18 solution q. E. D

Ans.
Ans:
s
C
=
78.0 mm

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949
© 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. 
18–38.
SOLUTION
System:
Ans.
Block:
Ans.
T
=
163 N
0
+
20(9.81)(0.30071)
-
T
(0.30071)
=
1
2
(20)(1)
2
T
1
+ ©
U
1
-
2
=
T
2
s
=
0.30071 m
=
0.301 m
[0
+
0]
+
0
=
1
2
(20)(1)
2
+
1
2
[50(0.280)
2
](5)
2
-
20(9.81) 
s
T
1
+
V
1
=
T
2
+
V
2
v
A
=
0.2
v
=
0.2(5)
=
1 m
>
s
The spool has a mass of 50 kg and a radius of gyration
If the 20-kg block A is released from rest,
determine the distance the block must fall in order for the
spool to have an angular velocity 
Also, what is
the tension in the cord while the block is in motion? Neglect
the mass of the cord.
v
=
5 rad
>
s.
k
O
=
0.280 m.
A
0.2 m
O
0.3 m
A
0.2 m
O
0.3 m
Ans:
s
=
0.301 m
T
=
163 N

---

950
© 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. 
18–39.
A
0.2 m
O
0.3 m
The spool has a mass of 50 kg and a radius of gyration
. If the 20-kg block A is released from rest,
determine the velocity of the block when it descends 0.5 m.
k
O
=
0.280 m
SOLUTION