MHT CET · Physics · Rotational Motion
The moment of inertia of a thin uniform rod of mass ' \(M\) ' and length ' \(L\) ' about an axis passing through a point at a distance \(\frac{L}{4}\) from one of its ends and perpendicular to the length of the rod is
- A \(\frac{\mathrm{ML}^2}{48}\)
- B \(\frac{7 \mathrm{ML}^2}{48}\)
- C \(\frac{5 \mathrm{ML}^2}{48}\)
- D \(\frac{9 \mathrm{ML}^2}{48}\)
Answer & Solution
Correct Answer
(B) \(\frac{7 \mathrm{ML}^2}{48}\)
Step-by-step Solution
Detailed explanation
The moment of inertia of the rod about an axis passing through the center and perpendicular to its length is given by
\(\mathrm{I}_0=\frac{\mathrm{ML}^2}{12}\)
Hence by parallel axis theorem,
\(\mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48}\)
\(\mathrm{I}_0=\frac{\mathrm{ML}^2}{12}\)
Hence by parallel axis theorem,
\(\mathrm{I}=\mathrm{I}_0+\mathrm{M}\left(\frac{\mathrm{L}}{4}\right)^2=\frac{\mathrm{ML}^2}{12}+\frac{\mathrm{ML}^2}{16}=\frac{7 \mathrm{ML}^2}{48}\)
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