A uniform circular disc of radius R and mass | vedclass

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(D) The moment of inertia of the original disc of mass $M$ and radius $R$ about the central axis is $I_1 = \frac{1}{2} MR^2$.The mass of the removed c

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The moment of inertia of the original disc of mass $M$ and radius $R$ about the central axis is $I_1 = \frac{1}{2} MR^2$.The mass of the removed circular part of radius $r = R/2$ is $M' = \frac{M}{\pi R^2} 	imes \pi (R/2)^2 = M/4$.The moment of inertia of this removed part about its own central axis is $I_{cm} = \frac{1}{2} M' r^2 = \frac{1}{2} (M/4) (R/2)^2 = \frac{1}{32} MR^2$.Using the parallel axis theorem,the moment of inertia of the removed part about the original central axis of the disc (at a distance $d = R/2$ from its own center) is $I_2 = I_{cm} + M' d^2 = \frac{1}{32} MR^2 + (M/4) (R/2)^2 = \frac{1}{32} MR^2 + \frac{1}{16} MR^2 = \frac{3}{32} MR^2$.The moment of inertia of the remaining part is $I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{3}{32} MR^2 = \frac{16-3}{32} MR^2 = \frac{13}{32} MR^2$.

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