A circular disc of radius R and thickness fra | vedclass

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(C) The mass of the disc and the sphere remains the same because the disc is melted and recast into a sphere.Volume of disc = Volume of sphere$\pi R^2

Vedclass · Accepted Answer

$\frac{I}{5}$

Vedclass · Answer

(C) The mass of the disc and the sphere remains the same because the disc is melted and recast into a sphere.Volume of disc = Volume of sphere$\pi R^2 t = \frac{4}{3} \pi R_s^3$Given $t = \frac{R}{6}$,so $\pi R^2 (\frac{R}{6}) = \frac{4}{3} \pi R_s^3$$\frac{R^3}{6} = \frac{4}{3} R_s^3 \implies R_s^3 = \frac{R^3}{8} \implies R_s = \frac{R}{2}$The moment of inertia of the disc is $I = \frac{1}{2} M R^2$,so $M R^2 = 2I$.The moment of inertia of the sphere about its diameter is $I_s = \frac{2}{5} M R_s^2$.Substituting $R_s = \frac{R}{2}$,we get $I_s = \frac{2}{5} M (\frac{R}{2})^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{M R^2}{10}$.Since $M R^2 = 2I$,then $I_s = \frac{2I}{10} = \frac{I}{5}$.

$A$ circular disc of radius $R$ and thickness $\frac{R}{6}$ has a moment of inertia $I$ about an axis passing through its center and perpendicular to its plane. It is melted and recast into a solid sphere. The moment of inertia of this sphere about its diameter is:

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