TS EAMCET · Physics · Rotational Motion
If the moment of inertia of a thin circular ring about an axis passing through its edge and perpendicular to its plane is \(I\), then the moment of inertia of the ring about its diameter is
- A \(\frac{\mathrm{I}}{4}\)
- B 4 I
- C \(\frac{I}{2}\)
- D 2 I
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{I}}{4}\)
Step-by-step Solution
Detailed explanation
\(I = MR^2_{center,\perp} + MR^2 = MR^2 + MR^2 = 2MR^2\) \(MR^2_{center,\perp} = 2I_{diameter} \implies MR^2 = 2I_{diameter}\) \(I_{diameter} = \frac{MR^2}{2} = \frac{1}{2}\left(\frac{I}{2}\right) = \frac{I}{4}\)
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