MHT CET · Physics · Rotational Motion
A solid sphere has mass ' \(M\) ' and radius ' \(R\) '. Its moment of inertia about a parallel axis passing through a point at a distance \(\frac{R}{2}\) from its centre is
- A \(\frac{8 \mathrm{MR}^2}{11}\)
- B \(\frac{11 \mathrm{MR}^2}{18}\)
- C \(\frac{7 \mathrm{MR}^2}{10}\)
- D \(\frac{13 \mathrm{MR}^2}{20}\)
Answer & Solution
Correct Answer
(D) \(\frac{13 \mathrm{MR}^2}{20}\)
Step-by-step Solution
Detailed explanation
Concept: Parallel axis theorem application.
The moment of inertia for a sphere about the central rotation axis is:
\(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{MR}^2\)
See the diagram below,
To find the moment of inertia about the new rotation axis we use parallel axis theorem:
\(I_{\text {new }}=I_{\text {sphere }}+{M b^2}^2=\frac{2}{5}{M R^2}+\frac{M R^2}{4}=\frac{(8+5)}{20} M R^2=\frac{13}{20} M R^2\)
The moment of inertia for a sphere about the central rotation axis is:
\(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{MR}^2\)
See the diagram below,
To find the moment of inertia about the new rotation axis we use parallel axis theorem:
\(I_{\text {new }}=I_{\text {sphere }}+{M b^2}^2=\frac{2}{5}{M R^2}+\frac{M R^2}{4}=\frac{(8+5)}{20} M R^2=\frac{13}{20} M R^2\)
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