Chapter 8 Torque and Angular Momentum
Review of Chapter 5
We had a table comparing parameters from linear and rotational motion. Today we fill in the
table. Here it is
Description
Linear
x
x
vx
x
v x ,av 
t
x
v x  lim
t 0 t
v
a x ,av  x
t
v
ax  lim x
t  0 t
m
F

p
position
displacement
Rate of change of position
Average rate of change of position
Instantaneous rate of change of position
Average rate of change of speed
Instantaneous rate of change of speed
Inertia
Influence that causes acceleration
Momentum
Rotational




t

  lim
t  0 t

 av 
t

  lim
t  0 t
I
 av 


L
The relations (often physical laws) for rotational motion can be found by a simple substitution of
rotational variables for the corresponding linear variables.
Rotational Kinetic energy
A wheel suspended at its axis can spin in space. Since
the points of the wheel are moving, the wheel has
kinetic energy.
All the pieces in a rigid body remain at the same
location relative to all the other pieces. For a rotating
object, the parts further away from the axis of rotation
are moving faster.
v  r
The total kinetic energy of all the pieces will be
K total  12 m1v1  12 m2 v2    12 mN vN
2
N
  12 mi vi
2
2
i 1
8-1
2
N
K total   12 mi ri  2
2
i 1
 N
2
 12   mi ri  2
 i 1

The quantity in parentheses is called the rotational inertia (or the moment of inertia)
N
I   mi ri
2
i 1
Finding the Rotational Inertia (page 262)
1. If the object consists of a small number of particles, calculate the sum directly.
2. For symmetrical objects with simple geometric shapes, calculus can be used to perform
the sum.
3. Since the rotational inertia is a sum, you can always mentally decompose the object into
several parts, find the rotational inertia of each part, and then add them.
The rotational inertia depends on the location of the rotation axis. The same object will have a
different rotational inertia depending on where it is rotating. Look at the formula for a thin rod
below.
8-2
The rotational kinetic energy of a rigid object rotating with angular velocity  is
K  12 I 2
Compare to the translational kinetic energy
K  12 mv2
Torque
A quantity related to force, called torque, plays the role in rotation that force itself plays in
translation. A torque is not separate from a force; it is impossible to exert a torque without
exerting a force. Torque is a measure of how effective a given force is at twisting or turning
something.
The torque due to a force depends of the magnitude of the applied force, the force’s point
of application, and the force’s direction.
First definition of torque
  rF
r
Because rotations have directions, we assign the + sign to torques that cause counterclockwise
rotations, and – sign to torques that cause clockwise rotations. What is the sign of the torque in
the figures above?
Torques are measured in the units of force times distance. This is the same dimensions as
work. However, torque has a different effect than work. To keep the two concepts distinct, we
measure work in joules and torque in newton-meters.
Second definition of torque
  r F
8-3
r
Find the lever arm (or moment arm) by extending the line of the force and drawing a line from
the axis of rotation so that is crosses the line of the force at a right angle. Finding the lever arm
is often the most difficult part of a torque problem.
Finding Torques Using the Lever Arm (p 269)
1. Draw a line parallel to the force through the force’s point of application; this line is called
the force’s line of action.
2. Draw a line from the rotation axis to the line of action. This line must be perpendicular
to both the axis and the line of action. The distance from the axis to the line of action
along this perpendicular line is the lever arm (r⊥). If the line of action of the force goes
through the rotation axis, the lever arm and the torque are both zero.
3. The magnitude of the torque is the magnitude of the force times the lever arm:
  r F
4. Determine the algebraic sign of the torque as before.
Center of Gravity
When the gravitational force acts on an object, all the small pieces of the object experience the
gravitational force. This very large number of forces taken around an axis will create a torque.
How do we deal with that?
Fortunately, we can greatly simplify the problem. The total force can be considered to
act a single point called the center of gravity. If the gravitational field is uniform in magnitude
and direction, the center of gravity is located at the center of mass.
becomes
8-4
Work done by a torque
The expression for the work done is
W  F s
 
  r 
r
  
Power is the rate of doing work
W
t


t

P
Rotational Equilibrium
We remember that for an object to remain at rest,
the net force acting on it must be equal to zero.
(Newton’s first law.) However, that condition is not
sufficient for rotational equilibrium. What happens
to the object to the right? The forces have the same
magnitude.

F2

F1
F1  F2
8-5
Conditions for equilibrium (both translational and rotational):

F
  0 and
  0
The obedient spool.
F1 and F2 make the spool roll to the left, F4 to the right, and F3 makes it slide.
Problem-Solving Steps in Equilibrium Problems (page 274)
1. Identify an object or system in equilibrium. Draw a diagram showing all the forces
acting on that object, each drawn at its point of application. Use the center of gravity
(CM) as the point of application of any gravitational forces.
2. To apply the force conditions, choose a convenient coordinate system and resolve each
force into its x- and y-components.
3. To apply the torque condition, choose a convenient rotation axis – generally one that
passes through the point of application of an unknown force. Then find the torque due to
each force. Use whichever method is easier: either the lever arm times the magnitude of
the force or the distance times the perpendicular component of the force. Determine the
direction of each torque; then either set the sum of all torques (with their algebraic signs)
equal to zero or set the magnitude of the CW torques equal to the magnitudes of the
CCW torques.
4. Not all problems require all three equilibrium equations (two force component equations
and one torque equation). Sometimes it is easier to use more than one torque equation,
with a different axis. Before diving in and writing down all the equations, think about
which approach is the easiest and most direct.
There are many good examples worked out for you in the text. See pages 274-279.
Example: What is the smallest angle a ladder can make so that it does not slide?
8-6
FW
mg
N

fs
Solution: We will use the condition for rotational equilibrium
  0
We can choose any axis about which to take torques. The axis I choose is where the ladder
touches the floor. The lever arms for the normal force and the frictional force will be zero and
their torques will also be zero. Recall that the torque is
   Fr
If the ladder has length L, the lever arm for the weight is the short horizontal line below
the floor in the diagram. The lever arm is “the perpendicular distance from the line of the force
to the point of rotation”. Here it is
L
cos 
2
r 
The lever arm for the force of the wall pushing against the ladder is
r  L sin 
Using the condition for rotational equilibrium
  0
 F   mg  0
 FW L sin   mg
The conditions for translational equilibrium are
8-7
L
cos   0
2
F
x
 0 and
F
y
0
Referring to the FBD, the x-components give
F
0
x
FW  f s  0
FW  f s
Again looking at the FBD, the y-components
F
0
y
N  mg  0
N  mg
Oh, no! Four equations:
 FW L sin   mg
L
cos   0
2
FW  f s
N  mg
f s  s N
Use the torque relation
 FW L sin   mg
L
cos   0
2
FW L sin   mg
L
cos 
2
mg
cos 
2
mg
tan  
2 FW
FW sin  
Substitute the two force equations from Newton’s second law
tan  
mg
2 FW

N
2 fs
8-8
fs 
N
2 tan 
Since this is a static friction force,
fs  s N
N
 s N
2 tan 
1  2  s tan 
tan  
1
2
If the coefficient of static friction is 0.4, the smallest angle is 51o.
Equilibrium in the Human Body
Forces act on the structures in the body.
Example 8.10: The deltoid muscle exerts Fm on the humerus as shown. The force does two
things. The vertical component supports the weight of the arm and the horizontal component
stabilizes the joint by pulling the humerus in against the shoulder.
There are three forces acting on the arm: its weight (Fg), the force due to the deltoid muscle (Fm)
and the force of the shoulder joint (Fs) constraining the motion of the arm.
Since the arm is in equilibrium, we use the equilibrium conditions. To use the torque
equation we use a convenient rotation axis. We choose the shoulder joint as the rotation axis as
that will eliminate Fs from consideration. (Why?)
  0
g m  0
 Fg rg  Fm  rm  0
8-9
 Fg rg  Fm sin 15rm  0
Fm 
Fg rg
rm sin 15

(30 N)(0.275 m)
 266 N
(0.12 m) sin 15
To support the 30 N arm a 270 N force is required. Highly inefficient!!
The Iron Cross.
Here is an interesting video: http://www.youtube.com/watch?v=Sd1LjYgIm_4
Because of the symmetry, half of the gymnast’s weight is supported by each ring. Consider the
FBD above.
  0
w m  0
1
2
Wrw  Fm r m  0
Fm 
Wrw
2r m
W (0.60 m)
2(0.045 m) sin 45
 9.4W

8-10
The force exerted by the latissimus dorsi and the pectoralis major on one side of the gymnast’s
body is more than nine times his weight!
“The structure of the human body makes large muscular forces necessary. Are there any
advantages to the structure? Due to the small lever arms, the muscle forces are much larger than
they would otherwise be, but the human body has traded this for a wide range of movement of
the bones. The biceps and triceps muscles can move the lower arms through almost 180º while
they change their lengths by only a few centimeters.” (p. 282)
A video of equilibrium and how easily it can be disrupted:
http://www.youtube.com/watch?v=K6rX1AEi57c
Rotational Form of Newton’s second law
  I
Very similar to the other second law.
Motion of Rolling Objects
A rolling object has rotational kinetic energy and translational kinetic energy.
K  Ktrans  K rot
 12 mvCM  12 I CM  2
2
Why does the object roll (and not slide)? Frictional forces exert a torque on the object.
8-11
Example 8.13 The acceleration of a rolling ball.
The rotational form of Newton’s second law is
  I
The torque on the ball is due to friction
  rf
So
  I
rf  I
I
f 
r
We can use Newton’s second law to find the linear acceleration of the ball. As we usually do,
take the +x-axis along the incline.
F
x
 max
mg sin   f  ma
Use the expression for the frictional force to find,
mg sin   f  ma
I
mg sin  
 ma
r
8-12
But the acceleration of the ball is related to its angular acceleration, a = r.
I
 ma
r
Ia
mg sin   2  ma
r
Ia
mg sin   2  ma
r
mg sin 
a
m  I r2
mg sin  
For a uniform, solid sphere, I = (2/5)MR2 and for a thin ring, I = MR2. Which has the larger
acceleration? Consider the effect of the rotational inertia on the acceleration.
Example: A solid sphere rolls down a hill that has a height h. What is its speed at the bottom?
Solution: Use conservation of energy. Since the ball rolls without slipping, the frictional force
doesn’t do any work. The displacement is zero in the definition W = F r cos .
U 1  K1  U 2  K 2
mgy1  0  0 

1
2
mv 2  12 I 2

The translational speed of the ball is related to its rotational speed, v = r.
mgy1  12 mv 2  12 I 2
mgh  12 mv 2  12 I (v / r ) 2
 12 (m  I r 2 )v 2
v
2mgh
m  I r2
For the solid sphere, I = (2/5)MR2
v
2mgh
2mgh
2 gh



2
m I r
m  (2 5)m
(7 5)
10
7
gh
This is less than the answer we found when we ignored rolling, 𝑣 = √2𝑔ℎ.
Angular momentum
We introduced the idea of linear momentum in chapter 7. We had
8-13
 dp
 F  dt
A similar expression exits for rotational motion
dL
  dt
The net external torque acting on a system is equal to the rate of change of the angular
momentum of the system. The angular momentum
L  I
is the tendency of a rotating object to continue rotating with the same angular speed about the
same axis. Angular momentum is measured in kgm2/s.
If the net torque is zero, we have the conservation of angular momentum
L  0
Li  L f
If the rotational inertia of the system changes, its angular speed will change to compensate.
Angular momentum is a vector. (So is the torque!) The direction is given by the righthand rule.
Our seasons are a consequence of the conservation of angular momentum.
8-14
We have completed our study of rotational motion. Try to see how it is analogous to (the more
familiar) linear motion. Here is a summary:
Description
position
displacement
Rate of change of position
Rate of change of velocity
Linear
x
x
vx
ax
x
v x ,av 
t
x
v x  lim
t 0 t
v
a x ,av  x
t
v
ax  lim x
t  0 t
Average rate of change of position
Instantaneous rate of change of position
Average rate of change of velocity
Instantaneous rate of change of velocity
Equations of uniform acceleration
Influence that causes acceleration
Newton’s second law
 f  i  t
x  v x t  12 a x (t ) 2
  t  12  (t ) 2
x  12 (v fx  vix )t
  12 ( f  i )t
v fx  vix  2ax
 f 2  i 2  2
2
m

F
 Fx  max
F
y
Work
Kinetic energy





 av 
t

  lim
t  0 t

 av 
t

  lim
t  0 t
v fx  vix  a x t
2
Inertia
Rotational
 may
W  F r cos 
2
1
2 mv
8-15
I   mi ri
2
i
  Fr  F r 
  I
W   
2
1
2 I


p  mv
 dp
 F  dt

F  0
Momentum
Newton’s second law
Condition for conservation of momentum


pi  p f
Conservation of momentum
8-16


L  Iω
dL
  dt
  0
Li  L f