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CHAPTER 17CHAPTER 17
3
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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
p
w
0 300
2
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 0
2
2
6
2
1
2
1
2
3000 10
1 48044 10
0
=
=
= ¥ ◊
=
w
p( )( )
. N m
Work: U M
M
M
1 2
1500 2
9424 8
Æ = -
= -
= -
q
p( )( )
.
rev rad/rev
Principle of work and energy: T U T
M
1 1 2 2
61 48044 10 9424 8 0
+ =
¥ - =
Æ
. .
Average friction couple: M = ◊157 08. N m M = ◊157 1. N m �
4
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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
p
w
0 3600
2
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 0
2
2
2
1
2
1
2
1 620 120
115 12
0
=
=
=
=
w
p( . )( )
. ,kJ
Work: U M1 2
3 5
Æ = -
= - ◊
q
q( . )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
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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 I
I I I
n
B
A
A B
=
= È
ÎÍ
˘
˚̇
=
= +
= +
1
2
3
3
1
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 2
2
0 4
1
2
= = =
=
Æ q p
w
( )rad
total
T U T M n I A1 1 2 2
4
2
20 4
1
2
1 3+ = + = +Æ : ( ) ( )p w
 w p2
2
4
8
1 3
=
+
M
n I A( )
For Point D on rim of disk B
 v nrD = ( )w2 or v n r
Mr
I
n
nD A
2 2 2
2
2
2 2
4
8
1 3
= = ◊
+
w p
6
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PROBLEM 17.3 (Continued)
Value of n for maximum final speed.
For maximum v d
dn
n
nD
:
2
41 3
0
+
Ê
ËÁ
ˆ
¯̃
=
1
1 3
12 1 3 2 0
12 2 6 0
2 3 1 0
4 2
2 3 4
5 5
4
( )
[ ( ) ( )( )]
( )
+
- + =
- - =
- =
n
n n n n
n n n
n n
n = 0 and n = ÊËÁ
ˆ
¯̃
=1
3
0 7598
0 25.
. n = 0 760. �
7
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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
=
= È
ÎÍ
˘
˚̇
=
1
2
3
3
1
2
3
4
4 4
4
pr
pr
( )( )
 I I I n IA B Atotal = + = +( )1 3
4 (1)
Angular velocity: w
w
p
1
2
0
600
20
=
=
=
rpm
rad/s
Rotation: q p= =4 8rev rad
Kinetic energy: T
T I
1
2 2
2
0
1
2
=
= total w
Work: U M1 2
20 8
502 65
Æ =
= ◊
=
q
p( )( )
.
N m rad
J
8
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PROBLEM 17.4 (Continued)
Principle of work and energy: T U T
I
I
1 1 2 2
20 502 65
1
2
20
0 25465
+ =
+ =
= ◊
Æ
. ( )
.
total
total
2kg m
p
But, I m rA A A=
=
= ◊
1
2
1
2
15 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
n
n
Radius of disk B: r nrB A= = ( . )( )0 79076 125 mm rB = 98 8. mm �
9
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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. w
p
w
p
1
1 1
2
2
3
300
10
1
2
1
2
108 10
53 296 10
=
=
=
=
= ¥
rpm
rad/s
J
T I( )( )
.
Position 2. T I2 2
2
2
21
2
54= =w w
Work. U1 2 2500Æ = - J
Principle of work and energy for punching.
T U T1 1 2 2
3
2
253 296 10 2500 54+ = ¥ - =Æ : . w
(a) w2
2 940 66= .
 w2 30 67= . rad/s w2 293= rpm �
 Principle of work and energy for speed recovery.
 
T U T
U
M
U 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
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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 p
w w
w 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
1
2
1 2 2
21
2
1
2
+ =
+ =
Æ
Æw w
(a) Solving for I I
U
=
-
-
Æ2 1 2
2
2
1
2w w
 = - -
-
= ◊( )( )
( ) ( . )
.
2 2250
12 11 4
32 475
2 2
2
p p
kg m I = ◊32 5. kg m2 �
 Principle of work and energy for speed recovery.
T U T2 2 3 3+ =Æ
 But, T T
U T T
T T
U
3 1
2 3 3 2
1 2
1 2
2250
=
= -
= -
= -
= ◊
Æ
Æ
N m
 Work, U M2 3Æ = q
(b) q =
=
ÆU
M
2 3
2250
27
 = 83 33. rad q = 13 26. rev �
11
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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 M
rF
r mgk
1 2Æ =
=
=
q
q
m q( )
Kinetic energy of disk A.
Angular velocity becomes constant when w
w
2
1
2 2
2
2
2
2
0
1
2
1
2
1
2
4
=
=
=
= ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
v
r
T
T I
mr v
r
mv
Principle of work and energy for disk A.
T U T r mg mvk1 1 2 2
2
0
4
+ = + =Æ : m q
Angle change. q
m
= v
r gk
2
4
rad q
p m
= v
r gk
2
8
rev �
12
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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 M
rF
r mgk
1 2Æ =
=
=
q
q
m q( )
Kinetic energy of disk A.
Angular velocity becomes constant when w
w
2
1
2 2
2
2
2
2
0
1
2
1
2
1
2
4
=
=
=
= ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
v
r
T
T I
mr v
r
mv
Principle of work and energy for disk A.
T U T r mg mvk1 1 2 2
2
0
4
+ = + =- : m q
Angle change q
m
= v
r gk
2
4
rad q
p m
= v
r gk
2
8
rev
Data: r
v
k
=
=
=
0 15
0 20
12
.
.
m
m/s
m
 q
p
=
( )
( . )( . )( . )
12
8 0 15 0 20 9 81
2m/s
m m/s2
 q = 19 47. rev �
13
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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
80
2 5
2 5
.
.
Kinetic energy. T I= 1
2
2w :
 Position 1. w p
w w p
C
A B
= =
= = =
100
10
3
250
25
3
rpm rad/s
rpm rad/s
 Gear A: ( ) ( . ) .T A1
3
2
1
2
8 64 10
25
3
2 9609= ¥ ÊËÁ
ˆ
¯̃
=- p J
 Gear B: ( ) ( . ) .T B1
3
2
1
2
8 64 10
25
3
2 9609= ¥ ÊËÁ
ˆ
¯̃
=- p J
 Gear C: ( ) ( ) .T C1
3
2
1
2
270 10
10
3
14 8044= ¥ ÊËÁ
ˆ
¯̃
=- p J
 System: T T T TA B C1 1 1 1 20 726= + + =( ) ( ) ( ) . J
 Position 2. w p
w w p
C
A B
= =
= =
450 15
37 5
rpm rad/s
rad/s.
 Gear A: ( ) ( . )( . ) .T A2
3 21
2
8 64 10 37 5 59 957= ¥ =- p J
14
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PROBLEM 17.9 (Continued)
 Gear B: ( ) ( . )( . ) .T B2
3 21
2
8 64 10 37 5 59 957= ¥ =- p J
 Gear C: ( ) ( )( ) .T C2
3 21
2
270 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 M
rt
A
A
= = 0 57142
0 08
.
.
 Ft = 7 14. N �
15
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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
80
2 5
2 5
.
.
Kinetic energy. T I= 1
2
2w :
 Position 1. w p
w w p
C
A B
= =
= = =
100
10
3
250
25
3
rpm rad/s
rpm rad/s
 Gear A: ( ) ( . ) .T A1
3
2
1
2
8 64 10
25
3
2 9609= ¥ ÊËÁ
ˆ
¯̃
=- p J
 Gear B: ( ) ( . ) .T B1
3
2
1
2
8 64 10
25
3
2 9609= ¥ ÊËÁ
ˆ
¯̃
=- p J
 Gear C: ( ) ( ) .T C1
3
2
1
2
270 10
10
3
14 8044= ¥ ÊËÁ
ˆ
¯̃
=- p J
 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
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PROBLEM 17.10 (Continued)
 Position 2. w p
w w p
C
A B
= =
= =
450 15
37 5
rpm rad/s
rad/s.
 Gear A: ( ) ( . )( . ) .T A2
3 21
2
8 64 10 37 5 59 957= ¥ =- p J
 Gear B: ( ) ( . )( . ) .T B2
3 21
2
8 64 10 37 5 59 957= ¥ =- p J
 Gear C: ( ) ( )( ) .T C2
3 21
2
270 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 5
15 959
.
.
. 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 M
rt
A
A
= = 1 4285
0 08
.
.
 Ft = 17 86. N �
17
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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 r
v r
v r
A B
A A
B B
= =
= =
= =
0 25 0 15
0 25
0 15
. , .
.
.
m m
w w
w w
s r
s r
A A
B B
= =
= =
q q
q q
0 25
0 15
.
.
(a) Cylinder A falls to ground. s
s
A
B
=
= ÊËÁ
ˆ
¯̃
=
0 9
0 9
0 25
0 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 2
2 2 20
1
2
1
2
1
2
= = + +; w
 
T m r m k m rA A C B B2
2 2 2 2 2 2
2
1
2
1
2
1
2
1
2
12 5 0 25 15 0 16
= + +
= +
w w w
( . ) ( . ) ( ) ( . )22 2 2
2
10 0 15
0 695125
+ÈÎ ˘̊
=
( ) ( . )
.
w
w
18
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it without permission.
PROBLEM 17.11 (Continued)
Principle of work and energy.
T U T1 1 2
20 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 B3
2 2
4
1
2
1
2
0= + =w ; : 
T m k m rC B B3
2 2 2 2
2 2
1
2
1
2
1
2
15 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 = - ¢ = - ¢.
Principle of work and energy.
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
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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 p
w
w
p
1
2
1 1
2
2
360 12
0
1
2
1
2
20 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 F
U M F
D f f
D f 6628 32
100 531
. )
.= - Ff
Principle of work and energy.
T U T Ff1 1 2 2 14 212 23 100 531 0+ = - =Æ : , . .
Ff = 141 371. N
Kinetic friction force. F N
N
F
f k
f
k
=
= = =
m
m
141 371
0 35
403 918
.
.
. 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
P
P
+ - =
=
( )( . ) ( )( . )
. N P = 417 N Ø �
20
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it without permission.
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 initialangular 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 p
w
w
p
1
2
1 1
2
2
360 12
0
1
2
1
2
20 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 F
U M F
D f f
D f 228 32
100 531
. )
.= - Ff
Principle of work and energy.
T U T Ff1 1 2 2 14 212 23 100 531 0+ = - =Æ : , . .
 Ff = 141 371. N
Kinetic friction. F N N
F
f k
f
k
= = = =m
m
:
.
.
.
141 371
0 35
403 918 N
Statics. + + = - - =SM P F NA f0 180 40 200 0: ( ) ( ) ( )mm mm mm
 
180 40 141 371 200 403 918 0
480 21
P
P
- - =
=
( )( . ) ( )( . )
. N P = 480 N Ø �
21
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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
0
2 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 4
0rp ( )
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
2
2
2 2
1
2
1
2
- = = =
= +
q p p
w
( )
( ) ( )
radians
shaft shaft
22 2
0
2
0
4
0
2
4
0
1
2
1
2
1
2
1
2
+
= + + Ê
ËÁ
ˆ
¯̃
+
( )
( ) ( )
I
I w I n I
n
n I
C C
A
A A
shaft w
w w
nn
I n
n
I n
n
A
A
2
2
0
2 2
2
0
2
2
1
2
2
1
1
2
1
Ê
ËÁ
ˆ
¯̃
= + +ÊËÁ
ˆ
¯̃
= +ÊËÁ
ˆ
¯̃
w
w
T U T M I n
nA1 1 2 2 0 0
2
2
0 2
1
2
1+ = + = +ÊËÁ
ˆ
¯̃- : p w
Angular velocity. w
p
A
n
M
I n
2 0
0 1
2
4 1=
+( )
 w
p
A
n
n
M
I
=
+
2
12
0
0
 �
22
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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
˜̃ +
Ê
ËÁ
ˆ
¯̃
+
Ê
ËÁ
ˆ
¯̃
È
Î
Í
Í
˘
˚
˙
˙
r r r
r
r r
rA A B
A
B
A C
A
C
A
4 2 4
2
2 4
2
2w w w
 T t r r r rA A A B C=
Ê
ËÁ
ˆ
¯̃
+ +ÈÎ ˘̊
1
2 2
2
2 2 2 2pr w (1)
23
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PROBLEM 17.15 (Continued)
Recall that m r tA A= p r
2 and write
Eq. (1) as: T r t r r r
T m r r
r
r
r
A A B C
A A
B
A
C
A
= ( ) + +( )
= +
Ê
ËÁ
ˆ
¯̃
+
Ê
ËÁ
ˆ
1
4
1
4
1
2 2 2 2
2
2
p r
( )
¯̃̄
È
Î
Í
Í
˘
˚
˙
˙
2
2w A
Data: w p pA
A A B Cm r r r
= ÊËÁ
ˆ
¯̃
=
= = = =
150
2
60
5
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
2
2 2
0 7 5
1
4
6 0 2 1
0 15
0 2
0 1
0 2
+ =
+ = + ÊËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
Æ
. ( )( . )
.
.
.
.
q
ÈÈ
Î
Í
Í
˘
˚
˙
˙
= + +È
ÎÍ
˘
˚̇
=
( )
. . ( )
. .
5
7 5 0 06 1
9
16
1
4
25
7 5 26 833
2
2
p
q p
q
 
q
p
=
Ê
ËÁ
ˆ
¯̃
=
3 5777
2
0 5694
.
.
rad
rev
rad
rev q = 0 569. rev �
24
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it without permission.
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
2
400 0 22 9 681
2 2kx ( )( . ) .N/m m J
Gravity: V Wh mgh
V 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
2
2 2
230 150 80 0 08
1
2
1
2
400 0 08 1 28
= - = =
= = =
mm mm mm m
N/m m
.
( )( . ) . JJ
Gravity: V Wh
V V V
g
e g
= =
= +
=
0
1 28
2
. J
25
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PROBLEM 17.16 (Continued)
Kinetic energy: v r
I mL
T mv
2 2 2
2 2 2
2 2
0 18
1
12
1
12
4 0 6 0 12
1
2
= =
= = = ◊
=
w w( . )
( )( . ) .
m
kg m kg m
22
2
2
2
2
2
2
2 2
2
1
2
1
2
4 0 18
1
2
0 12
0 1248
+
= +
=
I
T
w
w w
w
( )( . ) ( . )
.
kg
Conservation of energy:
 
T V T V1 1 2 2
2
2
2
2
2
0 16 743 0 1248 1 28
123 9
11 131
+ = +
+ = +
=
=
. . .
.
.
J J
rad/
w
w
w ss w2 11 13= . rad/s �
26
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SOLUTION
Position 1:
Spring: x CD
Ve
1 150 370 150 220 0 22= - = - = =( ) .mm mm m
Unstretched 
Length674 84
== = =1
2
1
2
400 0 22 9 681
2 2kx ( )( . ) .N/m m J
Gravity: V Wh mgh
V 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= .
Kinetic energy: T1 0=
Position 2:
Spring: x
V kxe
2
2
2 2
230 150 80 0 08
1
2
1
2
400 0 08 1 28
= - = =
= = =
mm mm mm m
N/m m
.
( )( . ) . JJ
Gravity: V Wh
V 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
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PROBLEM 17.17 (Continued)
Kinetic energy: v r
I mL
T mv
2 2 2
2 2 2
2 2
0 18
1
12
1
12
4 0 6 0 12
1
2
= =
= = = ◊
=
w w( . )
( )( . ) .
m
kg m kg m
22
2
2
2
2
2
2
2 2
2
1
2
1
2
4 0 18
1
2
0 12
0 1248
+
= +
=
I
T
w
w w
w
( )( . ) ( . )
.
kg
Conservation of energy:
 
T V T V1 1 2 2
2
2
2
2
2
0 2 617 0 1248 1 28
10 713
3 273
+ = +
+ = +
=
=
. . .
.
.
J J
rad/s
w
w
w w2 3 27= . rad/s �
28
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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 2
2
2
2
2
2
2
2
2
2
2
1
2
1
2
1
2 2
1
2
1
12
1
6
= +
= ÊËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
=
w
w w
w22
2
Work: U mg l1 2
2
Æ =
Principle of work and energy: T U T
mg l ml
1 1 2 2
2
2
20
2
1
6
+ =
+ =
Æ
w
(a) Expressions for angular velocity and reactions.
 w
w
2
2
2
2
3
2 2
3 3
2
=
= = ◊ =
g
l
a l l g
l
g
 w2
3= g
l
 �
+ S SF F A W ma= - =( )eff :
 
A mg m g
A mg
- =
=
3
2
5
2
 A = ≠5
2
W �
29
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PROBLEM 17.18 (Continued)
(b) Application of data:
 
W l
g
l
g
= =
= = =
10 1
3 3
1
29 432
2
N m
rad /s2 2
,
.w w2 5 42= . rad/s �
 A W= = =5
2
5
2
10 25( )N N A = 25 0. N ≠ �
30
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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, w
Elevation: h V mgh= = =0 0 1
Position 2. v b
I ml
T mv I
m b l
2 2
2
2 2
2
2
2
2 2
2
2
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 2
2 2
2
20 0
1
2
1
12
+ = + + = +ÊËÁ
ˆ
¯̃
-: w
 w2
2
2 1
12
2
2=
+
gb
b l
(a) Value of b for maximumw2.
 
d
db
b
b l
b l b b
b l
b
2 1
12
2
2 1
12
2
2 1
12
2
2
2
2
0
+
Ê
Ë
Á
ˆ
¯
˜ =
+( ) - ( )
+( )
= == 1
12
2l b l=
12
 �
(b) Angular velocity. w2
2 12
12 12
2
12
2 2
=
+
=
g
g
l
l
l l
 w2
1 412= / g
l
 w2 1 861= .
g
l
 �
31
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PROBLEM 17.19 (Continued)
 Reaction at C. a b
l g
l
g
n =
=
=
w2
2
12
12
+ 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
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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 m
2
Speeds: w1 10 0= =, v
Kinetic energy: T1 0=
(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 2
2 2
2
2
2 2
2 2
2
2
2
2
1
2
1
2
1
2
80
1
2
80 0 4
46 4
= +
= +
=
w
w w
w
( ) ( )( . )
.
 Principle of conservation of energy.
 T V T V1 1 2 2 2
20 784 8 46 4+ = + + =: . . w
 w2
2 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
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PROBLEM 17.20 (Continued)
 Kinematics:at = ( )( )1 a
an = = ¨( )( ) .1 16 91382
2w 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 3
246 4= . w
 Principle of conservation of energy.
T V T V1 1 3 3 3
20 784 8 46 4 784 8+ = + + = -: . . .w
 w3
2 = 33.828 rad /s2 2 w3 5 82= . rad/s �
 Kinematics: an = = ≠( )( . ) .1 33 828 33 828 m/s
2
 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
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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
2
30
2
30
2 2 2V W h W h
W l W l
AB AB BC BC
AB BC
( ) ( )
sin cos
TT I m m lB AB BC2 2
2 2
2
21
2
1
6
= = +w w( )
Position 3. q = ∞
= =
90
2
03 3V W
l TAB
Conservation of energy.
 T V T V2 2 3 3+ = + :
1
6 2
30
2
30 0
2
2
2
2( ) sin cosm m l W l W l W lAB BC AB BC AB+ + ∞ - ∞ = +w
 
w2
2 3 1 30 30
3
2
1 30 30
= ◊
- ∞ + ∞
+
= - ∞ + ∞
l
W W
m m
g
l
AB BC
AB BC
( sin ) cos
[ sin cos ]]
. .
.
.
.= = =2 049 2 049 9 81
0 4
50 25
g
l
 w2 7 09= . rad/s �
35
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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 =
2
w
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
2
2 2 2
2
2 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 20
1
6
1
2 2
+ = + + = +ÊËÁ
ˆ
¯̃
- -: 0 w
 w2
2 2 2 2
3 2
3
3 2
3
=
+
+
=
+
+
( ) ( )W d W L
m d m L
g m d m L
m 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 2
3 9 81
2 1 0 3 3 0 6
3 1 0 3 3 0 6
= +
+
È
Î
Í
˘
˚
˙( . )
( )( )( . ) ( . )
( )( . ) ( . )
 = 52 32. w = 7 23. rad/s �
36
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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 =
2
w
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
2
2 2 2
2
2 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
0
1
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
+ÊËÁ
ˆ
¯̃
- -w
 w2
2 2 2 2
3 2
3
3 2
3
=
+
+
=
+
+
( ) ( )W d W L
m d m L
g m d m L
m d m L
C C
C R
C R
C R
 (1)
Let x d
L
= .
 w2
2
3
2
3
= ◊
+
+
g
L
x m
m
x m
m
R
C
R
C
Data: m m
L
g
x
x
C R= =
= +
+
1 3
3
2 3
3 3
2
2
 kg, kg
w
37
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PROBLEM 17.23 (Continued)
L gw2 3/ is maximum. Set its derivative with respect to x equal to zero.
 
d
dx
L
g
x x x
x
x x
w2 2
2 2
2
3
3 3 2 2 3 6
3 3
0
6 18 6 0
Ê
ËÁ
ˆ
¯̃
= + - +
+
=
- - + =
( )( ) ( )( )
( )
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
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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= w
Position 1. At rest. T1 0=
Position 2. s v v
v
r
T mv I
mv mr
v
r
G
G
G
G
= = =
= +
= + ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
1 5
1
2
1
2
1
2
1
2
1
2
2
2 2
2 2
. m w
w
¯̃̄
= = =
2
2 2 23
4
3
4
20 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 2
20 135 15+ = + =Æ :
 vG
2 9= vG = Æ3 00. m/s
(b) Since the forces are constant, a a
a
v
s
G
G
G
= =
=
=
=
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
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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 v
r
= =w w
Position 1. At rest. T1 0=
Position 2. Cylinder has fallen through distance s.
 T mv I
mv mr v
r
mv
2
2 2
2 2
2
2
1
2
1
2
1
2
1
2
1
2
3
4
= +
= + ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
w
Work. U mgs1 2Æ =
Principle of work and energy.
 T U T mgs mv1 1 2 2
20
3
4
+ = + =Æ :
 v gs2 4
3
= v = 4
3
gs
 Ø �
40
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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 v
r
= =w w
Position 1. At rest. T1 0=
Position 2. Cylinder has fallen through distance s.
 T mv I
mv mr v
r
mv
2
2 2
2 2
2
2
1
2
1
2
1
2
1
2
= +
= + ÊËÁ
ˆ
¯̃
=
w
( )
Work. U mgs1 2Æ =
Principle of work and energy.
 T U T mgs mv1 1 2 2
20+ = + =Æ :
 v gs2 = v = Øgs �
41
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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
v
m
k
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 1
2
1
2
2 2 2
0
1
2
1
2
1
2
3 8 0 036
1
2
3 0 09 8
4 233
=
= +
= +
=
w
( )( . ) ( )( . ) ( )
. 66 J
Position 2. V Wh
mgh
T mv I
2
2 2
2
2
2
3 9 81 0 06
1 7658
1
2
1
2
1
2
3 0
=
=
=
=
= +
=
( )( . )( . )
.
( )( .
J
w
224
1
2
3 0 09
0 09855
2
2 2
2
2
2
2
w w
w
) ( )( . )
.
+
=
42
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PROBLEM 17.27 (Continued)
(a) Conservation of energy.
 
T V T V1 1 2 2
2
2
2
2
2
4 2336 0 0 09855 1 7658
25 041
5 004
+ = +
+ = +
=
=
. . .
.
.
J J
 
w
w
w rrad/s w2 5 00= . rad/s �
(b) Reaction at B.
 
+
 
ma m CG
F ma N m
n
y y
=
=
= Ø
= -
( )
(
.
w2
2
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
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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
1
2 2 2
1
2
1
2
1
2
1
2
1
2
1
2
2
1
2
v r v r
T mv mv I
m r m r
B A
B A
( ) ( )) ( )2 2 1
2
2
1
2
1
2
3
+
=
mr
mr
w
w
Position 2. Point B is at the bottom.
 w w w
w
w
= = =
= + +
= + +
2 2
2
2 2
2
2
2
2 2
0
1
2
1
2
1
2
0
1
2
1
2
 v v r
T mv mv I
m r mr
B A
B A
( ) ( ))
( )
w
w
w
w
2
2
2
2
2
2
1
2
2
1
2
3
9
=
=
=
mr
mr
mr
Work. U mg h mgr1 2 2Æ = =( )D
Principle of work and energy.
 T U T mr mgr m1 1 2 2
2
1
2
1
23 2 9+ = + =Æ : w w
 w1
2
3
= g
r
 w1 0 577= .
g
r
 �
44
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it without permission.
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 21
2= = -ÊËÁ
ˆ
¯̃
( )w
p
w
Moment of inertia: I mr m mr m r mr= - = - ÊËÁ
ˆ
¯̃
= -Ê
ËÁ
ˆ
¯̃
2 2 2
2
2
2
0 6
2
1
4
( . )
p p
Kinetic energy: T mv I
m r mr
2 2
2
2
2
2
2
2
2 2
2 2
2
1
2
1
2
1
2
1
2 1
2
1
4
1
= +
= -ÊËÁ
ˆ
¯̃
+ -Ê
ËÁ
ˆ
¯̃
=
w
p
w
p
w
22
1
4 4
1
4
1
2
2
4
2
2 2
2
mr
mr
- +Ê
ËÁ
ˆ
¯̃
+ -Ê
ËÁ
ˆ
¯̃
È
Î
Í
˘
˚
˙
= -ÊËÁ
ˆ
¯̃
p p p
p
Work: U W OG mg r mgr1 2
2 2
Æ = = =( ) p p
Principle of work and energy: T U T
mg r mr
g
r
1 1 2 2
2
2
2
2
2
2
0
2 1
2
2
4
2
1
1 7519
+ =
+ = -ÊËÁ
ˆ
¯̃
=
-( ) ◊ =
Æ
p p
w
w
p p
.
gg
r
45
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PROBLEM 17.29 (Continued)
(a) Angular velocity. w2 1 324= .
g
r
 �
(b) Reaction at A.
 Kinematics: Since O moves horizontally, ( )a y0 0=
 
a
r g
r
g
n =
= ÊËÁ
ˆ
¯̃
= ≠
( . )
.
.
0 6
2
1 7519
1 1153
2
2w
p
 Kinetics:
 + S SF F A mg mgy y= - =( ) .eff : 1 1153 A = ≠2 12. mg �
46
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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
A
AB
B
A A B
v
r
v r r
= =
= =
2
1
2
1
2
Moment of inertia. I W
g
r= 1
2
2
Kinetic energy.
 Cyl B: 1
2
1
2
1
2
1
4
2 2 2 2 2I W
g
r W
g
rB B Bw w w=
Ê
ËÁ
ˆ
¯̃
=
 Cyl A: 1
2
1
2
1
2
1
2
1
2
1
2
1
2
3
2 2
2
2
2
mv I W
g
r W
g
rA A B B+ =
Ê
ËÁ
ˆ
¯̃
+
Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
w w w
116
2 2W
g
r Bw
Total: T W
g
r B=
7
16
2 2w
(a) Distance h that cylinder A will rise.
 Conservation of energy for system.
 
T V T V W
g
r W
g
r Wh
h r
g
B B
B
1 1 2 2
2
1
2 2
2
2
2
1
7
16
0
7
16
7
16
+ = + + = +
=
: ( ) ( )
[( )
w w
w 22 2
2
2
2 27
16
0 1
9 81
30 5
0 3902
-
= ÊËÁ
ˆ
¯̃
-
=
( ) ]
( . )
( . )
( )
.
wB
 m h = 0 390. m �
47
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it without permission.
PROBLEM 17.30 (Continued)
(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 W
g
r Ph Wh W
g
r
P W Wr
g
B B1 1 2 2
2
1
2 2
2
2
2
3
16
2
3
16
1
2
3
32
+ = + - =
= -
Æ : ( ) ( )w w
hh
W W
W
B B( ) ( )
( )( . )
w w1
2
2
2
1
2
3
14
2
7
2
7
7 9 81
-ÈÎ ˘̊
= -
= = P = 19 62. N �
48
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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
Kinematics. v rAB B= w
Point C is the instantaneous center of cylinder A.
 
w w
w w
A
AB
B
A A B
v
r
v r r
= =
= =
2
1
2
1
2
Moment of inertia. I W
g
r= 1
2
2
Kinetic energy.
 Cyl B: 1
2
1
2
1
2
1
4
2 2 2 2 2I W
g
r W
g
rB B Bw w w=
Ê
ËÁ
ˆ
¯̃
=
 Cyl A: 1
2
1
2
1
2
1
2
1
2
1
2
1
2
3
2 2
2
2
2
mv I W
g
r W
g
rA A B B+ =
Ê
ËÁ
ˆ
¯̃
+
Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
w w w
116
2 2W
g
r Bw
Total: T W
g
r B=
7
16
2 2w
Position 1. Rest T V1 10 0= =
Position 2. Cylinder has fallen through. d = 1 m.
 T W
g
r
V Wd
B2
2 2
2
7
16
=
= -
w
49
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PROBLEM 17.31 (Continued)
Principle of conservation of energy: T V T V
W
g
r WdB
1 1 2 2
2 20 0
7
16
+ = +
+ = -w
where r
gd
rB
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
2
0 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 2
2
2Æ = +( )
 Principle of work and energy for cylinder A:
 
T U T P d Wd W
g
r
P W Wr
g d
P
B
B
1 1 2 2
2 2
2 2
0 2
3
16
1
2
3
32
1
2
7 9
+ = - + =
= -
=
Æ : ( )
( )(
w
w
.. )
( )( . )( . ) ( . )
( . )( )
. .
81
32
7 9 81 0 1 2242 29
9 81 1
34 335 14 715 1
2
- 3
= - = 99 620.
 
P = 19 62. N
 
�
50
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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 v
BE
v v v v
v v
v
CF
v
B AB A
B AB
A B AB
C AB C
C AB
= = = = =
= = =
w
w w
0 075
2 2
0 075
.
.
m
m
AAB = 0
Kinetic energy. T m v I m v m v IA A A A AB AB B B B B2
2 2 2 21
2
1
2
1
2
1
2
1
2
= + + + +w w
 = +
Ê
ËÁ
ˆ
¯̃
+
È1
2
2 2
1
2
6 0 15
0 075
52 2
2
2( )( ) ( )( . )
.
( )m/s kg m kgv v vAB AB AB
ÎÎ
Í
Í
+ + Ê
ËÁ
ˆ
¯̃
˘
˚
˙
˙
=
( . )( ) ( . )( . )
.
1 5
1
2
1 5 0 075
0 075
1
2
2
2
kg kg mv vAB AB
224 12 5 1 5 0 75
21 625
2
2
2
+ + + +[ ]
=
. .
.
v
T v
AB
AB
51
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PROBLEM 17.32 (Continued)
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
Principle of work and energy: T U T
v
v
v
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
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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, determinethe velocity of the cradle after it 
has moved 250 mm.
SOLUTION
Moments of inertia. I I Wr
gA B
= =
2
2
Kinematics. w wA B
C
A B C
v
r
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 C2
2 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
È
ÎÍ
˘
˚̇
= +
= +
=
2
2
2
2
1
2
3
1
2
3 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 2
20 7 5 13 5+ = + =Æ : . . vC = Æ0 745. m/s �
53
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PROBLEM 17.34
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 Wr
gA B
= =
2
2
Kinematics. w wA B
C
A B
v
r
v v
= =
= = 0
Kinetic energy. T1 0=
 
T m v I m v I m v
m m m
A A A A B B B B C C2
2 2 2 2 21
2
1
2
1
2
1
2
1
2
1
2
0
1
2
0
1
2
= + + + +
= + + + +
w w
CC C
C C
C
C
v
m m v
v
v
È
ÎÍ
˘
˚̇
= +
= +
=
2
2
2
2
1
2
1
2
6 9
7 5
( )
( )
.
Work. U Fs1 2 30 0 25 7 5Æ = = =( )( . ) .N m J
Principle of work and energy. T U T vC1 1 2 2
20 7 5 7 5+ = + =Æ : . . vC = Æ1 000. m/s �
54
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PROBLEM 17.35
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 Wr
gA B
= =
2
2
Kinematics. w wA B
C
A B C
v
r
v v v= = = =
2
1
2
Kinetic energy. T1 0=
 
T m v I m v I m v
m m m
A A A A B B B B C C2
2 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
2
0 75
1
2
0 75 6 9
6 75
2
2
2
2
m m v
m m v
v
v
C C
C C
C
C
+È
ÎÍ
˘
˚̇
= +
= +
=
( . )
[( . )( ) ]
.
Work. U Fs1 2 30 0 25 7 5Æ = = =( )( . ) .N m J
Principle of work and energy. T U T vC1 1 2 2
20 7 5 6 75+ = + =Æ : . . vC = Æ1 054. m/s �
55
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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: T
V 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 2
2
0 5196
10 9 81 0 5196
50 974
1
2
1
2
= -
= -
= -
= +
( . )
( )( . )( . )
.
m
kg
J
ww
w w
w
2
2
2
2 2
2
2
2
2
1
2
10 0 6
1
2
1
2
10 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
2
2
2
2
- = -
=
. . .
.
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
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it without permission.
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: T
V W
mg
1
1
0
0 2052
0 2052
=
= -
= -
( . )
( . )
m
Position 2: V W mg
T m I
m m
2
2 2
2
2
2
2
2
0 6 0 6
1
2
1
2
1
2
0 6
1
2
1
12
1
= - = -
= +
= +
( . ) ( . )
( . ) ( .
m
v w
w 22
0 24
2
2
2
2 2
2
)
.
Ê
ËÁ
ˆ
¯̃
=
w
wT m
Principle of conservation of energy. T V T V
mg m mg
g
1 1 2 2
2
2
2
2
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
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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. 
v
h 
elongation of spring. 
h
e e
1
1 1
0
0
=
= =
 T mv I
V ke Wh
1 1
2
1
2
1 1
2
1
1
2
1
2
0
1
2
0
= + =
= + =
w
Position 2. q = ∞30
 
e L L
e L
h L L
V ke Wh
2
2
2
2 2
2
2
30
1 30
2
30
1
4
1
2
1
+ ∞ =
= - ∞
= - ∞ = -
= + =
cos
( cos )
sin
22
1 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
2
2 2
2
2
( cos )
ÊÊ
ËÁ
ˆ
¯̃
= 1
6
2 2W
g
L w
58
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PROBLEM 17.38 (Continued)
Conservation of energy.
 
T V T V W
g
L kL WL
g
L W
1 1 2 2
2 2 2 2
20 0
1
6
1
2
1 30
1
4
3 3
+ = + + = + - ∞ -
= -
: ( cos )w
w kg (( cos )1 30 2- ∞
Data: W
g
L
k
=
=
=
=
( . )( . )
.
.
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 81
1 30
41 87
= - - ∞
=
( )( . )
( . )
( )( . )
( . )( . )
( cos )
. 003 w = 6 47. rad/s �
 vB = ∞( . )(cos )( . )0 6 30 6 4707 vB = 3 36. m/s �
59
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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 L
x L L
B
2 2
2
1
2
50
0 6 1 50
0 21433
=
=
= - ∞
= - ∞
=
w
w
cos
. ( cos )
.
m
m
Position 1. V W L kx
V
1 1
2
1
2
50
1
2
4 5 9 81
0 6
2
50
1
2
600
= - ∞ +
= - 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 2
2
2
2
2
2
2
0
1
2
1
2
1
2 2
1
2
1
12
= + =
= +
= ÊËÁ
ˆ
¯̃
+ ÊË
( ) ( )
w
w ÁÁ
ˆ
¯̃
= = =
w
w w w
2
2
2
2
2 2
2
2
2
21
6
1
6
4 5 0 6 0 27mL ( . )( . ) .kg m
60
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PROBLEM 17.39 (Continued)
Conservation of energy: T V T V1 1 2 2
2
20 3 6361 0 27
+ = +
+ ◊ =. .N m w
 
w
w
2
2
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
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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
T
V
A B
Position 2. q = ∞30
Kinematics. Locate the instantaneous center C. Triangle ABC is equilateral.
 v v L
v L
A B
G
= =
= ∞
w
w cos30
Moment of inertia. I ml= 1
12
2
Kinetic energy. T mv I T m L mL
ml
G2
2 2
2
2 2 2
2
1
2
1
2
1
2
30
1
2
1
12
5
12
= + = ∞ + ÊËÁ
ˆ
¯̃
=
w w w
w
: ( cos )
22
Potential energy. V mg L mgL2
2
30
1
4
= - ∞ = -sin
Conservation of energy.
T V T V mL mgL1 1 2 2
2 20 0
5
12
1
4
+ = + + = -: w
 
w
w
2 0 6
0 775
0 775
0 775
=
=
=
=
.
.
.
.
g
L
g
L
v gL
v gL
A
B vA gL= ¨0 775. �
 vB gL= 0 775. 60∞ �
62
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 17.41
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 v
h
mg R1 1
1
0
2
= = =
Position 2: Since instantaneous center is at B,
 
v R
T mv I
m R mR
2 2
2 2
2
2
2
2
2
2
2
2
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
2
2
2
2 2
mR
V Wh mg R
w
Position 3: V3 0=
Since both vA and vB are horizontal, w3 0= (1)
 T mv3 2
21
2
=
63
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 17.41 (Continued)
(a) From 1 to 2: Conservation of energy
 
T V T V mgR mR mgR
g
R
g
R
1 1 2 2
2
2
2
2
2
2
0
1
2
1
6
1
4
3
2
3
2
+ = + + = +
=
=
: w
w
w w2 1 225= .
g
R
 �
 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 3
2
3
2
0
1
2
1
2
+ = + + =
=
:
 v3 = ÆgR �
64
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 17.42
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
0
2
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 w
w
w
w
AB BD
D
G
v l
v l
=
=
=
= ∞
2
2
30cos
 Kinetic energy.
T I mv I T mL m lA G BD2 2
2 2
2
2
2
2
2
2
2
1
2
1
2
1
2
1
2
1
3
1
2
3= + + = ÊËÁ
ˆ
¯̃
+w w w w: ( cos 00 1
2
1
12
7
12
2 2
2
2
2
2
2
∞ + ÊËÁ
ˆ
¯̃
=
) mL
mL
w
w
 Potential energy. V mgh mghAB BD2 = +
V mg L mg L
mgL
2
2
30
3
2
30= - ∞ÊËÁ
ˆ
¯̃
+ - ∞ÊËÁ
ˆ
¯̃
= -
sin sin
65
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