Download Solutions #9

PHYSICS 171
AQ 2009
Solutions to Homework
#9
1. Giancoli Chapter 10, Problem 12
Convert the rpm values to angular velocities.
rev   2 rad   1 min 

0   130


  13.6 rad s
min   1 rev   60 sec 

rev   2 rad   1 min 

   280


  29.3 rad s
min   1 rev   60 sec 

(a) The angular acceleration is found from Eq. 10-3a.
  0 29.3rad s  13.6 rad s


 3.93rad s 2  3.9 rad s 2
t
4.0 s
(b)
To find the components of the acceleration, the instantaneous angular velocity is
needed.
  0   t  13.6 rad s   3.93rad s 2   2.0s   21.5 rad s
The instantaneous radial acceleration is given by a R   2 r.
aR   2 r   21.5rad s   0.35m   160 m s 2
2
The tangential acceleration is given by atan   r .

atan   r  3.93rad s2
  0.35m  1.4 m s
2
2. Giancoli Chapter 10, Problem 15
(a) The direction of 1 is along the axle of the wheel, to the left. That is the iˆ direction.
The
direction of 2 is also along its axis of rotation, so it is straight up. That is the
kˆ direction. That is also the angular velocity of the axis of the wheel.
(b) At the instant shown in the textbook, we have the vector
relationship
as shown in the diagram.
    
2
1
2
2
 44.0 rad s 
2
  35.0 rad s   56.2 rad s
2

35.0
  tan 2  tan 1
 38.5
2
44.0
1
(c) Angular acceleration is given by α 
35.0kˆ rad s , α 
dω1
dt
dω
dt
2
z


1
. Since ω  ω1  ω2 , and ω 2 is a constant
. ω1 is rotating counterclockwise about the z axis with the
x
angular velocity of 2 , and so if the figure is at t = 0, then


ω1  1  cos 2tˆi  sin 2tˆj .
α
dω
dt

d  ω1  ω2 
dt

dω1
dt



d 1  cos 2tˆi  sin 2tˆj 
dt

 12 sin 2tˆi  cos 2tˆj

 
α  t  0   12  ˆj    44.0 rad s  35.0 rad s  ˆj  1540 rad s 2 ˆj
3. Giancoli Chapter 10, Problem 22
We are given that   8.5t  15.0t 2  1.6t 4 .
d
 8.5  30.0t  6.4t 3 , where  is in rad/sec and t is in sec.
(a)  
dt
d
 30.0  19.2t 2 , where  is in rad sec2 and t is in sec.
(b)  
dt
3
(c)   3.0   8.5  30.0  3.0   6.4  3.0   91rad s
  3.0  30.0  19.2  3.0  140 rad s2
2
(d) The average angular velocity is the angular displacement divided by the elapsed time.
   3.0     2.0 
avg 

t
3.0s  2.0s
8.5  3.0   15.0  3.0  2  1.6  3.0  4   8.5  2.0   15.0  2.0  2  1.6  2.0  4 
 


1.0s
 38 rad s
(e) The average angular acceleration is the change in angular velocity divided by the elapsed
time.
 avg 

t

  3.0     2.0 
3.0s  2.0s
8.5  30.0  3.0   6.4  3.0 3   8.5  30.0  2.0   6.4  2.0 3 
 
  92 rad s2

1.0s
4. Giancoli Chapter 10, Problem 30
For each torque, use Eq. 10-10c. Take counterclockwise torques to be positive.
(a)
Each force has a lever arm of 1.0 m.
 about   1.0 m  56 N  sin 30  1.0 m  52 N  sin 60  17m N
C
(b)
The force at C has a lever arm of 1.0 m, and the force at the top has a lever arm
of 2.0 m.
 about    2.0 m  56 N  sin 30  1.0 m  65 N  sin 45  10 m N (2 sig fig)
P
The negative sign indicates a clockwise torque.
5. Giancoli Chapter 10, Problem 33
(a) The torque exerted by the frictional force is   rFfr sin  . The force of
friction is assumed to be tangential to the clay, and so   90.
 total  rFfr sin    12  0.12 m   1.5 N  sin 90  0.090 m N
(b) The time to stop is found from   o   t , with a final angular
velocity of 0. The angular acceleration can be found from
 total  I . The net torque (and angular acceleration) is negative
since the object is slowing.
  o   o 0  1.6 rev s  2 rad rev 
t


 12 s

 I
 0.090 m N   0.11kg m 2 
direction of
rotation
Ffr
6. Giancoli Chapter 10, Problem 40
(a)
To calculate the moment of inertia about the y axis (vertical), use the following.
2
2
2
2
I   M i Rix2  m  0.50 m   M  0.50 m   m 1.00 m   M 1.00 m 
  m  M   0.50 m   1.00 m     5.3kg   0.50 m   1.00 m    6.6 kg m 2
2
(b)
2
2
2
To calculate the moment of inertia about the x-axis (horizontal), use the
following.
I   M i Riy2   2m  2 M  0.25m   0.66 kg m2
2
(c) Because of the larger I value, it is ten times harder to accelerate the array about
the vertical axis .
7. Giancoli Chapter 10, Problem 46
(a) The free body diagrams are shown. Note that only the forces producing
torque are shown on the pulley. There would also be a gravity force on the pulley
(since it has mass) and a normal force from the pulley’s suspension, but they are not
shown.
(b) Write Newton’s second law for the two blocks, taking the positive x direction as shown
in the free body diagrams.
FNB
FTA
FTB
FNA
y
FTA
x
A
mAg
A
y
FTB
x
B
B
m Bg
mA :
F
x
 FTA  mA g sin  A  mA a 
FTA  mA  g sin  A  a 


  8.0kg   9.80 m s 2 sin 32  1.00 m s 2   49.55 N
 50 N
mB :
F
x
 2 sig fig 
 mB g sin  B  FTB  mBa 
FTB  mB  g sin  B  a 


 10.0kg   9.80 m s 2 sin 61  1.00 m s 2   75.71N
 76 N
(c) The net torque on the pulley is caused by the two tensions. We take clockwise torques
as positive.
   F
TB
 FTB  R   75.71N  49.55 N  0.15m   3.924m N  3.9m N
Use Newton’s second law to find the rotational inertia of the pulley. The tangential
acceleration of the pulley’s rim is the same as the linear acceleration of the blocks,
assuming that the string doesn’t slip.
a
  I  I R   FTB  FTB  R 
I
 FTB  FTB  R 2
a

 75.71N  49.55 N  0.15 m  2
1.00 m s
2
 0.59 kg m 2
  168.9  170 .
8. Giancoli Chapter 10, Problem 51
We assume that mB  mA , and so mB will accelerate down, mA will accelerate up, and the
pulley will accelerate clockwise. Call the direction of acceleration the positive direction for
each object. The masses will have the same acceleration since they are connected by a cord.
The rim of the pulley will have that same acceleration since the cord is making it rotate, and
so  pulley  a R . From the free-body diagrams for each object, we have the following.
F
F
yA
 FTA  mA g  mA a  FTA  mA g  mA a
yB
 mB g  FTB  mB a  FTB  mB g  mB a
  F
TB
r  FTA r  I   I

I
a
R
Substitute the expressions for the tensions into the torque equation, and
solve for the acceleration.
y
R
FTA
FTB
FTA
FTB
mA
mAg
mB  y
m Bg
FTB R  FTA R  I
a
m
a
R
 m B  mA 
A
 mB g  mBa  R   mA g  mA a  R  I

 mB  I R 2

a
R

g
If the moment of inertia is ignored, then from the torque equation we see that FTB  FTA , and
the acceleration will be a I 0 
 m B  mA 
g.
 mA  m B 
We see that the acceleration with the moment
of inertia included will be smaller than if the moment of inertia is ignored.
9. Giancoli Chapter 10, Problem 68
(a)
The kinetic energy of the system is the kinetic energy of the two masses, since
the rod is treated as massless. Let A represent the heavier mass, and B the lighter mass.
K  12 I AA2  12 I BB2  12 mA rA2A2  12 mB rB2A2  12 r 2 2  mA  mB 

1
2
 0.210 m 2  5.60 rad s 2  7.00 kg  
4.84 J
(b) The net force on each object produces centripetal motion, and so can be expressed as
mr .
2
FA  mA rAA2   4.00 kg  0.210 m  5.60 rad s   26.3 N
2
FB  mB rBB2   3.00 kg  0.210 m  5.60 rad s   19.8 N
2
These forces are exerted by the rod. Since they are unequal, there would be a net
horizontal force on the rod (and hence the axle) due to the masses. This horizontal
force would have to be counteracted by the mounting for the rod and axle in order for
the rod not to move horizontally. There is also a gravity force on each mass, balanced
by a vertical force from the rod, so that there is no net vertical force on either mass.
(c)
Take the 4.00 kg mass to be the origin of coordinates for determining the center
of mass.
m x  mB xB  4.00 kg  0   3.00 kg  0.420 m 
xCM  A A

 0.180 m from mass A
mA  mB
7.00 kg
So the distance from mass A to the axis of rotation is now 0.180 m, and the distance
from mass B to the axis of rotation is now 0.24 m. Re-do the above calculations with
these values.
K  12 I AA2  12 I BB2  12 mA rA2A2  12 mB rB2A2  12  2  mA rA2  mB rB2 
 5.60 rad s  2  4.00 kg  0.180 m  2   3.00 kg  0.240 m 2  
2
FA  mA rAA2   4.00 kg  0.180 m  5.60 rad s   22.6 N

1
2
4.74 J
FB  mB rBB2   3.00 kg  0.240 m  5.60 rad s   22.6 N
2
Note that the horizontal forces are now equal, and so there will be no horizontal force
on the rod or axle.