From a uniform circular thin disc of mass $9 \mathrm{M}$ and radius $\mathrm{R}$, a small disc of radius $\frac{\mathrm{R}}{3}$ is removed. The centre of the small disc is at a distance $\frac{2 \mathrm{R}}{3}$ from the centre of original disc. The moment of inertia of the remaining disc about an…

Question

From a uniform circular thin disc of mass $9 \mathrm{M}$ and radius $\mathrm{R}$, a small disc of radius $\frac{\mathrm{R}}{3}$ is removed. The centre of the small disc is at a distance $\frac{2 \mathrm{R}}{3}$ from the centre of original disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc of radius $\mathrm{R}$ is

Aicharya · Accepted Answer

The moment of inertia of the complete disc $\mathrm{I}_{1}=\frac{9 \mathrm{MR}^{2}}{2}$ The moment of inertia of the removed disc $=$ $\begin{aligned} I^{\prime} &=\frac{M}{2}\left(\frac{R}{3}ight)^{2}+M\left(\frac{2 R}{3}ight)^{2}=\frac{M R^{2}}{18}+\frac{4 M R^{2}}{9} \ &=\frac{M R^{2}}{2} \end{aligned}$ [Since mass is proportional to area, it is proportional to square of the radius. Since the radius of the removed disc is $\frac{R}{3}$, its mass will be $M$ ] $	herefore$ Moment of inertia of the remaining disc will be $I_{2}=I_{1}-I^{\prime}=\frac{9 m R^{2}}{2}-\frac{M R^{2}}{2}=4 M R^{2}$

From a uniform circular thin disc of mass $9 \mathrm{M}$ and radius $\mathrm{R}$, a small disc of radius…

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