# Total momentum of a rotating object

1. Oct 24, 2009

### Tominator

Hi
I am a bit puzzled by a specific type of collisions and distribution of momentum. Well, I know that the change in momentum of two coliding objects is equal and opposite. But I am not certain how is it in a case of collision of a directly moving object and a rotating one. Is the sum of a liner momentum and angular momentum, total momentum of a rotating object?

2. Oct 24, 2009

### Staff: Mentor

Linear and angular momentum are separately conserved. So the change in linear momentum of the colliding objects is equal and opposite, and so is the change in their angular momentum.

3. Oct 24, 2009

### Tominator

Well, but there are colisions, where the rotating object almost stops rotating thanks to colision wih another. So how is rotational movement translated to unidirectional in such a case?

4. Oct 24, 2009

### Staff: Mentor

What's conserved is total angular momentum, not just angular momentum about a body's center of mass. A body in pure translation will still have angular momentum about some point.

5. Oct 25, 2009

### Tominator

I must admitt I do not understand this very well. Are you saying that it is not possible to translate rotational movement to unidirectional and vice versa? Because for example if two balls rotating clockwise crash, their rotation can cause them to move even quicker than before the crash. In an empty space, their rotation would be slower and their unidirectional movement would be quicker. I do not understand what you mean by total angular momentum, can you explain that please?

6. Oct 25, 2009

### Staff: Mentor

However they move after the collision, their total linear momentum and total angular momentum will be the same.
Sure. Total angular momentum = Angular momentum due to rotation about the center of mass + angular momentum due to the translation of the center of mass.

That second part may be new to you. Here's an example. Say a particle is moving with speed v in the +x direction at some distance d from the x-axis. What is its angular momentum about the origin? L = r X mv, so its angular momentum = mvd.

If this "particle" were really an object rotating about its center of mass, then you'd have to add the angular momentum due to that rotation to get the total.

7. Oct 28, 2009