A uniform circular disc of radius $R$ and mass $M$ is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius $R/2$ is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
Answer: (D) $\dfrac{13}{32}MR^2$
The removed disc has radius $R/2$, so its area (and mass) is one quarter of the whole: $m = M/4$. Its centre is at distance $R/2$ from the axis.
Moment of inertia of the removed part about the axis (parallel axis theorem):
$$I_{rem} = \frac{1}{2}\cdot\frac{M}{4}\left(\frac{R}{2}\right)^2 + \frac{M}{4}\left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3}{32}MR^2$$
Remaining part:
$$I = \frac{1}{2}MR^2 - \frac{3}{32}MR^2 = \frac{13}{32}MR^2$$
Solution by Sreeraj P, M.Sc Physics