A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity (a) (ω M)/(M+2m) (b) (ω(M+2m))/(M) (c) (ω M)/(M+m) (d) (ω(M-2m))/(M+2m)

Answer: (a). Conservation of L: MR²ω = (M+2m)R²ω'. ω' = (ω M)/(M+2m) So the answer is (a) (ω M)/(M+2m).

A thin circular ring of mass M and radius R is rotating in a… | NEET 2009 (AIPMT)

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Angular Momentum & its Conservation

A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$ . If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

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Answer: (a)

Conservation of $L$ : $MR^2\omega = (M+2m)R^2\omega'$ .
$\omega' = \frac{\omega M}{M+2m}$
So the answer is (a) $\frac{\omega M}{M+2m}$ .

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