PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 1
Two balls, both with a mass of 3.5 kg are connected by a mass-less rigid rod. The centre of mass of the
system is as the axis of rotation. The balls are 1.0 m from the centre of mass and rotate about it with an
angular speed of 12 radians per second.
a) Find the moment of inertia of the system.
m1  m2  3.5 kg, r1  r2  1.0 m,   12 s1
terms:
I   mi ri2
i
 m1r12  m2 r22
  3.5 kg 1.0 m    3.5 kg 1.0 m 
2
2
 7.0 kg  m 2
The moment of inertia is 7.0 kg·m2.
b)
Find the rotational kinetic energy of the system.
K R  12 I  2

1
2
 7.0 kg  m 12 s 
 504 kg·
2
1
2
m2
s2
 5.0  102 J
The rotational kinetic energy of the system is 5.0102 J.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 2
Three 3.6 kg balls are equally space about a circle of radius of 1.0 m forming an equilateral triangle. Find
the moment of inertial of the system if
Four balls are equally space about a circle of radius of 1.0 m. All of the balls have a mass of 3.6 kg. Find
the moment of inertial of the system if
a)
all of the balls rotate at a radius of 1.0 m about the centre of the
system.
terms:
1.0 m
m1  m2  m3  3.6 kg,
r1  r2  r3  1.0 m
I   mi ri2
i
 m1r12  m2 r22  m3r32
  3.6 kg 1.0 m    3.6 kg 1.0 m    3.6 kg 1.0 m 
2
2
2
 10.8 kg  m 2
The moment of inertia of the system is 11 kg·m2.
b)
the balls rotate about the line joining two of the balls.
1.0 m
The 1.0 m radius in question a) forms the hypotenuse of a small
inner triangle of the equilateral triangle, shown at right. The new
radius for two of the balls is 0.866 m. The ball on the axis of
rotation (top ball) has a radius of zero.
terms:
30°
0.866 m
m1  m2  m3  3.6 kg,
r1  r2  0.866 m, r3  0
I   mi ri2
i
0.866 m
 m1r12  m2 r22  m3r32
  3.6 kg  0.866 m    3.6 kg  0.866 m    3.5 kg  0 
2
2
2
 5.4 kg  m 2
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
2
The moment of inertia of the system is 5.4 kg·m .
Here we have two identical systems rotated about two different axes. The moments of inertia for the
two identical systems are different. The moment of inertia depends on the axis of rotation.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 3
For a long thin rod with total mass of 5.0 kg and length 1.2 m, find the moment of inertia
a) about an axis of rotation in the centre with the rod perpendicular to the axis,
b) about an axis of rotation in the end with the rod perpendicular to the axis.
c) Find the moment of inertia of the rod in part b) if there is a ball attached to the rod 1.0 m
from the axis. The mass of the ball is 2.0 kg.
M  5.0 kg, L  1.2 m
terms:
a)
about an axis of rotation in the centre with the rod perpendicular to the axis.
I  121 ML2
 121  5.0 1.2 
2
L
 0.60 kg·m 2
The moment of inertia of the rod rotating about its centre is 0.60 kg·m2.
b)
about an axis of rotation in the end with the rod perpendicular to the axis.
I  13 ML2
 13  5.0 1.2 
2
L
 2.4 kg·m 2
The moment of inertia for the rod rotating about its end is 2.4 kg·m2.
c)
Find the moment of inertia of the rod in part b) if there is a ball attached to the rod 1.0 m
from the axis. The mass of the ball is 2.0 kg.
new terms:
m  2.0 kg, l  1.0 m
I   Ii
l
i

1 ML2
3
 mr
L
2
 2.4 kg  m 2   2.0 kg 1.0 m 
2
 4.4 kg  m 2
The moment of inertia of the rod / ball system is 4.4 kg·m2.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 4
a)
Use the parallel axis theorem to verify that the moment of inertia of a long thin rod rotating
about its end is I  1 ML2 given that the moment of inertia for the rod rotating about its
3
centre is I  1 ML2 .
12
For a uniform rod, the centre of mass is at the centre of the rod,
so the distance from the centre to the end is
d
L
2
centre of mass
d
L
According to the table, shape 5, the moment of inertia through
the centre of the rod is
1 ML2
ICM  12
According to the parallel axis theorem
I parallel axis  I CM  Md 2
1 ML2  M  L 
 12
 
2
2
1 ML2  1 ML2
 12
4
 13 ML2
The moment of inertia of a rod rotating about its end is I  1 ML2 .
3
b)
Show that the moment of inertia of a tire (thin walled cylinder) rotating about its edge is
I  2MR2 .
For a thin walled cylinder, the centre of mass is along the central axis
of the cylinder, so the distance from the centre to the rim is
d
R
dR
According to the table, shape 1, the moment of inertia through
the centre of the rod is
ICM  MR 2
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centre of mass.
inside
5
PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
According to the parallel axis theorem
I parallel axis  ICM  Md 2
 MR 2  MR 2
 2MR 2
The moment of inertia of a tire rotating about its rim end is I  2MR2 .
For a rod with total mass of 5.0 kg and length 1.2 m, find the moment of inertia
a) about an axis of rotation in the centre with the rod perpendicular to the axis,
M  5.0 kg, L  1.2 m
1 ML2
I  12
1 5.0 kg 1.2 m
 12



2
 0.60 kg  m 2
The moment of inertia is 0.60 kg·m2.
b)
about an axis of rotation in the end with the rod perpendicular to the axis.
M  5.0 kg, L  1.2 m
I  13 ML2

1
3
L
 5.0 kg 1.2 m 2
 2.4 kg  m 2
The moment of inertia is 2.4 kg·m2.
c)
Find the moment of inertia of the rod in part b) if there is a ball attached to the rod 1.0 m
from the axis. The mass of the ball is 2.0 kg.
M  5.0 kg, L  1.2 m, m  2.0 kg, r  1.0 m
L
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
We can write
I   mi ri2
i
as
I   Ii
i
The moment of inertia of a combined object is simply the sum of the moments of
inertia of the parts. For the combined object in question
I   Ii
i
 I rod  Iball
 13 ML2  mr 2
 13  5.0 kg 1.2 m    2.0 kg 1.0 m 
2
2
 4.4 kg  m 2
The moment of inertia is 4.4 kg·m2.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 5
1
a)
A force of 5.0 N is applied to a nut with a moment arm of 0.50 m. A second force of 2.0 N
is applied to the nut with moment arm of 1.2 m. Find the total torque on the following
system
assuming the torques produced are in the same direction.
F1  5.0 N, d1  0.50 m, F2  2.0 N, d2  1.2 m
 net  1   2
 F1d1  F2 d 2
  5.0 N  0.50 m    2.0 N 1.2 m 
 4.9 Nm
The net torque is 4.9 Nm.
b)
assuming the torques produced are in the opposite direction.
We will assume that the second torque is negative.
 net  1   2
 F1d1  F2 d 2
  5.0 N  0.50 m    2.0 N 1.2 m 
 0.1 Nm
The net torque is 0.1 Nm.
2
a)
Find the angular acceleration of the nut in the last question if the momentum of inertia is 4.0
kg·m2.
torques in same direction
I  4.0 kg·m 2 ,  net , a  4.9 Nm,  net ,b  0.1 Nm
 net , a  I


 net , a
I
4.9 Nm
The angular acceleration is 1.2 s-2.
4.0 kg·m 2
 1.23 s 2
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
b)
torques in opposite direction
 net ,b  I


 net ,b
I
0.1 Nm
The angular acceleration is 0.03 s-2.
4.0 kg·m 2
 0.025 s 2
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 6
A 240 kg crate sits on a horizontal surface. The coefficient of kinetic friction between the crate and floor
is 0.50. A rope runs horizontally from the crate, over a pulley, to a 175 kg hanging mass. The pulley has
a radius of 0.50 m and a moment of inertia of 32 kg·m2. Find the acceleration of the crate.
Sketch of system
+y
+x
+y
+x
The coordinate system for the hanging mass is rotated 90° to make its acceleration in the same direction
as the crate.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
Terms:
mC  240 kg, mH  175 kg,   0.50, r  0.50 m, I  32 kg·m2
+y
Draw FBDs
+
+x
r
T1
Ff
T2
+x
FN
T1
T2
FgH
FgC
Pulley
Crate
Hanging Mass
The FBDs give us the following.
The tensions are tangent to the pulley, perpendicular to the radius to the pulley axis.
 net  I
T2 r  T1r  I
ax
r
1
We need to find the friction force for the crate. The net force in the y-direction is
Fnet , y  0
FN  FgC  0
FN  mC g
The friction force is then
F f   FN
  mC g
The net force in the x-direction, the direction of acceleration is
Fnet , x  mC ax
T1  F f  mC ax
T1   mC g  mC ax
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 2
11
PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
The forces on the hanging mass obey the relation
Fnet , x  mH ax
FgH  T2  mH ax
 3
T2  mH g  mH ax
Equations 2 and 3 can inserted into equation 1 to get
T2 r  T1r  I
ax
r
 mH g  mH ax  r    mC g  mC ax  r  I
ax
r
Rearranging gives
mH gr  mH ax r   mC gr  mC a x r  I
mH ax r  mC ax r  I
ax 
ax
r
ax
 mH gr   mC gr
r
mH gr   mC gr
I
mH r  mC r 
r
Substituting values (make sure all values are in MKS),
a
175 9.8 0.50    0.50  240  9.8 0.50 
175 0.50    240  0.50  
32
0.50
 0.993 m/s 2
The acceleration of the system is 0.99 m/s2.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 7
A uniform rod of length L = 1.0 m and mass M = 0.30 kg is free to rotate about a frictionless pin through
one end. Released from rest in the horizontal position, the rod accelerates due to gravity.
a)
What is the angular speed when the rod reaches its vertical position?
rotation
Sketch:
pin
All terms are defined in the question.
The moment of inertia of the rod is
I  13 ML2
For the change in gravitational potential energy, we only need to look at change in height of the centre
of the mass of the rod. The centre of mass of the rod only falls a distance L/2 = 0.50 m.
Setting the final gravitational potential energy equal to 0, the initial gravitational energy is
L
2
U gi  Mg
Energy is conserved so
KRf  U Rf  KRi  U Ri
The gravitational potential energy in the final position is 0 and the initial kinetic energy is zero (from
rest), so the final rotational kinetic energy is
K Rf  U Ri
1
2

I  2  Mg
1 ML2
3

2
L
2
 MgL
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
Isolating the angular speed



3M g L
M L2
3g
L

3 9.8 m/s 2

1.0 m
 5.42 s 1
The angular speed of the rod is 5.4 s-1 or 5.4 radians per second at the vertical position.
b)
Does the angular speed in part a) depend on the mass of the rod?
No, the masses cancel out when we insert the expressions for moment of inertia and
gravitational potential energy into the conservation of energy equation.
Determining the dependence of one variable on another is yet another benefit of using algebra until the
very end of the question.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 8
Find the torque for the following
a)
F  2.6 N ˆi  5.0 N ˆj, r  1.8 m ˆi  1.1 m ˆj
Both vectors are in the x-y plane so we can use the simplified from
τ  rF


 rx Fy  ry Fx kˆ
  1.8 m  5.0 N    1.1 m  2.6 N   kˆ
 11.86 Nm kˆ
The torque is 12 Nm in the z-direction.
b)
F = 12.5 N, r = 3.4 m,   85 , magnitude only
Assume both vectors are in the x-y plane so the cross product is in the z-direction.
  rF sin 
  3.4 m 12.5 N  sin 85
 42.3 Nm
The torque is 42 Nm.
c)
F = 12.5 N, r = 3.4 m,   5
Assume both vectors are in the x-y plane so the cross product is in the z-direction.
  rF sin 
  3.4 m 12.5 N  sin 5
 3.70 Nm
The torque is 3.7 Nm.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 9
1.
a)
A skater, with a moment of inertia of 9.0 kg·m2, is rotating at 9.1 radians per second with
arms extended.
Find the skater’s angular momentum.
  9.1 s 1 , I  9.0 kg·m 2 , L  ?
L  I


 9.0 kg·m 2 9.1 s 1

m2
 81.9 kg
s
The angular momentum of the skater is 82 kg·m2/s.
b)
Find the skater’s angular speed if she pulls her arms and legs in, reducing her moment if
inertia to 3.2 kg·m2.
  ?, I  3.2 kg·m 2 , L  82 kg
m2
s
L  I

L
I
m2
s

3.2 kg·m 2
81.9 kg
 25.6 s 1
The angular speed of the skater after tucking her arms in is 26 radians per second, or
about 4.1 rotations per second.
Angular momentum is conserved if there are no outside forces. An object can greatly increase its
rotation speed by reducing its moment of inertia. This little trick is how a figure skater can speed up in
twirls (not sure of technical name) by pulling her/his arms in towards her/his body. Pulling in arms
reduces the skater’s moments of inertia. This trick is not available with linear momentum because the
mass of an object can be altered by simple changing its shape.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 10
A torque of 13 Nm is applied to an object for 4.2 seconds, increasing its angular speed from 1.1 radians
per second to 3.4 radians per second. Calculate the moment of inertia of the object.
I  ?,  net  13 Nm, t  4.2 s, i  1.1 s1 ,  f  3.4 s1
Torque is the rate of change of the angular momentum (just as force is the rate of change of the linear
momentum).
 net 

L
t
I  f  I i
I
t
 f  i
t
Rearranging
I

 net t
 f  i
13 Nm  4.2 s 
3.4 s 1  1.1 s 1
 23.7 kg·m 2
The moment of inertia is 24 kg·m2.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 11
A two dimensional tractor sits on level ground. The distance between the
front and back wheel is 2.7 m. The 1.0 x 104 N weight of the tractor has a
centre of mass 1.0 m from the back of the tractor. Calculate the weight on
each wheel (two variables).
We will assume that all vectors are vertical so we can use the scalar form.
FB  ?, FF  ?, FgCM  1.0 104 N, xF  2.7 m, xCM  1.0 m
1.
Sketch
FB
moment arms
xF
FF
xCM
axis of rotation
(non-rotation)
FgCM
This is a fairly simple system, the choice of the axis of rotation is not critical. We will put it at one of the
unknown forces. This removes that force from the torque equations.
Assume that the weight of the tractor and the two forces from the wheels lie along a single horizontal
line. The forces are vertical so the moment arms are horizontal regardless of the vertical location of the
forces. If you are not convinced, you can imagine that the centre of mass is 1.0 m above the two wheel
forces. The moment arms are still horizontal, their sizes unchanged.
moment arms
xCM
FB
axis of rotation
(non-rotation)
rCM
FF
FgCM
xF
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
2.
FBD for the beam
FB
+ FF
–
axis of
rotation
FgCM
xCM
xF
The torque directions are noted for each torque. These are the directions that he tractor
would rotate in the absence of other torques. By defining the positive and negative
directions for the torques in the FBD, we can treat all forces and distances as positive and
account for the direction with the + or – in front of the terms in the net torque equation. This
should remove the need for negative distances and forces.
3.
In the sketch, the axis of rotation is at the force from the back wheel. This choice
means that the torque from the back wheel is zero. The system is in static
equilibrium so the total torque on the beam gives us
 net  FB  0   FgCM xCM  FF xF


0  0  1.0  104 N 1.0 m   FF  2.7 m 
FF 
1.0  104 Nm
2.7 m
 3.7  103 N
4.
As far as the Fnet  0 analysis is concerned, we can ignore all of the distances on the FBD.
Fnet  FB  FgCM  FF

 
0  FB  1.0  104 N  3.7  103 N

FB  6.3  103 N
5.
We can now employ Newton’s third law (the weight on the wheels is equal and opposite to
the forces from the wheels. The weight on the back wheel is 6.3103 N and the weight on
the front wheel is 3.7103 N.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
PROBLEM 12
Anne (61 kg) stands 2.0 m up a 3.0 m ladder that leans against a wall, making an angle of 51° with the
ground. The 7.5 kg mass of the ladder is uniformly distributed along the ladder. Fiction between the
ladder and ground keeps the ladder from slipping. There is no friction between the ladder and the wall
so the wall force is perpendicular to the wall. Calculate the normal force on the ladder, the friction force
and the force from the wall.
1.
Sketch
FW
xL
xA
FgA
xl
FgL
FN
θ
axis of rotation
(non-rotation)
FF
The length of the ladder is xL  3.0 m .
The ladders mass is equally distributed along the ladder so its centre of mass is at the midpoint
xl 
xL
 1.5 m .
2
The other terms are
FN  ?, FF  ?, FW  ?, mA  61 kg, mL  1.0 104 N, xA  2.0 m,   51
We will use the point where the ladder touches the floor as our pivot. A pivot there means that the
torques from the normal force and friction force are zero, leaving the wall force as the only unknown in
the net torque equation.
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
2.
FBD for the beam
+
xL
FW
θ
–
xA
–
xl
FgL


FgA
  90  
FN
θ
axis of rotation
(non-rotation)
FF
θ
Using geometry, the angle wall force and the ladder is   51 , and the angles between the
ladder weight and Anne’s weight and the ladder are   90   .
The torque directions are noted for each torque. These are the directions that he ladder
would rotate in the absence of other torques. By defining the positive and negative
directions for the torques in the FBD, we can treat all forces and distances as positive and
account for the direction with the + or – in front of the terms in the net torque equation. This
should remove the need for negative distances and forces.
3.
In the sketch, the axis of rotation is at the base of the ladder. This axis location was chosen
to make the torque from the normal force and friction force zero. The system is in static
equilibrium so the total torque on the beam gives us
 net  FN  0   FF  0   FgL xl  FgA x A  FW xL
0  0  mL gxl  mA gx A  FW xL
The only unknown in this expression is the force from the wall. Isolating the force from the
wall gives
FW 

mL gxl  mA gx A
xL
 7.5 kg   9.8 m/s 2  1.5 m    61 kg  9.8 m/s 2   2.0 m 
3.0 m
 434 N
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PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
4.
As far as the Fnet  0 analysis is concerned, we can ignore all of the distances on the FBD.
Each force is in only one direction so we don’t have to break them into components.
For the x-direction
Fnet , x  FF  FW
0  FF   434 N 
FF  434 N
For the y-direction
Fnet , y  FN  FgA  FgL



0  FN   61 kg  9.8 m/s 2   7.5 kg  9.8 m/s 2

FN  671 N
5.
The wall force and friction forces are both 430 N. The normal force is 670 N.
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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22
PHYSICS SEMESTER ONE
UNIT 7: ANSWERS TO PROBLEMS
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23