18 solution q. E. D

18–67. Continued
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
(86.20
+
148.19
+
109.12)
=
4.50(
v
A
)
2
2
+
(0
+
58.86
+
49.05)
(
v
A
)
2
=
7.2357 m
>
s
=
7.24 m
>
s 
Ans.
Ans:
(
v
A
)
2
=
7.24 m
>
s

---

983
© 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. 
Solution
Kinetic Energy. Since the system is released from rest, T
1
=
0. Referring to the 
kinematics diagram of the rod at final position with IC so located, Fig. a, 
r
A
>
IC
=
2 cos 30
°
=
1.7321 m r
B
>
IC
=
2 cos 60
°
=
1.00 m
r
Gr
>
IC
=
2
1
2
+
1.00
2
-
2(1)(1.00) cos 60
°
=
1.00 m
Then
(
v
A
)
2
=
(
v
r
)
2
(r
A
>
IC
); (
v
A
)
2
=
(
v
r
)
2
(1.7321)
(
v
r
)
2
=
0.5774(
v
A
)
2
(
v
B
)
2
=
(
v
r
)
2
(r
B
>
IC
); (
v
B
)
2
=
[0.5774(
v
A
)
2
](1.00)
=
0.5774(
v
A
)
2
(
v
Gr
)
2
=
(
v
r
)
2
(r
Gr
>
IC
); (
v
Gr
)
2
=
[0.5774(
v
A
)
2
](1.00)
=
0.5774(
v
A
)
2
Since the disk rolls without slipping, 
(
v
B
)
2
=
v
d
r
d
; 0.5774(
v
A
)
2
=
(
v
d
)
2
(0.5)
(
v
d
)
2
=
1.1547(
v
A
)
2
Thus, the kinetic energy of the system at final position is
T
2
=
1
2
m
r
(
v
Gr
)
2
2
+
1
2
I
Gr
(
v
r
)
2
2
+
1
2
m
d
(
v
B
)
2
2
+
1
2
I
B
(
v
d
)
2
2
+
1
2
m
c
(
v
A
)
2
2
=
1
2
(12)[0.5774(
v
A
)
2
]
2
+
1
2
c
1
12
(12)
(
2
2
)
d
[0.5774(
v
A
)
2
]
2
+
1
2
(3.0)[0.5774(
v
A
)
2
]
2
+
1
2
c
1
2
(30)
(
0.5
2
)
d
[1.1547(
v
A
)
2
]
2
+
1
2
(5)(
v
A
)
2
2
=
12.6667(
v
A
)
2
2
Potential Energy. Datum is set as shown in Fig. a. Here, 
S
B
=
2 cos 30
°
-
2 cos 45
°
=
0.3178 m
Then 
(y
d
)
1
=
0.3178 sin 30
°
=
0.1589 m
(y
r
)
1
=
0.3178 sin 30
°
+
1 sin 75
°
=
1.1248 m
(y
r
)
2
=
1 sin 60
°
=
0.8660 m
(y
c
)
1
=
0.3178 sin 30
°
+
2 sin 75
°
=
2.0908 m
(y
c
)
2
=
2 sin 60
°
=
1.7321 m
*18–68.
The system consists of a 30-kg disk A, 12-kg slender rod BA, 
and a 5-kg smooth collar A. If the disk rolls without slipping, 
determine the velocity of the collar at the instant 
u
=
30
°
. 
The system is released from rest when 
u
=
45
°
.
0.5 m
2 m
A
C
B
30
u

---

984
© 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. 
Thus, the gravitational potential energies of the disk, rod and collar at initial and 
final position are 
(V
d
)
1
=
m
d
g
(y
d
)
1
=
30(9.81)(0.1589)
=
46.77 J
(V
d
)
2
=
m
d
g
(y
d
)
2
=
0
(V
r
)
1
=
m
r
g
(y
r
)
1
=
12(9.81)(1.1248)
=
132.42 J
(V
r
)
2
=
m
r
g
(y
r
)
2
=
12(9.81)(0.8660)
=
101.95 J
(V
c
)
1
=
m
c
g
(y
c
)
1
=
5(9.81)(2.0908)
=
102.55 J
(V
c
)
2
=
m
c
g
(y
c
)
2
=
5(9.81)(1.7321)
=
84.96 J
Conservation of Energy. 
T
1
+
V
1
=
T
2
+
V
2
0
+
(46.77
+
132.42
+
102.55)
=
12.6667(
v
A
)
2
2
+
(0
+
101.95
+
84.96)
(
v
A
)
2
=
2.7362 m
>
s
=
2.74 m
>
s 
Ans.
*18–68. Continued
Ans:
(
v
A
)
2
=
2.74 m
>
s