A circular hole of radius frac{R}{4} is made | vedclass

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(B) The moment of inertia of the complete disc about point $O$ is $I_{total} = \frac{M R^2}{2}$.The radius of the removed disc is $r = \frac{R}{4}$.Si

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$\frac{237 M R^2}{512}$

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(B) The moment of inertia of the complete disc about point $O$ is $I_{total} = \frac{M R^2}{2}$.The radius of the removed disc is $r = \frac{R}{4}$.Since the disc is uniform,the mass is proportional to the area $(M \propto R^2)$. Therefore,the mass of the removed disc is $m = M \left( \frac{r}{R} \right)^2 = M \left( \frac{R/4}{R} \right)^2 = \frac{M}{16}$.The moment of inertia of the removed disc about its own central axis passing through $O'$ is $I_{cm} = \frac{1}{2} m r^2 = \frac{1}{2} \left( \frac{M}{16} \right) \left( \frac{R}{4} \right)^2 = \frac{M R^2}{512}$.Using the parallel axis theorem,the moment of inertia of the removed disc about point $O$ is $I_{removed} = I_{cm} + m d^2$,where $d = \frac{3R}{4}$ is the distance between $O$ and $O'$.$I_{removed} = \frac{M R^2}{512} + \left( \frac{M}{16} \right) \left( \frac{3R}{4} \right)^2 = \frac{M R^2}{512} + \frac{9 M R^2}{256} = \frac{M R^2 + 18 M R^2}{512} = \frac{19 M R^2}{512}$.The moment of inertia of the remaining portion is $I_{remaining} = I_{total} - I_{removed} = \frac{M R^2}{2} - \frac{19 M R^2}{512} = \frac{256 M R^2 - 19 M R^2}{512} = \frac{237 M R^2}{512}$.

$A$ circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$,as shown in the figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is:

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