From a disc of radius R,a concentric circular | vedclass

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(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$ and its moment of iner

Vedclass · Accepted Answer

$ \frac{1}{2}M(R^2 + r^2) $

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(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$ and its moment of inertia is $I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} \sigma \pi R^4$.The mass of the removed circular portion of radius $r$ is $M_2 = \sigma \pi r^2$ and its moment of inertia is $I_2 = \frac{1}{2} M_2 r^2 = \frac{1}{2} (\sigma \pi r^2) r^2 = \frac{1}{2} \sigma \pi r^4$.The mass of the remaining annular disc is $M = M_1 - M_2 = \sigma \pi (R^2 - r^2)$,which implies $\sigma \pi = \frac{M}{R^2 - r^2}$.The moment of inertia of the annular disc is $I = I_1 - I_2 = \frac{1}{2} \sigma \pi (R^4 - r^4)$.Substituting $\sigma \pi$,we get $I = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^4 - r^4) = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^2 - r^2)(R^2 + r^2) = \frac{1}{2} M(R^2 + r^2)$.

From a disc of radius $R$,a concentric circular portion of radius $r$ is cut out so as to leave an annular disc of mass $M$. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

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