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# A circular ring of mass M and radius R is rotating about its axis with constant angular velocity $w$. Two particle each of mass $m$ are attached gently to the opposite ends of a diameter of the ting. The angular velocity of the ring will now become :

$\begin{array}{1 1} (1)\; \frac{m\omega}{M+2m} \\ (2)\; \frac{M\omega}{M-2m} \\ (3)\; \frac{M\omega}{M + 2m} \\ (4)\; \frac{M + 2m}{M\omega} \end{array}$

Thank u Priyanka ji

$\; \; \; \; \; \; \; I_{\omega} = I' \omega'$
$M r^2 \omega = (Mr^2 + 2mr^2)\omega'$
$\omega' = \frac{M \omega}{M + 2m}$