Last time we solved the general one-dimensional elastic collision problem.
m1
v1i
v2i
m2
v1f
m1
initial
m2
v2f
final
Conservation of momentum gives
m1v1i  m2v2i  m1v1 f  m2v2 f
Elastic collision means that kinetic energy is conserved
1
2
m1v1i  12 m2v2i  12 m1v1 f  12 m2v2 f
2
2
2
2
Cancelling the ½ and grouping like masses on the same side of the equation,
m1 (v1i  v1 f )  m2 (v2 f  v2i )
2
2
2
2
Doing the same with the linear momentum equation
m1 (v1i  v1 f )  m2 (v2 f  v2i )
Divide the two equations
m1 (v1i  v1 f )
2
2
m1 (v1i  v1 f )
m2 (v2 f  v2i )
2

2
m2 (v2 f  v2i )
v1i  v1 f  v2 f  v2i
This can be rewritten as
v1i  v2i  v2 f  v1 f
The equation states that m1 approaches m2 with the same speed as m2 moves away from m1 after
the collision. Important result. Solve the above for v2f
v2 f  v1 f  v1i  v2i
and substitute into the momentum equation
m1v1i  m2v2i  m1v1 f  m2v2 f
 m1v1 f  m2 (v1 f  v1i  v2i )
 m1v1 f  m2v1 f  m2v1i  m2v2i
(m1  m2 )v1i  2m2v2i  (m1  m2 )v1 f
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
Using very similar reasoning, we can find an expression for the final speed of m2
 2m1 
 m  m1 
v1i   2
v2i
v2 f  
 m1  m2 
 m1  m2 
Example: The masses are m1 = 2 kg and m2 = 3 kg and they collide elastically in one dimension.
Before the collision, m1 is moving at 4 m/s to the right and m2 is moving at 1 m/s to the left.
What are the velocities of m1 and m2 after the collision?
Solution: Use the equations derived above,
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
 2 kg  3 kg 
 2(3 kg) 
(4 m/s)  
(1m/s)
 
 2 kg  3 kg 
 2 kg  3 kg 
 2 m/s
 2m1 
 m  m1 
v1i   2
v2i
v2 f  
 m1  m2 
 m1  m2 
 2(2 kg ) 
 3 kg  2 kg 
(4 m/s)  
(1m/s)
 
 2 kg  3 kg 
 2 kg  3 kg 
 3 m/s
m1
4 m/s
1 m/s
To review relative velocity, the velocity of 2 relative to 1 is



v 21  v 2G  v G1
m2
Taking components,
v21x  v2Gx  vG1x  1 m/s  (4 m/s)  5 m/s
As seen from m1, m2 is moving to the left at 5 m/s. After the collision,
2 m/s
m1
m2
3 m/s
v21x  v2Gx  vG1x  3 m/s  (2 m/s)  5 m/s
The masses approach at 5 m/s and separate at 5 m/s.
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
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
Linear
x
x
vx
x
t
x
v x  lim
t 0 t
v
a x ,av  x
t
v
ax  lim x
t  0 t
m
F

p
v x ,av 
Rotational




 av 
t

  lim
t 0 t

 av 
t

  lim
t 0 t
I


L
The relations (often physical laws) for rotational motion are found by a simple substitution of
rotational variables for the corresponding linear variables.
Rotational Kinetic energy
An 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
Ktotal  12 m1v1  12 m2v2    12 mN vN
2
N
2
  12 mi vi
2
2
i 1
N
  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
above.
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

F

F
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 figure?
Torques are measured in the units of force times distance. This is the same dimensions as
work. However, the 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
r
r

F
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.
Work done by a torque
The expression for the work done is
W   
Fig. 08.16
Power is the rate of doing work
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 below?
Conditions for equilibrium (both translational and rotational):

F
  0 and
  0
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 (tow 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.
We will have a ViaResponse question similar to Example 8.7. It will be open book and open
note. You will have 15 minutes to complete the quiz. We will start class Thursday with the
ViaResponse quiz.