Q62 / 120 · NEET 2010 (Mains) · Physics

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with 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 angular velocity given by

  • A
    $\frac{(M+2 m) \omega}{2 m}$
  • B
    $\frac{2 \mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}}$
  • C
    $\frac{(M+2 m) \omega}{M}$
  • Correct answer
    D
    $\frac{M \omega}{M+2 m}$

Solution

In the absence of external torque, angular momentum remain constant $$ \begin{aligned} \mathrm{L} & =\mathrm{I} \omega=\mathrm{I}^{\prime} \omega^{\prime} \\ \therefore \quad \mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\ \omega^{\prime} & =\frac{\mathrm{M} \omega}{(\mathrm{M}+2 \mathrm{~m})} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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