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[10(9.81)(1.5 sin 60
°
)]
=
254.87 J
(V
g
)
2
=
2mgy
2
=
0
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
254.87
=
30.0(
v
BC
)
2
2
+
0
(
v
BC
)
2
=
2.9147 rad
>
s
=
2.91 rad
>
s 
Ans.
(
v
AB
)
2
=
(
v
BC
)
2
=
2.91 rad
>
s 
Ans.
Ans:
(
v
BC
)
2
=
2.91 rad
>
s
(
v
AB
)
2
=
2.91 rad
>
s

---

960
© 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. 
18–49.
The assembly consists of two 10-kg bars which are pin 
connected. If the bars are released from rest when 
u
=
60
°
, determine their angular velocities at the instant
u
=
30
°
. The 5-kg disk at C has a radius of 0.5 m and rolls 
without slipping.
A
3 m
3 m
C
B
u
u
Solution
Kinetic Energy. Since the system is released from rest, T
1
=
0. Referring to the 
kinematics diagram of bar BC at final position with IC so located, Fig. a, 
r
B
>
IC
=
r
C
>
IC
=
3 m r
G
>
IC
=
3 sin 60
°
=
1.5
2
3 m
Thus,
(
v
B
)
2
=
(
v
BC
)
2 
r
B
>
IC
; (
v
B
)
2
=
(
v
BC
)
2
(3)
(
v
C
)
2
=
(
v
BC
)
2 
r
C
>
IC
; (
v
C
)
2
=
(
v
BC
)
2
(3)
(
v
G
)
2
=
(
v
BC
)
2 
r
G
>
IC
;
(
v
G
)
2
=
(
v
BC
)
2
(
1.5
2
3
)
Then for rod AB,
(
v
B
)
2
=
(
v
AB
)
2 
r
AB
; (
v
BC
)
2
(3)
=
(
v
AB
)
2
(3)
(
v
AB
)
2
=
(
v
BC
)
2
For the disk, since it rolls without slipping, 
(
v
C
)
2
=
(
v
d
)
2
r
d
; (
v
BC
)
2
(3)
=
(
v
d
)
2
(0.5)
(
v
d
)
2
=
6(
v
BC
)
2
Thus, the kinetic energy of the system at final position is
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
I
C
(
v
d
)
2
2
+
1
2
m
d
(
v
C
)
2
2
=
1
2
c
1
3
(10)
(
3
2
)
d
(
v
BC
)
2
2
+
1
2
c
1
12
(10)
(
3
2
)
d
(
v
BC
)
2
2
+
1
2
(10)
c
(
v
BC
)
2
(
1.5
2
3
)
d
2
+
1
2
c
1
2
(5)
(
0.5
2
)
d
[6(
v
BC
)
2
]
2
+
1
2
(5)[(
v
BC
)
2
(3)]
2
=
86.25(
v
BC
)
2
2
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[10(9.81)(1.5 sin 60
°
)]
=
254.87 J
(V
g
)
2
=
2mgy
2
=
2[10(9.81)(1.5 sin 30
°
)]
=
147.15 J

---

961
© 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
+
254.87
=
86.25(
v
BC
)
2
2
+
147.15
(
v
BC
)
2
=
1.1176 rad
>
s
=
1.12 rad
>
s 
Ans.
(
v
AB
)
2
=
(
v
BC
)
2
=
1.12 rad
>
s 
Ans.
18–49. Continued
Ans:
(
v
AB
)
2
=
(
v
BC
)
2
=
1.12 rad
>
s

---

962
© 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. 
18–50.
SOLUTION
Set 
,
Substituting and solving yields,
Ans.
v
A
=
0.1(14.04)
=
1.40 m
>
s
v
=
14.04 rad
>
s
s
B
=
0.06 m
s
A
=
0.2 m
s
B
=
0.3
s
A
u
=
s
B
0.03
=
s
A
0.1
[0
+
0
+
0]
+
[0
+
0]
=
1
2
[3(0.045)
2
]
v
2
+
1
2
(2)(0.03
v
)
2
+
1
2
(2)(0.1
v
)
2
-
2(9.81)
s
A
+
2(9.81)
s
B
T
1
+
V
1
=
T
2
+
V
2
The compound disk pulley consists of a hub and attached
outer rim. If it has a mass of 3 kg and a radius of gyration
determine the speed of block A after A
descends 0.2 m from rest. Blocks A and B each have a mass
of 2 kg. Neglect the mass of the cords.
k
G
=
45 mm,
B
100 mm
30 mm
A
Ans:
v
A
=
1.40 m
>
s

---

963
© 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. 
18–51.
The uniform garage door has a mass of 150 kg and is guided
along smooth tracks at its ends. Lifting is done using the two
springs, each of which is attached to the anchor bracket at A
and to the counterbalance shaft at B and C. As the door is
raised, the springs begin to unwind from the shaft, thereby
assisting the lift. If each spring provides a torsional moment of
, where is in radians, determine the angle
at which both the left-wound and right-wound spring
should be attached so that the door is completely balanced by
the springs, i.e., when the door is in the vertical position and is
given a slight force upwards, the springs will lift the door along
the side tracks to the horizontal plane with no final angular
velocity. Note: The elastic potential energy of a torsional
spring is 
, where 
and in this case
.
k
=
0.7 N
#
m
>
rad
M
=
k
u
V
e
=
1
2
k
u
2
u
0
u
M
=
(0.7
u
) N
#
m
SOLUTION
Datum at initial position.
Ans.
u
0
=
56.15 rad
=
8.94 rev.
0
+
2
c
1
2
(0.7)
u
0
2
d
+
0
=
0
+
150(9.81)(1.5)
T
1
+
V
1
=
T
2
+
V
2
3 m
4 m
C
A
B
Ans:
u
0
=
8.94 rev

---

964
© 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. 
*18–52.
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
=
0
°
. Neglect 
the mass of the roller at C.
A
C
B
2 m
k
20 N
/
m
2 m
u
Solution
Kinetic Energy. Since the system is released from rest, T
1
=
0 . Referring to the 
kinematics diagram of rod BC at the final position, Fig. a we found that IC is located 
at C. Thus, (
v
C
)
2
=
0. Also, 
(
v
B
)
2
=
(
v
BC
)
2 
r
B
>
IC
; (
v
B
)
2
=
(
v
BC
)
2
(2)
(
v
G
)
2
=
(
v
BC
)
2 
r
C
>
IC
; (
v
G
)
2
=
(
v
BC
)
2
(1)
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)]
2
=
16.0(
v
BC
)
2
2