18 solution q. E. D

W
P
h
=
0.3 sin 45°
W
s
P
=
0.6 m
P
u
=
45°
W
P
N
B
N
A
T
1
=
0
=
3.6
v
2
2
=
1
2
(15)[
w
2
(0.6708)]
2
+
1
2
(0.45) 
v
2
2
T
2
=
1
2
m
(
v
G
)
2
2
+
1
2
I
G
v
2
2
=
1
12
(15)(0.6
2
)
=
0.45 kg
#
m
2
I
G
=
1
12
ml
2
(
v
G
)
2
=
v
2
r
G
>
IC
=
v
2
(0.6708)
r
G
>
IC
=
3
0.3
2
+
0.6
2
=
0.6708 m
r
A
>
IC
=
0.6 tan 45°
=
0.6 m
Ans:
v
2
=
4.97 rad
>
s

---

932
© 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. 
18–21.
SOLUTION
Ans.
s
=
0.304 ft
0
+
(0.3)(
s
)
=
1
2
a
0.3
32.2
b
(1.40)
2
+
1
2
c
(0.06)
2
a
0.3
32.2
b d
(70)
2
T
1
+ ©
U
1
-
2
=
T
2
v
G
=
(0.02)70
=
1.40 ft
>
s
A yo-yo has a weight of 0.3 lb and a radius of gyration
If it is released from rest, determine how far it
must descend in order to attain an angular velocity
Neglect the mass of the string and assume
that the string is wound around the central peg such that the
mean radius at which it unravels is 
r
=
0.02 ft.
v
=
70 rad
>
s.
k
O
=
0.06 ft.
O
r
Ans:
s
=
0.304 ft

---

933
© 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. 
18–22.
SOLUTION
K
i
net
i
c Ener
g
y and Work
:
Since the windlass rotates about a fixed axis,
or 
. The mass moment of inertia of the windlass about its
mass center is
Thus, the kinetic energy of the system is
Since the system is initially at rest,
. Referring to Fig. a,
W
A
,
A
x
,
A
y
, and 
R
B
do no work, while 
W
C
does positive work. Thus, the work done by 
W
C
, when it
displaces vertically downward through a distance of 
, is
Pr
i
nc
ip
le of Work and Ener
g
y
:
Ans.
v
C
=
19.6 ft
>
s
0
+
500
=
1.2992v
C
2
T
1
+ ©
U
1
-
2
=
T
2
U
W
C
=
W
C
s
C
=
50(10)
=
500 ft
#
lb
s
C
=
10 ft
T
1
=
0
=
1.2992v
C
2
=
1
2
(0.2614)(2v
C
)
2
+
1
2
a
50
32.2
b
v
C
2
=
1
2
I
A
v
2
+
1
2
m
C
v
C
2
T
=
T
A
+
T
C
I
A
=
1
2
a
30
32.2
b
A
0.5
2
B
+
4
c
1
12
a
2
32.2
b
A
0.5
2
B
+
2
32.2
A
0.75
2
B
d
=
0.2614 slug
#
ft
2
v
A
=
v
C
r
A
=
v
C
0.5
=
2v
C
v
C
=
v
A
r
A
If the 50-lb bucket is released from rest, determine its
velocity after it has fallen a distance of 10 ft. The windlass A
can be considered as a 30-lb cylinder, while the spokes are
slender rods, each having a weight of 2 lb. Neglect the
pulley’s weight.
4 ft
0.5 ft
0.5 ft
3 ft
B
A
C
Ans:
v
C
=
19.6 ft
>
s

---

934
© 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. 
18–23.
The coefficient of kinetic friction between the 100-lb disk
and the surface of the conveyor belt is 
0.2. If the
conveyor belt is moving with a speed of 
when
the disk is placed in contact with it, determine the number
of revolutions the disk makes before it reaches a constant
angular velocity.
v
C
=
6 ft
>
s
m
A
SOLUTION
Equation of Motion
: In order to obtain the friction developed at point A of the
disk, the normal reaction N
A
must be determine first.
Work
:
e
l
p
u
o
c
t
n
a
t
s
n
o
c
a
s
p
o
l
e
v
e
d
n
o
i
t
c
i
r
f
e
h
T
t
n
i
o
p
t
u
o
b
a
f
o
t
n
e
m
o
m
O
when the disk is brought in
contact with the conveyor belt. This couple moment does positive work of
when the disk undergoes an angular displacement . The normal
reaction N, force F
OB
and the weight of the disk do no work since point O does not
displace.
Principle of Work and Energy
: The disk achieves a constant angular velocity
when the points on the rim of the disk reach the speed of that of the conveyor
belt,
i.e;
. This constant angular velocity is given by
. The mass moment inertia of the disk about point O is 
. Applying Eq.18–13, we have
Ans.
u
=
2.80 rad
*
1 rev
2
p
rad
=
0.445 rev
0
+
10.0
u
=
1
2
(0.3882) 
A
12.0
2
B
0
+
U
=
1
2
I
O
v
2
T
1
+
a
U
1
-
2
=
T
2
I
O
=
1
2
mr
2
=
1
2
a
100
32.2
b
A
0.5
2
B
=
0.3882 slug
#
ft
2
v
=
y
C
r
=
6
0.5
=
12.0 rad
>
s
y
C
=
6 ft
>
s
u
U
=
10.0(
u
)
M
=
20.0(0.5)
=
10.0 lb
#
ft
F
f
=
m
k
N
=
0.2(100)
=
20.0 lb
+ c ©
F
y
=
m
(
a
G
)
y
;
N
-
100
=
0
N
=
100 lb
C
= 6 ft/s
v
A
B
0.5 ft
Ans:
u
=
0.445 rev

---

935
© 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. 
*18–24.
The 30-kg disk is originally at rest, and the spring is 
unstretched. A couple moment of M
=
80 N
#
m is then 
applied to the disk as shown. Determine its angular velocity 
when its mass center G has moved 0.5 m along the plane. 
The disk rolls without slipping.
0.5 m
G
M
80 N 
m
k
200 N
/
m
A
Solution
Kinetic Energy. Since the disk is at rest initially, T
1
=
0. The disk rolls without 
slipping. Thus, 
v
G
=
v
r
=
v
(0.5). The mass moment of inertia of the disk about its 
center of gravity G is I
G
=
1
2
mr
=
1
2
(30)
(
0.5
2
)
=
3.75 kg
#
m
2
. Thus, 
T
2
=
1
2
I
G
v
2
+
1
2
M
v
G
2
=
1
2
(3.75)
v
2
+
1
2
(30)[
v
(0.5)]
2
=
5.625 
v
2
Work. Since the disk rolls without slipping, the friction F
f
does no work. Also when 
the center of the disk moves S
G
=
0.5 m, the disk rotates 
u
=
s
G
r
=
0.5
0.5
=
1.00 rad. 
Here, couple moment M does positive work whereas the spring force does negative 
work. 
U
M
=
M
u
=
80(1.00)
=
80.0 J
U
Fsp
=
-
1
2
kx
2
=
-
1
2
(200)
(
0.5
2
)
=
-
25.0 J
Principle of Work and Energy.
T
1
+
Σ
U
1
-
2
=
T
2
0
+
80
+
(
-
25.0)
=
5.625 
v
2
v
=
3.127 rad
>
s
=
3.13 rad
>
s 
Ans.
Ans:
v
=
3.13 rad
>
s

---

936
© 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. 
18–25.
The 30-kg disk is originally at rest, and the spring is 
unstretched. A couple moment M
=
80 N
#
m is then 
applied to the disk as shown. Determine how far the center 
of mass of the disk travels along the plane before it 
momentarily stops. The disk rolls without slipping.
0.5 m
G
M
80 N 
m
k
200 N
/
m
A
Solution
Kinetic Energy. Since the disk is at rest initially and required to stop finally, 
T
1
=
T
2
=
0.
Work. Since the disk rolls without slipping, the friction F
f
does no work. Also, when 
the center of the disk moves s
G
, the disk rotates 
u
=
s
G
r
=
s
G
0.5
=
2 s
G
. Here, couple 
moment M does positive work whereas the spring force does negative work. 
U
M
=
M
u
=
80(2 s
G
)
=
160 s
G
U
Fsp
=
-
1
2
kx
2
=
-
1
2
(200) s
G
2
=
-
100 s
G
2