# Angular momentum help!

1. Dec 9, 2012

### NasuSama

1. The problem statement, all variables and given/known data

Suppose the system is rotating clockwise when the metal disk is dropped on it. The direction of the frictional force is...

zero on the disk and the system
counter-clockwise on the disk, clockwise on the system
clockwise on the disk, zero on the system
counter-clockwise on the disk zero on the system
zero on the disk, clockwise on the system
clockwise on the disk and the system
counter-clockwise on the disk and the system
clockwise on the disk, counter-clockwise on the system
zero on the disk, counter-clockwise on the system

2. The attempt at a solution

I believe that the answer is:

"counter-clockwise on the disk, clockwise on the system"

But it is not right. I thought that a system keeps turning clockwise, and that when a disk is place on the system, then the frictional force acts counter-clockwise on the disk.

2. Dec 9, 2012

### haruspex

If that is the entire description you're given, I agree with you. Is there more?

3. Dec 9, 2012

### NasuSama

See:

http://www.physics.qc.edu/files

Check under PH. 145.1 Then, check Trial 1 under the Procedures.

That describes the experiment, which is related to the question I'm asked.

4. Dec 9, 2012

### haruspex

This is strange. I read the OP several times before saying I agree with you, but reading it again I don't! Must have got something backwards before.
When the disk is dropped on it is not rotating. Friction will lead to its rotating in the same direction as the system, clockwise. So the force must be clockwise on the disc (and the other way on the system).

5. Dec 9, 2012

### NasuSama

That means frictional force goes clockwise while the system goes counterclockwise?

6. Dec 9, 2012

### haruspex

In the lab described, the system will tend to slow down when the disk is dropped on it. Therefore friction acts on the system counter to the system's motion.