- #1

Waxbear

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Imagine a rod standing on a table, the base of the rod is attached to the table with a hinge, so that the rod is able to swing between standing position and parallel with the table. The rod has the length L.

The rod is standing upright, so that it's in equilibrium. A ball comes flying through the air, and hits the rod at 1/2 L, after which the ball drops vertically to the table. Consider the transfer of kinetic energy to be instant.

The equivalence between kinetic energy in the ball and in the rotational motion of the rod should be: [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]=[tex]\frac{1}{2}[/tex]I[tex]\omega[/tex][tex]^{2}[/tex] right?

Now consider the ball having the same mass and the same speed as before, but now it hits the rod at L. wouldn't the initial rotational velocity of the rod be larger? if so, why? the kinetic energy in the ball hasn't changed.

It's probably something trivial and stupid that I've overlooked. So if anyone would care to explain what I've missed, it would be much appreciated :-)