18 solution q. E. D

Principle of Work and Energy.
T
1
+
Σ
U
1
-
2
=
T
2
0
+
160 s
G
+
(
-
100 s
G
2
)
=
0
160 s
G
-
100 s
G
2
=
0
s
G
(160
-
100 s
G
)
=
0
Since s
G
≠
0, then
160
-
100 s
G
=
0
s
G
=
1.60 m 
Ans.
Ans:
s
G
=
1.60 m

---

937
© 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. 
18–26.
A
B
1.5 ft
3 ft
u
Two wheels of negligible weight are mounted at corners A
and B of the rectangular 75-lb plate. If the plate is released
from rest at 
, determine its angular velocity at the
instant just before 
.
u
=
0°
u
=
90°
SOLUTION
Kinetic Energy and Work: Referring Fig. a,
The mass moment of inertia of the plate about its mass center is
. Thus, the final
kinetic energy is
Since the plate is initially at rest,
. Referring to Fig. b,
and 
do no work,
while 
does positive work. When 
,
displaces vertically through a distance
of 
, Fig. c. Thus, the work done by 
W is 
Principle of Work and Energy:
Ans.
v
2
=
5.37 rad
>
s
0
+
125.78
=
4.3672
v
2
2
T
1
+ ©
U
1
-
2
=
T
2
U
W
=
Wh
=
75(1.677)
=
125.78 ft
#
lb
h
=
2
0.75
2
+
1.5
2
=
1.677 ft
W
u
=
0°
W
N
B
N
A
T
1
=
0
=
4.3672
v
2
2
=
1
2
a
75
32.2
b
(1.677
v
2
)
2
+
1
2
I
G
(2.1836)
v
2
2
T
2
=
1
2
m
(
v
G
)
2
2
+
1
2
v
2
2
I
G
=
1
12
m
(
a
2
+
b
2
)
=
1
12
a
75
32.2
b
(1.5
2
+
3
2
)
=
2.1836 slug
#
ft
2
(
v
G
)
2
=
v
r
A
>
IC
=
v
a
2
0.75
2
+
1.5
2
b
=
1.677
v
2
Ans:
v
2
=
5.37 rad
>
s

---

938
© 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. 
18–27.
The link AB is subjected to a couple moment of 
M
=
40 N
#
m. If the ring gear C is fixed, determine the 
angular velocity of the 15-kg inner gear when the link has 
made two revolutions starting from rest. Neglect the mass 
of the link and assume the inner gear is a disk. Motion 
occurs in the vertical plane.
150 mm
200 mm
M
40 N 
m 
A
B
C
Solution
Kinetic Energy. The mass moment of inertia of the inner gear about its center 
B
is I
B
=
1
2
mr
2
=
1
2
(15)
(
0.15
2
)
=
0.16875 kg
#
m
2
. Referring to the kinematics 
diagram of the gear, the velocity of center B of the gear can be related to the gear’s 
angular velocity, which is
v
B
=
v
r
B
>
IC
;
v
B
=
v
(0.15)
Thus,
T
=
1
2
I
B
v
2
+
1
2
M
v
G
2
=
1
2
(0.16875) 
v
2
+
1
2
(15)[
v
(0.15)]
2
=
0.253125 
v
2
Since the gear starts from rest, T
1
=
0.
Work. Referring to the FBD of the gear system, we notice that M does positive 
work whereas W does no work, since the gear returns to its initial position after 
the link completes two revolutions. 
U
M
=
M
u
=
40[2(2
p
)]
=
160
p
J
Principle of Work and Energy. 
T
1
+
Σ
U
1
-
2
=
T
2
0
+
160
p
=
0.253125 
v
2
v
=
44.56 rad
>
s
=
44.6 rad
>
s 
Ans.
Ans:
v
=
44.6 rad
>
s

---

939
© 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. 
*18–28.
SOLUTION
Free Body Diagram
: The spring force F
sP
does negative work since it acts in the
opposite direction to that of its displacement s
sp
, whereas the weight of the
cylinder acts in the same direction of its displacement s
w
and hence does positive
work. Also, the couple moment 
M does positive work as it acts in the same
direction of its angular displacement . The reactions A
x
and A
y
do no work since
point A does not displace. Here,
and
.
Principle of Work and Energy: The mass moment of inertia of the cylinder about
point A is 
.
Applying Eq.18–13, we have
Ans.
v
=
4.60 rad
>
s
0
+
10(9.81)(0.1373)
+
15
a
p
3
-
p
6
b
-
1
2
(40)
A
0.2745
2
B
=
1
2
(1.875) 
v
2
0
+
Ws
W
+
M
u
-
1
2
ks
P
2
=
1
2
I
A
v
2
T
1
+
a
U
1
-
2
=
T
2
I
A
=
1
12
ml
2
+
md
2
=
1
12
(10)
A
0.75
2
B
+
10
A
0.375
2
B
=
1.875 kg
#
m
2
s
W
=
0.375 cos 30°
-
0.375 cos 60°
=
0.1373 m
s
sp
=
0.75 sin 60°
-
0.75 sin 30°
=
0.2745 m
u
The 10-kg rod AB is pin-connected at A and subjected to
a couple moment of M
15 N m. If the rod is released
from rest when the spring is unstretched at 
30 ,
determine the rod’s angular velocity at the instant 
60 .
As the rod rotates, the spring always remains horizontal,
because of the roller support at C.
u
u
.
C
A
B
k
= 40 N/m
M
= 15 N · m
0.75 m
Ans:
v
=
4.60 rad
>
s

---

940
© 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. 
18–29.
SOLUTION
Kinematics:
Ans.
v
G
=
11.9 ft
>
s
0
+
10(10.5)(1
-
cos 45°)
=
1
2
a
10
32.2
b
v
2
G
+
1
2
c
2
5
a
10
32.2
b
(0.5)
2
d a
v
G
0.5
b
2
T
1
+ ©
U
1
-
2
=
T
2
v
G
=
0.5
v
G
The 10-lb sphere starts from rest at 
0° and rolls without
slipping down the cylindrical surface which has a radius of
10 ft. Determine the speed of the sphere’s center of mass
at the instant 
45°.
u
u
10 ft
0.5 ft
θ
Ans:
v
G
=
11.9 ft
>
s

---

941
© 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. 
18–30.
M
C
P
750 N
3 m
4 m
A
u
B
Motor M exerts a constant force of 
on the rope.
If the 100-kg post is at rest when 
, determine the
angular velocity of the post at the instant 
. Neglect
the mass of the pulley and its size, and consider the post as a
slender rod.
u
=
60°
u
=
0°
P
=
750 N
SOLUTION
Kinetic Energy and Work: Since the post rotates about a fixed axis,
The mass moment of inertia of the post about its mass center is
. Thus, the kinetic energy of the post is
This result can also be obtained by applying , where 
.
Thus,
Since the post is initially at rest,
. Referring to Fig. a,
,
, and 
do no
work, while 
does positive work and 
does negative work. When 
,
displaces , where 
and 
. Thus,
. Also,
displaces
vertically upwards through a distance of 
. Thus, the work
done by and 
is
Principle of Work and Energy:
Ans.
v
=
2.50 rad
>
s
0
+
[2210.14
-
1274.36]
=
150
v
2
T
1
+ ©
U
1
-
2
=
T
2
U
W
= -
Wh
= -
100
(9.81)(1.299)
= -
1274.36 J
U
P
=
Ps
P
=
750(2.947)
=
2210.14 J
W
P
h
=
1.5
sin
60°
=
1.299 m
W
s
P
=
5
-
2.053
=
2.947 m
A
¿
C
=
2
4
2
+
3
2
=
5 m
AC
=
2
4
2
+
3
2
-
2(4)(3)
cos
30°
=
2.053 m
s
P
=
A
¿
C
-
AC
P
u
=
60°
W
P
R
C
B
y
B
x
T
1
=
0
T
=
1
2
I
B
v
2
=
1
2
(300)
v
2
=
150
v
2
1
12
(100)(3
2
)
+
100
(1.5
2
)
=
300 kg
#
m
2
I
B
=
T
=
1
2
I
B
v
2
=
150
v
2
=
1
2
(100)[
v
(1.5)]
2
+
1
2
(75)
v
2
T
=
1
2
mv
G
2
+
1
2
I
G
v
2
I
G
=
1
12
(100)(3
2
)
=
75 kg
#
m
2
v
G
=
v
r
G
=
v
(1.5).