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MHT CET · Physics · Rotational Motion

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?

  1. A \(\frac{2}{3} \mathrm{MR}^2\)
  2. B \(\frac{5}{2} \mathrm{MR}^2\)
  3. C \(\frac{5}{21} \mathrm{MR}^2\)
  4. D \(\frac{21}{5} \mathrm{MR}^2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{21}{5} \mathrm{MR}^2\)

Step-by-step Solution

Detailed explanation

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,
\(\begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\
\mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\
& =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\
& =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\
& =\frac{21}{5} \mathrm{MR}^2
\end{aligned}\)
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