PROBLEM 17.9
The 10-in.-radius brake drum is attached to a larger flywheel which
is not shown. The total mass moment of inertia of the flywheel and
drum is 16 lb ft s2 and the coefficient of kinetic friction between
the drum and the brake shoe is 0.40. Knowing that the initial angular
velocity is 240 rpm clockwise, determine the force which must be
exerted by the hydraulic cylinder if the system is to stop in 75
revolutions.
SOLUTION
Kinetic energies.
1
240 rpm 8 rad/s
I
16 lb ft s 2
T1
2
Angular displacement.
Work.
U1
1
I
2
0
1
(16)(8 ) 2
2
T2 0
2
1
75 rev
2
M
F
Principle of work and energy.
T1
k N:
75(2 ) 150 rad
10
ft
12
(150 rad)
U1
T2 :
5053 392.7 F
F
5053 ft lb
12.868
2
F
0
(0.40) N
N
392.7 F
12.868 lb
32.17 lb
Free body brake arm:
MA
0: B (6 in.)
F (6 in.)
N (18 in.)
0
B(6 in.) (12.868 lb)(6 in.) (32.17 lb)(18 in.) 0
B 83.64 lb
B
83.6 lb
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PROBLEM 17.11
Each of the gears A and B has a mass of 2.4 kg and a radius of
gyration of 60 mm, while gear C has a mass of 12 kg and a radius
of gyration of 150 mm. A couple M of constant magnitude 10 N m
is applied to gear C. Determine (a) the number of revolutions of
gear C required for its angular velocity to increase from 100 to
450 rpm, (b) the corresponding tangential force acting on gear A.
SOLUTION
Moments of inertia.
mk 2
(2.4)(0.06)2
Gears A and B:
IA
IB
Gear C:
IC
(12)(0.15)2
Kinematics.
Kinetic energy.
Position 1.
rA
A
rB
A
B
A
B
T
1
I
2
C
A
rC
B
200
80
2.5 C
2
270 10 3 kg m2
C
C
2.5
C
:
100 rpm
B
10
3
250 rpm
rad/s
25
3
rad/s
(T1 ) A
1
25
(8.64 10 3 )
2
3
2
Gear A:
(T1 ) B
1
25
(8.64 10 3 )
2
3
2
Gear B:
(T1 )C
1
10
(270 10 3 )
2
3
2
Gear C:
System:
T1
Position 2.
C
A
(T1 ) A
8.64 10 3 kg m2
(T1 ) B
(T1 ) C
2.9609 J
2.9609 J
14.8044 J
20.726 J
450 rpm 15 rad/s
B
37.5 rad/s
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PROBLEM 17.11 (Continued)
1
(8.64 10 3 )(37.5 )2
2
(T2 ) A
Gear A:
59.957 J
Gear B:
(T2 ) B
1
(8.64 10 3 )(37.5 )2
2
Gear C:
(T2 )C
1
(270 10 3 )(15 )2
2
299.789 J
System:
T2
(T2 ) A
419.7 J
Work of couple.
U1
2
M
C
(T2 ) B
10
59.957 J
(T2 ) C
C
Principle of work and energy for system.
T1
(a)
U1
2
T2 : 20.726 10
C
419.7
C
39.898 radians
Rotation of gear C.
C
Rotation of gear A.
A
6.35 rev
(2.5)(39.898)
99.744 radians
Principle of work and energy for gear A.
(T1 ) A
MA
A
(T2 ) A : 2.9609
M A (99.744)
MA
(b)
Tangential force on gear A.
Ft
59.957
0.57142 N m
MA
rA
0.57142
0.08
Ft
7.14 N
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PROBLEM 17.14
The double pulley shown has a mass of 15 kg and a
centroidal radius of gyration of 160 mm. Cylinder A
and block B are attached to cords that are wrapped on
the pulleys as shown. The coefficient of kinetic
friction between block B and the surface is 0.2.
Knowing that the system is at rest in the position
shown when a constant force P 200 N is applied to
cylinder A, determine (a) the velocity of cylinder A as
it strikes the ground, (b) the total distance that block
B moves before coming to rest.
SOLUTION
Kinematics. Let rA be the radius of the outer pulley and rB that of the inner pulley.
vA
rA
C
vB
rB
sA
rA
C
sB
rB
sA
rA
C
rB
vA
rA
Use the principle of work and energy with position 1 being the initial rest position and position 2 being when
cylinder A strikes the ground.
T1
U1
2
T2 :
where
T1
0
and
T2
1
mA v A2
2
with
mA
5 kg, mB
T2
mC kC2
15 kg, IC
1
mA
2
1
5 kg
2
mB rB2
IC
rA2
rA2
1
mB vB2
2
(15 kg)(0.160 m)2
1
IC
2
2
C
0.384 kg m2
v A2
(15 kg)(0.150 m) 2
(0.250 m) 2
0.384 kg m 2 2
vA
(0.250 m)2
(8.272 kg)v A2
Principle of work and energy applied to the system consisting of blocks A and B and the double pulley C.
Work.
where
U1
2
sA
Ps A
mA g sA
FF s B
m B g s B sin 30
1m
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PROBLEM 17.14 (Continued)
and
rB
sA
rA
sB
0.150 m
(1 m)
0.250 m
0.6 m
To find Ff use the free body diagram of block B.
60° F
0:
NB
m B g cos 30
0
NB
m B g cos 30
(15 kg)(9.81 m/s) cos 30
Ff
U1
k NB
(0.2)(127.44 N)
127.44 N
25.487 N
(200 N)(1 m) (5 kg)(9.81 m/s 2 )(1 m)
2
(25.487 N)(0.6 m) (15 kg)(9.81 m/s 2 )(0.6 m)sin 30
189.613 J
Work-energy:
(a)
0 189.613 J
Velocity of A.
vA
(8.272 kg)vA2
4.7877 m/s
vA
4.79 m/s
when the cylinder strikes the ground,
vB
C
rB
vA
rA
vA
rA
0.150 m
(4.7877 m/s)
0.250 m
4.7877 m/s
0.250 m
2.8726 m/s
19.1508 rad/s
After the cylinder strikes the ground use the principle of work and energy applied to a system
consisting of block B and double pulley C.
Let T3 be its kinetic energy when A strikes the ground.
T3
1
1
mB vB2
I C C2
2
2
1
(15 kg)(2.8726 m/s) 2
2
132.305 J
When the system comes to rest,
U3
4
T4
1
(0.384 kg m 2 )(19.1508 rad/s)2
2
0
(25.487 N) sB
(15 kg)(9.81 m/s 2 )( sB sin 30 )
(99.062 N) sB
where s B is the additional travel of block B.
T3
U3
4
T4 : 132.305 J
(99.062 N) s B
sB
(b)
Total distance:
0
1.3356 m
sB
sB
1.936 m
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PROBLEM 17.18
A slender 9 lb rod can rotate in a vertical plane about a pivot at B. A spring
of constant k 30 lb/ft and of unstretched length 6 in. is attached to the rod
as shown. Knowing that the rod is released from rest in the position shown,
determine its angular velocity after it has rotated through 90°.
SOLUTION
Position 1:
Unstretched
Length
Spring:
Gravity:
Kinetic energy:
x1
CD ( 6 in. ) 14.866 6 8.8661 in. 0.73884 ft
Ve
1 2
kx1
2
Vg
Wh
V1
Ve
T1
0
x2
9 in. 6 in. 3 in. 0.25 ft
Ve
1 2
kx2
2
1
(30 lb/ft)(0.73884) 2
2
8.1882 lb ft
7
ft
5.25 lb ft
12
8.1882 lb ft 5.25 lb ft 13.438 lb ft
(9 lb)
Vg
Position 2:
Spring:
1
(30 lb/ft)(0.25 ft) 2
2
0.9375 lb ft
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PROBLEM 17.18 (Continued)
Gravity:
Vg
Wh
0
V2
Ve Vg
0.9375 lb ft
Kinetic energy:
v2
I
T2
r
2
7
ft
12
1
1 9 lb
mL2
(2 ft) 2
12
12 32.2
1
1 2
mv22
I 2
2
2
1 9 lb
2 32.2
T2
2
0.094138
7
ft
12
2
2
0.093168 slug ft 2
1
(0.093168)
2
2
2
2
2
Conservation of energy:
T1 V1 T2 V2
0 13.438 0.094138
2
2
2
2
132.79
2
11.524 rad/s
0.9375
2
11.52 rad/s
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