18 solution q. E. D

Ans:
v
=
14.0 rad
>
s

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914
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
18–3.
The wheel is made from a 5-kg thin ring and two 2-kg slender
rods. If the torsional spring attached to the wheel’s center has
a stiffness 
so that the torque on the center
of the wheel is 
where 
is in radians,
determine the maximum angular velocity of the wheel if it is
rotated two revolutions and then released from rest.
u
M
=
1
2
u
2
N
#
m,
k
=
2 N
#
m
>
rad,
SOLUTION
Ans.
v
=
14.1 rad/s
(4
p
)
2
=
0.7917
v
2
0
+
L
4
p
0
2
u
d
u
=
1
2
(1.583) 
v
2
T
1
+ ©
U
1
-
2
=
T
2
I
o
=
2
c
1
12
(2)(1)
2
d
+
5(0.5)
2
=
1.583
M
O
0.5 m
Ans:
v
=
14.1 rad
>
s

---

915
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*18–4.
A force of P
=
60 N is applied to the cable, which causes 
the 200-kg reel to turn since it is resting on the two rollers 
A
and B of the dispenser. Determine the angular velocity of 
the reel after it has made two revolutions starting from rest. 
Neglect the mass of the rollers and the mass of the cable. 
Assume the radius of gyration of the reel about its center 
axis remains constant at k
O
=
0.6 m.
Solution
Kinetic Energy. Since the reel is at rest initially, T
1
=
0. The mass moment of inertia 
of the reel about its center O is I
0
=
mk
0
2
=
200
(
0.6
2
)
=
72.0 kg
#
m
2
. Thus,
T
2
=
1
2
I
0 
v
2
=
1
2
(72.0)
v
2
=
36.0 
v
2