# the questions in PPT

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still don't understand why the angular momentum is not change but the angular speed is decreasing in first two questions. and why does dropping the block in different ways make different angular momentum?

asked Oct 28, 2013

This is all depends on the angular momentum of the block before it is dropped on the disk. In the first two cases, the angular momentum of the block is 0. The path of the block goes through the rotational axis, therefore the momentum would be zero -- because $R_\perp =0$.
So in the first two cases, before and after collision, the total angular momentum is the same. But after collision the moment of inertia increases, therefore, the angular speed would be smaller $I_i\omega_i=I_f\omega_f$.
In the third case, the path of the block doesn't go through the rotational axis and the block has a momentum in the clockwise direction opposite to the disk's angular momentum. Therefore after the collision the angular momentum of the disk and the block together will be smaller than the angular momentum of the disk only before collision. The total angular momentum will still be the same before and after collision. $L_{Di}+L_{Bi}=L_{(D+B)f}$.