Question

A unifonn circular ring of mass M and radius r is rotating with an angular speed $\omega$ about an axis passing through its centre and perpendicular to the plane of the ring. Two identical beads, each of mass m, somehow get attached at two diametrically opposite points. The rotational speed of the ring will become

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

Answer 1

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

Since the two beads are diametrically opposite, resultant force on the system is zero.

Since the ring is rotating at constant angular velocity. Since there is no external torque, the angular momentum of the system must be conserved.

$\therefore I\omega =\text{Constant}$

$Mr\omega =(M+2m)r{\omega }^{\prime }$

$\therefore {\omega }^{\prime }=\frac{M\omega }{M+2m}$