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 data**

The 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 inertia**

- The 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, $I=\underset{}{\u0101\x88\x91m{x}^{2}}$, 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: Diagram**

**Step 5: Finding the radius of the disc**

According 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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