18 solution q. E. D

Potential Energy: With reference to the datum established in Fig. a, the gravitational
potential energy of block A at position 1 and 2 are
Kinetic Energy: Since the spool rotates about a fixed axis,
.
Here, the mass moment of inertia about the fixed axis passes through point O is
. Thus, the kinetic energy of the system is
Since the system is at rest initially,
Conservation of Energy:
Ans.
=
1.29 m
>
s
v
A
=
1.289 m
>
s
0
+
0
=
59
v
A
2
+
(
-
98.1)
T
1
+
V
1
=
T
2
+
V
2
T
1
=
0
=
1
2
(3.92)(5
v
A
)
2
+
1
2
(20)
v
A
2
=
59
v
A
2
T
=
1
2
I
O
v
2
+
1
2
m
A
v
A
2
I
O
=
mk
O
2
=
50
(0.280)
2
=
3.92 kg
#
m
2
v
=
v
A
r
A
=
v
A
0.2
=
5
v
A
V
2
=
(
V
g
)
2
= -
W
A
y
2
= -
20
(9.81)(0.5)
= -
98.1 J
V
1
=
(
V
g
)
1
=
W
A
y
1
=
20
(9.81)(0)
=
0
Ans:
v
A
=
1.29 m
>
s

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951
© 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. 
*18–40.
An automobile tire has a mass of 7 kg and radius of gyration
If it is released from rest at A on the incline,
determine its angular velocity when it reaches the horizontal
plane. The tire rolls without slipping.
k
G
=
0.3 m.
SOLUTION
Datum at lowest point.
Ans.
v
=
19.8 rad s
0
+
7(9.81)(5)
=
1
2
(7)(0.4
v
)
2
+
1
2
[7 (0.3)
2
]
v
2
+
0
T
1
+
V
1
=
T
2
+
V
2
n
G
=
0.4
v
0.4 m
30
°
5 m
G
A
B
0.4 m
Ans:
v
=
19.8 rad
>
s

---

952
© 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. 
18–41.
The spool has a mass of 20 kg and a radius of gyration of 
k
O
=
160 mm. If the 15-kg block A is released from rest, 
determine the distance the block must fall in order for the 
spool to have an angular velocity 
v
=
8 rad
>
s. Also, what is 
the tension in the cord while the block is in motion? Neglect 
the mass of the cord.
Solution
Kinetic Energy. The mass moment of inertia of the spool about its center 
O
is I
0
=
mk
0
2
=
20
(
0.16
2
)
=
0.512 kg
#
m
2
. The velocity of the block is 
v
b
=
v
s
r
s
=
v
s
(0.2). Thus,
T
=
1
2
I
0 
v
2
+
1
2
m
b
v
b
2
=
1
2
(0.512)
v
s
2
+
1
2
(15)[
v
s
(0.2)]
2
=
0.556 
v
s
2
Since the system starts from rest, T
1
=
0. When 
v
s
=
8 rad
>
s,
T
2
=
0.556
(
8
2
)
=
35.584 J
Potential Energy. With reference to the datum set in Fig. a, the initial and final 
gravitational potential energy of the block are 
(V
g
)
1
=
m
b
g y
1
=
0
(V
g
)
2
=
m
b
g (
-
y
2
)
=
15(9.81)(
-
s
b
)
=
-
147.15 s
b
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
0
=
35.584
+
(
-
147.15 s
b
)
s
b
=
0.2418 m
=
242 mm 
Ans.
Principle of Work and Energy. The final velocity of the block is 
(
v
b
)
2
=
(
v
s
)
2
r
s
=
8(0.2)
=
1.60 m
>
s. Referring to the FBD of the block, Fig. b and 
using the result of S
b
,
T
1
+
Σ
U
1
-
2
=
T
2
0
+
15(9.81)(0.2418)
-
T
(0.2418)
=
1
2
(15)
(
1.60
2
)
T
=
67.75 N
=
67.8 N 
Ans.
Ans:
s
b
=
242 mm
T
=
67.8 N
200 mm
A
O

---

953
© 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. 
18–42.
The spool has a mass of 20 kg and a radius of gyration of 
k
O
=
160 mm. If the 15-kg block A is released from rest, 
determine the velocity of the block when it descends 
600 mm.
Solution
Kinetic Energy. The mass moment of inertia of the spool about its center O 
is I
0
=
mk
0
2
=
20
(
0.16
2
)
=
0.512 kg
#
m
2
. The angular velocity of the spool is 
v
s
=
v
b
r
s
=
v
b
0.2
=
5
v
b
. Thus, 
T
=
1
2
I
0 
v
2
+
1
2
m
b
v
b
2
=
1
2
(0.512)(5
v
b
)
2
+
1
2
(15)
v
b
2
=
13.9
v
b
2
Since the system starts from rest, T
1
=
0.
Potential Energy. With reference to the datum set in Fig. a, the initial and final 
gravitational potential energies of the block are 
(V
g
)
1
=
m
b
g y
1
=
0
(V
g
)
2
=
m
b
g y
2
=
15(9.81)(
-
0.6)
=
-
88.29 J
Conservation of Energy.
T
1
+
V
1
=
T
2
+
V
2
0
+
0
=
13.9
v
b
2
+
(
-
88.29)
v
b
=
2.5203 m
>
s
=
2.52 m
>
s 
Ans.
200 mm
A
O
Ans:
v
b
=
2.52 m
>
s

---

954
© 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. 
18–43.
SOLUTION
Potent
i
al Ener
g
y
:
Datum is set at point A. When the ladder is at its initial and final
position, its center of gravity is located 5 ft and 
above
the datum. Its
initial and final gravitational potential energy are 
and
respectively. Thus, the initial and final potential
energy are
K
i
net
i
c Ener
g
y
:
The mass moment inertia of the ladder about point A is
. Since the ladder is initially at
rest, the initial kinetic energy is 
. The final kinetic energy is given by
Conservat
i
on of Ener
g
y
:
Applying Eq. 18–18, we have
Equat
i
on of Mot
i
on
:
The mass moment inertia of the ladder about its mass center is
. Applying Eq. 17–16, we have
If the ladder begins to slide, then 
. Thus, for 
u
>0,
Ans.
u
=
48.2°
45.0 sin 
u
(1 
- 
1.5 cos 
u
) = 0
=
45.0 sin 
u
(1 
- 
1.5 cos 
u
) = 0
-
24.15 sin 
u
cos 
u
)
A
x
= -
30
32.2
(48.3 sin 
u
-
48.3 sin
u
cos
u
+
30
32.2
[4.83 sin 
u
(5)] cos 
u
+c ©
F
x
=
m
(
a
G
)
x
;
A
x
= -
30
32.2
[9.66(1
-
cos 
u
)(5)] sin 
u
a
=
4.83 sin 
u
+ ©
M
A
= ©
(
M
k
)
A
;
-
30 sin 
u
(5)
= -
7.764
a
-
a
30
32.2
b
[
a
(5)](5) 
I
G
=
1
12
a
30
32.2
b
(10
2
)
=
7.764 slug
#
ft
2
v
2
=
9.66(1
-
cos 
u
)
0
+
150
=
15.53
v
2
+
150 cos 
u
T
1
+
V
1
=
T
2
+
V
2
T
2
=
1
2
I
A
v
2
=
1
2
(31.06)
v
2
=
15.53
v
2
T
1
=
0
I
A
=
1
12
a
30
32.2
b
(10
2
)
+
a
30
32.2
b
(5
2
)
=
31.06 slug
#
ft
2
V
1
=
150 ft
#
lb
V
2
=
150 cos 
u
ft
#
lb
30(5 cos 
u
)
=
150 cos 
u
ft
#
lb,
30(5)
=
150 ft 
#
lb
(5 cos 
u
) ft
A
10 ft
u
A uniform ladder having a weight of 30 lb is released from 
rest when it is in the vertical position. If it is allowed to fall 
freely, determine the angle 
u
at which the bottom end A 
starts to slide to the right of A. For the calculation, assume 
the ladder to be a slender rod and neglect friction at A. 
A
x
=
0