A thin circular ring of mass M and radius R i | vedclass

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(C) Since there is no external torque acting on the system,the angular momentum $L$ is conserved.Initial angular momentum $L_1 = I_1 \omega_1$,where $

Vedclass · Accepted Answer

$\frac{\omega M}{M + 2m}$

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(C) Since there is no external torque acting on the system,the angular momentum $L$ is conserved.Initial angular momentum $L_1 = I_1 \omega_1$,where $I_1 = MR^2$ is the moment of inertia of the ring.Final angular momentum $L_2 = I_2 \omega_2$,where $I_2 = MR^2 + 2(mR^2) = (M + 2m)R^2$ is the moment of inertia of the ring with the two masses attached.By the law of conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.Substituting the values: $(MR^2) \omega = (M + 2m)R^2 \omega_2$.Solving for $\omega_2$: $\omega_2 = \frac{MR^2 \omega}{(M + 2m)R^2} = \frac{M \omega}{M + 2m}$.

Column-$I$	Column-$II$
$a$. Bicuspid valve	$p$. Brain
$b$. Nephron	$q$. Liver
$c$. Alveoli	$r$. Heart
$d$. Cerebrum	$s$. Kidney
	$t$. Lungs

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

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