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Two spheres each of mass ' $M$ ' and radius $\frac{R}{2}$ are connected at the ends of massless rod of length ' $2 \mathrm{R}$ '. What will be the moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod?
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Verified Answer
The correct answer is:
$\frac{21}{5} \mathrm{MR}^2$
From parallel axis theorem,
$\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2$
Let the moment of inertia of sphere 1 be
$\mathrm{I}_1=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2$
and,
Let the moment of inertia of sphere 2 be
$\mathrm{I}_2=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2$
Moment of inertia of the rod $\mathrm{I}_3=0$
$\therefore \quad$ Moment of inertia of the system,
$\begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\
\mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\
& =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\
& =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\
& =\frac{21}{5} \mathrm{MR}^2
\end{aligned}$
$\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2$
Let the moment of inertia of sphere 1 be
$\mathrm{I}_1=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2$
and,
Let the moment of inertia of sphere 2 be
$\mathrm{I}_2=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2$
Moment of inertia of the rod $\mathrm{I}_3=0$
$\therefore \quad$ Moment of inertia of the system,
$\begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\
\mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\
& =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\
& =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\
& =\frac{21}{5} \mathrm{MR}^2
\end{aligned}$
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