MHT CET · Physics · Rotational Motion
The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity \(\omega\). Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is
- A \(\frac{\mathrm{I} \omega^2}{16}\)
- B \(\frac{\mathrm{I} \omega^2}{8}\)
- C \(\frac{\mathrm{I} \omega^2}{4}\)
- D \(\frac{\mathrm{I} \omega^2}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{I} \omega^2}{4}\)
Step-by-step Solution
Detailed explanation
\(I_{initial} = I\), \(I_{final} = 2I\) \(I_{initial}\omega_{initial} = I_{final}\omega_{final}\)
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