# Cross Product, Torque, and Angular Momentum Physics I Class 15 ```Physics I
Class 15
Cross Product, Torque, and
Angular Momentum
15-1
Review of Angular and
Linear Acceleration
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.
15-2
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.
15-3
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!)
15-4
The Vector Cross Product
15-5
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 ).
15-6
Example of Torque - A Rod
Suspended by a String Off Center
 

|  |  | r | | F | sin( )

|  | r m g
Forces 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?
15-7
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.
15-8
Class #15
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 .
15-9
Activity #15
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.)
15-10
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