# Does moment of inertia about an axis depend on whether the object is rotating around it or not?

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.

• It would be useful to post an image or the full problem, cause I don't get what are the impacts. Mar 2, 2023 at 7:55
• I think connecting it to P makes it to start rotating around P instead of rotating around its center. Mar 2, 2023 at 8:07
• 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. Mar 2, 2023 at 8:13

$$I_0 = \frac{1}{2}mR^2$$
$$I_0 + mR^2 = \frac{3}{2}mR^2$$
• 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? Mar 6, 2023 at 12:35