Question

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with an angular speed $\omega$. Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:

• A. $\omega \frac{M}{M+2m}$
  
• B. $\omega \frac{M}{M+m}$
  
• C. $\omega \frac{M+2m}{M}$
  
• D. $\omega \frac{M-2m}{M+2m}$
  

Answer 1

The correct option is A $\omega \frac{M}{M+2m}$


External torque is zero on the system. ${\tau }_{net}=0$, so angular momentum is conserved.

By angular momentum conservation:
$I_i{\omega }_i=I_f{\omega }_f$

$\left(M{R}^2\right)\omega =(M{R}^2+2m{R}^2){\omega }_f$

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

${\omega }_f=\frac{M\omega }{M+2m}$