Question 542: Centre of Mass and Rotational Motion

Question

A solid sphere of mass ' $m$ ', radius ' $R$ ', having moment of inertia about an axis passing through center of mass as 'I' is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains 'I'. Then the radius of disc is

Accepted Answer

The moment of inertia of a solid sphere
$$=\frac{2}{5} \mathrm{MR}^2$$
Moment of inertia of a disc through its rim,
$$=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$$
Since both the moment of inertias are equal, $	herefore \quad \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2$, where r is the radius of the disc
$	herefore \quad r=\frac{2 R}{\sqrt{15}}$

Answer

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

Answer

$\left(\sqrt{\frac{2}{15}}\right) \mathrm{R}$

Answer

$\frac{4 \mathrm{R}}{\sqrt{15}}$

Answer

$\frac{\mathrm{R}}{4}$

Centre of Mass and Rotational Motion

A solid sphere of mass ' m ', radius ' R ', having moment of inertia about an axis passing through center of mass as 'I' is recast into a disc of thickness ' t ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains 'I'. Then the radius of disc is

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