

4 08

1 2 8 0 3 1 9222

.

. ( )( . ) .

kg m

kg m

2

2I

Equation (2): 8 4 084 08

1 921 1 51

2 2( ) ..

.( . )vr = -Ê

ËÁˆ¯̃

( ) .vr 12 21 29094= m /s2 ( ) .vr 1 1 136= m/s �

138


PROBLEM 17.91

A 3 kg collar C is attached to a spring and can slide on rod AB, which in turn

can rotate in a horizontal plane. The mass moment of inertia of rod AB with

respect to end A is 0.5 kg ◊ m2. The spring has a constant k = 3000 N/m and an

undeformed length of 250 mm. At the instant shown the velocity of the collar

relative to the rod is zero and the assembly is rotating with an angular velocity

of 12 rad/s. Neglecting the effect of friction, determine (a) the angular velocity of

the assembly as the collar passes through a point located 180 mm from end A of

the rod, (b) the corresponding velocity of the collar relative to the rod.

SOLUTION

Potential energy of spring: undeformed length = 250 mm = 0.25 m

Position 1: Position 2:

D

D

= - =

= =

= ◊

0 65 0 25 0 4

1

2

1

23000 0 4

240

12 2

. . .

( )( . )

m m m

N m

V k

D

D

= - =

= =

=

0 30806 0 25 0 05806

1

2

1

23000 0 05806

5 0564

22 2

. . .

( )( . )

.

m

N

V k

◊◊ m

Kinematics:

Kinetics: Since moments of all forces about shift at A are zero, ( ) ( )H HA A1 2=

I m v r I m v r

I m r I m rR C R C C

R C R C

w w

w w1 0 1 0 2 2

12

1 23

2

+ = +

+( ) = +( )( ) ( )

139



Data: I mr rR C= ◊ =

= = =0 5 3

0 6 0 18 121 2 1

. ,

. .

kg m kg

m, m, rad/s

2

w

0 5 3 0 6 12 0 5 3 0 18

18 96 0 597

2 22. ( )( . ) ( ) . ( )( . )

. .

+ÈÎ ˘̊ = +ÈÎ ˘̊

=

rad/s w

22 31 7482 2w w; .= rad/s

(a) Angular velocity. w2 31 7= . rad/s �

Kinetic energy. T I m v m vA C D C R1 12

12

121

2

1

2

1

2= + +w ( ) ( )

= + +

= += ◊

=

1

20 5 12

1

23 0 6 12 0

36 77 76

113 76

1

2

2 2

1

2

( . )( ) ( )( . ) ( )

.

.

2

N mT

T II m v m vR B Rw22

22

2 22

2

1

2

1

2

1

20 5 31 748

1

23 0 18

+ +

=

+

( ) ( )

( . )( . )

( )( . ) (

2

331 7481

23

300 9695 1 5

222

2 22

. ) ( )( )

. . ( )

+

= +

v

T v

R

r

Principle of conservation of energy: T V T V1 1 2 2+ = +

Recall: V V1 2240 5 0564= ◊ = ◊N m and N m.

113 76 240 300 9695 1 5 5 0564

47 7341 1 5

22

22

. . . ( ) .

. .

+ = + +

=

v

vr

r

(b) Velocity of collar relative to rod. ( ) .vr 2 5 64= m/s �

140


PROBLEM 17.92

A uniform rod AB, of mass 7.5 kg and length 1 m is attached to the 12.5-kg

cart C. Knowing that the system is released from rest in the position shown and

neglecting friction, determine (a) the velocity of Point B as rod AB passes through

a vertical position (b) the corresponding velocity of cart C.

SOLUTION

Kinematics v vC A=

v vC ABÆ = Æ + ¨( . )0 5 m w

v vC AB= - 0 5. w (1)

AB = 1 m

Mass. m mC AB= =12 5 7 5. .kg, kg

Kinetics

Linear momentum

+ 0 = +m v m vC C AB AB

vmm

v v v vABC

ABC C AB C= - = - = -

( . ),

12 5 5

3

kg

(7.5 kg) (2)

Substitute into Eq. (1): v vC C= - -5

30 5. w

8

30 5vC = - . w vC = - 0 1875. w (3)

Substitute into Eq. (2): vAB = - -5

30 1875( . )w

vAB = 0 3125. w (4)

141



Kinetic and potential energies.

TV W b

V

T m

AB

1

1

2

2

0

7 5 9 81 0 5 1 30

4 9286

0

1

2

== = - ∞= ◊=

=

( . )( . )( . )( cos )

. N m

CC C AB AB ABv m v I2 2 2

2

1

2

1

2

1

212 5 0 1875

1

27 5 0 3125

+ +

= - +

w

w( . )( . ) ( . )( . ww w

w

) ( . )( )

( . . . )

.

2 2 2

2

1

2

1

127 5 1

0 21973 0 36621 0 3125

0 8

+ ÈÎÍ

˘˚̇

= + +

= 99844 2w


0 4 9286 0 89844

5 4857 2 3422

2

2

+ =

= =

. .

. .

w

w w rad/s

(b) Velocity of C: Eq. (3) vC = - 0 1875. w vC = ¨0 439. m/s �

(a) Velocity of B: vB Cv= + Æ = ¨ + Æ( ) ( . ) [( )( . )] ]1 0 43916 1 2 3422w m/s

vB = ¨ + Æ0 43916 2 3422. . vB = Æ1 903. m/s �

142


PROBLEM 17.93

In Problem 17.83, determine the velocity of rod AB relative to cylinder DE as end B of the rod strikes the end

E of the cylinder.

SOLUTION

Kinematics and geometry.

vv

1 1

1

0 04 0 4

1 6

= ==

( . ( .

.

m) m)(40 rad/s)

m/s

w v2 0 28= ( . m) 2w

Initial position Final position


+ Moments about C: I AB = = ◊1

123 0 9 0 16( )( . . kg m) kg m2 2

I mv I I mv IAB DE AB DEw w w w1 1 1 2 2 20 04 0 28

0 16

+ + = + +

◊

( . ( .

( .

m) m)

kg m2 ))( ( ( . ( .40 8 0 04 0 025 rad/s) kg)(1.6 m/s) m) kg m )(40 rad2+ + ◊ //s)

kg m kg)(0.28 kg m22

2= ◊ + + ◊( . ) ( )( . ) ( . )

(

0 16 3 0 28 0 025

6

2 2w w w.. . . ) ( . . . )

. . ; .

4 0 192 1 00 0 16 0 2352 0 025

7 592 0 4202 18

2

2 2

+ + = + += =

ww w 0068 18 072 rad/s: rad/sw = .

Conservation of energy ( ) .vr = 0 075 m/s

V V

T I I m v m vDE AB AB AB r

1 2

1 12

12

12

12

0

1

2

1

2

1

2

1

2

1

20 025

= =

= + + +

=

w w ( )

( . kgg m rad/s kg m rad/s

kg m/s

2 2◊ + ◊

+ +

)( ) ( . )( )

( )( . )

401

20 16 40

1

23 1 6

2 2

2 11

33 0 075( )( .kg m/s)2

143



Tv

1

2 2

20 120 3 84 0 008 151 85

0 28 0 28 18 06

= + + + == +

J J J J J

m m

. . .

( . ) ( . )( .w 88 5 059

1

2

1

2

1

2

1

22 2

222

22

22

rad/s m/s) .

( )

=

= + + +T I I m v m vDE AB AB AB rw w

== ◊

+ ◊

1

20 025 18 068

1

20 16 18 068

2( . )( . )

( . )( .

kg m rad/s

kg m rad/s

2

2 ))

( )( . ) ( )( )

. . .

2

222

2

1

23 5 059

1

23

4 081 26 116 38 3

= +

= + +

kg m/s kg

J J

v

T

r

991 1 5

68 587 1 5

151 85 0 68

22

2 22

1 1 2 2

J

J

J

+

= +

+ = + + =

. ( )

. . ( )

: .

v

T v

T V T V

r

r

.. . ( )

. . ( )

587 1 5

83 263 1 5

22

22

J +

=

v

vr

r

Velocity of rod relative to cylinder. ( ) .vr 2 7 45= m/s �

144


PROBLEM 17.94

In Problem 17.81 determine the velocity of the tube relative to the rod as the

tube strikes end E of the assembly.

PROBLEM 17.81 A 1.6-kg tube AB can slide freely on rod DE which in turn

can rotate freely in a horizontal plane. Initially the assembly is rotating with

an angular velocity w = 5 rad/s and the tube is held in position by a cord. The

moment of inertia of the rod and bracket about the vertical axis of rotation is

0 30. kg m2◊ and the centroidal moment of inertia of the tube about a vertical

axis is 0 0025 2. . kg m◊ If the cord suddenly breaks, determine (a) the angular

velocity of the assembly after the tube has moved to end E, (b) the energy lost

during the plastic impact at E.

SOLUTION

Let Point C be the intersection of axle C and rod DE. Let Point G be the mass center of tube AB.

Masses and moments of inertia about vertical axes.

m I IAB AB DCE= = ◊ = ◊1 6 0 0025 0 30. . , .kg, kg m kg m2 2

State 1. ( ) ( ) . ( )/r vG A r1 1 1

1

2125 62 5 5 0= = = =mm, rad/s,w

State 2. ( ) . . ,/rG A 2 2500 62 5 437 5= - = =mm, w w v vr r= =( )2 0

Kinematics. ( ) /v v rG G Cq q w= =


145



Moments about C:

I I m v r I I m v rAB DCE AB G C AB DCE AB G Cw w w wq q1 1 1 1 2 2 20+ + + = + +( ) ( ) ( ) ( )/ / 22

12

1 22

2

0

I I m r I I m rAB DCE AB G C AB DCE AB G C+ +ÈÎ ˘̊ = + +ÈÎ ˘̊( ) ( )

[ .

/ /w w

00025 0 30 1 6 0 0625 5 0 0025 0 30 1 6 0 43752 2+ + = + +. ( . )( . ) ]( ) [ . . ( . )( . ) ]www w

2

2 20 30875 5 0 60875 2 5359( . )( ) . .= = rad/s

Kinetic energy. T I I m vAB DCE AB= + +1

2

1

2

1

2

2 2 2w w

= + + +( )= +

1

2

1

2

1

2

1

20 0025 5

1

20

2 2 2 2 2

12

I I m r v

T

AB DCE AB G C rw w w/

( . )( ) ( .. )( ) ( . )( . ) .

( . )( . )

3 51

21 6 0 0625 0 3 859375

1

20 0025 2 5359

2 2

2

+ + =

=

J

T 22 2

2 2

1

20 30 2 5359

1

21 6 0 0625 2 5359

1

21 6

+

+ +

( . )( . )

( . )( . ) ( . ) ( . )( )vr 222

221 95737 0 8= +. . ( )vr

Work. The work of the bearing reactions at C is zero. Since the sliding contact between the rod and the tube

is frictionless, the work of the contact force is zero.

U1 2 0Æ =

Principle of work and energy. T U T1 1 2 2+ =Æ

3 859375 0 1 95737 0 8 22. . . ( )+ = + vr

Velocity of the tube relative to the rod. ( ) .vr 2 1 542= m/s �

146


PROBLEM 17.95

The 3-kg steel cylinder A and the 5-kg wooden cart B are at rest

in the position shown when the cylinder is given a slight nudge,

causing it to roll without sliding along the top surface of the cart.

Neglecting friction between the cart and the ground, determine

the velocity of the cart as the cylinder passes through the lowest

point of the surface at C.

SOLUTION

Kinematics (when cylinder is passing C )

+ v v r vB C A= = -w

w =+v vr

A B


Syst. of Momenta1 + Syst. Ext. Imp.1Æ2 = Syst. Momenta2

+ x components: m v m vv v v v

A A B B

A B B A

- == =

0

3 5 0 6; .

Work: U W TA1 2 10 15 3 9 81 0 15 4 4145 0Æ = = = ◊ =( . ) ( )( . )( . ) .m kg m/s m N m;2

Kinetic energy: T m v I m vA A B22 2

321

2

1

2

1

2= + +w

v v v vr

v vr

vr

T v r

B AA B A A A

A

= =+

=+

=

= ( ) +

0 60 6 1 6

1

23

1

2

3

22

22

. ;. .

( )

w

kgkgÊÊ

ËÁˆ

¯̃ÊËÁ

ˆ¯̃

+

= + +

1 6 1

25 0 6

1 5 1 92 0 9

22

2 2 2

.( )( . )

. . .

vr

v

v v v

AA

A A A

kg

== 4 32 2. vA

Principle of work and energy: T U T1 1 2 2+ =Æ

0 4 4145 4 32

1 021875 1 01088

2

2

+ =

= = Æ

. .

. .

v

vA

A Av m/s

v vB A= =0 6 0 6 1 01088. . ( . ) vB = ¨0 607. m/s �

147


PROBLEM 17.96

A bullet weighing 40 gm is fired with a horizontal velocity of 550 m/s into

the lower end of a slender 7.5-kg bar of length L = 800 mm. Knowing that

h = 300 mm and that the bar is initially at rest, determine (a) the angular velocity

of the bar immediately after the bullet becomes embedded, (b) the impulsive

reaction at C, assuming that the bullet becomes embedded in 0.001 s.

SOLUTION

Bar: L m

I mL

= = =

= = = ◊

800 0 8 7 5

1

12

1

127 5 0 8 0 42 2

mm m kg

kg m2

. .

( . )( . ) .

Bullet: m0 40 0 04= =g kg.

Support location: h = =300 0 3mm m.

Kinematics. v L h

v L h

B

G

= - = - =

= -ÊËÁ

ˆ¯̃

= - =

( ) ( . . ) .

( . . ) .

w w w

w w w

0 8 0 3 0 5

20 4 0 3 0 1

Kinetics.


Moments about C: m v L h m v L h mv L h IB G0 0 02

( ) ( )- = - + -ÊËÁ

ˆ¯̃

+ w

( . )( )( . ) ( . )( . )( . ) ( . )( . )( . ) ( . )0 04 550 0 5 0 04 0 5 0 5 7 5 0 1 0 1 0 4= + +w w w

148



(a) 11 0 01 0 075 0 4 0 485 22 6804= + + = =. . . . .w w w w wor w = 22 7. rad/s �

vv

B

G

= == =

( . )( . ) .

( . )( . ) .

0 5 22 6804 11 3402

0 1 22 6804 2 26804

m/s

m/s

+ Horizontal components:

- + = - - = - -= -

m v C t m v mv C t m v v mvC t

B G B0 0 0 0 0 0

0 04 550

( ) : ( ) ( )

( ) ( . )(

D DD 111 3402 7 5 2 26804

4 53609

. ) ( . )( . )

.

-= ◊N s

(b) C C tt

= =DD

4 53609

0 001

.

. C = Æ4540 N �

149


PROBLEM 17.97

In Problem 17.96, determine (a) the required distance h if the impulsive reaction

at C is to be zero, (b) the corresponding angular velocity of the bar immediately

after the bullet becomes embedded.

SOLUTION

Bar: L m

I mL

= = =

= = = ◊

800 0 8 0 75

1

12

1

127 5 0 8 0 42 2

mm m kg

kg m2

. .

( . )( . ) .

Bullet: m0 0 04= . kg

Kinematics. v L h h

v L h h

B

G

= - = -

= -ÊËÁ

ˆ¯̃

= -

( ) ( . )

( . )

w w

w w

0 8

20 4

Kinetics.


moment about B: 0 02

+ = - ÊËÁ

ˆ¯̃

I mv LGw

0 0 0 4 7 5 0 4 0 4+ = - -. ( . )( . ) ( . )w wh

Divide by w 0 0 4 3 0 4= - -. ( . )h

150



(a) h = =4

150 26667m m. h = 267 mm �

v

v

B

G

= -ÊËÁ

ˆ¯̃

=

= -ÊËÁ

ˆ¯̃

=

0 84

15

8

15

0 44

15

2

15

.

.

w w

w w

+ Horizontal components: m v mv m vG B0 0 00+ = +

( . )( ) ( . ) ( . )0 04 550 0 7 52

150 04

8

15+ = Ê

ËÁˆ¯̃

+ ÊËÁ

ˆ¯̃

w w

(b) w = 21 541. w = 21 5. rad/s �

151


PROBLEM 17.98

A 45-g bullet is fired with a velocity of 400 m/s at q = ∞30 into a 9-kg

square panel of side b = 200 mm. Knowing that h = 150 mm and that the

panel is initially at rest, determine (a) the velocity of the center of the panel

immediately after the bullet becomes embedded, (b) the impulsive reaction

at A, assuming that the bullet becomes embedded in 2 ms.

SOLUTION

m m I m bB P G P= = = = = ◊0 045 91

6

1

69 0 200 0 062 2. ( )( . ) .kg kg kg m2

Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity w = w .

vGb= Æ2

w .

Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.


(a) + Moments about A:

( cos ) sin

cos

m v h m v b I m v b

m v h b

B B G P G

B

0 0

0

30 302

02

30

∞ + ∞ÊËÁ

ˆ¯̃

+ = +

∞ +

w

2230

1

4

0 045 400 0 150 30 0

2sin

( . )( )( . cos .

∞ÊËÁ

ˆ¯̃

= +ÊËÁ

ˆ¯̃

∞ +

I m bG P w

1100 30

0 061

49 0 2 0 15

21 588

2

sin )

. ( )( . ) .

.

∞

= +ÈÎÍ

˘˚̇

=

=

w w

w rad/s

vB = =( . )( . ) .0 100 21 556 2 1556 m/s vG = Æ2 16. m/s �

152



(b) + Linear momentum: m v A t m v

AB x P G

x

0 30

0 045 400 30 0 002 9 2 1

cos ( )

( . )( cos ) ( . ) ( )( .

∞ + =

∞ + =

D

5556)

A X = 1920 N A x = Æ1920 N

Linear momentum: - ∞ + =m v A tB y0 30 0sin ( )D

- ∞ + =( . )( )sin ( . )0 045 400 30 0 002 0Ay

Ay = 4500 N A y = ≠4500 N

` A = = = = ∞4892 4 8924500

192066 9N kW. tan .b b

A = 4 87. kN 66.9∞�

153


PROBLEM 17.99

A 45-g bullet is fired with a velocity of 400 m/s at q = ∞5 into a 9-kg square

panel of side b = 200 mm. Knowing that h = 150 mm and that the panel

is initially at rest, determine (a) the required distance h if the horizontal

component of the impulsive reaction at A is to be zero, (b) the corresponding

velocity of the center of the panel immediately after the bullet becomes

embedded.

SOLUTION

m m I m bB P G P= = = = = ◊0 045 91

6

1

69 0 200 0 062 2. ( )( . ) .kg kg kg m2

Kinematics. After impact, the plate is rotating about the fixed Point A with angular velocity w = w .

vGb= Æ2

w .

Principle of impulse and momentum. To simplify the analysis, neglect the mass of the bullet after impact.

Also A tX ( ) .D = 0


+ Linear momentum: m v m v m bB P G P0 5 0

2cos ∞ + = = Ê

ËÁˆ¯̃

w

( . )( cos ) ( )( . ) .0 045 400 5 9 0 100 19 9239∞ = =w w rad/s

vG = =( . )( . ) .0 100 19 9239 1 99239 m/s (1)

+ Moments about A: ( cos ) ( sin )m v h m v b I m v bB B G P G0 05 5

2 2∞ + ∞ = +w

m v b b I m b

h

B G P025

25

1

4

0 045 400

cos sin

( . )( )( cos

∞ + ∞ÊËÁ

ˆ¯̃

= +ÊËÁ

ˆ¯̃

w

55 0 100 5 0 061

49 0 100 19 9239

17 9315

2∞ + ∞ = +ÈÎÍ

˘˚̇

+

. sin ) . ( )( . ) ( . )

. h 11 5688 2 9886. .=

(a) h = 0 07918. m h = 79 2. mm �

(b) From Eq. (1), vG = Æ1 992. m/s �

154


PROBLEM 17.100

A 8-kg wooden panel is suspended from a pin support at A and is initially

at rest. A 2-kg metal sphere is released from rest at B and falls into a

hemispherical cup C attached to the panel at a point located on its top edge.

Assuming that the impact is perfectly plastic, determine the velocity of the

mass center G of the panel immediately after the impact.

SOLUTION

Mass and moment of inertia

m m

I m

s P

P

= =

= = = ◊

2 8

1

60 5

1

68 0 5 0 33332 2

kg kg

m kg m2( . ) ( )( . ) .

Velocity of sphere at C. ( ) ( . )( . ) .v gyC 1 2 2 9 81 0 25 2 2147= = =m/s m m/s2

Impact analysis.

Kinematics Immediately after impact in terms of w2

vvC

2 2

2 2

0 25

0 2

==

.

( ) .

ww



155



+ Moments about A:

m v m v I m vs C s C P( ) ( . ) ( ) ( . ) ( . )

( )( .

1 2 2 20 2 0 0 2 0 25

2 2 2147

m m m

kg m

+ = + +w

//s m kg m kg m)( . ) ( )( . )( . ) . ( )( . )

.

0 2 2 0 2 0 2 0 33333 8 0 25

0 8

2 22

2= + +w w w88589 0 08 0 33333 0 500 2= + +( . . . )w

w2 0 96995 0 96995= =. . rad/s rad/s2w

Velocity of the mass center

v2 20 25 0 25 0 96995= =( . ) ( . )( . )m m rad/sw

v2 0 24249= . m/s v2 242= Æmm/s �

156


PROBLEM 17.101

An 8-kg wooden panel is suspended from a pin support at A and is initially

at rest. A 2-kg metal sphere is released from rest at B¢ and falls into a

hemispherical cup C¢ attached to the panel at the same level as the mass

center G. Assuming that the impact is perfectly plastic, determine the

velocity of the mass center G of the panel immediately after the impact.

SOLUTION

Mass and moment of inertia. mm

I m

S

P

P

==

=

=

= ◊

2

8

1

60 5

1

68 0 5

0 3333

2

2

2

kg

kg

m

kg m

( . )

( )( . )

.

Velocity of sphere at C ¢. ( )

( . )( . )

.

v gyC ¢ =

==

1

2

2

2 9 81 0 5

3 1321

m/s m

m/s

Impact analysis.

Kinematics Immediately after impact in terms of w2.

ACACC

¢ = + == ¢=

¢

( . ) ( . ) .

( )

.

0 2 0 25 0 32016

0 32016

2 2

2 2

2

m

v ww

v2 20 25= . w q (perpendicular to .)AC


Syst. Momenta1 + Syst. Ext. Imp.1-2 = Syst. Momenta2

157



+ Moments about A:

m v m v AC I m vS C S C P( ) ( . ) ( ) ( ) ( . )

( )( .

¢ ¢+ = ¢ + +1 2 2 20 2 0 0 25

2 3 1321

m m

kg

wmm/s m kg m

kg

)( . ) ( )( . )( . )

. ( )( .

0 2 2 0 32016 0 32016

0 3333 8 0 25

2

2

=

+ +

w

w mm

rad/s

)

. ( . . . )

.

22

2

2

1 25284 0 2050 0 3333 0 500

1 2066

ww

w= + +=

Velocity of the mass center. v2 20 25

0 25 1 2066

==

( . )

( . )( . )

m

m rad/s

w

= 0 3016. m/s v2 302= ¨mm/s �

158


PROBLEM 17.102

The gear shown has a radius R = 150 mm and a radius of gyration

k = 125 mm. The gear is rolling without sliding with a velocity v1of magnitude 3 m/s when it strikes a step of height h = 75 mm.

Because the edge of the step engages the gear teeth, no slipping

occurs between the gear and the step. Assuming perfectly plastic

impact, determine the angular velocity of the gear immediately

after the impact.

SOLUTION

Kinematics. Just before impact, the contact point with the rack is the instantaneous center of rotation of the

gear.

v1 1= ¨Rw

Just after impact, Point S is the instantaneous center of rotation

v2 2= Rw q ( )perpendicular to GS


+ Moments about S: mv R h I mv R I2 1 2 2( )- + = +w w

m R R h mk m R R mk

R R h k R k

R

( )( ) ( )

[ ( ) ] ( )

w w w w

w w

w

12

1 22

2

21

2 22

2

- + = +

- + = +

=22 2

2 2 1 2 2 2 11+ -

+= -

+ÈÎÍ

˘˚̇

k RhR k

RhR k

w w w (1)

Data: R k v h= = = =150 125 3 751mm mm m/s mm, , ,

w11 3

0 15020= = =

vR

m/s

mrad/s

.

Angular velocity.

From (1), w2 2 21

150 75

150 12520 0 7049 20= -

+È

ÎÍ

˘

˚˙ =( )( )

( )( ) . ( )rad/s w2 14 10= . rad/s �

159


PROBLEM 17.103

A uniform slender rod AB of mass m is at rest on a frictionless

horizontal surface when hook C engages a small pin at A.

Knowing that the hook is pulled upward with a constant

velocity v0, determine the impulse exerted on the rod (a) at A,

(b) at B. Assume that the velocity of the hook is unchanged

and that the impact is perfectly plastic.

SOLUTION

At the given instant, vB = 0.

Moment of inertia. I mL= 1

12

2

Kinematics. (Rotation about B). w =vL0

v L vG = =2

1

20w

Kinetics.


Moments about B: 02

+ Ú = +L Adt mv L IG w

(a) Ú = + = + =Adtmv mL mv mv mvG

2 12 4 12 3

0 0 0w Ú = ≠Adt

mv0

3 �

Moments about A: 02

+ Ú = -L Bdt mv L IG w

(b) Ú = - = - =Bdtmv mL mv mv mvG

2 12 4 12 6

0 0 0w Ú = ≠Bdt

mv0

6 �

160


PROBLEM 17.104

A uniform slender bar of length L and mass m is supported

by a frictionless horizontal table. Initially the bar is spinning

about its mass center G with a constant angular velocity w1.

Suddenly latch D is moved to the right and is struck by end A

of the bar. Assuming that the impact of A and D is perfectly

plastic, determine the angular velocity of the bar and the

velocity of its mass center immediately after the impact.

SOLUTION


12

2

Before impact. ( )v LA 1 1

2= Øw

Impact condition. ( ) ( )v e v eLA A2 1 1

1

2= - = ≠w

Kinematics after impact. v v L eL LA2 2 2 1 22

1

2

1

2= + = +( ) w w w

Principle of impulse-momentum at impact.


Moments about D: I I mv L

I I m eL L L

mL mL

w w

w w w w

w

1 2 2

1 2 1 2

21

02

1

2

1

2 2

1

12

1

12

+ = +

= + +ÊËÁ

ˆ¯̃

= 222

21

22

2 1

2 1

1

4

1

4

1

41 3

1

2

1

2

1

41

w w w

w w

w

+ +

= -

= + ÊËÁ

ˆ¯̃

-

mL e mL

e

v Le L

( )

( 331

811 1e e L) ( )w w= +

For perfectly plastic impact, e = 0 w2 1

1

4= w �

v2 1

1

8= ≠Lw �

161


PROBLEM 17.105

Solve Problem 17.104, assuming that the impact of A and D

is perfectly elastic.

PROBLEM 17.104 A uniform slender bar of length L and

mass m is supported by a frictionless horizontal table.

Initially the bar is spinning about its mass center G with a

constant angular velocity w 1. Suddenly latch D is moved

to the right and is struck by end A of the bar. Assuming that

the impact of A and D is perfectly plastic, determine the

angular velocity of the bar and the velocity of its mass center

immediately after the impact.

SOLUTION


12

2

Before impact. ( )v LA 1 1

2= Øw

Impact condition. ( ) ( )v e v eLA A2 1 1

1

2= - = ≠w

Kinematics after impact. v v L eL LA2 2 2 1 22

1

2

1

2= + = +( ) w w w



Moments about D: I I mv L

I I m eL L L

mL mL

w w

w w w w

w

1 2 2

1 2 1 2

21

02

1

2

1

2 2

1

12

1

12

+ = +

= + +ÊËÁ

ˆ¯̃

= 222

21

22

2 1

2 1

1

4

1

4

1

41 3

1

2

1

2

1

41

w w w

w w

w

+ +

= -

= + ÊËÁ

ˆ¯̃

-

mL e mL

e

v Le L

( )

( 331

811 1e e L) ( )w w= +

162



For perfectly elastic impact, e = 1

w w w2 1 1

1

41 3

1

2= - = -( ) w2 1

1

2= w �

v L L L2 1 1 1

1

2

1

2

1

2

1

4= + -Ê

ËÁˆ¯̃

=w w w v2 1

1

4= ≠Lw �

163


PROBLEM 17.106

A uniform slender rod of length L is dropped onto rigid supports at A

and B. Since support B is slightly lower than support A, the rod strikes

A with a velocity v1 before it strikes B. Assuming perfectly elastic

impact at both A and B, determine the angular velocity of the rod and

the velocity of its mass center immediately after the rod (a) strikes

support A, (b) strikes support B, (c) again strikes support A.

SOLUTION


12

2

(a) First Impact at A.


Condition of impact: e vA= = ≠1 2 1: ( )v

Kinematics: v L v L vA2 2 12 2

= - = -w w( )

Moments about A: mv L mv L I1 2 22

02

+ = + w

= -ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

m L v L mL2 2

1

121

22w w w2

13=

vL

�

v L vL

v v21

1 12

3 1

2= Ê

ËÁˆ¯̃

- = v2 1

1

2= Øv �

( ) ( )v L v v v vB A2 2 1 1 13 2= - = - = Øw

(b) Impact at B.


Condition of impact. e vB= = ≠1 23 1: ( )v

Kinematics: v v L v LB3 2 1

22

2= - = -( ) w w

164



Moments about B: - + + = -mv L I mv L I2 2 3 32

02

w w

- ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

+ = -ÊËÁ

ˆ¯̃

-m v L mL vL

m v L L1

2 2

1

12

30 2

2 21

2 11 3w 11

12

23mLÊ

ËÁˆ¯̃

w w313

=vL

�

v v L vL

v3 11

122

3 1

2= - Ê

ËÁˆ¯̃

= v3 1

1

2= ≠v �

( ) ( )v L v v v vA B3 3 1 1 13 2= - = - = Øw

(c) Second Impact at A.


Condition of impact. e vA= = ≠1 4 1: ( )v

Kinematics: v v L v LA4 4 4 1 4

2 2= + = +( ) w w

Moments about A: mv L I mv L I3 3 4 42

02

+ + = +w w

m v L mL vL

m v L L1

2 2

1

12

30

2 2

1

11

2 11 4

ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

+ = +ÊËÁ

ˆ¯̃

+w22

24mLÊ

ËÁˆ¯̃

w w4 0= �

v v4 1 0= + v4 1= ≠v �

165


PROBLEM 17.107

A uniform slender rod AB is at rest on a frictionless horizontal

table when end A of the rod is struck by a hammer which delivers

an impulse that is perpendicular to the rod. In the subsequent

motion, determine the distance b through which the rod will

move each time it completes a full revolution.

SOLUTION


12

2

Kinetics.


Moments about A: 0 02

+ + +I mv Lw

v ImL

mLmL

L= = =2 2 1

6

112

2w ww

Motion after impact. q w qw

pw

= = =t t 2

b vt L= = ÊËÁ

ˆ¯̃

1

6

2w pw

b L= p3

�

166


PROBLEM 17.108

A uniform sphere of radius r rolls down the incline shown without slipping.

It hits a horizontal surface and, after slipping for a while, it starts rolling

again. Assuming that the sphere does not bounce as it hits the horizontal

surface, determine its angular velocity and the velocity of its mass center

after it has resumed rolling.

SOLUTION


5

2

Kinematics.

Before After

Before impact (rolling).

v r1 1= w

After slipping has stopped.

v r2 2= w

Kinetics.


Moments about C: I mv r I mv r

I mr I mr

w b w

w w b w w1 1 2 2

12

1 22

2

0+ + = +

+ = +

cos

cos

w b wb

w2

2

2 1

25

2 2

25

2 2 1= ++

=+

+I mr

I mrmr mr

mr mrcos cos

w2 1

1

72 5= +( cos )b w �

v r r2 2 1

2 5

7= = +w b wcos

v2 1

1

72 5= + ¨( cos )b v �

167


PROBLEM 17.109

The slender rod AB of length L forms an angle b with the vertical as it strikes the

frictionless surface shown with a vertical velocity v1 and no angular velocity.

Assuming that the impact is perfectly elastic, derive an expression for the angular

velocity of the rod immediately after the impact.

SOLUTION


12

2

Perfectly elastic impact. e v e v evv v v

A y A y

A A x A y A x

= = - = ≠

= + =

1 2 1 1 [( ) ] [( ) ]

( ) ( ) ( )v i j i ++ v1j

Kinetics.


+ horizontal components: 0 0 0+ = =mv mvx x

Kinematics. v v vG A G A y A xv v v L= + ≠ = ≠ + Æ + ÈÎÍ/ [ ] [ ] [( ) ] 1

2w b˘

˚̇

Velocity components : v v Ly = -1

2w bsin

Moments about A: mv L mv L I

mv L m L v L

y1

1 1

20

2

2 2 2

sin sin

sin sin sin

b b w

b w b b

+ = - +

= -ÊËÁ

ˆ¯̃ ++ 1

12

2mL w

1

12

1

4

2 2 21mL mL mv L+Ê

ËÁˆ¯̃

=sin sinb w b w bb

=+12

1 3 2

1sin

sin

vL

�

168


PROBLEM 17.110

Solve Problem 17.109, assuming that the impact between rod AB and the

frictionless surface is perfectly plastic.

SOLUTION


12

2

Perfectly plastic impact. e v e vv v v

A y A y

A A x A y A x

= = - =

= + =

0 02 1 [( ) ] ( )

( ) ( ) ( )v i j i

Kinetics.


+ horizontal components: 0 0 0+ = =mv mvx x

Kinematics. v v vG A G A y A xv v L= + ≠ = Æ + + ÈÎÍ/ [ ] [( ) ]

2w b˘

˚̇

Velocity components ≠: v Ly = -

2w bsin

Moments about A: mv L mv L I

mv L m L L

y1

1

20

2

2 2 2

1

1

sin sin

sin sin sin

b b w

b w b b

+ = - +

= ÊËÁ

ˆ¯̃

+22

2mL w

1

12

1

4

1

2

2 2 21mL mL mv L+Ê

ËÁˆ¯̃

=sin sinb w b w bb

=+6

1 3 2

1sin

sin

vL

�

169


PROBLEM 17.111

A uniformly loaded rectangular crate is released from rest in the

position shown. Assuming that the floor is sufficiently rough

to prevent slipping and that the impact at B is perfectly plastic,

determine the smallest value of the ratio a/b for which corner A will remain in contact with the floor.

SOLUTION

We consider the limiting case when the crate is just ready to rotate about B. At that instant the velocities must

be zero and the reaction at corner A must be zero. Use the principle of impulse and momentum.


+ Moments about B:

I mv b mv ayw1 1 2 1

2 20 0+ - + =( ) ( ) (1)

Note: sin , cosf f=+

=+

b

a b

a

a b2 2 2 2

v AG v a b1 12 2

1

1

2= = +( ) w

Thus: ( ) ( )sinmv mv m a b b

a bmbx1 1

2 21

2 21

2

1

2= = +

+=f w w

Also, ( ) ( )cosmv mv may1 1

1

2= =f w

I m a b= +1

12

2 2( )

From Eq. (1) 1

12

1

2 2

1

2 202 2

1 1 1m a b mb b ma a( ) ( ) ( )+ + - =w w w

1

3

1

602

12

1mb ma vw - = ab

ab

ab

2

22 2 1 414= = = . �

170


PROBLEM 17.112

A uniform slender rod AB of length L is falling freely

with a velocity v0 when cord AC suddenly becomes

taut. Assuming that the impact is perfectly plastic,

determine the angular velocity of the rod and the

velocity of its mass center immediately after the cord

becomes taut.

SOLUTION

Immediately after impact

tan .q q= = ∞1

226 565

Due to constraint of inextensible cord,

vA Av= q

Kinematics: v v vG A G A= + /

= vA q w+ ØL2

(1)



Components q : mv mv mLA0 0

2cos cosq w q+ = +

v L vA + ÊËÁ

ˆ¯̃

=1

20cos ( ) cosq w q (2)

+ Moments about A: mv L mv L mL L IA0

20

2 2 2+ = + Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

+( cos )q w w

L v L L L vA2

1

4

1

12 2

2 20cosq wÊ

ËÁˆ¯̃

+ +ÊËÁ

ˆ¯̃

= (3)

Solving Eqs. (2) and (3) simultaneously (with q = ∞26 565. )

v v L vA = =0 55902 0 750 0. .w

Angular velocity after impact. w = 0 750 0.vL

�

171



Velocity of mass center. v v vG A G A= + /

vG v= 0 55902 0. q + Ø1

20 75 0( . )v

= ¨ + Ø + Ø0 25 0 5 0 3750 0 0. . .v v v

= ¨ + Ø0 25 0 8750 0. .v v vB v= 0 910 0. 74 1. ∞ �

172


PROBLEM 17.113

A uniform slender rod AB of length L is falling freely with a velocity v0

when cord AC suddenly becomes taut. Assuming that the impact is perfectly

plastic, determine the angular velocity of the rod and the velocity of its mass

center immediately after the cord becomes taut.

SOLUTION

Immediately after impact,

tan .q q= = ∞1

226 565

Due to constraint of inextensible cord,

vA Av= q

Kinematics. v v vG A G A= + /

= vA q w+ ÆL2



Components q : mv mv mLA0 0

2cos sinq w q+ = -

v L vA - ÊËÁ

ˆ¯̃

=1

20sin ( ) cosq w q (2)

173



+ Moments about A: 0 0

2 2 2+ = - + Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

+( sin )mv L mL L IA q w w

- ÊËÁ

ˆ¯̃

+ +ÊËÁ

ˆ¯̃

=L v L LA2

1

4

1

1202 2sinq w (3)

Solving Eqs. (2) and (3) simultaneously (with q = ∞26 565. ),

v v L vA = =1 05527 0 705880 0. .w

Angular velocity after impact. w = 0 706 0.vL

�

Velocity of mass center. v v vG A G A= + /

vG v= 1 055227 0. q + Æ1

20 70588 0( . )v

= ¨ + Ø + Æ0 47059 0 94118 0 352940 0 0. . .v v v

= ¨ + Ø0 11765 0 941180 0. .v v vG v= 0 949 0. 82 9. ∞ �

174


PROBLEM 17.114

A slender rod of length L and mass m is released from rest in the position

shown. It is observed that after the rod strikes the vertical surface it

rebounds to form an angle of 30° with the vertical. (a) Determine the

coefficient of restitution between knob K and the surface. (b) Show that

the same rebound can be expected for any position of knob K.

SOLUTION

For analysis of the downward swing of the rod before impact and for the upward swing after impact use the

principle of conservation of energy.

Before impact.

V

V W L mg L

T

T I mv mL v

1

2

1

2 22

22 2

22

0

2 2

0

1

2

1

2

1

2

1

12

1

2

=

= - = -

=

= + = ÊËÁ

ˆ¯̃

+w mm mL1

2

1

62

22

22w wÊ

ËÁˆ¯̃

=

T V T V mL mg L gL1 1 2 2

222

22

23+ = + = = - =: 0

1

6w w; w2 1 73205= .

gL

After impact.

V W L mg L3

2 2= - = -

175



V W L

T I mv

mL m

4

3 32

32

232

3

230

1

2

1

2

1

2

1

12

1

2

1

2

= - ∞

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

cos

w

w w ˆ̂¯̃

=

=

+ = + - = - ∞

2

232

4

3 3 4 42

32

1

6

0

1

6 20

230

mL

T

T V T V mL m L mg L

w

w q: cos

w32 1 30= - ∞( cos )

gL

w3 0 63397= .gL

Analysis of impact.

Let r be the distance BK.

Before impact, ( ) .vk b b gL3 2 1 73205= Æ = Æw

After impact, ( ) .vk b b gL4 3 0 63397= ¨ = ¨w

Coefficient of restitution. evv

k n

k n=

| |

|( |

( )

)

4

3

e = 0 63397

1 73205

.

. e = 0 366. �

(b) Clearly the answer is independent of b.

176


PROBLEM 17.115

The uniform rectangular block shown is moving along a

frictionless surface with a velocity v1 when it strikes a

small obstruction at B. Assuming that the impact between

corner A and obstruction B is perfectly plastic, determine

the magnitude of the velocity v1 for which the maximum

angle q through which the block will rotate is 30∞.

SOLUTION

Let m be the mass of the block.

Dimensions: ab

= == =

250 0 25

125 0 125

mm m

mm m

.

.

Moment of inertia about the mass center.

I m a b= +1

12

2 2( )

Let d be one half the diagonal. d a b= + = =1

2

5

160 139752 2 m m.

Kinematics. Before impact v v1 1 1 0= Æ =, w

After impact, the block is rotating about corner at B.

w2 2= w v d2 2= w



+ Moments about B: mv b I mdv

mv b m a b md

m a b

12 2

12 2

22

2

2 22

20

1

2

1

12

1

3

+ = +

= + +

= +

w

w w

w

( )

( )

177



Angular velocity after impact w21

2 2

3

2=

+v b

a b( ) (1)

The motion after impact is a rotation about corner B.

Position 2 (immediately after impact). v d2 2= w

Position 3 ( ).q = ∞30 b

b

= = = ∞

= + ∞ = ∞ =

- -tan tan . .

sin( ) . sin .

1 1 0 5 26 565

30 0 13975 56 565

ba

h d 00 11663

0 03 3

. m

w = =v

Potential energy: V mgb V mgh2 32

= =

Kinetic energy: T I mv I md

m a b T

2 22

22 2

22

2 222

3

1

2

1

2

1

2

1

60

= + = +

= + =

w w

w

( )

( )


T V T V

m a b mgb mgh

g h ba b

2 3 3 3

2 222

22

2 2

1

6 20

3 2 3

+ = +

+ + = +

= -+

=

( )

( )

( )

( )

w

w (( . )( . . )

( . ) ( . )

9 81 0 23325 0 125

0 25 0 1252 2

-+

= 40 7794. (rad/s)2 w2 6 3859= . rad/s

Magnitude of initial velocity.

Solving Eq. (1) for v1 v a bb1

2 222

3=

+( )w

v1

2 22 0 25 0 125 6 3859

3 0 125=

+ÈÎ ˘̊( ) ( . ) ( . ) ( . )

( )( . ) v1 2 66= . m/s �

178


PROBLEM 17.116

A slender rod of mass m and length L is released from rest in the position

shown and hits edge D. Assuming perfectly plastic impact at D, determine

for b L= 0 6. , (a) the angular velocity of the rod immediately after the impact,

(b) the maximum angle through which the rod will rotate after the impact.

SOLUTION

For analysis of the falling motion before impact use the principle of conservation of energy.

Position 1: T V mg L1 10

4= =,

Position 2: V2 0=

T m L mL

T mL

T V T V

2 2

22

22

22

22

1 1 2 2

1

2 2

1

2

1

12

1

6

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

+ = +

w w

w

: 004

1

6

3

2

222

2+ = =mg L mL gL

w w

Analysis of impact. Kinematics

Before impact, rotation is about Point A. v L2 2

2= w

After impact, rotation is about Point D. v L3 3

10= w

Principle of impulse-momentum.


+ Moments about D: I mv L I mv Lw w2 2 3 3

10 10- Ê

ËÁˆ¯̃

= + ÊËÁ

ˆ¯̃ (1)

1

12 2 10

1

12 10 10

1

12

1

20

22 2

23 3mL m L L mL m L Lw w w w- Ê

ËÁˆ¯̃

= + ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ̂¯̃

= +ÊËÁ

ˆ¯̃

w w2 3

1

12

1

100

179



(a) Angular velocity. w w3 2

5

14

5

14

3

2= = g

L w3 0 437= .

gL

�

For analysis of the rotation about Point D after the impact use the principle of conservation of energy.

Position 3. (Just after impact)

v L V

T m L mL mL

3 3 3

3 3

22

32 2

100

1

2 10

1

2

1

12

14

300

= =

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

w

w w w332

2

214

300

5

14

2

2 112=

Ê

ËÁˆ

¯̃=mL g

LmgL

Position 4. q = maximum rotation angle.

¢ =

= ¢ =

= = =

+ = +

h L

V mgh mgL

v T

T V T V mgL

10

10

0 0 0

11

4

4 4 4

3 3 4 4

sin

sin

, ,

;

q

q

w

220 0

10+ = + mgL

sinq

(b) Maximum rotation angle. sinq = 10

112 q = ∞5 12. �

180


PROBLEM 17.117

A 30-g bullet is fired with a horizontal velocity of 350 m/s into the 8-kg wooden

beam AB. The beam is suspended from a collar of negligible mass that can

slide along a horizontal rod. Neglecting friction between the collar and the rod,

determine the maximum angle of rotation of the beam during its subsequent

motion.

SOLUTION

Mass of bullet. ¢ = =m g30 0.03 kg

Mass of beam AB. m = 8 kg

Mass ratio. b b= ¢ = ¢ =mm

m m0 00375.

Since b is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining

the motion after the impact.


12

2

Impact kinetics.


+ linear components: - + = =b bmv mv v v0 2 2 00

Moments about B: 0 02

2+ = -I mv Lw

wb

wb

= =

=

mv LI

m v LmL

vL

2 0

2

0

2

12

2

6

181



Motion during rising. Position 2. Just after the impact.

V mg L A

T mv I

m v

2

2 22

22

02

2

1

2

1

2

1

2

1

2

1

12

= -

= +

= +

( )

( )

datum at level

w

b mmLv

L

mv

2 02

202

6

2

ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

b

b

Position 3. w q q= =0, . m


V mg L

T mv

m3

3 32

2

1

2

= -

=

cosq

+ linear components: mv mv v v v2 3 3 2 00+ = = = b


T V T V mv mg L m v mg L

vgL

m2 2 3 32

02

02

202

22

1

2 2

31

+ = + - = -

= -

: ( ) cos

cos

b b q

bqq

qb

m

mv

gLcos

( )( . ) ( )

( . )( . )

= -

= -

13

13 0 00375 350

9 81 1 2

202

2 2

= 0 56099. qm = ∞55 9. �

182


PROBLEM 17.118

For the beam of Problem 17.117, determine the velocity of the 30-g bullet for

which the maximum angle of rotation of the beam will be 90∞.

PROBLEM 17.117 A 30-g bullet is fired with a horizontal velocity of 350 m/s

into the 8-kg wooden beam AB. The beam is suspended from a collar of negligible

weight that can slide along a horizontal rod. Neglecting friction between the

collar and the rod, determine the maximum angle of rotation of the beam during

its subsequent motion.

SOLUTION

Mass of bullet. ¢ = =m g30 0 03. kg

Mass of beam AB. m = 8 kg

Mass ratio. b b= ¢ = ¢ =mm m m0 00375.

Since b is so small, the mass of the bullet will be neglected in comparison with that of the beam in determining

the motion after the impact.


12

2

Impact Kinetics.


+ linear components: - + = =b bmv mv v v0 2 2 00

Moments about B: 0 02

2+ = -I mv Lw

wb

wb

= =

=

mv LI

m v LmL

vL

2 0

2

0

2

12

2

6

183



Motion during rising. Position 2. Just after the impact.

V mg L A

T mv I

m v

2

2 22

22

02

2

1

2

1

2

1

2

1

= -

= +

= +

datum at level ( )

( )

w

b22

1

12

6

2

2 02

202

mLv

L

mv

ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

b

b

Position 3. w q q= =0, . &m


V mg L

T mv

m3

3 32

2

1

2

= -

=

cosq

+ linear components: mv mv v v v2 3 3 2 00+ = = = b


T V T V mv mg L m v mg L

v gL

m2 2 3 32

02

02

0

22

1

2 2

1

31

+ = + - = -

= -

: ( ) cos

( cos

b b q

b qqm )

( . )( . )( cos )

.

= ÊËÁ

ˆ¯̃

- ∞

=

1

39 81 1 2 1 90

1 98091 m/s

v0

1 98091

0 00375= .

. v0 528= m/s �

184


PROBLEM 17.119

A uniformly loaded square crate is released from rest with

its corner D directly above A; it rotates about A until its

corner B strikes the floor, and then rotates about B. The floor

is sufficiently rough to prevent slipping and the impact at B

is perfectly plastic. Denoting by w 0 the angular velocity of

the crate immediately before B strikes the floor, determine

(a) the angular velocity of the crate immediately after B strikes

the floor, (b) the fraction of the kinetic energy of the crate lost

during the impact, (c) the angle q through which the crate will

rotate after B strikes the floor.

SOLUTION

Let m be the mass of the crate and c be the length of an edge.

Moment of inertia I m c c mc= + =1

12

1

6

2 2 2( )


Kinematics: v r c

v r c

G A

G B

0 0 0

1

22

1

22

= =

= =

/

/

w w

w w

Moments about B: I I r mv

mc mc c m c m

G Bw w

w w w

0

20

2

0

1

60

1

6

1

22

1

22

2

3

+ = +

+ = + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

/

cc2w

(a) Solving for w , w = 1

40w �

Kinetic Energy.

Before impact: T I mv

mc m c

mc

1 02

02

202

0

2

20

1

2

1

2

1

2

1

6

1

2

1

22

1

3

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

w

w w

w22

185



After impact: T I mv mc m c

mc m

22 2 2 2

2

2 2

1

2

1

2

1

2

1

6

1

2

1

22

1

3

1

3

= + = ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= =

w w w

w cc mc20

22

0

1

4

1

48w wÊ

ËÁˆ¯̃

=

(b) Fraction of energy lost: T T

T1 2

1

13

148

13

11

16

-=

-= -

15

16 �

Conservation of energy during falling. T V T V0 0 1 1+ = + (1)

Conservation of energy during rising. T V T V3 3 2 2+ = + (2)

Conditions: T T T T0 3 2 10 01

16= = =,

V mg c V V mg c V mgh0 1 2 3 3

1

22

1

2= Ê

ËÁˆ¯̃

= = ÊËÁ

ˆ¯̃

=

From Equation (1), T V V mgc1 0 1

1

22 1= - = -( )

From Equation (2), T V V mgh mgc2 3 2 3

1

2= - = -

h ch c3

12

12

32 1

1

16

1

2

1

162 1

--

= = + -( )ÈÎÍ

˘˚̇

(c) From geometry, h c3

1

22 45= + ∞sin ( )q

Equating the two expressions for h3,

sin ( )452 1

2

12

116

12

∞ + =+ -( )

q

45 46 503∞ + = ∞q . q = ∞1 50. �

186


PROBLEM 17.120

A uniform slender rod AB of length L = 800 mm is placed

with its center equidistant from two supports that are located

at a distance b = 200 mm from each other. End B of the rod is

raised a distance h0 100= mm and released; the rod then rocks

on the supports as shown. Assuming that the impact at each

support is perfectly plastic and that no slipping occurs between

the rod and the supports, determine (a) the height h1 reached

by end A after the first impact, (b) the height h2 reached by

end B after the second impact.

SOLUTION


12

2

Let q be the angle of inclination of the bar and h the elevation of the center of gravity.

Position 0 Position¢

Position 0. V mgh mgbh

L bT0 0

00 0= =

+=

Position¢. ¢ = ¢ ¢ =v b V1

20w

¢ = ¢ + ¢ = ¢ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃ ¢

=

T m v I m b mL1

2

1

2

1

2

1

2

1

2

1

12

1

2 22

2 2( ) ( ) ( )w w w

22432 2 2m L b( )( )+ ¢w

Conservation of energy. T V T VmgbhL b

m L b0 00 2 2 20

1

243+ = ¢ + ¢ +

+= + ¢: ( )( )w

( )( )( )

¢ =+ +

=w 2 0

2 2 1 0

24

3

gbhL b L b

C h (1)

where C gbL b L b1 2 2

24

3=

+ +( )( ) (2)

187



Impact at D.

Syst. Momenta¢ + Syst. Ext. Imp.¢Æ≤ = Syst. Momenta≤

Kinematics. ¢ = ¢ ¢¢ = ¢¢v b v b1

2

1

2w w

Moments about D: mv b I mv b I¢ - ¢ + = ¢¢ + ¢¢2

02

w w

- ¢ÊËÁ

ˆ¯̃

+ ¢ + = ¢¢ÊËÁ

ˆ¯̃

+ ¢¢m b b mL m b b mL1

2 2

1

120

1

2 2

1

12

2 2w w w w

¢¢ = -+

¢ = ¢w w wL bL b

C2 2

2 2 2

3

3 (3)

where C L bL b2

2 2

2 2

3

3= -

+ (4)

Position 1 Maximum elevation of end A.

Conservation of energy. T V T V1 1+ = ¢¢ + ¢

Result ( )¢¢ =w 21 1C h

hC

CC

C CC

h C h1

2

1

22 2

1

22

1

11 2

21= ¢¢ =

¢= =( ) ( )w w

Data: L b= = = =800 0 8 200 0 2 mm m, mm m. .

C

C

2

2 2

2 2

22

0 8 3 0 2

0 8 3 0 2

13

190 68421

169

3610 4

= -+

= =

= =

( . ) ( . )

( . ) ( . ).

. 668144

(a) h h1 00 468144 0 468144 100 46 8144= = =. ( . )( . mm) mm h1 46 8= . mm �

(b) h h2 10 468144 0 468144 46 8144 21 916= = =. ( . )( . ) . mm h2 21 9= . mm �

188


PROBLEM 17.121

A small plate B is attached to a cord that is wrapped around a uniform 4 kg disk

of radius R = 200 mm. A 1.5 kg collar A is released from rest and falls through

a distance h = 300 mm before hitting plate B. Assuming that the impact is

perfectly plastic and neglecting the weight of the plate, determine immediately

after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.

SOLUTION

The collar A falls a distance h. From the principle of conservation of energy.

v gh1 2=

Impact analysis. Kinematics. Plastic impact. e = 0

Collar A and plate B move together. The cord is inextensible.

v R vR2 22= =w wor

Let m = mass of collar A and M = mass of disk

Moment of inertia of disk: I MR= 1

2

2


I w1 0=

Syst Momenta. 1 + Syst Ext Imp. . .1 2Æ = Syst Momenta. 2

189



+ Moments about C: mv R I mv R1 2 2= +w (1)

mv R MR vR

mv R

mv Mv mv

v mm M

v

12 2

2

1 2 2

2 1

1

2

1

2

2

2

= ÊËÁ

ˆ¯̃

+

= +

=+

Data: mMhR

v

=== == =

==

1 5

4

300 0 3

200 0 2

2 9 81 0 3

2 42

1

.

.

.

( . )( . )

.

kg

kg

mm m

mm m

661 m/s

(a) Velocity of A. v v v2 1 1

2 1 5

2 1 5 4

3

7=

+=( )( . )

[( )( . ) ]

v2

3

72 4261 1 03976= =( . ) . m/s v2 1 040= Ø. m/s �

(b) Angular velocity. w2

1 03976

0 25 198= =.

.. 8 rad/s w2 5 20= . rad/s �

190


PROBLEM 17.122

Solve Problem 17.121, assuming that the coefficient of restitution between A

and B is 0.8.

PROBLEM 17.121 A small plate B is attached to a cord that is wrapped

around a uniform 4-kg disk of radius 200 mm. A 1.5 kg collar A is released

from rest and falls through a distance h = 300 mm before hitting plate B.

Assuming that the impact is perfectly plastic and neglecting the weight of the

plate, determine immediately after the impact (a) the velocity of the collar,

(b) the angular velocity of the disk.

SOLUTION

mR

I m R

mh

D

D D

A

==

= = = ◊

==

4

0 2

1

2

1

24 0 2 0 08

1 5

300

2 2

kg

m

kg m

kg

mm

2

.

( )( . ) .

.

== 0 3. m

Collar A falls through distance h. Use conservation of energy.

TV W h

T m v

V

T V T V W h m v

v m

A

A A

A A A

AA

1

1

22

2

1 1 2 22

2

0

1

2

0

01

20

2

==

=

=

+ = + + = +

=

:

hhW

ghA

A

=

=

=

= Ø

2

2 9 81 0 3

5 886

2 4261

( )( . )( . )

.

.

m /s

m/s

2 2

v

Impact. Neglect the mass of plate B. Neglect the effect of weight during the duration of the impact.

191



Kinematics. ¢ =w w ¢ = Ø = ¢ ØvB Rw w0 2.

Conservation of momentum.

+ Moments about D: m v R m v R I m v RA A A A D B B+ = ¢ + ¢ + ¢0 w

( . )( . )( . ) ( . )( . ) ( . )

. .

1 5 2 4261 0 2 1 5 0 75 0 08 0

1 125 0 08

= ¢ + ¢ +

¢ + ¢

v

vA

A

w

ww = 0 72783. (1)

Coefficient of restitution. ¢ - ¢ = -v v e v vB A A B( )

0 2 0 8 2 4261 0. . ( . )¢ - ¢ = -w vA (2)

- ¢ + ¢ =vA 0 2 1 94088. .w

Solving Eqs. (1) and (2) simultaneously

(a) Velocity of A. ¢ = -vA 0 0318. m/s ¢ = ≠vA 0 0318. m/s �

(b) Angular velocity. ¢ =w 9 545. rad/s w¢ = 9 55. rad/s �

192


PROBLEM 17.123

A slender rod AB is released from rest in the position shown. It

swings down to a vertical position and strikes a second and identical

rod CD which is resting on a frictionless surface. Assuming that

the coefficient of restitution between the rods is 0.5, determine the

velocity of rod CD immediately after the impact.

SOLUTION


12

2 for each rod.

Rod AB swings to vertical position.


Position 1. V T1 10 0= =

Position 2. V mg L2

2= -

T mv I

m L mL

mL

2 22

22

2

22

22

222

1

2

1

2

1

2 2

1

2

1

12

1

6

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

w

w w

w

Conservation of energy. T V T V mL mg L1 1 2 2

2220 0

1

6 2+ = + + = -: w

w2

3= gL

v2

2

2

3

3

= Æ

= =

L gL

v L gLB( ) w

193



Impact condition: ( ) ( ) ( )

( )

( )

v v e v

v L e gL

v L e gL

C B B

C

C

3 3 2

3 3

3 3

3

3

- =

- =

= +

w

w



Moments about B: mv L I mv L I m v LC2 2 3 3 32

02

+ + = + +w w ( )

m L gL

L mL gL

m L L mL m L e g2

3

2

1

12

30

2 2

1

1232

32

3 3

Ê

ËÁˆ

¯̃+ + = Ê

ËÁˆ¯̃

+ + +w w w LL L

e gL

v e gL e gL

v

C

C

( )

= -ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

+

=

w3

3

3

1

4

3

4

3

1

4

3

43 3

1

4

( )

( ) (11 3+ e gL)

For e = 0 5. ( ) ( . )v gLC 3

1

41 0 5 3= + ( ) .vC gL3 0 650= Æ �

194


PROBLEM 17.124

Solve Problem 17.123, assuming that the impact between the rods

is perfectly elastic.

PROBLEM 17.123 A slender rod AB is released from rest in the

position shown. It swings down to a vertical position and strikes

a second and identical rod CD which is resting on a frictionless

surface. Assuming that the coefficient of restitution between the

rods is 0.5, determine the velocity of rod CD immediately after

the impact.

SOLUTION


12

2 for each rod.

Rod AB swings to vertical position.


Position 1 V T1 10 0= =

Position 2 V mg L

T mv I

m L mL

2

2 22

22

2

22

22

2

1

2

1

2

1

2 2

1

2

1

12

= -

= +

= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

=

w

w w

11

6

222mL w

Conservation of energy. T V T V mL mg L1 1 2 2

2220 0

1

6 2+ = + + = -: w

w2

3= gL

v2

2

2

3

3

= Æ

= =

L gL

v L gLB( ) w

195



Impact condition: - - =

- =

= +

( ) ( ) ( )

( )

( )

v v e v

v L e gL

v L e gL

C B B

C

C

3 3 2

3 3

3 3

3

3

w

w



Moments about B: mv L I mv L I m v LC2 2 3 3 32

02

+ + = + +w w ( )

m L gL

L mL gL

m L L mL m L e g2

3

2

1

12

30

2 2

1

1232

32

3 3

Ê

ËÁˆ

¯̃+ + = Ê

ËÁˆ¯̃

+ + +w w w LL L

e gL

v e gL e gL

v

C

C

( )

= -ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

+

=

w3

3

3

1

4

3

4

3

1

4

3

43 3

1

4

( )

( ) (11 3+ e gL)

For perfectly elastic impact, e = 1. ( ) ( )v gLC 3

1

41 1 3= + ( ) .vC gL3 0 866= Æ �

196


PROBLEM 17.125

The plank CDE has a mass of 15 kg and rests on a small pivot

at D. The 55-kg gymnast A is standing on the plank at C when

the 70-kg gymnast B jumps from a height of 2.5 m and strikes the

plank at E. Assuming perfectly plastic impact and that gymnast A

is standing absolutely straight, determine the height to which

gymnast A will rise.

SOLUTION

Moment of inertia. I m L m LP P= =1

122

1

3

2 2( )

Velocity of jumper at E. ( )v gh1 12= (1)



Kinematics: v L v LC D= =w w

Moments about D: m v L m v L m v L I

m L m L m L

mm m m

v

E E E C C

E C P

E

E C P

1

2 2 2

13

1

0

1

3

+ = + +

= + +

=+ +

w

w w w

wLL

v L m vm m mC

E

E C P= =

+ +w 1

13

(2)

Gymnast (flier) rising. hv

gCC=2

2 (3)

Data: m mm mm

h

E B

C A

P

= == ===

70

55

15

2 51

kg

kg

kg

m.

197



From Equation (1) v1 2 9 81 2 5

7 0036

==

( )( . )( . )

. m/s

From Equation (2) vC =+ +

=

( )( . )

.

70 7 0036

70 55 5

3 7712 m/s

From Equation (3) h2

23 7712

2 9 81= ( . )

( )( . )

= 0 725. m h2 725= mm �

198


PROBLEM 17.126

Solve Problem 17.125, assuming that the gymnasts change places so

that gymnast A jumps onto the plank while gymnast B stands at C.

PROBLEM 17.125 The plank CDE has a mass of 15 kg and rests

on a small pivot at D. The 55-kg gymnast A is standing on the plank

at C when the 70-kg gymnast B jumps from a height of 2.5 m and

strikes the plank at E. Assuming perfectly plastic impact and that

gymnast A is standing absolutely straight, determine the height to

which gymnast A will rise.

SOLUTION

Moment of inertia. I m L m LP P= =1

122

1

3

2 2( )

Velocity of jumper at E. ( )v gh1 12= (1)



Kinematics: v L v LC D= =w w

Moments about D: m v L m v L m v L I

m L m L m L

mm m m

v

E E E C C

E C P

E

E C P

1

2 2 2

13

1

0

1

3

+ = + +

= + +

=+ +

w

w w w

wLL

v L m vm m mC

E

E C P= =

+ +w 1

13

(2)

Gymnast (flier) rising. hv

gCC=2

2 (3)

199



Data: m mm mm

h

E A

C B

P

= == ===

55

70

15

2 51

kg

kg

kg

m.

From Equation (1) v1 2 9 81 2 5

7 0036

==

( )( . )( . )

. m/s

From Equation (2) vC =+ +

=

( )( . )

.

55 7 0036

55 70 5

2 9631 m/s

From Equation (3) h2

22 9631

2 9 81= ( . )

( )( . )

= 0 447. m h2 447= mm �

200


PROBLEM 17.127

Member ABC has a mass of 2.4 kg and is attached to a pin support

at B. An 800-g sphere D strikes the end of member ABC with

a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and

that the coefficient of restitution between the sphere and member

ABC is 0.5, determine immediately after the impact (a) the angular

velocity of member ABC, (b) the velocity of the sphere.

SOLUTION

mL

L

m

D

AC

==

=

=

0 800

0 750

0 1875

2 4

.

.

.

.

kg

m

1

4m

kg

Let Point G be the mass center of member ABC.

I m LG AC=

=

= ◊

1

12

1

122 4 0 750

0 1125

2

2( . )( . )

. kg m2

Kinematics after impact. ¢ = ¢w w , ¢ = ¢ ≠vGL4

w , ¢ = ¢ ØvAL4

w

Conservation of momentum.

+ Moments about B: m v L m v L I m v L

m v L m v L I m L

D D D D G AC G

D D D D G AD

20

2 4

4 4 4

+ = ¢ + ¢ + ¢

= ¢ + + ÊËÁ

ˆ¯

w

˜̃È

ÎÍÍ

˘

˚˙˙

¢2

w

( . )( )( . ) ( . )( . ) [ . ( . )( .0 800 3 0 1875 0 800 0 1875 0 1125 2 4 0 1875= ¢ + +vD )) ]

. . .

2

0 45 0 15 0 196875

¢= ¢ + ¢

wwvD (1)

201



Coefficient of restitution. ¢ - ¢ = ¢ - ¢

= - -

v v v L

e v v

D A D

D A

4w

( )

¢ - ¢ = - -vD 0 1875 0 5 3 0. ( . )( )w (2)

Solving Eqs. (1) and (2) simultaneously.

(a) Angular velocity. ¢ =w 3 ¢ =w 3 00. rad/s �

(b) Velocity of D. ¢ = -vD 0 9375. ¢ = ≠vD 0 938. m/s �

202


PROBLEM 17.128

Member ABC has a mass of 2.4 kg and is attached to a pin support

at B. An 800-g sphere D strikes the end of member ABC with

a vertical velocity v1 of 3 m/s. Knowing that L = 750 mm and

that the coefficient of restitution between the sphere and member

ABC is 0.5, determine immediately after the impact (a) the

angular velocity of member ABC, (b) the velocity of the sphere.

SOLUTION

Let M be the mass of member ABC and I its moment of inertia about B.

M I M L= =2 41

122 2. ( )kg

where L = =750 0 75mm m.

Let m be the mass of sphere D. m = =800 0 8g kg.

Impact kinematics and coefficient of restitution.

( sin ) ( ) : ( ) ( sin )v e L v v L v eD n D n1 2 2 1q w w q= - = - (1)



+ Moments about E: mv L I m v L

mv L M L m L v e L

D n1 2

12

2 2 1

1

122

sin ( )

sin ( ) [ ( sin ) ]

q w

q w w q

= +

= + -

mv ML mL m v e

m e vL

M m

1 2 2 1

12

1

3

11

3

sin ( sin )

( ) sin

q w w q

q w

= - -

+ = +ÊËÁ

ˆ¯̃

203



(a) Angular velocity. wq

213 1

3=

++

( )( ) sine mvM m

w2

3 1 5 0 8 3 60

2 4 2 4 0 75

2 5981

= ∞+

=

( )( . )( . )( )sin

( . . )( . )

. w2 2 60= . rad/s �

(b) Velocity of D.

From Eq. (1), ( ) ( . )( . ) ( sin )( . )

.

( ) cos

v

v v

D n

D t

= - ∞==

0 75 2 5981 3 60 0 5

0 64976

61

m/s

00

3 60

1 5

∞= ∞=

cos

. m/s

( ) .vD n = 0 64976 m/s 30∞

( ) .vD t = 1 5 m/s 30∞

vD = +=

=

= ∞= ∞

( . ) ( . )

.

tan.

.

.

0 64976 1 5

1 63468

0 64976

1 5

23 4

30

2 2

m/s

q

qq == ∞53 4. vD = 1 635. m/s 53 4. ∞ �

204


PROBLEM 17.129

A slender rod CDE of length L and mass m is attached to a pin

support at its midpoint D. A second and identical rod AB is

rotating about a pin support at A with an angular velocity w1

when its end B strikes end C of rod CDE. Denoting by e the

coefficient of restitution between the rods, determine the angular

velocity of each rod immediately after the impact.

SOLUTION

Rod AB.

Kinematics. w1 1= w

w2 2= w

( )vABL

1 12

= Øw

( )v LAB 2 2

2= Øw



+ Moments about A: I m v L F t L I m v L

AB AB ABw w1 1 2 22 2

+ - = +( ) ( ) ( ) ( )D

1

12 2 2

1

12 2 2

1

3

21 1

22 2mL m L L F t L mL m L L

mL

AB ABw w w w+ ÊËÁ

ˆ¯̃

- = +( ) ( ) ( )D

221

22

21 2

1

3

1

3

w w

w w

- =

= -

( ) ( )

[ ( ) ]

F t L mL

F t mL

AB

AB

D

D (1)

205



Rod CE.



+ Moments about D: ( ) ( )

( ) ( )

F t L I

F t L mL

CE

CE

D

D

2

2

1

12

2

22

=

=

w

w

Substitute for ( )F tD from (1) 1

3 2

1

12

21 2

22mL L mLAB CE[ ( ) ] ( )w w w- =

w w w1 2 2

1

2- =( ) ( )AB CE (2)

Condition of impact. e =coefficient of restitution.

( ) ( ) ( )

( ) ( )

v e v v

L e L L

B C B

CE AB

1 2 2

1 2 22

= -

= -w w w

( ) ( )w w wAB CE e2 2 1

1

2= - (3)

From Eq. (2) w w w w1 2 1 2

1

2

1

2- -È

ÎÍ˘˚̇

=( ) ( )CE CEe

w w1 21( ) ( )+ =e CE ( ) ( )wCE e2 1 1= +w �

From Eq. (3) ( ) ( )w w w w w wAB e e e e2 1 1 1 1 1

1

21

1

2

1

2- + - = + - ( ) ( )w AB e2 1

1

21= -w �

206


PROBLEM 17.130

The 2.5-kg slender rod AB is released from rest in the position shown

and swings to a vertical position where it strikes the 1.5-kg slender rod

CD. Knowing that the coefficient of restitution between the knob K

attached to rod AB and rod CD is 0.8, determine the maximum angle

qm through which rod CD will rotate after the impact.

SOLUTION

Let b = = =mm

CD

AB

1 5

2 50 6

.

..

kg

kg

Let m mAB= .

Then m mCD = b .

Moments of inertia. I I mL

I mL

AB

CD

= =

=

1

12

1

12

2

2b

Rod AB falls to vertical position.

Position 0. V T0 00 0= =

Position 1. V mg L

v L

T m v I

mL

AB AB

AB AB

AB

1

1 1

1 12

12

2

2

2

1

2

1

2

1

6

= -

=

= +

=

( ) ( )

( ) ( )

( )

w

w

w 112

207



Conservation of energy. T V T V mL mgLAB0 0 1 12

120 0

1

6

1

2+ = + + = -: ( )w

( )w ABgL1

2 3= (1)

Impact.

Syst Momenta. 1 + Syst Ext Imp. . .1 2® == Syst Momenta. 2

Kinematics ( ) ( )

( ) ( )

v L

v L

AB AB

AB AB

1 1

2 2

2

2

=

=

w

w

Moments about B: m v L I b Kdt m v L IAB AB AB AB( ) ( ) ( )1 1 2 22 2

+ - Ú = ( ) ÊËÁ

ˆ¯̃

+w w

1

3

1

3

21

22mL b Kdt mLAB AB( ) ( )w w- Ú = (2)

Syst Momenta. 1 + Syst Ext Imp. . .1 2® == Syst Momenta. 2

Kinematics ( ) ( )v LCD CD2 2

2= w

Moments about C: 02

2 2+ Ú = +b Kdt m v L ICD CDb b w( ) ( )

b Kdt mL CDÚ = 1

3

22b w( ) (3)

208



Add Equations (1) and (2) to eliminate b KdtÚ .

1

3

1

3

1

3

21

22

22

2 2 1

mL mL mLAB AB CD

CD AB AB

( ) ( ) ( )

( ) ( ) ( )

w w b w

b w w w

= +

+ = (4)

Condition of impact. b b ebCD AB AB( ) ( ) ( )w w w2 2 1- =

( ) ( ) ( )w w wCD AB ABe2 2 1- = (5)

Add Equations (4) and (5) to eliminate ( )w AB 2 .

( )( ) ( )( )

( ) ( )

1 1

1

1

2 1

2 1

+ = +

= ++

ÊËÁ

ˆ¯̃

b w w

wb

w

CD AB

CD AB

e

e (6)

Rod CD rises to maximum height.

Position 2. V mg L

T m v I

mL

CD CD

CD

2

2 22

22

222

2

1

2

1

2

1

6

= -

= +

=

b

b b w

b w

( ) ( )

( )

Position 3. V mg L

T

m3

3

2

0

= -

=

b qcos

209




T V T V mL mgL mgL

Lg

CD m

m

2 2 3 32

221

6

1

20

1

2

13

+ = + - = -

- =

: b w b b q

q

( ) cos

cos (ww

bw

b

q

CD

AB

m

Lg

e

Lg

e gL

)

( )

cos

22

2

12

2

3

1

1

3

1

1

3

1

= ++

ÊËÁ

ˆ¯̃

= ++

ÊËÁ

ˆ¯̃

= -- ++

ÊËÁ

ˆ¯̃

1

1

2eb

= - ++

ÊËÁ

ˆ¯̃

11 0 8

1 0 6

2.

. qm = ∞105 4. �

210


PROBLEM 17.131

Sphere A of mass m and radius r rolls without slipping with

a velocity v1 on a horizontal surface when it hits squarely an

identical sphere B that is at rest. Denoting by mk the coefficient

of kinetic friction between the spheres and the surface, neglecting

friction between the spheres, and assuming perfectly elastic

impact, determine (a) the linear and angular velocities of each

sphere immediately after the impact, (b) the velocity of each

sphere after it has started rolling uniformly.

SOLUTION


5

2

Analysis of impact. Sphere A.


Kinematics: Rolling without slipping in Position 1.

w Avr

= 1

Moments about G: I Ivr

A

A

w w

w w

1

11

0+ =

= =

+ Linear components: mv Pdt mvA1 - =Ú (1)

Analysis of impact. Sphere B.


+ Linear components: 0 + Ú =Pdt mvB (2)

211



Add Equations (1) and (2) to eliminate Ú Pdt.

mv mv mv v v vA B B A1 1= + + =or (3)

Condition of impact. e = 1. v v ev vB A- = =1 1 (4)

Solving Equations (4) and (5) simultaneously,

v v vA B= =0 1,

Moments about G: 0 0 0+ = =I B Bw w

(a) Velocities after impact. vA Avr

= =0 1; w ; vB Bv= Æ =1 0; w �

Motion after Impact. Sphere A.

Syst Momenta. 1 + Syst Ext Imp. . .1 2® = Syst Momenta. 2

Condition of rolling without slipping: ¢ = ¢v rA Aw

Moments about C: I I mv rA A Aw w+ + ¢ + ¢0

2

50

2

5

2

7

2 1 2

1

mr vr

mr m r r

vr

v

A A

A

A

ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

+ = ÊËÁ

ˆ¯̃ ¢ + ¢

¢ =

¢

w w

w

( )

== 2

71v

Motion after impact. Sphere B.


212



Condition of rolling without slipping: ¢ = ¢v rB Bw

Moments about C: mv r I mv r

mv r mr m r r

vr

v

B B B

B B

B

+ = ¢ + ¢

+ = ÊËÁ

ˆ¯̃ ¢ + ¢

¢ =

¢

0

02

5

5

7

12

1

w

w w

w

( )

BB v= 5

71

(b) Final Rolling Velocities. ¢ = Æ ¢ = Æv vA Bv v2

7

5

71 1; �

213


PROBLEM 17.132

A small rubber ball of radius r is thrown against a rough floor with a

velocity vA of magnitude υ0 and a backspin w A of magnitude w 0 . It

is observed that the ball bounces from A to B, then from B to A, then

from A to B, etc. Assuming perfectly elastic impact, determine the

required magnitude w 0 of the backspin in terms of υ0 and r.

SOLUTION


5

2 Ball is assumed to be a solid sphere.

Impact at A.


For the velocity of the ball to be reversed on each impact,

¢ = =¢ = =

v v vA A

A A

0

0w w w

This is consistent with the assumption of perfectly elastic impact.

Moments about C: mv r I I mv rA A A Acos cos60 0 60∞ - + = ¢ - ¢ ∞w w

mv r mr mr mv r

r v

02

02

0 0

0 0

602

50

2

560

2

560

cos cos

cos

∞ - + = - ∞

= ∞

w w

w w005

4=

vr

�

214


PROBLEM 17.133

In a game of pool, ball A is rolling without slipping with a velocity v0 as it hits

obliquely ball B, which is at rest. Denoting by r the radius of each ball and by mk

the coefficient of kinetic friction between the balls, and assuming perfectly elastic

impact, determine (a) the linear and angular velocity of each ball immediately

after the impact, (b) the velocity of ball B after it has started rolling uniformly.

SOLUTION


5

2

(a) Impact analysis.

Ball A:


Kinematics of rolling: w00=

vr

+ Linear components: mv Pdt m vA x0 cos ( )q - =Ú (1)

+ Linear components: mv m vA y0 0sin ( )q + = (2)

Moments about y axis: I I Aw q w b0 0cos cos+ = (3)

Moments about x axis: - + = -I I Aw q w b0 0sin sin (4)

Ball B:


+ Linear components: 0 + =Ú Pdt m vB x( ) (5)

+

Linear components: 0 0+ = m vB y( ) (6)

215



Moments about y axis: 0 0+ = I Bw gcos (7)

Moments about x axis: 0 0+ = I Bw gsin (8)

Adding Equations (1) and (5) to eliminate PdtÚ ,

mv m v m vA x B x0 0cos ( ) ( )q + = +

or ( ) ( ) cosv v vB x A x+ = 0 q (9)

Condition of impact. e v v ev vB x A x= - = =1 0 0( ) ( ) cos cosq q (10)

Solving Equations (9) and (10) simultaneously,

( ) , ( ) cosv v vA x B x= =0 0 q

From Equations (2) and (6), ( ) sin , ( )v v vA y B y= =0 0q v vA = ( sin )0 q j �

v vB = ( cos )0 q i �

From Equations (3) and (4) simultaneously,

b q w w= = =, Avr00 w A

vr

= - +0 ( sin cos )q qi j �

From Equations (7) and (8) simultaneously,

wB = 0 w B = 0 �

(b) Subsequent motion of ball B.


Kinematics of rolling without slipping. ¢ = ¢v rB Bw

Moments about C: mv r I mv r

mr m r r

vr

vr

B B B

B B

BB

+ = ¢ + ¢

= ¢ + ¢

¢ =¢

=

0

2

5

5

7

5

7

2

1

w

w w

wq

( )

cos

¢ =v vB5

71 cosq ¢ =v iB v5

70( cos )q �

216


PROBLEM 17.134

Each of the bars AB and BC is of length L = 400 mm and mass m = 1 kg. Determine the

angular velocity of each bar immediately after the impulse QDt = ◊( . )1 5 N s i is applied at C.

SOLUTION


Bar BC:


+ Moments about B: 02

1

12 2 2

2

+ = +

= + +ÊËÁ

ˆ¯̃

( )

( )

Q t L I mv L

Q t L mL m L L L

BC BC

BC AB BC

D

D

w

w w w

Q t mL mLAB BCD = +1

2

1

3w w (1)

+ x components: Q t B t mvx BCD D- =

Q t B t m Lx AB BCD D- = +ÊËÁ

ˆ¯̃

w w1

2 (2)

Bar AB:


217



+ Moments about A: 02

1

12

1

2 2

2

+ = +

= + ÊËÁ

ˆ¯̃

( )

( )

B t L I mv L

B t L mL m L

x AB AB

x AB AB

D

D

w

w w

B t mLx ABD = 1

3w (3)

Add Eqs. (2) and (3): Q t mL mLAB BCD = +4

3

1

2w w (4)

Subtract Eq. (1) from Eq. (4): 05

6

1

6= +mL mLAB BCw w

w wBC AB= -5 (5)

Substitute for wBC in Eq. (1): Q t mL mL

mL

AB AB

AB

D = + -

= -

1

2

1

35

7

6

w w

w

( )

w ABQ tmL

= - 6

7

D (6)

Substituting into Eq. (5): wBCQ tmL

= - -ÊËÁ

ˆ¯̃

56

7

D

wBCQ tmL

= 30

7

D (7)

Given data: LQ t

m

= == ◊=

400 0 4

1 5

1

mm m

N s

kg

.

.D

Angular velocity of bar AB. w ABQ tmL

= - = -6

7

6

7

1 5

1 0 4

D ( . )

( )( . ) w AB = 3 21. rad/s �

Angular velocity of bar BC. wBCQ tmL

= =30

7

30 1 5

7 1 0 4

D ( )( . )

( )( )( . ) w BC = 16 07. rad/s �

218


PROBLEM 17.135

The motion of the slender 250-mm rod AB is guided by pins at A and B that

slide freely in slots cut in a vertical plate as shown. Knowing that the rod has

a mass of 2 kg and is released from rest when q = 0, determine the reactions

at A and B when q = ∞90 .

SOLUTION

Let Point G be the mass center of rod AB.

mL

I mLG

==

= = ◊

2

0 025

1

120 01046672

kg

m

kg m2

.

.

Kinematics. q

b b

= ∞

= == =

= = = ∞

= =

=

90

0 125

0 25

1

230

20 125

0

AD RAB L

RL

AG L

BG

.

.

sin

.

m

m

m

..125 m

Point E is the instantaneous center of rotation of bar AB.

v L

v Lv L

G

A

B

= =

= ∞ == ∞ =

20 125

30 0 21651

30 0 125

w w

w ww w

.

( cos ) .

( sin ) .

Use principle of conservation of energy to obtain the velocities when q = ∞90 :

Use level A as the datum for potential energy.

Position 1. q = =

= -

= -= -

0 0

2

2 9 81 0 125

2 4525

1

1

T

V mg L

( )( . )( . )

. J

219




T I mv

V

G G22 2

2 2

2

2

1

2

1

2

1

20 0104667

1

20 125

0 0208583

= +

= +

=

w

w w

w

( . ) ( . )

.

== - +ÊËÁ

ˆ¯̃

= - + ∞= -

mg R L2

2 9 81 0 125 0 125 30

4 5764

cos

( )( . )( . . cos )

.

b

J

TT V T V1 1 2 22

2

2 4525 0 0208583 4 5764

101 826

+ = + - = -

=

: 0

rad /s2 2

. . .

.

w

www =

===

10 091

0 21651 10 091

2 1848

0 125 1

.

( . )( . )

.

( . )(

rad/s

m/s

v

v

A

G 00 091

1 2614

. )

.= m/s

More kinematics: For Point A moving in the curved slot,

a i j

i j

i j

A C xA

C x

C x

a vR

a

a

= +

= +

= +

( )

( )( . )

.

( ) .

2

22 1847

0 125

38 1833

For the rod AB, a = =ak v j, B Bv

r i j

i j

r r

A B

G B A B

L L/

/ /

= - ∞ + ∞= - +

=

= -

sin cos

. .

.

30 30

0 125 0 21651

1

2

0 06255 0 108253

0 125 0 21651

2

i j

a a r r

j k i j

+

= + + -= + ¥ - +

.

( . . )

A B A B A B

Baa / /wa

-- - += - - +

( . ) ( . . )

. . .

10 091 0 125 0 21651

0 125 0 21651 12 728

2 i j

j j iaB a a 55 22 0468i j- .

220



Matching vertical components of aA

38 1833 0 125 22 0468

0 125 60 2301

. . .

. .

= - -= += +

= +

aa

B

B

G B G B

B

aa

a

a a a

a k

/

¥¥ -= + + ¥ - +

-

r

j k i jG B G Br/ /w

a a

2

0 125 60 2324 0 0625 0 108253

1

( . . ) ( . . )

( 00 091 0 0625 0 108253

0 125 60 2301 0 0625 0 1

2. ) ( . . )

. . . .

- += + - -

i j

j j ja a 008253

6 3643 11 0232

0 108253 6 3643 0 0625

a

a a

i

i j

a i

+ -= - + + +

. .

( . . ) ( .G 449 2069. )j

Kinetics: Use rod AB as a free body.

+ S SME EM= ( ) :eff

- = + ¥ - ∞mg L IG G E G2

2 9 81 0 125 30sin ( )( . )( . )sinbk r a ka /

= + +- = +

0.0104667a ( . . )( )

. . .

0 0625 0 108253

1 22625 0 0104667 0

i j am G

a 003125 4 7730

0 0417167 5 999

143 808

21 9

aa

a

+= -

= -

= -

.

. .

.

( .

R

aG

rad/s2

333 40 2189 m/s m/s2 2) ( . )i j+

+ S SF F m a Bx x G x= = - = - =( ) ( ) ( )( . ) .eff : N2 21 932 43 864 B = Æ43 9. N �

+

S SF F m a A mgy y G y= = - = =( ) ( ) ( )( . ) .eff : 2 40 4289 80 4378

A = + =( )( . ) . .2 9 81 80 4378 100 058 A = ≠100 1. N �

Check by considering

+ S SM MG G= eff : SM A BG = - = ◊( . ) . .0 0625 0 108253 1 5052 N m

S( ) ( ) ( . )( . ) .M IG Geff N m= - = = ◊a 0 0104667 143 808 1 5052

221


PROBLEM 17.136

A uniform disk of constant thickness and initially at rest is placed

in contact with the belt shown, which moves at a constant speed

n = 25 m/s. Knowing that the coefficient of kinetic friction between

the disk and the belt is 0.15, determine (a) the number of revolutions

executed by the disk before it reaches a constant angular velocity,

(b) the time required for the disk to reach that constant angular velocity.

SOLUTION

Kinetic friction. F Nf k= =m 0 15. N

+ S≠ = ∞ - ∞ - =

∞ - ∞ =

=∞

F N F mgN mg

N mg

y f

k

cos sin

(cos sin )

cos

25 25 0

25 25

25

m

-- ∞==

=

0 15 25

1 18636

0 15 1 18636

0 177954

. sin

.

( . )( . )

.

mgF mg

mgf

Final angular velocity. w2 = vr


2

2

(a) Principle of work and energy.

T W T TW F r mgr

T I mr

f

1 1 2 2 1

1 2

2 22 2

0

0 177954

1

2

1

2

1

2

+ = == =

= = ÊËÁ

ˆ

Æ

Æ

:

.q q

w¯̃̄

ÊËÁ

ˆ¯̃

=vr

mv2

21

4

0 0 1779541

4

1 40486

1 40486 25

9 81 0 1

2

2

2

+ =

=

=

.

.

( . )( )

( . )( .

mgr mv

vgr

q

q

220)

= 745 87. radians q = 118 7. rev �

222



(b) Principle of impulse-momentum.


Moments about A: 0 2+ =F tr If w

t IF r

mr

mgrvg

f

vr

=

=( )( )

=

w2

12

2

0 177954

2 8097

.

.

= ( . )( )

.

2 8097 25

9 81 t = 7 16. s �

223


PROBLEM 17.137

Solve Problem 17.136, assuming that the direction of motion of the belt

is reversed.

SOLUTION

While slipping occurs:

+

SF N N mgy k= + - =0 0: cos sinb m b

N mg

k=

+cos sinb m b (1)

For cylinder slipping occurs until w =

= =

vr

M Fr F AF Moment of about .

Work: U M FrNrF

k

1 2Æ = ¥=

q qm q

Kinetic energy: T T I mr vr

mv1 22 2

220

1

2

1

2

1

2

1

2= = = Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

=: w


Nr mrk

1 1 2 2

201

4

+ =

+ =

Æ

m q

qm

mb m q

qm

b m

= ◊

= ◊+

= ◊ +

1

4

1

1

4

1

4

2

2

2

mvr N

mvr mg

vrg

k

k

k

kk

cos sin

(cos sinqq ) (2)

Principle of impulse-momentum

224



+ Moments about A: Ftr I

Ntr mv vrk

=

= ÊËÁ

ˆ¯̃

w

m 1

2

2

Substituting for N: mb m ak

k

mg tr mrvcos sin+

ÊËÁ

ˆ¯̃

= 1

2

t vgk

k= ◊ +1

2 mb m b(cos sin ) (3)

Data: mbk

vr

== ∞==

0 15

25

25

.

m/s

0.12 m

(a) From Eq. (2), q = ∞ + ∞[ ](

( . )( . )cos ( . )sin

25

4 0 15 0 1225 0 15 25

m/s)

m)(9.81 m/s

2

2

q = 858 05. rad q = 136 6. revolutions �

(b) From Eq. (3), t = ∞ + ∞2525 0 15 25

m/s

2(0.15)(9.81 m/s2 )[cos ( . )sin ] t = 3 24. s �

225


PROBLEM 17.138

A uniform slender rod is placed at corner B and is given a slight

clockwise motion. Assuming that the corner is sharp and becomes

slightly embedded in the end of the rod, so that the coefficient of static

friction at B is very large, determine (a) the angle b through which

the rod will have rotated when it loses contact with the corner, (b) the

corresponding velocity of end A.

SOLUTION

Position 1. T

V mgh mgL1

1 1

0

2

=

= =

Position 2. V mgh mgL

T I mL

2 2

2 22 2

22

2

1

2

1

2

1

3

= =

= = ÊËÁ

ˆ¯̃

cos b

w w


T V T VmgL mL mgL

gL

1 1 2 2

222

22

02

1

2

1

3 2

31

+ = +

+ = ÊËÁ

ˆ¯̃

+

= -

w b

w b

cos

( cos ) (1)

Normal acceleration of mass center.

a L gn = = -2

3

212

2w b( cos )

+ S SF F man= + =eff

mg mgcos ( cos )b b= -3

21

(a) Angle b. 5

2

3

20 6cos cos .b b= = b = ∞53 1. �

From (1) w w22

2

31 0 6 1 2 1 09545= - = =g

LgL

gL

( . ) . .

(b) Velocity of end A v tA = w2 vA gL= 1 095. 53 1. ∞ �

226


PROBLEM 17.139

A 35-g bullet B is fired with a velocity of 400 m/s into the side of a

3-kg square panel suspended as shown from a pin at A. Knowing that

the panel is initially at rest, determine the components of the reaction at

A after the panel has rotated 90∞.

SOLUTION

Masses and moment of inertia. mm

b

I m b

B

P

G P

=== =

= = =

0 035

3

500

1

6

1

63 0 5 0 122 2

.

( )( . ) .

kg

kg

mm 0.5 m

55 kg m2◊

Note: The mass of the bullet is neglected in comparison with that of the plate after impact.

Analysis of impact: Use principle of impulse and momentum.

Kinematics: After impact the plate rotates about the pin at A.

v bG = =

2

0 5

2w w.

+ Moments about A: m v b I m v b

G G p G02

02

+ = +w

1

2

1

2

1

20 035 400 0 5 0 125

1

23 0 5

02

2

m v b I m bG G P= +

= +È

( )

( . )( )( . ) . ( )( . )

w

ÎÎÍ˘˚̇

=

= =

w

w 7

0 5

27 2 4749

rad/s

m/svG.

( ) .

227



Corresponding kinetic energy. T I m v

T

G P G12 2

12 2

1

2

1

2

1

20 125 7

1

23 2 4749

12 25

= +

= +

=

w

( . )( ) ( )( . )

. J

Plate rotates through 45∞.

Position 1: q = ∞0

Use Point A as the datum for potential energy

V m g bP1

2

3 9 810 5

2

10 4051

= -

= -

= -

( )( . ).

. J

Position 2: q = ∞90

V G A

T I m vG P G

2

2 22

22

0

1

2

1

2

1

20 125

=

= +

=

since is at level .

( )

( . )

w

w222

2

21

23

0 5

2

0 25

+ ÊËÁ

ˆ¯̃

=

( ).

.

w

w22


T V T V1 1 2 2

22

22

2

12 25 10 4051 0 25 0

7 3796

+ = +

- = +

==

. . .

. )

J J

(rad/s2

w

ww 22 7165. rad/s

Analysis at 90∞ rotation. a = a

Kinematics: ( ).a b

G t = =2

0 5

2a a ( ) .aG t = Ø0 35355a

( )

( . )( . )

a bG n =

=

2

0 5 7 3796

2

2w

( ) .aG n = Æ2 6091 m/s2

228



Kinematics: Use the free body diagram of the plate.

+ S SM M m g b I m a bA A P G P G t= = +( ) ( )eff :

2 2a

= +ÊËÁ

ˆ¯̃

= +ÈÎÍ

˘˚̇

I m bG P1

2

3 9 81 0 5

20 125

1

23 0 5

2

2

a

a( )( . )( . ). ( )( . )

a = 20 810. rad/s2

( ) .aG t = Ø7 3574 m/s2

+ S SF F A m ax x x P G n= = =( ) ( ) ( )( . )eff : 3 2 6091 Ax = Æ7 83. N �

+ S SF F A m g m ay y y P P G y= - = -( ) : ( )eff

Ay - = -( )( . ) ( )( . )3 9 81 3 7 3574 A y = ≠7 35. N �

229


PROBLEM 17.140

A square block of mass m is falling with a velocity v1 when it strikes a small obstruction

at B. Assuming that the impact between corner A and the obstruction B is perfectly

plastic, determine immediately after the impact (a) the angular velocity of the block

(b) the velocity of its mass center G.

SOLUTION

Moment of inertia. I mb= 1

6

2

Kinematics. Before impact, block is translating.

v1 1 1 0= Ø =v w

After impact, block is rotating about edge A.

v2 22

= b w 45∞



+ Moments about A. mv b I mv AG

mb m b b

mb

1 2 2

22 2

22

2

1

6 2 2

2

3

= +

= + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=

w

w w

w

( )

(a) Angular velocity. w213

4=

vb

w210 750= .

vb

�

(b) Velocity of the mass center. v b v2 2 12

3

4 2= =w v2 10 530= . v 45∞ �

230


PROBLEM 17.141

Solve Problem 17.140, assuming that the impact between corner A and the obstruction

B is perfectly elastic.

SOLUTION

Moments of inertia. I mb= 1

6

2

Kinematics. Before impact, block is translating.

v1 1 1 0= Ø =v w

After impact with e = 1: vA v= ≠1

v v v2 = +A G A/

= ≠ + ÈÎÍ

[ ]v b1 2

2w 45∞] (1)


Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta2

+ Moments about A:

mv b I mv b m b b

I mb Ag b

mv bb

mb mv b m

1 1 1 2

2

12

2 1

2 2 2 2

1

6 2

2

1

2

= - +

= =

= - +

w w

w

( )

bb

mv b b

2

2

3

2

2

12

2

ÊËÁ

ˆ¯̃

=

w

w

231



(a) Angular velocity. w213

2=

vb

w2 1 500= .vb

�

(b) Velocity of the mass center.

From Eq. (1), v v b vb2 11

2

3

2= ≠ + ◊È

ÎÍ[ ] 45∞

˘

˚˙

= ≠ + ∞ ØÈÎÍ

˘˚̇

+ ∞ ÆÈÎÍ

˘˚̇

= ≠ + ØÈÎ

[ ] sin cos

[ ]

v v v

v v

1 1 1

1 1

3

2 245

3

2 245

3

4ÍÍ˘˚̇

+ ÆÈÎÍ

˘˚̇

= ≠ÈÎÍ

˘˚̇

ÆÈÎÍ

˘˚̇

3

4

1

4

3

41 1v v v

v2 10 791= . v 18.4° �

232


PROBLEM 17.142

A 3-kg bar AB is attached by a pin at D to a 4-kg square plate, which can rotate

freely about a vertical axis. Knowing that the angular velocity of the plate is

120 rpm when the bar is vertical, determine (a) the angular velocity of the plate

after the bar has swung into a horizontal position and has come to rest against

pin C, (b) the energy lost during the plastic impact at C.

SOLUTION

Moments of inertia about the vertical centroidal axis.

Square plate. I mL= = = ◊1

12

1

124 0 500 0 0833332 2( )( . ) . kg m2

Bar AB vertical. I = approximately zero

Bar AB horizontal. I mL= = = ◊1

12

1

123 0 500 0 06252 2( )( . ) . kg m2

Position 1. Bar AB is vertical. I1 0 083333= ◊. kg m2

Angular velocity. w p1 120 4= = rpm rad/s

Angular momentum about the vertical axis.

( ) ( . )( ) .H IO 1 1 1 0 083333 4 1 04720= = = ◊w p kg m /s2

Kinetic energy. T I1 1 12 21

2

1

20 083333 4 6 5797= = =w p( . )( ) . J

Position 2. Bar AB is horizontal. I2 0 145833= ◊. kg m2

( ) .H IO 2 2 2 20 145833= =w w

Conservation of angular momentum. ( ) ( ) :H HO O1 2=

1 04720 0 145833 7 18082 2. . .= =w w rad/s

(a) Final angular velocity. w2 68 6= . rpm �

(b) Loss of energy.

T T T I1 2 1 2 22 21

26 5797

1

20 145833 0 71808- = - = -w . ( . )( . )

T T1 2 2 82- = . J �

233


PROBLEM 17.143

A 300 40¥ 0 mm-rectangular plate is suspended by pins at A and B. The

pin at B is removed and the plate swings freely about pin A. Determine

(a) the angular velocity of the plate after it has rotated through 90°,

(b) the maximum angular velocity attained by the plate as it swings freely.

SOLUTION

Let m be the mass of the plate.

Dimensions: a b= =0 4 0 3. .m m

Moment of inertia about A I m a bA = +1

3

2 2( )

Position 1. Initial position. w1 0=

Position 2. Plate has rotated about A through 90°.

Position 3. Mass center is directly below pivot A.

Potential energy. Use level A as datum.

V mab V mga V mgd1 2 32 2

= - = - = -

where d a b= + =1

20 252 2 . m

Kinetic energy. T T I T IA A1 2 22

3 320

1

2

1

2= = =w w

234



(a) 90° rotation. Conservation of energy.

T V T V mgb m a b mga

g a ba b

1 1 2 22 2

22

22

2 2

02

1

2

1

3 2

3

+ = + - = ◊ + +

= -+

=

: ( )

( ) (

w

w 33 9 81 0 4 0 3

0 4 0 311 772

2 2

2)( . )( . . )

( . ) ( . ). ( )

-+ÈÎ ˘̊

= rad/s

w2 3 43= . rad/s �

(b) w is maximum. Conservation of energy.

T V T V mgb m a b mgd

g d ba b

1 1 3 32 2

32

32

2 2

02

1

2

1

3

6 3

+ = + - = ◊ + -

= ++

=

: ( )

( ) (

w

w 99 81 1 5 0 9

0 4 0 323 544

2 2

. )( . . )

( . ) ( . ).

-+ÈÎ ˘̊

= (rad/s)2

w3 4 85= . rad/s �

235


PROBLEM 17.144

Disks A and B are made of the same material and are of the same thickness;

they can rotate freely about the vertical shaft. Disk B is at rest when it is

dropped onto disk A, which is rotating with an angular velocity of 500 rpm.

Knowing that disk A has a mass of 10 kg, determine (a) the final angular

velocity of the disks, (b) the change is kinetic energy of the system.

SOLUTION

Disk A: m rA A= = =10 150 0 15kg mm m.

I m rA A A= = = ◊1

2

1

210 0 15

9

80

2 2( )( . ) kg m2

Disk B: rB = =100 0 1mm m.

m m rr

I m r

B AB

A

B B B

=ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

=

= = ÊË

2 2

2

10100

150

40

9

1

2

1

2

40

9

( ) kg

ÁÁˆ¯̃

= ◊( . )0 11

45

2 kg m2



+ Moments about B: I I IA A Bw w w1 2 20 0+ + = +

w w w w2 1 1 1

81

970 83505=

+= =

II I

A

A B.

Initial angular velocity of disk A: w1 500 52 36= =rpm rad/s.

236



(a) Final angular velocity of system: w2 0 83505 52 36= ( . )( . )

w2 43 723= . rad/s w2 418= rpm �

Initial kinetic energy: T I A1 121

2= w

T121

2

9

8052 36 154 213= Ê

ËÁˆ¯̃

= ◊( . ) . N m

Final kinetic energy: T I IA B2 221

2= +( )w

T221

2

9

80

1

4543 723 128 774= +Ê

ËÁˆ¯̃

= ◊( . ) . N m

(b) Change in energy: T T2 1 25 439- = - ◊. N m DT = - ◊25 4. N m �

237


PROBLEM 17.145

At what height h above its center G should a billiard ball of radius r be

struck horizontally by a cue if the ball is to start rolling without sliding?

SOLUTION


5

2



Kinematics. Rolling without sliding. Point C is the instantaneous center of rotation.

+ Linear components: 0 2+ =P t mvD

= mrw2

Moments about G: 0 2+ =hP t ID w

02

52

22+ = Ê

ËÁˆ¯̃

h mr mr( )w w h r= 2

5 �

238


PROBLEM 17.146

A large 1.5 kg sphere with a radius r = 100 mm is thrown into a light

basket at the end of a thin, uniform rod weighing 1 kg and length

L = 250 mm as shown. Immediately before the impact the angular

velocity of the rod is 3 rad/s counterclockwise and the velocity of

the sphere is 0.5 m/s down. Assume the sphere sticks in the basket.

Determine after the impact (a) the angular velocity of the bar and

sphere, (b) the components of the reactions at A.

SOLUTION

Let Point G be the mass center of the sphere and Point C be that of the rod AB.

Rod AB: m

I m L

AB

AB AB

=

= =

= ◊

= ¥ ◊-

1

1

12

1

121 0 25

1

192

5 2083 10

2 2

3

kg

kg m

kg

2

( )( . )

. mm2

Sphere: m

I m r

S

G S

=

= = = = ¥ ◊-

1 5

2

5

2

51 5 0 1

3

5006 102 2 3 2

.

( . )( . )

kg

kg m

Impact. Before impact, bar AB is rotating about A with angular velocity w0 0= w ( )w0 3= rad/s and the

sphere is falling with velocity v0 0 0 0 5= Ø =v v( . ).m/s After impact, the rod and the sphere move together,

rotating about A with angular velocity w = w .

Geometry. R L rrL

= + = + =

= = = ∞

2 2 2 20 25 0 1 0 26926

0 1

0 2521 8

( . ) ( . ) .

tan.

..

m

q q

Kinematics: Before impact, v LC = = =

20 125 3 0 3750w ( . )( ) . m/s≠

After impact, v vC GL R= ¢ Ø = ¢2

w w, q

Principle of impulse and momentum. Neglect mass of the rod and sphere over the duration of the impact.

239



(a) +

Moments about A:

m v L I m v L I m v R I m v LS AB AB C G S G AB AB C0 0

20

2- - + = ¢ + ¢ + ¢ +w w w

or m v L I m v L I m R I m LS AB AB C G S AB AB0 02 2

2

1

4- - = + + +Ê

ËÁˆ¯̃ ¢w w (1)

( . )( . )( . ) ( ) ( )( . )( . )

( .

1 5 0 5 0 251

1923 1 0 375 0 125

3

5001 5

- ÊËÁ

ˆ¯̃

-

= + ))( . ) ( )( . )0 269261

192

1

41 0 252 2+ + Ê

ËÁˆ¯̃

ÈÎÍ

˘˚̇

0 125 0 13558 0 9219. . .= ¢ ¢ =w w 3 rad/s w¢ = 0 922. rad/s �

Normal accelerations at C and G.

( ) ( ) ( . )( . ) .aC nL= ¢ = = ¨2

0 125 0 92193 0 106242 2w m/s2

( ) ( ) ( . )( . ) .a RG n = ¢ = =w 2 20 26926 0 92193 0 22886 m/s2 21.8°

Tangential accelerations at C and G. a = a

( ) . ( ) .a L a RC t G t= = Ø = =2

0 125 0 26926a a a a 21.8°

(b) Kinetics. Use bar AB and the sphere as a free body.

+ S SM MA A= ( ) :eff

W L W L I L m a I m a R

I m L I m R

AB S AB AB C t G S G t

AB AB G S

2 2

1

4

2 2

+ = + + +

= + + +

a a( ) ( )

ÊÊËÁ

ˆ¯̃

+ = + ÊËÁ

ˆ¯̃

a

( )( . )( . ) ( . )( . )( . ) (1 9 81 0 125 1 5 9 81 0 251

192

1

41))( . ) ( . )( . )0 25

3

5001 5 0 269262 2+ +È

ÎÍ˘˚̇a

4 905 0 13558. .= a a = 36 1766. rad/s2

240



( ) ( . )( . ) . , ( ) ( . )( .a aC t G t= = Ø =0 125 36 1766 4 522 0 26926 36 17621 m/s 66 9 7409) .= m/s2 21.8°

+ S SF Fx x= ( ) :eff

A m a m a m aA

x AB C n S G n S G t

x

= - - ∞ + ∞= -

( ) ( ) cos . ( ) sin .

( )( .

21 8 21 8

1 0 106244 1 5 0 22886 21 8 1 5 9 7409 21 8) ( . )( . )cos . ( . )( . )sin .- ∞ + ∞

Ax = 5 2137. N Ax = Æ5 21. N �

+ S SF F A W W m a m a m ay y y AB S AB C t S G t S G n= - - = - - ∞ -( ) : ( ) ( ) cos . ( ) sieff 21 8 nn .21 8∞Ay - - = - -( )( . ) ( . )( . ) ( )( . ) ( . )( . )cos .1 9 81 1 5 9 81 1 4 5221 1 5 9 7409 21 8∞∞ - ∞( . )( . )sin .1 5 0 22886 21 8

Ay = 6 309. N A y = ≠6 31. N �

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