Q132 / 150 · JEE Main 2006 · Physics

A thin circular ring of mass $\mathrm{m}$ and radius $\mathrm{R}$ is rotating about its axis with a 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 an angular velocity $\omega^{\prime}=$

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

Solution

$L_i=L_f$ $m R R^2 \omega=\left(m R^2+2 M R^2\right) \omega^{\prime}$ $\omega^{\prime}=\left(\frac{m \omega}{m+2 M}\right)$

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

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