PROBLEM 17.

15
Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B
has a mass of 4 kg and a radius of gyration of 75 mm; gear C has a
mass of 9 kg and a radius of gyration of 100 mm. The system is at rest
when a couple M0 of constant magnitude 4 N · m is applied to gear C.
Assuming that no slipping occurs between the gears, determine the
number of revolutions required for disk A to reach an angular velocity
of 300 rpm.

SOLUTION

Moments of inertia: I mk 2
Gear A: IA (1 kg)(0.030 m)2 0.9 u 103 kg ˜ m2

Gear B: IB (4 kg)(0.075 m)2 22.5 u 103 kg ˜ m2

Gear C: IC (9 kg)(0.100 m)2 90 u 103 kg ˜ m2

Let rA be the radius of gear A, r1 the outer radius of gear B, r2 the inner radius of gear B, and rC the radius of
gear C.
rA 50 mm, r1 100 mm, r2 50 mm, rC 150 mm
At the contact point between gears A and B,
rA
r1Z B rAZ A : ZB ZA 0.5Z A
r1
At the contact point between gear B and C.
r2
rC ZC r2ZB : ZC ZB 0.33333ZB
rC

ZC 0.16667Z A

1 1 1
Kinetic energy: T I AZ A2  I BZB2  I C ZC2
2 2 2
1
T [0.9 u 10 Z A  (22.5 u 103 )(0.5Z A ) 2  (90 u 103 )(0.16667Z A )2 ]
3 2
2
(4.5125 u 103 kg ˜ m 2 )Z A2

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1717
 PROBLEM 17.15 (Continued)

Use the principle of work and energy applied to the system of all three gears with position 1 being the initial
rest position and position 2 being when Z A 300 rpm.

2S rad 1min rev

ZA ˜ ˜ 300 31.416 rad/s
rev 60 s min
T1 0
T2 (4.5125 u 103 kg ˜ m 2 )(31.416 rad/s)2 4.4565 J

U1o2 M TC (4 N ˜ m)TC
Principle of work and energy.
T1  U1o2 T2 : 0  4.4565 J 4(N ˜ m)TC

TC 1.11413 rad
TC
TA 6TC 6.6848 rad
0.16667
6.6848 rad
TA T A 1.063 rev W
2S rad/rev

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1718
 PROBLEM 17.21
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.
L
Rod vR Z
2
Collar vC dZ
Position 1. Z 0
T1 0 V1 0

1 1 1
Position 2. T2 mR vR2  I RZ 2  mC vC2
2 2 2
2
1 §L · 1§ 1 · 1
mR ¨ Z ¸  ¨ mR L2 ¸ Z 2  mC d 2Z 2
2 ©2 ¹ 2 © 12 ¹ 2
1 1
mR L2Z 2  mC d 2Z 2
6 2
L
V2 WC d  WR
2
§1 1 · L
T1  V1 T2  V2 : 0  0 ¨ mR L2  mC d 2 ¸ Z 2  WC d  WR
©6 2 ¹ 2
3(2WC d  WC L) 3g (2mC d  mR L)
Z2 (1)
3mC d  mR L
2 2
3mC d 2  mR L2

Data: mC 1 kg, d 0.3 m, mR 3 kg, L 0.6 m

ª (2)(1)(0.3)  3(0.6) º
From Eq. (1), Z2 3(9.81) « 2»
¬ 3(1)(0.3)  3(0.6) ¼
2

52.32 Z 7.23 rad/s W

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1728
 PROBLEM 17.72
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
1
Moment of inertia. I = mA r 2
2
1
= (8 kg)(0.24 m)2
2
= 0.2304 kg ◊ m2
Cylinder alone:

Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2

Moments about C: 0 + 0 = I w - mA v A r

or 0 = 0.2304w - (8)(0.24) v A (1)

Cylinder and carriage:

Syst. Momenta1 + Syst. Ext. Imp. 1Æ2 = Syst. Momenta 2

0 + Pt = mA v A + mB v B
+

Horizontal components:

or 0 + (10)(1.2) = 8 v A + 3v B (2)

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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1812
 PROBLEM 17.72 (Continued)

Kinematics. v A = v B - rw
v A = v B - 0.24w (3)

Solving Equations (1), (2) and (3) simultaneously gives ␻ = 5.68 rad/s

(a) Velocity of the carriage. vB = 2.12 m/s Æ 䉳

(b) Velocity of the center of the cylinder. vA = 0.706 m/s Æ 䉳

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
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1813
 PROBLEM 17.83

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 Z 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 ˜ m 2 and the centroidal moment of inertia of the tube about a vertical
axis is 0.0025 kg ˜ m 2 . 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.
mAB 1.6 kg
I AB 0.0025 kg ˜ m 2
I DCE 0.30 kg ˜ m 2

1
State 1. (rG/A )1 (125)
2
62.5 mm
Z1 5 rad/s

State 2. (rG/A )2 500  62.5

437.5 mm
Z Z2

Kinematics. (vG )T vT rG/C Z

Syst. Momenta1  Syst. Ext. Imp. 1o2 Syst. Momenta2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1832
 PROBLEM 17.83 (Continued)

Moments about C:
I ABZ1  I DCE Z1  m AB (vT )1 (rG/C )1  0 I ABZ2  I DCE Z2  mAB (vT )2 (rG/C )2
ª I AB  I DCE  m AB (rG/G )12 º Z1 ª I AB  I DCE  m AB (rG/C )22 º Z2
¬ ¼ ¬ ¼
[0.0025  0.30  (1.6)(0.0625)2 ](5) [0.0025  0.30  (1.6)(0.4375) 2 ]Z2
(0.30875)(5) 0.60875Z2
Z2 2.5359 rad/s

(a) Angular velocity after the plastic impact. 2.54 rad/s W

1 1 1
Kinetic energy. T I ABZ 2  I DCE Z 2  mAB v 2
2 2 2
1 1 1
T1 (0.0025)(5)2  (0.30)(5)2  (1.6)(0.0625)2 (5)2
2 2 2
3.859375 J
1 1 1
T2 (0.0025)(2.5359) 2  (0.30)(2.5359) 2  (1.6)(0.4375)2 (2.5359) 2
2 2 2
1.9573 J

(b) Energy lost. T1  T2 1.902 J W

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1833
 PROBLEM 17.116
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.

V1 0
L L
V2 W  mg
2 2
T1 0
2
1 1 1§ 1 2· 2 1 §1 · 1 2 2
T2 I Z22  mv22 ¨ mL ¸ v2  m ¨ Z2 ¸ mL Z2
2 2 2 © 12 ¹ 2 ©2 ¹ 6
1 2 2 L g g
T1  V1 T2  V2 : 0 mL Z2  mg ; Z22 3 Ȧ2 1.73205
6 2 L L
After impact.

L L
V3 W mg
2 2

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1889
 PROBLEM 17.116 (Continued)

L
V4 W cos 30q
2
1 2 1
T3 I Z3  mv32
2 2
2
1§ 1 · 1 §1 ·
mL2 ¸ Z32  m ¨ Z3 ¸
2 ¨© 12 ¹ 2 ©2 ¹
1 2 2
mL Z3
6
T4 0
1 2 2 L L
T3  V3 T4  V4 : mL Z3  mg 0  mg cos30q
6 2 2
g g
Z32 3(1  cos 30q) Z3 0.63397
L L
Analysis of impact.
Let r be the distance BK.
g
Before impact, ( v k )3 rZ3 1.73205r
L

g
After impact, ( v k )4 rZ4 0.63397r
L
|(vk )4 n |
Coefficient of restitution. e
|(vk )3n |

0.63397
e e 0.366 W
1.73205

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1890
 PROBLEM 17.145
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.

1 1
Square plate. I mL2 (4)(0.500)2 0.083333 kg ˜ m 2
12 12
Bar AB vertical. I approximately zero

1 1
Bar AB horizontal. I mL2 (3)(0.500)2 0.0625 kg ˜ m 2
12 12

Position 1. Bar AB is vertical. I1 0.083333 kg ˜ m 2

Angular velocity. Z1 120 rpm 4S rad/s

Angular momentum about the vertical axis.

( H O )1 I1Z1 (0.083333)(4S ) 1.04720 kg ˜ m 2 /s

1 1
Kinetic energy. T1 I1Z12 (0.083333)(4S )2 6.5797 J
2 2

Position 2. Bar AB is horizontal. I2 0.145833 kg ˜ m 2

( H O )2 I 2Z2 0.145833Z2

Conservation of angular momentum. ( H O )1 ( H O )2 :

1.04720 0.145833Z2 Z2 7.1808 rad/s

(a) Final angular velocity. Z2 68.6 rpm W

(b) Loss of energy.
1 1
T1  T2 T1  I 2Z22 6.5797  (0.145833)(7.1808) 2
2 2
T1  T2 2.82 J W

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1952