18 solution q. E. D

Potential Energy. With reference to the datum set in Fig. b, the initial and final 
gravitational potential energies of the system are
(V
g
)
1
=
2mgy
1
=
2[12(9.81)(1 sin 60
°
)]
=
203.90 J
(V
g
)
2
=
2mgy
2
=
0
The stretch of the spring when the system is at initial and final position are
x
1
=
2(2 cos 60
°
)
-
1.5
=
0.5 m
x
2
=
4
-
1.5
=
2.50 m
Thus, the initial and final elastic potential energies of the spring is
(V
e
)
1
=
1
2
kx
2
1
=
1
2
(20)
(
0.5
2
)
=
2.50 J
(V
e
)
2
=
1
2
kx
2
2
=
1
2
(20)
(
2.50
2
)
=
62.5 J

---

965
© 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. 
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
(203.90
+
2.50)
=
16.0(
v
BC
)
2
2
+
(0
+
62.5)
(
v
BC
)
2
=
2.9989 rad
>
s
=
3.00 rad
>
s 
Ans.
18–52. Continued
Ans:
(
v
BC
)
2
=
3.00 rad
>
s

---

966
© 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. 
18–53.
The two 12-kg slender rods are pin connected and released 
from rest at the position 
u
=
60
°
. If the spring has an 
unstretched length of 1.5 m, determine the angular velocity 
of rod BC, when the system is at the position 
u
=
30
°
.
Solution
Kinetic Energy. Since the system is released from rest, T
1
=
0 . Referring to the 
kinematics diagram of rod BC at final position with IC so located, Fig. a 
r
B
>
IC
=
r
C
>
IC
=
2 m r
G
>
IC
=
2 sin 60
°
=
2
3 m
Thus,
(V
B
)
2
=
(
v
BC
)
2 
r
B
>
IC
;
(V
B
)
2
=
(
v
BC
)
2
(2)
(V
C
)
2
=
(
v
BC
)
2 
r
C
>
IC
;
(V
C
)
2
=
(
v
BC
)
2
(2)
(V
G
)
2
=
(
v
BC
)
2 
r
G
>
IC
;
(V
G
)
2
=
(
v
BC
)
2
(
2
3
)
Then for rod AB,
(V
B
)
2
=
(
v
AB
)
2 
r
AB
;
(
v
BC
)
2
(2)
=
(
v
AB
)
2
(2)
(
v
AB
)
2
=
(
v
BC
)
2
Thus,
T
2
=
1
2
I
A
(
v
AB
)
2
2
+
1
2
I
G
(
v
BC
)
2
2
+
1
2
m
r
(V
G
)
2
2
=
1
2
c
1
3
(12)
(
2
2
)
d
(
v
BC
)
2
2
+
1
2
c
1
12
(12)
(
2
2
)
d
(
v
BC
)
2
2
+
1
2
(12)[(
v
BC
)
2
1
3]
2
=
28.0(
v
BC
)
2
2
Potential Energy. With reference to the datum set in Fig. b, the initial and final 
gravitational potential energy of the system are
(V
g
)
1
=
2mgy
1
=
2[12(9.81)(1 sin 60
°
)]
=
203.90 J
(V
g
)
2
=
2mgy
2
=
2[12(9.81)(1 sin 30
°
)]
=
117.72 J
The stretch of the spring when the system is at initial and final position are
x
1
=
2(2 cos 60
°
)
-
1.5
=
0.5 m
x
2
=
2(2 cos 30
°
)
-
1.5
=
1.9641 m
Thus, the initial and final elastic potential energy of the spring are
(V
e
)
1
=
1
2
kx
2
1
=
1
2
(20)
(
0.5
2
)
=
2.50 J
(V
e
)
2
=
1
2
kx
2
2
=
1
2
(20)
(
1.9641
2
)
=
38.58 J
A
C
B
2 m
k
20 N
/
m
2 m
u

---

967
© 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. 
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
(203.90
+
2.50)
=
28.0(
v
BC
)
2
2
+
(117.72
+
38.58)
v
BC
=
1.3376 rad
>
s
=
1.34 rad
>
s 
Ans.
18–53. Continued
Ans:
v
BC
=
1.34 rad
>
s

---

968
© 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. 
18–54.
k 
75 lb
/
ft
0.75 ft
0.375 ft
O
If the 250-lb block is released from rest when the spring is
unstretched, determine the velocity of the block after it has
descended 5 ft. The drum has a weight of 50 lb and a radius
of gyration of 
about its center of mass O.
k
O
=
0.5 ft
SOLUTION
Potential Energy: With reference to the datum shown in Fig. a, the gravitational
potential energy of the system when the block is at position 1 and 2 is
When the block descends , the drum rotates through an angle of
. Thus, the stretch of the spring is 
. The elastic potential energy of the spring is
Since the spring is initially unstretched,
. Thus,
Kinetic Energy: Since the drum rotates about a fixed axis passing through point O,
. The mass moment of inertia of the drum about its mass
center is 
.
Since the system is initially at rest,
.
Ans.
T
v
b
=
15.5 ft
>
s
0
+
0
=
4.2271
v
b
2
-
1015.625
T
1
+
V
1
=
T
2
+
V
2
T
1
=
0
=
4.2271
v
b
2
=
1
2
(0.3882)(1.333
v
b
)
2
+
1
2
a
250
32.2
b
v
b
2
T
=
1
2
I
O
v
2
+
1
2
m
b
v
b
2
I
O
=
mk
O
2
=
50
32.2
a
0.5
2
b
=
0.3882 slug
#
ft
2
v
=
v
b
r
b
=
v
b
0.75
=
1.333
v
b
V
2
=
(
V
g
)
2
+
(
V
e
)
2
= -
1250
+
234.375
= -
1015.625 ft
#
lb
V
1
=
(
V
g
)
1
+
(
V
e
)
1
=
0
(
V
e
)
1
=
0
(
V
e
)
2
=
1
2
kx
2
=
1
2
(75)(2.5
2
)
=
234.375 ft
#
lb
r
sp
u
+
0
=
0.375(6.667)
=
2.5 ft
x
=
s
+
s
0
=
u
=
s
b
r
b
=
5
0.75
=
6.667 rad
s
b
=
5 ft
(
V
g
)
2
= -
W
(
y
G
)
2
= -
250(5)
= -
1250 ft
#
lb
(
V
g
)
1
=
W
(
y
G
)
1
=
250(0)
=
0
Ans:
v
b
=
15.5 ft
>
s

---

969
© 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. 
18–55.
The slender 15-kg bar is initially at rest and standing in the 
vertical position when the bottom end A is displaced slightly 
to the right. If the track in which it moves is smooth, 
determine the speed at which end A strikes the corner D. 
The bar is constrained to move in the vertical plane. Neglect 
the mass of the cord BC.
Solution
x
2
+
y
2
=
5
2
x
2
+
(7
-
y
)
2
=
4
2
Thus, 
y
=
4.1429 m
x
=
2.7994 m
(5)
2
=
(4)
2
+
(7)
2
-
2(4)(7) cos 
f
f
=
44.42
°
h
2
=
(2)
2
+
(7)
2
-
2(2)(7) cos 44.42
°
h
=
5.745 m
T
1
+
V
1
=
T
2
+
V
2
0
+
147.15(2)
=
1
2
c
1
12
(15)(4)
2
d
v
2
+
1
2
(15)(5.745
v
)
2
+
147.15 
a
7
-
4.1429
2
b
v
=
0.5714 rad
>
s
v
A
=
0.5714(7)
=
4.00 m
>
s 
Ans.
Ans:
v
A
=
4.00 m
>
s
4 m
4 m
5 m
A
D
B
C

---

970
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*18–56.
A
B
6 ft
4 ft
0.5 ft
If the chain is released from rest from the position shown,
determine the angular velocity of the pulley after the end B
has risen 2 ft.The pulley has a weight of 50 lb and a radius of
gyration of 0.375 ft about its axis. The chain weighs 6 lb/ft.
SOLUTION
Potential Energy:
,
,
, and 
. With
reference to the datum in Fig. a, the gravitational potential energy of the chain at
position 1 and 2 is
Kinetic Energy: Since the system is initially at rest,
. The pulley rotates about
a fixed axis, thus,
. The mass moment of inertia of
the pulley about its axis is 
. Thus, the
final kinetic energy of the system is
T
=
1
2
I
O
v
2
2
+
1
2
m
1
(
V
G
1
)
2
2
+
1
2
m
2
(
V
G
2
)
2
2
I
O
=
mk
O
2
=
50
32.2
(0.375
2
)
=
0.2184 slug
#
ft
2
(
V
G
1
)
2
=
(
V
G
2
)
2
=
v
2
r
=
v
2
(0.5)
T
1
=
0
= -
6(2)(1)
-
6(8)(4)
= -
204
ft
#
lb
V
2
=
(
V
g
)
2
= -
W
1
(
y
G
1
)
2
+
W
2
(
y
G
2
)
2
= -
6(4)(2)
-
6(6)(3)
= -
156
ft
#
lb
V
1
=
(
V
g
)
1
=
W
1
(
y
G
1
)
1
-
W
2
(
y
G
2
)
1
(
y
G
2
)
2
=
4 ft
(
y
G
1
)
2
=
1
ft
(
y
G
2
)
1
=
3 ft
(
y
G
1
)
1
=
2 ft
+
1
2
c
6(0.5)(
p
)
32.2
d
[
v
2
(0.5)]
2
=
1
2
(0.2184)
v
2
2
+
1
2
c
6(2)
32.2
d
[
v
2
(0.5)]
2
+
1
2
c
6(8)
32.2
d
[
v
2
(0.5)]
2