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I got this tricky question, need help.

A uniform rod of mass $M$ and length $L$ is attached to an axis at its top, a bullet with mass $m$ traveling at speed $U$ (horizontal) hits the rod at $2L/3$ from the axis and then LEAVES the rod at speed $U/2$.

I need to find the angular speed of the rod right after collision assuming collision time is very short.

If the bullet stayed in I would just do conservation of angular momentum, find the new moment of inertia with the bullet using Steiner's law and solve for $\Omega$.

But what do I do when the bullet leaves the rod?

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closed as off-topic by John Rennie, Emilio Pisanty, Alexander, centralcharge, Brandon Enright Feb 3 '14 at 16:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

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You do exactly the same! You just have that the final angular momentum is the sum of the one in rod and the remaining in the bullet.

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so, Li = mu*(2/3)L, and Lfinal = Iw + m*(0.5u)*(2/3)L ? while Iw would be (1/3ML^2) which is rhe rod in relation to its axis – user2970357 Feb 3 '14 at 10:07
@user2970357 Do not confuse angular momentum and moment of inertia. Your expression for the rod's term of Lfinal is not correct. – DarioP Feb 3 '14 at 10:51

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