From a disc of mass M and radius R,a circular | vedclass

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(A) The moment of inertia of the remaining part of the disc $(I_r)$ is given by the principle of superposition:$I_r = I_{	ext{disc}} - I_{	ext{hole}

Vedclass · Accepted Answer

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

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(A) The moment of inertia of the remaining part of the disc $(I_r)$ is given by the principle of superposition:$I_r = I_{	ext{disc}} - I_{	ext{hole}}$where $I_{	ext{disc}}$ is the moment of inertia of the original disc about the central axis,and $I_{	ext{hole}}$ is the moment of inertia of the removed circular part about the same axis.$1$. Moment of inertia of the original disc:$I_{	ext{disc}} = \frac{1}{2} MR^2$$2$. Properties of the removed hole:Radius of the hole $r = \frac{R}{2}$.Since the surface mass density $\sigma = \frac{M}{\pi R^2}$ is uniform,the mass of the hole $M_h$ is:$M_h = \sigma \cdot \pi r^2 = \left(\frac{M}{\pi R^2}ight) \cdot \pi \left(\frac{R}{2}ight)^2 = \frac{M}{4}$.$3$. Moment of inertia of the hole about the central axis of the original disc:Using the parallel axis theorem,$I_{	ext{hole}} = I_{	ext{cm}} + M_h d^2$,where $I_{	ext{cm}} = \frac{1}{2} M_h r^2$ and $d = \frac{R}{2}$ is the distance between the center of the hole and the center of the original disc.$I_{	ext{hole}} = \frac{1}{2} \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2 + \left(\frac{M}{4}ight) \left(\frac{R}{2}ight)^2$$I_{	ext{hole}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.$4$. Moment of inertia of the remaining part:$I_r = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

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

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