A solid sphere of mass 'M' and radius 'R' has moment of inertia 'l' about its diameter. It is recast into a…
A solid sphere of mass 'M' and radius 'R' has moment of inertia 'l' about its diameter.
It 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'. Radius of the disc will
be
$\mathrm{R}/\sqrt{19}$
$R / \sqrt{15}$
$2 \mathrm{R} / \sqrt{15}$
$2 R / \sqrt{19}$
Solution
Moment of inertia of solid sphere is
$\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}$
Moment of inertia of the disc is given by about the given axis is
$\begin{aligned}
\mathrm{I}_{d} &=\frac{3}{2} \mathrm{Mr}^{2} \\
\therefore \mathrm{r}^{2} &=\frac{4}{15} \mathrm{R}^{2} \quad \therefore \frac{3}{2} \mathrm{Mr}^{2}=\frac{2}{5} \mathrm{MR}^{2} \\
\therefore \mathrm{r} &=\frac{2 \mathrm{R}}{\sqrt{15}}
\end{aligned}$