Angular Momentum and Coefficient of Restitution

[DeadEyeWilly]

Here is my problem.

A rod spinning about it's cg of angular speed omega. So there is no linear momentum, I believe that's correct. Just angular momentum. A ball, v_i = 0, impact the rod at radius, r. The coefficient of restitution is e. How do I use this to find the final velocity of the ball? I tried saying

e = (v_f-r*omega_f)/(vi-r*omega_i)

and conservation of momentum:

Irod*omega_i = Irod*omega_f + m_ball * v_f * r

but I don't think the coeff. of restitution equation is correct. I don't think you can say the tip speed slows by this much. Perhaps in a golf swing analysis where you have a large mass at radius r moving basically linearly you can say this but not with a spinning rod.

Thoughts?

 

[AlephZero]

You didn't say if the rod is fixed at the CG, but that's what you assumed in your equations.

Either you have some signs wrong in your coeff of restitution equation, or you are taking e to be negative. Is that the problem?

e = relative speed of separation / relative speed of approach.

where all three terms are positive quantities.

If the mass of the ball is very small, then omega will be almost unchanged and the speed of the ball will be about 2 r omega.

If the mass of the ball is very large, then omega_f = -e omega_i approximately, and the velocity of the ball is almost zero.

 

[DeadEyeWilly]

Yep. That was it; wrong sign in the coefficient of restitution equations (and no I was not assuming a negative coefficient of restitution)

And just to clarify, yes the rod is pinned in the center.

Thank you so much for your help and time.

Lee