Q110 / 200 · NEET 2009 (Screening) · Physics

A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

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

Solution

Key Idea Apply law of conservation of angular momentum. $$ \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 $$ In the given case $$ \begin{aligned} \mathrm{I}_1 & =\mathrm{MR}^2 \\ \mathrm{I}_2 & =\mathrm{MR}^2+2 \mathrm{mR}^2 \\ \omega_1 & =\omega \end{aligned} $$ Then $$ \omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega $$

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

Relevant Questions