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PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
HOMEWORK PROBLEMS Chapter 8:
ROTATIONAL EQUILIBRIUM AND
ROTATIONAL DYNAMICS
PART-A: Hand in your answers in class on scantron on Monday 08 November-2010.
Write
your name, class (1401) and HW # 8 on the scantron.
1. A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because
of the cramped
space, the force must be exerted upward at an angle of 60.0° with respect to
a line from the bolt
through the end of the wrench. How much torque is applied to the nut? (a)
34.6 N · m (b) 4.56
N · m (c) 11.8 N · m (d) 14.2 N · m (e) 20.0 N · m
  rF sin    0.500 m   80.0 N  sin  60.0°   36.4 N  m which is choice
(a).
2. A horizontal plank 4.00 m long and having mass 20.0 kg rests on two
pivots, one at the left
end and a second 1.00 m from the right end. Find the magnitude of the force
exerted on the
plank by the second pivot. (a) 32.0 N (b) 45.2 N (c) 112 N (d) 131 N (e) 98.2
N
Using the left end of the plank as a pivot and requiring that
  0 gives mg  2.00 m   F2  3.00 m   0 , or

2  20.0 kg  9.80 m s2
2mg

3
3
so choice (d) is the correct response.
F2 

 131 N
3. What is the magnitude of the angular acceleration of a 25.0-kg disk of
radius 0.800 m when a
torque of magnitude 40.0 N · m is applied to it? (a) 2.50 rad/s2 (b) 5.00
rad/s2 (c) 7.50 rad/s2 (d)
10.0 rad/s2 (e) 12.5 rad/s2
Assuming a uniform, solid disk, its moment of inertia about a perpendicular
axis through its center is
I  MR 2 2 , so   I  gives
2  40.0 N  m 
2

 5.00 rad s2
2
2
MR
25.0
kg
0.800
m



and the correct answer is (b).
 
4. Estimate the rotational kinetic energy of Earth by treating it as a solid
sphere with uniform
density. (a) 3 X 1029 kg · m2/s2 (b) 5 X 1027 kg · m2/s2 (c) 7 X 1030 kg ·
m2/s2 (d) 4 X 1028 kg ·
m2/s2 (e) 2 X 1025 kg · m2/s2
For a uniform, solid sphere,
I  2 MR 2 5 and  E 
2 rad 
1d

 7.27  10 5 rad s

4
1 d  8.64  10 s 
so



24
6
1
1 2 5.98  10 kg 6.38  10 m
KEr  I  E2  
2
2
5

2 
yielding KEr  3  1029 J , making (a) the correct choice.
1
 7.27  10 5


rad s
2
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
5. Two forces are acting on an object. Which of the following statements is
correct? (a) The
object is in equilibrium if the forces are equal in magnitude and opposite in
direction. (b) The
object is in equilibrium if the net torque on the object is zero. (c) The
object is in equilibrium if
the forces act at the same point on the object. (d) The object is in
equilibrium if the net force
and the net torque on the object are both zero. (e) The object cannot be in
equilibrium because
more than one force acts on it.
In order for an object to be in equilibrium, it must be in both translational
equilibrium and rotational
equilibrium. Thus, it must meet two conditions of equilibrium, namely Fnet  0
and  net  0 . The
correct answer is therefore choice (d).
6. A disk rotates about a fixed axis that is perpendicular to the disk and
passes through its
center. At any instant, does every point on the disk have the same (a)
centripetal acceleration,
(b) angular velocity, (c) tangential acceleration, (d) linear velocity, or
(e) total acceleration?
In a rigid, rotating body, all points in that body rotate about the axis at
the same rate (or have the same
angular velocity). The centripetal acceleration, tangential acceleration,
linear velocity, and total
acceleration of a point in the body all vary with the distance that point is
from the axis of rotation. Thus,
the only correct choice is (b).
7- BLANK
8. A block slides down a frictionless ramp, while a hollow sphere and a solid
ball roll without
slipping down a second ramp with the same height and slope. Rank the arrival
times at the
bottom from shortest to longest. (a) sphere, ball, block (b) ball, block,
sphere (c) ball, sphere,
block (d) block, sphere, ball (e) block, ball, sphere.
When objects travel down ramps of the same length, the one with the greatest
translational kinetic
energy at the bottom will have the greatest final translational speed (and,
hence, greatest average
translational speed). This means that it will require less time to travel the
length of the ramp. Of the
objects listed, all will have the same total kinetic energy at the bottom,
since they have the same
decrease in gravitational potential energy (due to the ramps having the same
vertical drop) and no energy
has been spent overcoming friction. All of the block’s kinetic energy is in
the form of translational
kinetic energy. Of the rolling bodies, the fraction of their total kinetic
energy that is in the translational
form is
1
M v2
KEt
1
1
f 
 1 22


1
2
2
KEt  KEr
Mv  2 I
1  I MR2
1   I M    v
2
Since the ratio I/MR2equals 2/5 for a solid ball and 2/3 for a hollow sphere,
the ball has the larger
translational kinetic energy at the bottom and will arrive before the hollow
sphere. The correct rankings
of arrival times, from shortest to longest, is then block, ball, sphere, and
choice (e) is the correct
response.


9. A solid disk and a hoop are simultaneously released from rest at the top
of an incline and roll
down without slipping. Which object reaches the bottom first? (a) The one
that has the largest
mass arrives first. (b) The one that has the largest radius arrives first.
(c) The hoop arrives first.
(d) The disk arrives first. (e) The hoop and the disk arrive at the same
time.
Please read the answer to Question 8 above, since most of what is said there
also applies to this question.
The total kinetic energy of either the disk or the hoop at the bottom of the
ramp will be KEtotal = Mgh,
where M is the mass of the body in question and h is the vertical drop of the
ramp. The translational
kinetic energy of this body will then be KEt = f KEtotal = fMgh, where f is
the fraction discussed in
2
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
Question 8. Hence, M v2 2  f M gh and the translational speed at the bottom
is v  2 fgh .
Since f = 1/(1+1/2) = 2/3 for the disk and f = 1/(1+1) = 1/2 for the hoop, we
see that the disk will have
the greater translational speed at the bottom, and hence, will arrive first.
Notice that both the mass and
radius of the object has canceled in the calculation. Our conclusion is then
independent of the object’s
mass and/or radius. Therefore, the only correct response is choice (d)
10. A solid cylinder of mass M and radius R rolls down an incline without
slipping. Its moment
of inertia about an axis through its center of mass is MR2/2. At any instant
while in motion, its
rotational kinetic energy about its center of mass is what fraction of its
total kinetic energy? (a)
12 (b) 14 (c) 13 (d) 25 (e) None of these
The ratio of rotational kinetic energy to the total kinetic energy for an
object that rolls without slipping is
1
I2
KEr
KEr
1
1
2

 1


1
2
2
2
2
MR
KEtotal
KEt  KEr
M
v

I

M  v
2
2
1
1
I
I   
For a solid cylinder, I = MR2/2 and this ratio becomes
KEr
1
1


KEtotal
21
3
so the correct answer is (c). (But c = 13, not 1/3. Therefore, correct answer
is e)
11. The cars in a soapbox derby have no engines; they simply coast downhill.
Which of the
following design criteria is best from a competitive point of view? The car’s
wheels should (a)
have large moments of inertia, (b) be massive, (c) be hoop-like wheels rather
than solid disks,
(d) be large wheels rather than small wheels, or (e) have small moments of
inertia.
If a car is to reach the bottom of the hill in the shortest time, it must
have the greatest translational speed
at the bottom (and hence, greatest average speed for the trip). To maximize
its final translational speed,
the car should be designed so as much as possible of the car’s total kinetic
energy is in the form of
translational kinetic energy. This means that the rotating parts of the car
(i.e., the wheels) should have as
little kinetic energy as possible. Therefore, the mass of these parts should
be kept small, and the mass
they do have should be concentrated near the axle in order to keep the moment
of inertia as small as
possible. The correct response to this question is (e).
12. Consider two uniform, solid spheres, a large, massive sphere and a
smaller, lighter sphere.
They are released from rest simultaneously from the top of a hill and roll
down without
slipping. Which one reaches the bottom of the hill first? (a) The large
sphere reaches the
bottom first. (b) The small sphere reaches the bottom first. (c) The sphere
with the greatest
density reaches the bottom first. (d) The spheres reach the bottom at the
same time. (e) The
answer depends on the values of the spheres’ masses and radii.
Please review the answers given above for questions 8 and 9. In the answer to
question 9, it is shown
that the translational speed at the bottom of the hill of an object that
rolls without slipping is
v  2 fgh where h is the vertical drop of the hill and f is the ratio of the
translational kinetic energy to
the total kinetic energy of the rolling body. For a solid sphere, I  2 MR 2 5
, so the ratio f is
f 
1
1 I

MR2


1
1

1 25
1.4
and the translational speed at the bottom of the hill is v  2gh 1.4 Notice
that this result is the same
for all uniform, solid, spheres. Thus, the two spheres have the same
translational speed at the bottom of
3
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
the hill. This also means that they have the same average speed for the trip,
and hence, both make the
trip in the same time. The correct answer to this question is (d).
13. A mouse is initially at rest on a horizontal turntable mounted on a
frictionless, vertical axle.
As the mouse begins to walk clockwise around the perimeter, which of the
following
statements must be true of the turntable? (a) It also turns clockwise. (b) It
turns
counterclockwise with the same angular velocity as the mouse. (c) It remains
stationary. (d) It
turns counterclockwise because angular momentum is conserved. (e) It turns
counterclockwise
because mechanical energy is conserved.
Since the axle of the turntable is frictionless, no external agent exerts a
torque about this vertical axis of
the mouse-turntable system. This means that the total angular momentum of the
mouse-turntable system
will remain constant at its initial value of zero. Thus, as the mouse starts
walking around the axis (and
developing an angular momentum, Lmouse  I mm , in the direction of its
angular velocity), the
turntable must start to turn in the opposite direction so it will possess an
angular momentum,
Ltable  I t t , such that Ltotal  Lmouse  Ltable  I mm  I t t  0 .
Thus, the angular velocity of
the table will be t    I m I t   m . The negative sign means that if the
mouse is walking around the
axis in a clockwise direction, the turntable will be rotating in the opposite
direction, or counterclockwise.
The correct choice for this question is (d).
14. According to the manual of a certain car, a maximum torque of magnitude
65.0 N · m
should be applied when tightening the lug nuts on the vehicle. If you use a
wrench of length
0.330 m and you apply the force at the end of the wrench at an angle of 75.0°
with respect to a
line going from the lug nut through the end of the handle, what is the
magnitude of the
maximum force you can exert on the handle without exceeding the
recommendation?
(a) 122 N
(b) 184 N
(c) 204 N
(d) 254 N
The torque of the applied force is  = rF sin. Thus, if r = 0.330 m,  =
75.0º, and the torque
has the maximum allowed value of max = 65.0 N  m, the applied force is
F 
 max
65.0 N  m

 204 N
r sin 
 0.330 m  sin 75.0
15. A student gets his car stuck in a snowdrift. Not at a loss, having
studied physics, he attaches
one end of a stout rope to the car and the other end to the trunk of a nearby
tree, allowing for a
small amount of slack. The student then exerts a force on the center of the
rope in the
direction perpendicular to the car-tree line as shown in Figure P8.25. If the
rope is inextensible
and the magnitude of the applied force is 475 N, what is the force on the
car? (Assume
equilibrium conditions.)
(a) 2.56 kN
(b) 2.86 kN
(c) 3.04 kN
4
(d) 4.04 kN
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
Consider the free-body diagram of the
material making up the center point in
the rope given at the right. Since this
material is in equilibrium, it is necessary
to have Fx = 0 and Fy = 0, giving
T2 sin   T1 sin   0
Fx  0:
or T2 = T1, meaning that the rope has a uniform tension T throughout its
length.
0.500 m
Fy  0:
T cos   T cos   475 N  0 where cos  
 6.00 m 2   0.500 m 2
and the tension in the rope (force applied to the car) is
 475 N 
475 N
T 

2 cos
 6.00 m 2   0.500 m 2
2  0.500 m 
 2.86  103 N  2.86 kN
PART-B: Hand in your solutions to the following questions in class, on Monday
08 November2010
A uniform beam of length 7.60 m and weight 4.50 X 102 N is carried by two
workers,
Sam and Joe, as shown in Figure P8.8. (a) Determine the forces that each
person exerts on
the beam. (b) Qualitatively, how would the answers change if Sam moved closer
to the
midpoint? (c) What would happen if Sam moved beyond the midpoint
8.
(a)
Since the beam is in equilibrium, we choose the center
as our pivot point and require that
 center   FSam  2.80 m   FJoe 1.80 m   0
or
FJoe  1.56 FSam
[1]
Also,
Fy  0

FSam  FJoe  450 N [2]
Substitute Equation [1] into [2] to get the following:
450 N
 176 N
2.56
 1.56  176 N   274 N .
FSam  1.56FSam  450 N
Then, Equation [1] yields FJoe
(b)
FSam 
or
If Sam moves closer to the center of the beam, his lever arm about the beam
center decreases, so
the force FSam must increase to continue applying a clockwise torque capable
of offsetting Joe’s
counterclockwise torque. At the same time, the force FJoe would decrease
since the sum of the
two upward forces equal the magnitude of the downward gravitational force.
(c)
If Sam moves to the right of the center of the beam, his torque about the
midpoint would then be
counterclockwise. Joe would have to hold down on the beam in order to exert
an offsetting
clockwise torque.
5
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
22. A hungry 700-N bear walks out on a beam in an attempt to retrieve some
“goodies”
hanging at the end (Fig. P8.22). The beam is uniform, weighs 200 N, and is
6.00 m long;
the goodies weigh 80.0 N. (a) Draw a free-body diagram of the beam. (b) When
the bear is
at x = 1.00 m, find the tension in the wire and the components of the
reaction force at the
hinge. (c) If the wire can withstand a maximum tension of 900 N, what is the
maximum
distance the bear can walk before the wire breaks?
(a)
See the diagram below
(b) If x = 1.00 m, then
 left end  0    700 N  1.00 m    200 N   3.00 m 
  80.0 N   6.00 m    T sin 60.0°   6.00 m   0
giving T = 434 N.
Then, Fx  0  H  T cos 60.0°  0 , or H   343 N  cos 60.0°  172 N
(c)
and Fy  0  V  980 N +  343 N  sin 60.0°  0 , or V = 683 N.
When the wire is on the verge of breaking, T = 900 N and
  left end    700 N  xmax   200 N   3.00 m 
  80.0 N   6.00 m     900 N  sin 60.0°   6.00 m   0
which gives xmax = 5.14 m
31. Four objects are held in position at the corners of a rectangle by light
rods as shown in
Figure P8.31. Find the moment of inertia of the system about (a) the x-axis,
(b) the y-axis,
and (c) an axis through O and perpendicular to the page.
The moment of inertia for rotations about an axis is I   mi ri2 , where ri
is the distance mass mi is from that axis.
(a) For rotation about the x-axis,
I x   3.00 kg   3.00 m    2 .00 kg   3.00 m 
2
2
  2.00 kg   3.00 m    4.00 kg   3.00 m   99.0 kg  m 2
2
(b)
2
When rotating about the y-axis,
6
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
I y   3.00 kg   2 .00 m    2 .00 kg   2 .00 m 
2
HW-8
2
  2.00 kg   2.00 m    4.00 kg   2.00 m   44.0 kg  m 2
2
(c)
is
2
For rotations about an axis perpendicular to the page through point O, the
distance ri for each mass
 2 .00 m 2   3.00 m 2
ri 
Thus,

13.0 m


IO   3.00  2.00  2.00  4.00  kg  13.0 m2  143 kg  m2
40.
An Atwood’s machine consists of blocks of masses m1 = 10.0 kg and m2 = 20.0
kg
attached by a cord running over a pulley as in Figure P8.40. The pulley is a
solid cylinder
with mass M = 8.00 kg and radius r = 0.200 m. The block of mass m2 is allowed
to drop,
and the cord turns the pulley without slipping. (a) Why must the tension T2
be greater than
the tension T1? (b) What is the acceleration of the system, assuming the
pulley axis is
frictionless? (c) Find the tensions T1 and T2.
(a)
It is necessary that the tensions T1 and T2 be different in order to
provide a net torque about the axis of the pulley and produce and angular
acceleration of the pulley. Since intuition tells us that the system will
accelerate in the directions shown in the diagrams at the right when m2 > m1,
it is necessary that T2 > T1.
(b)
We adopt a sign convention for each object with the positive direction being
the indicated direction of the
acceleration of that object in the diagrams at the right. Then, apply
Newton’s second law to each object:
Fy  m1a

For m2 :
Fy  m2 a
 m2 g  T2  m2 a m2
For M :
  I 
For m1 :
Or
T1  m1  g  a  [1]
T1  m1g  m1a
 rT2  rT1  I 
or
T2  m2  g  a 
[2]
or
T2  T1  I  r
[3]
Substitute Equations [1] and [2], along with the relations I  Mr 2 2 and  
a r , into Equation [3] to obtain
m2  g  a   m1  g  a  
and
a 
(c)
 m2
 m1  g
m1  m2  M 2
From Equation [1]: T1
From Equation [2]: T2
Mr 2  a 
Ma





2r
r
2
or
M

 m1  m2  2  a   m2  m1  g
 20.0 kg  10.0 kg   9.80 m s2 
 2.88 m s2
20.0 kg  10.0 kg +  8.00 kg  2
  10.0 kg   9.80 m s2  2.88 m s2   127 N .
  20.0 kg   9.80 m s2  2.88 m s2   138 N .

7
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
52. Use conservation of energy to determine the angular speed of the spool
shown in Figure
P8.52 after the 3.00-kg bucket has fallen 4.00 m, starting from rest. The
light string attached
to the bucket is wrapped around the spool and does not slip as it unwinds.
As the bucket drops, it loses gravitational potential energy. The spool
gains rotational kinetic energy and the bucket gains translational kinetic
energy. Since the string does not slip on the spool,   r where r is the
radius of the spool. The moment of inertia of the spool is I  21 Mr 2 ,
where M is the mass of the spool. Conservation of energy gives
 KE
t
 KEr  PEg

f

 KEt  KEr  PEg
1
1
m 2  I  2  mgy f  0  0  mgyi
2
2
or
1
11

2
m  r    Mr 2   2  mg yi  y f

2
22
This gives

 

2mg yi  y f

 m  21 M  r 2


i


2  3.00 kg  9.80 m s2
  4.00 m 
 3.00 kg+ 21  5.00 kg    0.600 m 
2
 10.9 rad s
58. Halley’s comet moves about the Sun in an elliptical orbit, with its
closest approach to the
Sun being 0.59 A.U. and its greatest distance being 35 A.U. (1 A.U. is the
Earth– Sun
distance). If the comet’s speed at closest approach is 54 km/s, what is its
speed when it is
farthest from the Sun? You may neglect any change in the comet’s mass and
assume that its
angular momentum about the Sun is conserved.
Using conservation of angular momentum, Laphelion  Lperihelion .
 
 
 mra2  a ra   mrp2   p
Thus, mra2  a  mrp2  p . Since  = 1/r at both aphelion and perihelion,
this is equivalent to
 rp 
rp , giving
 0.59 A.U. 
a     p  
 54 km s   0.91 km s
 35 A.U. 
 ra 
8
PHYS-1401: College Physics-I
CRN 55178
Khalid Bukhari
HW-8
60. A playground merry-go-round of radius 2.00 m has a moment of inertia I =
275 kg · m2
and is rotating about a frictionless vertical axle. As a child of mass 25.0
kg stands at a
distance of 1.00 m from the axle, the system (merry-go-round and child)
rotates at the rate
of 14.0 rev/min. The child then proceeds to walk toward the edge of the
merry-go-round.
What is the angular speed of the system when the child reaches the edge?

From conservation of angular momentum: Ichild  I m-g-r
where I m-g -r  275 kg 
m2

f


 f  Ichild  I m-g-r i
i
is the constant moment of inertia of the merry-go-round.
Treating the child as a point object, I child  mr 2 where r is the distance
the child is from the rotation
axis. Conservation of angular momentum then gives
  25.0 kg   1.00 m 2  275 kg  m 2 
 mri2  I m -g -r 

  14.0 rev min 
f   2


 i
  25.0 kg   2.00 m 2  275 kg  m 2 
 mrf  I m -g -r 


or
rev  2 rad   1 min 
 f  11.2
 1.17 rad s
min  1 rev   60.0 s 
87. A 4.00-kg mass is connected by a light cord to a 3.00-kg mass on a smooth
surface (Fig.
P8.87). The pulley rotates about a frictionless axle and has a moment of
inertia of 0.500 kg
· m2 and a radius of 0.300 m. Assuming that the cord does not slip on the
pulley, find (a)
the acceleration of the two masses and (b) the tensions T1 and T2.
(a) Free-body diagrams for each block and the pulley are given at the right.
Observe that the angular acceleration of the pulley will be clockwise in
direction
and has been given a negative sign. Since   I  , the positive sense for
torques and angular acceleration must be the same (counterclockwise).
For m1: Fy  may  T1  m1g  m1  a 
T1  m1  g  a 
[1]
For m2: Fx  max  T2  m2 a
For the pulley:   I   T2r  T1r  I  a r  or
[2]
 I 
T1  T2   2  a [3]
r 
Substitute Equations [1] and [2] into Equation [3] and solve for a to obtain
m1g
a 
2
I r  m1  m2


or
a 
(b)
 4.00 kg   9.80 m s2 
 3.12 m s2
2
2
0.500
kg

m
0.300
m

4.00
kg

3.00
kg



Equation [1] above gives: T1   4.00 kg   9.80 m s2  3.12 m s2 
and Equation [2] yields: T2   3.00 kg   3.12 m s2   9.37 N .
9
 26.7 N ,
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