Physics I
Class 14
Cross Product, Torque, and
Angular Momentum
Rev. 23-Feb-04 GB
14-1
Review of Angular and
Linear Acceleration
MUST express angles in radians.
s  r
v  r
a tangential   r
a centripetal
v 2 2 r 2


 2 r
r
r
The radial direction is defined to be +
outward from the center.
a radial  a centripetal
14-2
Review of Torque
For linear motion, we have “F = m a”. For rotation, we have


  I
The symbol “” is torque. We will define it more precisely today.




 When the rotation is speeding up,
and
are in the
same direction.


 When the rotation is slowing down,  and are in
opposite directions.
Torque and angular acceleration are
always in the same direction in Physics 1.
14-3
Motivation for the Concept of
Torque
We know that when we try to throw an object by applying a force
off center, that we end up imparting an angular velocity to the object
as well as linear velocity. For a given force, the greater the distance
it is applied off center, the greater the resulting angular acceleration.
We need a mathematical technique to calculate the “off-center”
effect of a force and relate it to the direction and magnitude of the
resulting angular acceleration.
14-4
The Vector Cross Product
We learned how to “multiply” two vectors to get a scalar.
That was the “dot”
 product:



d  a  b | a | | b | cos( )
Now we will “multiply”
two vectors
to get another vector:




 
c  a  b ; | c |  | a | | b | sin( )
The direction comes from the right-hand rule. It is at a right angle to

the plane formed by a and b . In other words, the cross product is

at right angles to both a and b . (3D thinking required!)
14-5
The Vector Cross Product
14-6
Torque as a Cross Product
  
  rF
 

|  |  | r | | F | sin( )

r is the vector from the axis of rotation to where the force is applied.
The torque can be zero in
 three different ways:
1. No force is applied (| F |  0 ).

|
2. The
 force is applied at the axis of rotation ( r |  0 ).

3. F and r in the same or opposite directions (sin( )  0 ).
14-7
Example of Torque - A Rod
Suspended by a String Off Center
 

|  |  | r | | F | sin( )

|  | r m g
Gravity force on “extended” bodies
can be viewed as acting on a point
mass (with the same total mass)
at the object’s center of mass
(balancing point).
(out of page)
r
F=W
What would be the direction of the rod’s angular acceleration?
14-8
Angular Momentum of a Particle
center of rotation (defined)

r


p  mv
Angular momentum of a particle
once a center is defined:
  
l  r p
(What is the direction of angular
momentum here?)
Once we define a center (or axis) of rotation, any object with a
linear momentum that does not move directly through that point
has an angular momentum defined relative to the chosen center.
14-9
Class #14
Take-Away Concepts




I

1. : Speeding up, slowing down.
.
2.
Definition of vector cross product:
 
   
c  a  b ; | c |  | a | | b | sin( )
  
3. Torque as a cross product:   r F .
 
4. Angular momentum of a particle: l  r  p .
14-10
Class #14
Problems of the Day
__1. One of your classmates in Physics 1 is sitting on a stool that rotates on
a vertical axle. She is facing north and puts out her right arm toward
where you are standing to the east of her. You push her arm forward
(north) and she begins to spin counter-clockwise as seen from above.
The direction of the torque vector from your push is
A)
B)
C)
D)
E)
F)
G)
H)
North.
South.
East.
West.
Up.
Down.
Clockwise.
Counter-clockwise.
14-11
Answer to Problem 1 for Class #14
The answer is E, up.
The direction of the angular velocity vector is along the axis of
rotation. The axis of rotation is vertical, which means that the
angular velocity must be up or down. The angular velocity is
increasing, which means the angular acceleration is in the same
direction. Torque is in the same direction as angular acceleration.
By the right-hand rule, the direction of angular velocity is up.
Another way to figure this out: r vector (E or +X) cross F vector
(N or +Y) is in the +Z or up direction.
14-12
Class #14
Problems of the Day
2. A force of 5 N is applied to the rim of a wheel at an angle
of 30º with respect to the outward radial direction. (The force
is in the plane of the wheel.) The radius of the wheel is 0.3 m.
What is the magnitude of the torque on the wheel?
30º
F
14-13
Answer to Problem 2 for Class #14
We aren’t worried about direction here, just magnitude.
The magnitude is r F sin(30) = 0.3 x 5 x 0.5 = 0.75 N-m.
14-14
Activity #14
Torque and Angular Momentum
Objective of the Activity:
1.
2.
Think some more about rotation concepts.




I

Determine the direction of torque from
.
 
3. Determine the direction of torque from   r  F .
 
4. Calculate momentum of a particle from l  r  p .
(No data measurements today, only paper and pencil.)
14-15