Can a linear momentum generate angular momentum at collision?

I'm trying to get the facts straight here.

Suppose I'm throwing a ball with no angular momentum. It collides with the ground and Newton's third law tells us that a force opposite to the gravity will be applied to the ball at collision. During collision, the momentum will be converted to potential energy and then become kinetic energy again.

Am I right to say that some of the kinetic energy can be converted to angular momentum, thus making the ball rebounce not straight up? (assuming some friction)

• If you're throwing the ball straight down, I don't think so. By cylnidrical symmetry, the ball could equivalently go in any direction... so it will only bounce straight back along the same line. However, if you throw the ball at a slant, the friction will cause some trade-off between just "slipping" and bouncing vs a little bit of "rolling" and the ball will acquire some angular momentum. However, I wonder if you're referring to something else. – Siva May 22 '13 at 20:15
• Yes that's what I was thinking. Imagine a a point on the ball tangent to the velocity vector, and the point of impact. The tangent point will still have a velocity while the point of impact none right? Shouldn't this create some angular momentum? – Antz May 22 '13 at 21:08
• I'd like to stress that for all conservation law calculations in collisions with the ground you need to include the Earth's energy/momentum/angular momentum in the equation. Otherwise the system is open and no conservation law applies. – Emilio Pisanty May 22 '13 at 21:16

Am I right to say that some of the kinetic energy can be converted to angular momentum[?]

No, angular momentum is a conserved quantity. In any isolated interaction you get out exactly as much as you put in.

But you may have intended to ask

Can a ball that is not spinning when I toss it at the ground come off with spin?

to which the answer is "Yes".

There are two ways that can happen:

1. You've thrown the ball at an angle so there is already angular momentum in the Earth--ball system, and some of it ends up spinning the ball around it's own CoM.
2. The ball hits a slope and changes the angular momentum of the Earth a little bit.

Both cases look like angular momentum being created out of nothing because (1) you don't see the action from far enough away to see the initial angular momentum in the first place or (2) you are completely unable to measure the change in the Earth angular momentum imparted. Or both.

• So imagine spinning a basketball on your finger by rapidly brushing your hand against the side of the ball to cause it to spin. Are you saying that ultimately the angular momentum will be sapped from the Earth's rotation? – Brandon Enright May 22 '13 at 20:56
• You can look at this on several levels. At the lowest level there was a pre-existing angular momentum between the hand and the ball and some is transferred. At the next level out you had to push on the ground to move the hand and that does modify the Earth's angular momentum at some level. But not necessarily "sapped" it depends what direction the pushes are in. The guarantee it that $L_{\mathrm{Earth},i} + L_{\mathrm{you},i}+ L_{\mathrm{ball},i} = L_{\mathrm{Earth},f} + L_{\mathrm{you},f}+ L_{\mathrm{ball},f}$ (all measurements around some inertial center). – dmckee May 22 '13 at 21:04
• It is also worth noting that if you look only at one part of a system you can give it some new angular momentum $L$ by doing some work (but you are ignoring the $-L$ that you have imparted to the other part of the system). – dmckee May 22 '13 at 21:06
• thank you very much, I learn something new every day! I didn't realize that linear momentum couldn't be converted to angular momentum (or vice-versa). My choice of the word "sap" was poor. I realized that I could just as easily be adding it depending on how my body pushed against the Earth as I sped the basketball up. – Brandon Enright May 22 '13 at 21:13
• Thank you I hadn't considered Earth's angular momentum, that indeed explains a lot of things :) However I was thinking that if we consider a point on the ball tangent to the velocity vector and the point of impact, the tangent point has velocity while the point of impact has none. Shouldn't this create some kind of "spin"? – Antz May 22 '13 at 21:20