# I Conservation of angular momentum in a two ball collision

1. Feb 19, 2017

### Tom Westbrook

I have two balls spinning with v1, omega1 and v2, omega2. They collide elastically with no tangential slip, resulting in new values for v1, omega1 and v2, omega2. I have the two components v1 & v2 figured out in the plane of contact, where angular momentum does not come into play. But I am still struggling with the tangential component that has effect on the final omegas. I have 4 unknowns, and can only come up with three equations. 1) linear momentum, 2) angular momentum, and 3) energy conservation.

I have decided to break the problem into three stages, initial, intermediate and final. The first stage would allow the calculation of an inelastic problem where v1tangential and omega of the intermediate system could be calculated, being they are the same for both balls. Then, the energy lost could be theoretical added back into each final system as energy stored in a clock spring to calculate the final omega1, omega2, and y1tangential, y2tangential. I'm still struggling with this last part. Anyone have any ideas/comments?

Thank you

2. Feb 19, 2017

### mathman

Momentum is a vector, so conservation of linear momentum (in a plane) has two equations.

3. Feb 19, 2017

### Tom Westbrook

I assume you are referring to the two orthogonal directions, say X and Y. I have solved the X direction (through to center points of the two balls, and the center point of contact. I am now attempting to solve in the Y direction (tangential to the ball surface). But just in this direction there are still 4 variables. The linear V of the two balls as they spread apart, and the angular spin of the two balls.

4. Feb 20, 2017

### BvU

is another equation.

You are dealing with a fairly complicated problem. A clear drawing might help a little. And a complete problem statement: the orientation of the spinning axis, for example -- though I think we may implicitly assume that axis is vertical (?)

5. Feb 20, 2017

### Tom Westbrook

The spinning axis would be out of the plane, in the z direction. Two circles in the x, y plane, both spinning in the z direction,

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6. Feb 20, 2017

### Tom Westbrook

I have broken to problem into two parts. The first part is to assume a non-elastic collision with energy lost to a theoretical clock spring that will be recovered later. Conservation of linear and angular momentum are conserved resulting in a Vfs (s for system), the same for both balls, and ωfs for the system. This is what I came up with so far (no mass, because it's the same for both balls):

(also, to simplify I am assuming point masses located at the radius. Like a dumbbell)

Linear momentum:
v1 + v2 = 2 vfs

Vfs = 1/2(v1 + v2)

Angular momentum:
-v1 r + ω1r2 + v2r + ω2r2 = -Vfs + ωfsr2 + Vfsr + ωfsr2

ωfs = 1/2(ω1 + ω2 + v2/r - V1/r)

Energy lost to a theoretical clock spring (to be recovered and entered back into the system later):

SE/m = 1/2(v12 + v22 + ω1r2 + ω22r2) - vfs2 - ωfs2r2

still working on part 2...

Last edited: Feb 20, 2017
7. Feb 20, 2017

### BvU

How does that show in the outcome ? Or the picture ?

PS would it be easier to work in the center-of mass system ? there you have $\sum \vec p = 0$

8. Feb 20, 2017

### Tom Westbrook

Not sure I understand the question. When the two balls come into contact, it would be as if they are locking gears with no slip. If, for example, the first ball is spinning and the second one is not, it would transfer some of it's angular momentum to the spin of the other ball, with no energy loss. Let's say both of the balls are spinning in same direction, which means when they bump into each other their tangential velocities would appose each other. Some of the angular momentum would be converted to translational momentum of the two balls moving away from each other. Again with no energy loss.

I think that is what I am doing, at least that is what I am attempting to do. Maybe it's not clear from the drawing, but if the balls are touching, both when they hit, and after they begin to separate, the center of mass would be at the point of contact. This is the point that I am using to calculate momentum.