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 $M_1 = \sigma \pi R^2$. The mass of the remov

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Column-$I$	Column-$II$
$(1)$ Moment of inertia of a solid sphere about its diameter	$(a)$ $\frac{2}{3}MR^2$
$(2)$ Moment of inertia of a solid sphere about a tangent	$(b)$ $\frac{2}{5}MR^2$
	$(c)$ $\frac{7}{5}MR^2$

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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Match the items in Column-$I$ with those in Column-$II$.
Column-$I$ Column-$II$
$(1)$ Moment of inertia of a solid sphere about its diameter $(a)$ $\frac{2}{3}MR^2$
$(2)$ Moment of inertia of a solid sphere about a tangent $(b)$ $\frac{2}{5}MR^2$
$(c)$ $\frac{7}{5}MR^2$

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