The moment of inertia of circular disc of mass M and radius R about an axis passing through the cent

Question

The moment of inertia of circular disc of mass M and radius R about an axis passing through the centre of mass is $\mathrm{I}_{0}$. The moment of inertia of another circular disc of same mass and thickness but half the density about the same axis is\

Professor PC Thomas Classes · Answer

(D) For circular disc 1\ mass M , radius $\mathrm{R}_{1}=\mathrm{R}$\ moment of inertia $\mathrm{I}_{1}=\mathrm{I}_{0}$\ For circular disc 2, of same thickness $t$, mass $=\mathrm{M}$\ But density $=\frac{1}{2} 	imes$ density of circular disc 1 .\ Let radius $=\mathrm{R}_{2}$\ then $\pi \mathrm{R}_{2}^{2} \mathrm{t} 	imes \frac{ho}{2}=\pi \mathrm{R}_{1}^{2} \mathrm{t} 	imes ho=\mathrm{M}$ $$ \begin{aligned} & R_{2}^{2}=2 R_{1}^{2} \ & R_{2}=\sqrt{2} R_{1}=\sqrt{2} R \end{aligned} $$ $\because$ the given axis passes through the centre of mass.\ $	herefore$ moment of inertia $\mathrm{I} \propto(	ext { Radius })^{2}$ $$ \begin{aligned} & \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}ight)^{2} \ & \quad \frac{\mathrm{I}_{0}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}ight)^{2} \ & \Rightarrow \mathrm{I}_{2}=2 \mathrm{I}_{0} \end{aligned} $$ OR\