A disc of radius R and thickness frac{R}{6} h | vedclass

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(A) The moment of inertia of a disc about its central axis is $I = \frac{MR^2}{2}$.Since the mass remains constant during recasting,$M' = M$.The volum

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$\frac{I}{5}$

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(A) The moment of inertia of a disc about its central axis is $I = \frac{MR^2}{2}$.Since the mass remains constant during recasting,$M' = M$.The volume of the disc is $V = \pi R^2 	imes 	ext{thickness} = \pi R^2 	imes \frac{R}{6} = \frac{\pi R^3}{6}$.The volume of the solid sphere is $V' = \frac{4}{3} \pi R'^3$.Equating volumes: $\frac{\pi R^3}{6} = \frac{4}{3} \pi R'^3$,which simplifies to $R'^3 = \frac{R^3}{8}$,so $R' = \frac{R}{2}$.The moment of inertia of a solid sphere about its diameter is $I' = \frac{2}{5} M' R'^2$.Substituting $M' = M$ and $R' = \frac{R}{2}$: $I' = \frac{2}{5} M \left(\frac{R}{2}ight)^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{1}{5} \left(\frac{MR^2}{2}ight)$.Since $I = \frac{MR^2}{2}$,we get $I' = \frac{I}{5}$.

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

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