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In this solved problem in my textbook:

A disc is freely rotating with an angular speed on a smooth horizontal plane. It is hooked at a rigid peg P and rotates about P without bouncing. What will be its angular speed after the impacts?

The book takes the moment of inertia about P after the collision to be $\frac{3mR^2}{2}$ after the collision, which is reasonable. However, it takes the moment of inertia just before the collision about P to be $\frac{mR^2}{2}$, which I do not understand. Clearly, the theorem of parallel axes tells us the same thing just before and after the collision. Also, isn't angular velocity about P zero? The body surely isn't rotating about P, it is moving in a straight line towards it.

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  • $\begingroup$ It would be useful to post an image or the full problem, cause I don't get what are the impacts. $\endgroup$
    – geofisue
    Mar 2, 2023 at 7:55
  • $\begingroup$ I think connecting it to P makes it to start rotating around P instead of rotating around its center. $\endgroup$
    – geofisue
    Mar 2, 2023 at 8:07
  • $\begingroup$ P is located in the circumference, that's why it says "impact". Parallel axes tells us the moment around the new axes is 1/2mR^2 + mR^2, which is 3/2mR^2. $\endgroup$
    – geofisue
    Mar 2, 2023 at 8:13

1 Answer 1

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When it is anchored to the fixed point P, the disk can no longer rotate about itself, so it will rotate around the new axis passing through P (since the angular momentum is conserved). Since we are talking about impact, we understand that P is on the circumference of the disk, so the distance to the center of the disk will be R. If the initial moment of inertia was:

$I_0 = \frac{1}{2}mR^2$

The theorem of parallel axes tells us that the new moment is:

$I_0 + mR^2 = \frac{3}{2}mR^2$

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  • $\begingroup$ Why can't we take the moment of inertia route for the first calculation? When is angular momentum $mvr+Iω$ and when is it $I'ω$, where $I'$ is the moment of inertia about P? $\endgroup$ Mar 6, 2023 at 12:35

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