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

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(A) The volume of the disc is $V = \pi R^2 	imes \frac{R}{6} = \frac{\pi R^3}{6}$.Given the moment of inertia of the disc is $I = \frac{1}{2} M R^2$.

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$\frac{I}{5}$

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(A) The volume of the disc is $V = \pi R^2 	imes \frac{R}{6} = \frac{\pi R^3}{6}$.Given the moment of inertia of the disc is $I = \frac{1}{2} M R^2$.When the disc is recast into a solid sphere of radius $r$,the volume remains constant:$\frac{\pi R^3}{6} = \frac{4}{3} \pi r^3$.Solving for $r$:$r^3 = \frac{3}{4} 	imes \frac{R^3}{6} = \frac{R^3}{8} \Rightarrow r = \frac{R}{2}$.The moment of inertia of the solid sphere about its diameter is $I_{sphere} = \frac{2}{5} M r^2$.Substituting $r = \frac{R}{2}$:$I_{sphere} = \frac{2}{5} M \left(\frac{R}{2}ight)^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{1}{10} M R^2$.Since $I = \frac{1}{2} M R^2$,we have $M R^2 = 2I$.Therefore,$I_{sphere} = \frac{1}{10} (2I) = \frac{I}{5}$.

$A$ disc of radius $R$ and thickness $\frac{R}{6}$ has a 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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