1. No category

PROBLEM 17.1
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 then coasts to rest after 5000 revolutions. Knowing that the kinetic friction of the rotor produces a
couple of magnitude 4 N · m, determine the centroidal radius of gyration of the rotor.
SOLUTION
ω1 = 3600
Angular velocities:
rev 1min 2π rad
⋅
⋅
= 120π rad/s
min 60 s
rev
ω2 = 0
Angular displacement:
5000 rev = 10000 π rad
Principle of work and energy: T1 + U1→2 = T2 :
1
1
I ω12 = I (120π ) 2 = 71.061 × 103 I
2
2
1
T2 = I ω22 = 0
2
U1→2 = − M θ = −(4 N ⋅ m)(10000π rad) = −40000π N ⋅ m
T1 =
71.061 × 103 I − 40000π = 0
I = 1.76839 kg ⋅ m 2
I = mk 2
Centroidal radius of gyration.
k =
I
1.76839 kg ⋅ m 2
=
= 0.1881 m
m
50 kg
k = 188.1 mm 
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1700
PROBLEM 17.19
A slender 9 lb rod can rotate in a vertical plane about a pivot at B. A spring of
constant k = 30 lb/ft and of unstretched length 6 in. 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:
CD = 142 + 52 = 14.866 in.
Unstretched
Length
Spring:

x1 = CD − ( 6 in. ) = 14.866 − 6 = 8.8661 in. = 0.73884 ft
Ve =
Gravity:
Kinetic energy:
1 2 1
kx1 = (30 lb/ft)(0.73884)2 = 8.1882 lb ⋅ ft
2
2
 7 
Vg = Wh = 9 lb =  − ft  = −5.25 lb ⋅ ft
 12 
V1 = Ve + Vg = 8.1882 lb ⋅ ft − 5.25 lb ⋅ ft = 2.9382 lb ⋅ ft
T1 = 0
Position 2:
Spring:
x2 = 9 in. − 6 in. = 3 in. = 0.25 ft
Ve =
1 2 1
kx2 = (30 lb/ft)(0.25 ft) 2 = 0.9375 lb ⋅ ft
2
2
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1727
PROBLEM 17.19 (Continued)
Gravity:
Vg = Wh = 0
V2 = Ve + Vg
= 0.9375 lb ⋅ ft
Kinetic energy:
 7 
v2 = rω2 =  ft  ω2
 12 
1
1  9 lb 
I = mL2 = 
(2 ft)2 = 0.093168 slug ⋅ ft 2
12
12  32.2 
1
1
T2 = mv22 + I ω22
2
2
2
=
1  9 lb    7  
1
2

   ft  ω2  + (0.093168)ω2
2  32.2    12  
2
T2 = 0.094138ω22
Conservation of energy:
T1 + V1 = T2 + V2
0 + 2.9382 = 0.094138ω22 + 0.9375
ω22 = 21.253
ω2 = 4.6101 rad/s
ω2 = 4.61 rad/s

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1728
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 = rω
Position 1. At rest.
ω=
v
r
T1 = 0
Position 2. Cylinder has fallen through distance s.
T2 =
1
1
mv 2 + I ω 2
2
2
1
11
 v 
mv 2 +  mr 2  
2
2 2
 r 
3
= mv 2
4
2
=
Work.
U1→2 = mgs
Principle of work and energy.
3
mv 2
4
4
gs

v2 =
3
T1 + U1→ 2 = T2: 0 + mgs =
v=
4 gs

3
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1737
PROBLEM 17.55
Two disks of the same thickness and same material are attached to a
shaft as shown. The 8-lb disk A has a radius rA = 3 in., and disk B has a
radius rB = 4.5 in. Knowing that a couple M of magnitude 20 lb ⋅ in. 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
2
r 
WB =  B  WB
 rA 
Weight of disk B.
2
 4.5 in. 
=
 (8 lb)
 3 in. 
= 18 lb
I = I A + IB
Moment of inertia.
2
1 8 lb  3 
1 18 lb  4.5 
ft  +
ft
=

2 32.2  12 
2 32.2  12 
2
= 0.04707 lb ⋅ ft ⋅ s 2
Angular velocity.
ω2 = 960 rpm = 100.53 rad/s
Moment.
M = 20 lb ⋅ in. = 1.667 lb ⋅ ft
Principle of impulse and momentum.
Syst. Momenta1 + Syst. Ext. Imp.1→2 =
Moments about C:
Required time.
Syst. Momenta2
0 + Mt = I ω2
I ω2
M
(0.04707 lb ⋅ ft ⋅ s2 )(100.53 rad/s)
=
1.667 lb ⋅ ft
t=
t = 2.839 s
t = 2.84 s 
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1791
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
rC = 0.150 m
For the double pulley,
rB = 0.080 m
k = 0.100 m
Principle of impulse and momentum.
Syst. Momenta1
+
Syst. Momenta 2
v = rC ω
Kinematics. Point C is the instantaneous center.
Moments about C:
=
Syst. Ext. Imp. 1→ 2
0 + Pt (rC + rB ) − mgtrC = I ω + mvrC
= mk 2ω + m(rC ω )rC
ω=
Pt (rC + rB ) − mgtrC
(
m k 2 + rC2
)
(24)(1.5)(0.230) − (3)(9.81)(1.5)(0.150)
3(0.1002 + 0.1502 )
= 17.0077 rad/s
=
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1813
PROBLEM 17.71 (Continued)
(a)
v = (0.150) (17.0077) = 2.55115 m/s
Linear components:
v = 2.55 m/s 
0 + Pt − mgt + Qt = mv
mv
+ mg − P
t
(3)(2.55115)
=
+ (3)(9.81) − 24
1.5
Q=
(b)
Q = 10.53 N 
Tension in cord C.
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1814
PROBLEM 17.80
A 2.5-lb disk of radius 4 in. is attached to the yoke BCD by means
of short shafts fitted in bearings at B and D. The 1.5-lb yoke has a
radius of gyration of 3 in. about the x axis. Initially the assembly is
rotating at 120 rpm with the disk in the plane of the yoke (θ = 0).
If the disk is slightly disturbed and rotates with respect to the yoke
until θ = 90°, where it is stopped by a small bar at D, determine
the final angular velocity of the assembly.
SOLUTION
2
 1.5   3 
−3
2
I C = mkC2 = 
  12  = 2.9115 × 10 lb ⋅ s ⋅ ft
32.2

 
Moment of inertia of yoke:
Moment of inertia of disk:
1
θ = 0: I A = mr 2
4
=
1  2.5  4 

4  32.2 
 12 
2
= 2.15666 × 10−3 lb ⋅ s 2 ⋅ ft
1
θ = 90°: I A = mr 2
2
=
1  2.5  4 

2  32.2 
 12 
2
= 4.3133 × 10−3 lb ⋅ s 2 ⋅ ft
Total moment of inertia about the x axis:
θ = 0: ( I x )1 = I C + I A
= 5.0682 × 10−3 lb ⋅ s 2 ⋅ ft
θ = 90°: ( I x )2 = I C + I A
= 7.2248 × 10−3 lb ⋅ s 2 ⋅ ft
Angular momentum about the x axis:
θ = 0: H1 = ( I x )1ω1
= 5.0682 × 10−3ω 1
θ = 90°: H 2 = ( I x ) 2 ω2
= 7.2248 × 10−3ω 2
Conservation of angular momentum.
H1 = H 2 : 5.0682 × 10−3 ω1 = 7.2248 × 10−3 ω2
ω2 = 0.7015ω1 = (0.7015)(120 rpm)
ω2 = 84.2 rpm 
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1831