# Linear momentum and angular momentum

1. May 7, 2007

### celestra

We see a lot of examples of converting between linear motion and angular motion in our daily life.
But, is the conversion between linear momentum and angular momentum possible?
If so, where have the conservation law of linear momentum and the conservation law of angular momentum gone?
If not so, how can I understand the phenomenon such as this.
A billiard ball that has high angular velocity and low linear velocity is collide with the cushion and bounce off with low angular velocity and high linear velocity.
Thanks.

2. May 7, 2007

### cesiumfrog

Looks like you haven't actually solved enough textbook problems. Basically, linear momentum can be expressed as angular momentum, and vice versa, and so both are always conserved individually.. it just makes problems easier sometimes if you consider just one first, and the other later.

3. May 8, 2007

### celestra

Thank you for your reply. But, I'm still in confusion. Where exactly originated the increased linear momentum of the billiard ball from? If it's from the angular momentum of the ball before the collision, then each conservations doesn't seem to work to me. If one of linear momentum(p) and angular momentum(L) can be converted into the other one according to L = r x p, then I think they cannot be conserved individually. Where did I mistake?

4. May 8, 2007

### cesiumfrog

Regards your question, the billiard ball exerts a force on the cushion. This exchanges linear and angular momentum with the table (and ground), which you didn't notice because the ball has a much smaller mass.

5. May 8, 2007

### celestra

L = angular momentum
P = linear momentum
b = ball
t = table
1 = before collision
2 = after collision
And let's try scalar form only instead of vectors.
Then,
Lb1 + Lt1 = Lb2 + Lt2
Pb1 + Pt1 = Pb2 + Pt2
According to the condition of the above problem,
Lt1 = Pt1 = 0
Lb1 > Lb2
Pb1 < Pb2
Then,
Lb1 - Lb2 = Lt2 > 0 (Surely, this must have been consumed by the cushion as it has been twisted)
But,
Pb1 - Pb2 = Pt2 < 0 (Again, this is magnitude only! It's not related with direction. And this must be supplied from somewhere, not consumed)
My question is if I can consider that the Pt2 is transformed from Lt2. If I can do so, it seems that angular momentum CAN be converted into linear momentum. And it sounds somewhat weird.

6. May 8, 2007

### cesiumfrog

You've made more mistakes there. Your first two inequalities are the wrong way around. (Consequently, Lt2 should be negative, and Pt2 +ve. Absolutely no idea what you meant by "consumed".) No you cannot consider Pt2 transformed from Lt2, as I stated previously.

It shouldn't be surprising. The effect occurs where the spinning ball starts exerting a frictional force on the surrounds. Newton's 3rd law states that the force and momentum imparted on the surrounds is equal and opposite to the force and additional momentum that friction simultaneously imparts on the ball, always giving zero change in (any form of) total momentum. You might say rotational kinetic energy is converted to translational kinetic energy.

By the way, welcome to PF

7. May 8, 2007

### celestra

Lt2 = Lb2 - Lb1 < 0
Pt2 = Pb1 + Pb2 > 0
Now I accept from your explanation that above two relations are totally independent from each other. So, one of them cannot influence the other. And I understand that I cannot predict the movement of the ball unless considering the kinetic energy. Am I right?
Thanks a lot.

8. May 8, 2007

### celestra

I think I had another mistake in the above mention. Let me ask the final question that IF I MAY SAY that some portion of the angular momentum(L) was CHANGED to linear momentum(p) according to the relation L=rxp(where r is the radius of the ball), AS I CAN SAY that some portion of the angular kinetic energy was CHANGED to linear kinetic energy in the above problem?

9. May 13, 2007

### Brock

I you have a ball flying though space it will have 100% KE but if it hits and connects to a bar that's on a stable axis then it will slow down some because some of the KE will transfer to the spring tension of the molocules in the bar and it will be PE. I guess you could say that the PE is actually KE in the swivel of the planet that the axis is connected too.

Lorentz Law is very interesting because if the charge turns then it should also slow down a little bit.