18 solution q. E. D

Ans:
v
=
2.50 rad
>
s

---

942
© 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. 
18–31.
The linkage consists of two 6-kg rods AB and CD and a 
20-kg bar BD. When 
u
=
0
°
, rod AB is rotating with an 
angular velocity 
v
=
2 rad
>
s. If rod CD is subjected to a 
couple moment of M
=
30 N
#
m, determine 
v
AB
at the 
instant 
u
=
90
°
.
Solution
Kinetic Energy. The mass moment of inertia of each link about the axis of rotation 
is I
A
=
1
12
(6)
(
1
2
)
+
6
(
0.5
2
)
=
2.00 kg
#
m. The velocity of the center of mass of the 
bar is 
v
G
=
v
r
=
v
(1). Thus, 
T
=
2
a
1
2
I
A
v
2
b
+
1
2
M
b
v
G
2
=
2
c
1
2
(2.00)
v
2
d
+
1
2
(20)[
v
(1)]
2
=
12.0 
v
2
Initially, 
v
=
2 rad
>
s. Then
T
1
=
12.0
(
2
2
)
=
48.0 J
Work. Referring to the FBD of the assembly, Fig. a, the weights 
W
b
, W
c
and couple moment M do positive work when the links 
undergo an angular displacement 
u
. When 
u
=
90
°
=
p
2
rad, 
U
W
b
=
W
b
s
b
=
20(9.81)(1)
=
196.2 J
U
W
c
=
W
c
s
c
=
6(9.81)(0.5)
=
29.43 J
U
M
=
M
u
=
30
a
p
2
b
=
15
p
J
Principle of Work and Energy. 
T
1
+
Σ
U
1
-
2
=
T
2
48.0
+
[196.2
+
2(29.43)
+
15
p
]
=
12.0 
v
2
v
=
5.4020 rad
>
s
=
5.40 rad
>
s 
Ans.
Ans:
v
=
5.40 rad
>
s
1.5 m
1 m
1 m
u
v
M
30 N 
m
B
C
A
D

---

943
© 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. 
*18–32.
The linkage consists of two 6-kg rods AB and CD and a 
20-kg bar BD. When 
u
=
0
°
, rod AB is rotating with an 
angular velocity 
v
=
2 rad
>
s. If rod CD is subjected to a 
couple moment M
=
30 N
#
m, determine 
v
at the instant
u
=
45
°
.
Solution
Kinetic Energy. The mass moment of inertia of each link about the axis of rotation 
is I
A
=
1
12
(6)
(
1
2
)
+
6
(
0.5
2
)
=
2.00 kg
#
m
2
. The velocity of the center of mass of 
the bar is 
v
G
=
v
r
=
v
(1). Thus, 
T
=
2
a
1
2
I
A
v
A
2
b
2
+
1
2
m
b
v
G
2
=
2
c
1
2
(2.00)
v
2
d
+
1
2
(20)[
v
(1)]
2
=
12.0 
v
2
Initially, 
v
=
2 rad
>
s. Then
T
1
=
12.0
(
2
2
)
=
48.0 J
Work. Referring to the FBD of the assembly, Fig. a, the weights 
W
b
, W
c
and couple moment M do positive work when the links 
undergo an angular displacement 
u
. when 
u
=
45
°
=
p
4
rad, 
U
W
b
=
W
b
s
b
=
20(9.81)
(
1
-
cos 45
°
)
=
57.47 J
U
W
c
=
W
c
s
c
=
6(9.81)[0.5(1
-
cos 45
°
)]
=
8.620 J
U
M
=
M
u
=
30
a
p
4
b
=
7.5
p
J
Principle of Work and Energy.
T
1
+
Σ
U
1
-
2
=
T
2
48.0
+
[57.47
+
2(8.620)
+
7.5
p
]
=
12.0 
v
2
v
=
3.4913 rad
>
s
=
3.49 rad
>
s
Ans.
1.5 m
1 m
1 m
u
v
M
30 N 
m
B
C
A
D
Ans:
v
=
3.49 rad
>
s

---

944
© 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. 
18–33.
The two 2-kg gears A and B are attached to the ends of a 
3-kg slender bar. The gears roll within the fixed ring gear C,
which lies in the horizontal plane. If a 
torque is
applied to the center of the bar as shown, determine the
number of revolutions the bar must rotate starting from rest
in order for it to have an angular velocity of 
For the calculation, assume the gears can be approximated by
thin disks.What is the result if the gears lie in the vertical plane?
v
AB
=
20 rad
>
s.
10-N
#
m
SOLUTION
Energy equation (where G refers to the center of one of the two gears):
,
,
g
n
i
s
U
and 
,
When 
Ans.
=
0.891 rev, regardless of orientation
u
=
5.60 rad
v
AB
=
20 rad
>
s,
10
u
=
0.0225
a
200
150
b
2
v
2
AB
+
2(0.200)
2
v
2
AB
+
0.0200
v
2
AB
v
gear
=
200
150
v
AB
I
AB
=
1
12
(3)(0.400)
2
=
0.0400 kg
#
m
2
I
G
=
1
2
(2)(0.150)
2
=
0.0225 kg 
#
m
2
m
gear
=
2 kg
10
u
=
2
a
1
2
I
G
v
2
gear
b
+
2
a
1
2
m
gear
b
(0.200
v
AB
)
2
+
1
2
I
AB
v
2
AB
M
u
=
T
2
400 mm
150 mm
M
= 10 N · m
150 mm
A
B
C
Ans:
u
=
0.891 rev,
regardless of orientation

---

945
© 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. 
18–34.
SOLUTION
T
1
+
Σ
U
1
-
2
=
T
2
2 
c
1
2
e
1
3
a
8
32.2
b
(2)
2
f
(2)
2
d
+
1
2
a
10
32.2
b
(4)
2
+
c
20(2)
+
15 
a
p
2
b
-
2(8)(1)
-
10(2)
d
=
2 
c
1
2
e
1
3
a
8
32.2
b
(2)
2
f
v
2
d
+
1
2
a
10
32.2
b
(2
v
)
2
v
=
5.74 rad
>
s 
Ans.
The linkage consists of two 8-lb rods 
AB
and 
CD
and 
a 10-lb bar 
AD
. When 
u
=
0
°
, rod 
AB
is rotating with an 
angular velocity 
v
AB
=
2 rad
>
s. If rod 
CD
is subjected to a 
couple moment 
M
=
15 lb
#
ft and bar 
AD
is subjected to a 
horizontal force 
P
=
20 lb as shown, determine 
v
AB
at the 
instant 
u
=
90
°
.
Ans:
v
=
5.74 rad
>
s
3 ft
2 ft
2 ft
M
15 lb · ft
B
C
A
D
P
20 lb
u
v
AB

---

946
© 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. 
18–35.
SOLUTION
T
1
+
Σ
U
1
-
2
=
T
2
2 
c
1
2
e
1
3
a
8
32.2
b
(2)
2
f
(2)
2
d
+
1
2
a
10
32.2
b
(4)
2
+
c
20(2 sin 45
°
)
+
15
a
p
4
b
-
2(8)(1
-
cos 45
°
)
-
10(2
-
2 cos 45
°
)
d
=
2
c
1
2
e
1
3
a
8
32.2
b
(2)
2
f
v
2
d
+
1
2
a
10
32.2
b
(2
v
)
2
v
=
5.92 rad
>
s 
Ans.
The linkage consists of two 8-lb rods 
AB
and 
CD
and 
a 10-lb bar 
AD
. When 
u
=
0
°
, rod 
AB
is rotating with an 
angular velocity 
v
AB
=
2 rad
>
s. If rod 
CD
is subjected to a 
couple moment 
M
=
15 lb
#
ft and bar 
AD
is subjected to a 
horizontal force 
P
=
20 lb as shown, determine 
v
AB
at the 
instant 
u
=
45
°
.
3 ft
2 ft
2 ft
M
= 15 lb · ft
AB
B
C
A
D
P
= 20 lb
Ans:
v
AB
=
5.92 rad
>
s

---

947
© 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. 
*18–36.
The assembly consists of a 3-kg pulley A and 10-kg pulley B. 
If a 2-kg block is suspended from the cord, determine the 
block’s speed after it descends 0.5 m starting from rest. 
Neglect the mass of the cord and treat the pulleys as thin 
disks. No slipping occurs.
A
B
30 mm
100 mm
Solution
T
1
+
V
1
=
T
2
+
V
2
[0
+
0
+
0]
+
[0]
=
1
2
c
1
2
(3)(0.03)
2
d
v
2
A
+
1
2
c
1
2
(10)(0.1)
2
d
v
2
B
+
1
2
(2)(
v
C
)
2
-
2(9.81)(0.5)
v
C
=
v
B
(0.1)
=
0.03
v
A
Thus, 
v
B
=
10
v
C
v
A
=
33.33
v
C
Substituting and solving yields, 
v
C
=
1.52 m
>
s 
Ans.
Ans:
v
C
=
1.52 m
>
s

---

948
© 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. 
18–37.
The assembly consists of a 3-kg pulley A and 10-kg pulley B. 
If a 2-kg block is suspended from the cord, determine the 
distance the block must descend, starting from rest, in order 
to cause B to have an angular velocity of 6 rad
>
s. Neglect 
the mass of the cord and treat the pulleys as thin disks. No 
slipping occurs.
A
B
30 mm
100 mm
Solution
v
C
=
v
B
(0.1)
=
0.03 
v
A
If 
v
B
=
6 rad
>
s then
v
A
=
20 rad
>
s
v
C
=
0.6 m
>
s
T
1
+
V
1
=
T
2
+
V
2
[0
+
0
+
0]
+
[0]
=
1
2
c
1
2
(3)(0.03)
2
d
(20)
2
+
1
2
c
1
2
(10)(0.1)
2
d
(6)
2
+
1
2
(2)(0.6)
2
-
2(9.81)s
C
s
C
=
78.0 mm