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Physics II
Homework I
CJ
Chapter 13; 4, 11, 13, 23, 55, 67, 71
13.4. Model: The angular velocity and angular acceleration graphs correspond to a rotating rigid body.
Solve:
(a) The -versus-t graph has a positive slope of 5 rad/s2 from t  0 s to t  2 s and a negative slope of 5 rad/s2 from t  2 s to t
 4 s.
(b) The angular velocity is the area under the -versus-t graph:

d
    ( x) dt  0  area under  graph.
dt
13.11. Visualize:
Solve: Torque by a force is defined as   Frsin where  is measured counterclockwise from the r vector to the
F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force:
(30 N)r1 sin1  (20 N)r2 sin2  (30 N)(0.02 m) sin (  90)  (20 N)(0.02 m) sin (90)
 (  0.60 N m)  (0.40 N m)  0.20 N m
Assess: A negative torque causes a clockwise motion of the pulley.
13.13. Visualize:
Solve:
The net torque on the spark plug is
  Fr sin  38 N m  F(0.25 m)sin(120)  F  175.5 N
That is, you must pull with a force of 175.5 N to tighten the spark plug.
Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net
torque.
13.23. Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular
to the disk.
Solve: To determine the torque () needed to take the plastic disk from i  0 rad/s to f  1800 rpm  (1800)(2)/
60 rad/s  60 rad/s in tf – ti  4.0 s, we need to determine the angular acceleration () and the disk’s moment of inertia (I) about the
axle in its center. The radius of the disk is R  10.0 cm. We have
1
1
MR2  (0.200 kg)(0.10 m)2  1.0  103 kg m2
2
2
   i 60 rad/s  0 rad/s
 f   i   (t f  t i )    f

 15 rad/s2
tf  ti
4.0 s
I
Thus,   I  (1.0  103 kg m2)(15 rad/s2)  0.0471 N m.
13.55. Visualize:
Solve: We can divide the face of the square rod into small areas dA  dx dy with mass dm. Because the mass of the
rod is uniformly distributed, dm is the same fraction of M as dA is of A  L2. That is
dA dm
 dx dy 

 dm  M  2 
L2
M
 L 
The moment of inertia is
I   r 2 dm 
L/2
M L/2 L/2 2
M L/2
M L/2 L/2
( x  y2 ) dx dy  2  x 2dx  dy  2  dx  y 2dy
2 

L L / 2 L / 2
L L / 2
L L / 2 L / 2
L / 2
M x3
 2
L 3
L/2
L/2
L/2
M
y3
y
 2x
L L / 2 3
L / 2
L / 2
L/2

L / 2
1
ML2
6
13.67. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and
that the string does not slip.
Visualize:
Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same.
Solve: Applying Newton’s second law to m1, m2, and the pulley yields the three equations:
T1  w1  m1a1
 w2  T2  m2a2
T2 R  T1R  0.50 N m  I
Noting that –a2  a1  a, I  21 mp R2 , and   a/R, the above equations simplify to
T1  m1g  m1a
m2g  T2  m2a
0.50 N m
1
 a  1 0.50 N m 1
T2  T1   mp R2   
 mpa 
2
R
R
R
2
0.060 m

 
Adding these three equations,
1 

(m2  m1 )g  a  m1  m2  mp   8.333 N
2


(m2  m1 )g  8.333 N (4.0 kg  2.0 kg)(9.8 m/s2 )  8.333 N
a

 1.610 m/s2
m1  m2  21 mp
2.0 kg  4.0 kg  (2.0 kg/2)
We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:
1
1
y1  y0  v0 (t1  t0 )  a2 (t1  t0 )2  0  1.0 m  0  (1.610 m/s2 )(t1  0 s)2
2
2
 t1 
2(1.0 m)
 1.11 s
(1.610 m/s2 )
13.71. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without
slipping. We also assume that the coefficient of rolling friction is zero.
Visualize:
The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational
potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy.
Solve: The conservation of energy equation Kf  Ugf  Ki  Ugi is
1
1
1
1
M (v1 )2cm  I cm (1 )2  Mgy1  M (v0 )2cm  I cm ( 0 )2  Mgy0
2
2
2
2
( v )2
1
1 2
1
1

0 J  0 J  Mgy1  M (v0 )2cm   MR2  ( 0 )2cm  0 J  Mgy1  M (v0 )2cm  MR 2 0 2cm
2
2 3
2
3
R

5
( v )2
5
5 (5.0 m/s)2
 gy1  (v0 )2cm  y1  6 0 cm 
 2.126 m
6
g
6 9.8 m/s2
The distance traveled along the incline is
s
y1
2.126 m

 4.25 m
sin30
0.5
Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of
the incline is approximately 10 mph.