(A) The initial moment of inertia of the ring about its axis is $I_i = Mr^2$. The initial angular momentum is $L_i = I_i \omega = Mr^2 \omega$. When t

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(A) The initial moment of inertia of the ring about its axis is $I_i = Mr^2$. The initial angular momentum is $L_i = I_i \omega = Mr^2 \omega$. When two particles of mass $m$ are attached at diametrically opposite points at a distance $r$ from the axis,the new moment of inertia becomes $I_f = Mr^2 + mr^2 + mr^2 = (M+2m)r^2$. Since there is no external torque acting on the system,the angular momentum is conserved,so $L_i = L_f$. $Mr^2 \omega = (M+2m)r^2 \omega'$. Solving for the new angular speed $\omega'$,we get $\omega' = \frac{Mr^2 \omega}{(M+2m)r^2} = \frac{\omega M}{M+2m}$.

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 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

MHT CET 2020 All MHT CET

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

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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 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

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