# Angular Momentum

1. A mass m is moving with velocity v. it then collids with the end of a rigid rod perpendicular to its intial path. They stick together. The other end of the rod is attached to a frictionless hinge which allows it to rotate in any direction. rod mass = M length = l
What is the angular velocity of the particle rod system after impact?

3. Initally the only angular momentum would be L = mvl
angular momentum is conserved so the intial L = final L
I am a bit confused about how to find the final L?
L=rmv
L = Iw(for the rod) +mwl for the particle?
I for the rod = MR^(2) / 3

Doc Al
Mentor
Once they collide and stick together they become one body. What's the rotational inertia of "rod + mass"?

Is it just I = (m+M)R^(2)

Doc Al
Mentor
Is it just I = (m+M)R^(2)
No. You already know the rotational inertia of the stick. Just add the rotational inertia of a point mass at its end.

So the final angular momentum would equal
L = [Ml^(2) / 3]w + [Ml^2]w?
where w = vl?

Doc Al
Mentor
So the final angular momentum would equal
L = [Ml^(2) / 3]w + [Ml^2]w?
Almost. That second mass should be m, not M.
where w = vl?
No. (If that were true, you could just write down the answer!) Set initial angular momentum equal to final angular momentum and solve for w.