MHT CET · Physics · Motion In Two Dimensions
A thin circular ring of mass 'M' and radius 'r' is rotating about its axis with an angular speed ' \(\omega\) '. Two particles each of mass 'm' are now attached at diametrically opposite points. The angular speed of the ring will become
- A \(\frac{\omega M}{M+2 m}\)
- B \(\frac{\omega M}{M+m}\)
- C \(\frac{\omega(M-2 m)}{M}\)
- D \(\frac{\omega(M-2 m)}{M+2 m}\)
Answer & Solution
Correct Answer
(A) \(\frac{\omega M}{M+2 m}\)
Step-by-step Solution
Detailed explanation
\(L_1 = I_1 \omega = Mr^2 \omega\) \(L_2 = I_2 \omega' = (Mr^2 + 2mr^2) \omega' = (M+2m)r^2 \omega'\)
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