A solid sphere of radius R has a moment of in | vedclass

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(A) $1$. Moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5}MR^2$.$2$. Volume of the sphere is $V = \frac{4}{3}\pi R^3$. The vo

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$\frac{2}{\sqrt{15}} R$

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(A) $1$. Moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5}MR^2$.$2$. Volume of the sphere is $V = \frac{4}{3}\pi R^3$. The volume remains constant when recast into a disc,so $V = \pi r^2 t$. Thus,$t = \frac{4R^3}{3r^2}$.$3$. The moment of inertia of a disc about its central axis perpendicular to its plane is $I_{cm} = \frac{1}{2}Mr^2$.$4$. Using the parallel axis theorem,the moment of inertia about an axis tangent to the edge and perpendicular to the plane is $I_{tangent} = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.$5$. Given $I_{tangent} = I$,we have $\frac{3}{2}Mr^2 = \frac{2}{5}MR^2$.$6$. Simplifying,$r^2 = \frac{4}{15}R^2$,which gives $r = \frac{2}{\sqrt{15}}R$.

$A$ solid sphere of radius $R$ has a moment of inertia $I$ about its geometric axis. It is melted and recast into a disc of radius $r$ and thickness $t$. If the moment of inertia of this disc about an axis tangent to its edge and perpendicular to its plane is also $I$,find the value of $r$.

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