Question

# A solid sphere of mass $m$, radius$r$, and having a moment of inertia about an axis passing through the center of mass as$I$ is recast into a disc of thickness $t$, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. Then, the radius of the disc will be

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Solution

## Step 1: Given dataThe mass of the sphere is $m$.The radius of the sphere is $r$.The moment of inertia of the sphere is ${I}_{s}=I.$The moment of inertia of the disc is ${I}_{d}=I.$.Step 2: Moment of inertiaThe moment of inertia of a body about a given axis in space is the sum of the products of the mass and square of the distance from the axis for each particle comprising the body.The moment of inertia is defined by the form, $I=\underset{}{ām{x}^{2}}$, where, m is the mass of the body and x is the distance of the body from the axis of rotation.Step 3: Moment of inertia of a solid sphere and a disc.The moment of inertia of a sphere about an axis passing through the diameter is ${I}_{s}=\frac{2}{5}m{r}^{2}$, where, m is the mass of the sphere and r is the radius of the solid sphere.The moment of inertia of a disc about an axis passing through its edge and perpendicular to its plane is ${I}_{d}=\frac{3}{2}m{R}^{2}$, where, m is the mass of the sphere and r is the radius of the solid sphere. Step 4: DiagramStep 5: Finding the radius of the discAccording to the question, in both cases, the moment of inertia is the same, i.e, ${I}_{s}={I}_{d}$.So, $\frac{2}{5}m{r}^{2}=\frac{3}{2}m{R}^{2}\phantom{\rule{0ex}{0ex}}or{R}^{2}=\frac{4{r}^{2}}{15}\phantom{\rule{0ex}{0ex}}orR=\sqrt{\frac{4{r}^{2}}{15}}=\frac{2r}{\sqrt{15}}\phantom{\rule{0ex}{0ex}}orR=\frac{2r}{\sqrt{15}}.$Therefore, the radius of the disc is $\frac{2r}{\sqrt{15}}$.

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