Re: Conservation of angular momentum

On Jun 1, 4:37 pm, "Timo A. Nieminen" <t...@xxxxxxxxxxxxxxxxx> wrote:
On Fri, 1 Jun 2007, Peter wrote:
On Jun 1, 12:14 pm, Ben C <spams...@xxxxxxxxx> wrote:

On that basis, working with a rod of 100g mass and a collision half way
up one half of the rod that stops the ball, the "missing energy" in the
impact comes to at most 0.03J. That doesn't sound like very much to be
absorbed by the rod and/or spindle, although I'm not sure how to
estimate how much to expect.

Your figures are good. The point is that, assuming no energy losses of
any kind, forangularmomentumto be conserved, the final kinetic
energy of the lever would have to be only 1/8 of the initial kinetic
energy of the cylinder. That means that 88% of the initial kinetic
energy was lost, How to account for such a result?

You have, of course, done something like dropping the puck onto the rod
with the rod supported at both ends and seen how high the puck bounces
back? In this case, the rod gains no KE, and the height to which the puck
rebounds tells you what the final KE is.

The coefficient of restitution in steel with steel collisions is very
high, but that is not the problem. Doing the test with a lever of mass
m and 2.99m (and the puck of mass m), both angular momentum and
kinetic energy are conserved. The problem starts with levers of mass
3.01m or more.

Peter




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