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(D) The moment of inertia of the original disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $

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$\frac{13 MR^{2}}{32}$

Vedclass · Answer

(D) The moment of inertia of the original disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{	ext{total}} = \frac{1}{2} MR^2$.The area of the original disc is $A = \pi R^2$. The diameter of the hole is $R$,so its radius is $r = R/2$. The area of the hole is $A_{	ext{hole}} = \pi (R/2)^2 = \frac{\pi R^2}{4} = \frac{A}{4}$.Since the mass is proportional to the area,the mass of the removed part is $M_{	ext{removed}} = \frac{M}{4}$.The center of the hole is at a distance $d = R/2$ from the center of the original disc. Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through the center of the original disc is:$I_{	ext{removed}} = I_{	ext{cm}} + M_{	ext{removed}} d^2$$I_{	ext{removed}} = \left[ \frac{1}{2} M_{	ext{removed}} r^2 ight] + M_{	ext{removed}} (R/2)^2$$I_{	ext{removed}} = \left[ \frac{1}{2} \left( \frac{M}{4} ight) \left( \frac{R}{2} ight)^2 ight] + \left( \frac{M}{4} ight) \left( \frac{R}{2} ight)^2$$I_{	ext{removed}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.The moment of inertia of the remaining part is:$I_{	ext{remaining}} = I_{	ext{total}} - I_{	ext{removed}}$$I_{	ext{remaining}} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2 = \frac{16 MR^2 - 3 MR^2}{32} = \frac{13 MR^2}{32}$.

From a disc of mass $M$ and radius $R$,a circular hole of diameter $R$ is cut such that its rim passes through the center of the disc. The moment of inertia of the remaining part of the disc about a perpendicular axis passing through the center of the original disc is

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