Question 569: Centre of Mass and Rotational Motion

Question

A disc of radius $$R$$ and thickness $$\frac{R}{6}$$ has moment of inertia I about an axis passing through its centre and perpendicular to its plane. Dise is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is

Accepted Answer

M.I. of disc, $$I=\frac{1}{2} ~\mathrm{MR}_{\mathrm{d}}^2$$ ..... (i)
M.I. of sphere, $$\mathrm{I}_{	ext {sphere }}=\frac{2}{5}~ \mathrm{MR}_{\mathrm{s}}^2$$ .... (ii)
$$\because \quad$$ volume of disc = volume of sphere
$$\begin{array}{ll}
	herefore & \pi \mathrm{R}_{\mathrm{d}}^2\left(\frac{\mathrm{R}_{\mathrm{d}}}{6}ight)=\frac{4}{3} \pi \mathrm{R}_{\mathrm{s}}^3 \
	herefore & \mathrm{R}_{\mathrm{d}}^3=8 \mathrm{R}_{\mathrm{s}}^3 \
	herefore & \mathrm{R}_{\mathrm{S}}=\frac{\mathrm{R}_{\mathrm{d}}}{2} \quad 	ext{.... (iii)}
\end{array}$$
Substitute equation (iii) in equation (ii)
$$\begin{aligned}
	herefore \quad I_{	ext {sphere }} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}_{\mathrm{d}}}{2}ight)^2=\frac{2}{5} 	imes \frac{1}{4} \mathrm{MR}_{\mathrm{d}}^2 \
& =\frac{1}{5}\left(\frac{1}{2} \mathrm{MR}_{\mathrm{d}}^2ight)=\frac{1}{5} \quad 	ext{.... [from (i)]}
\end{aligned}$$

Answer

$$\frac{\mathrm{I}}{5}$$

Answer

$$\frac{\mathrm{I}}{6}$$

Answer

$$\frac{\mathrm{I}}{32}$$

Answer

$$\frac{\mathrm{I}}{64}$$

Centre of Mass and Rotational Motion

A disc of radius $R and thickness \frac{R}{6}$ has moment of inertia I about an axis passing through its centre and perpendicular to its plane. Dise is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is

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