A thin circular ring of mass ' $\mathrm{M}^{\prime}$ and radius ' $\mathrm{R}^{\prime}$ is rotating about a…

A thin circular ring of mass ' $\mathrm{M}^{\prime}$ and radius ' $\mathrm{R}^{\prime}$ 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?
  1. $\frac{M \omega}{M+2 m}$
  2. $\frac{M(\omega)}{M+m}$
  3. $\frac{(M+2 m) \omega}{M}$
  4. $\frac{(\mathrm{M}-2 \mathrm{~m}) \omega}{\mathrm{M}+2 \mathrm{~m}}$

Solution

Initial angular momentum of the ring $\quad=I \omega=M R^{2} \omega$ If the new angular velocity is $\omega^{\prime}$ then the final angular momentum $=I^{\prime} \omega^{\prime}$ where $I^{\prime}=M R^{2}+2 m R^{2}=(M+2 m) R^{2}$. By law of conservation of momentum $\begin{array}{c} \mathrm{I}^{\prime} \omega^{\prime}=\mathrm{I} \omega \\ (\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^{2} \omega^{\prime} \\ \therefore \omega^{\prime}=\frac{\mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}} \end{array}=\mathrm{MR}^{2} \omega$ .

Asked in: MHT CET 2020 (14 Oct Shift 2)

Practice more Rotational Motion questions on Aicharya