Gre 08 q97 Angular Momentum

[pluspolaritons]

Hi I attempted to solve this question but got a different answer, I don't know what's wrong with my solution.

Homework Statement

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Homework Equations

The Attempt at a Solution

At point P before collision, the angular momentum from the top disc has two components, rotational and translational.

For the translational L=rxp=-MRv_0=-0.5*M*R^2*w_0. This value is negative as I choose out of the page as my positive axis.

For the rotational part, I applied parallel-axis theorem, thus I=(0.5MR^2+MR^2)=1.5MR^2. And the angular momentum is L=1.5*M*R^2*w_0

Adding these two components up I obtain: 0.5*M*R^2*w_0. The answer does not have this choice. The given solution is (A).

If I do not apply parallel-axis theorem, I will get the correct answer. In fact in most solutions I looked up, they do not apply parallel-axis theorem, which I think is incorrect since the axis of rotation is not the center of mass of the top disc but rather at point P, so we have to apply parallel-axis theorem to shift the rotational angular momentum axis of rotation to point P. Can someone please help to see where I did wrong and why my argument is invalid?

Thanks in advance.

 

[jbsteele]

Your answer would be correct if the system was not moving in a straight line before the collision. The point P is not the center of mass of the disc, so you must use parallel axis theorem to get the rotational angular momentum. However, since the system was moving in a straight line before the collision, the translational velocity is equal to the rotational velocity and thus you do not need to use the parallel axis theorem. The total angular momentum in this case is equal to just the translational angular momentum and thus is equal to -0.5 MR^2 w0.