Question

A thin circular ring of mass $M$ and radius $R$ 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

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

Answer 1

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

As we can see, there is no external torque on the system. Hence angular momentum of the system will be conserved about the rotational axis.
Given, mass of circular ring $=M$
Radius of circular ring $=R$
Angular velocity about rotational axis $=\omega$
MOI of circular ring $I=M{R}^2$
Hence, initial angular momentum $L_i=I\omega$
$L_i=\left(M{R}^2\right)\omega$
MOI of system (ring + objects)
$I^{\prime }=m{R}^2+m{R}^2+M{R}^2$
$I^{\prime }=(M+2m)R^2$
Hence, final angular momentum
$L_f=(M+2m)R^2{\omega }^{\prime }$
As we know, $L_i=L_f$
$\left(M{R}^2\right)\omega =(M+2m)R^2{\omega }^{\prime }$


i.e ${\omega }^{\prime }=\left(\frac{M\omega }{M+2m}\right)$