(A) The initial moment of inertia of the ring about the transverse axis is $I = M R^2$. The initial angular momentum is $L_i = I \omega = M R^2 \omega

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(A) The initial moment of inertia of the ring about the transverse axis is $I = M R^2$. The initial angular momentum is $L_i = I \omega = M R^2 \omega$.When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia becomes $I' = M R^2 + m R^2 + m R^2 = (M + 2m) R^2$.Since no external torque acts on the system,the angular momentum is conserved,so $L_i = L_f$.$M R^2 \omega = (M + 2m) R^2 \omega'$$\omega' = \frac{M R^2 \omega}{(M + 2m) R^2} = \frac{M \omega}{M + 2m}$.

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

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$A$ thin circular ring of mass $M$ and radius $R$ is rotating about a transverse axis passing through its centre with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

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A
$\frac{M \omega}{M+2 m}$
B
$\frac{M \omega}{M+m}$
C
$\frac{(M+2 m) \omega}{M}$
D
$\frac{(M-2 m) \omega}{M+2 m}$

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

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$A$ thin circular ring of mass $M$ and radius $R$ is rotating about a transverse axis passing through its centre with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

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