Question 525: Centre of Mass and Rotational Motion

Question

A solid metallic sphere of radius ' $R$ ' having moment of inertia '$I$' about diameter is melted and recast into a solid disc of radius ' $r$ ' of a uniform thickness. The moment of inertia of a disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is

Accepted Answer

M.I. of the solid sphere about a diameter
$\mathrm{I}=\frac{2}{5} \mathrm{MR}^2\quad	ext{.... (i)}$
M.I. of the disc about an axis through its edge and perpendicular to its plane is
$$\mathrm{I}=\frac{3}{2} \mathrm{Mr}^2\quad	ext{... (ii)}$$
$$\begin{array}{ll}
	herefore & \frac{2}{5} M R^2=\frac{3}{2} M r^2 \quad	ext{.... [From (i) and (ii)]}\
	herefore & r=\frac{2}{\sqrt{15}} \mathrm{R}
\end{array}$$

Answer

$\frac{1}{\sqrt{2}}$

Answer

$\frac{2}{\sqrt{5}}$

Answer

$\frac{2}{\sqrt{10}}$

Answer

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

Centre of Mass and Rotational Motion

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