Question

A uniform thin circular ring of mass 'M' and radius 'R' is rotating about its fixed axis, passing through its centre and perpendicular to its plane of rotation, 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 of :

• A. $\frac{\omega M}{(M+m)}$
• B. $\frac{\omega M}{(M+2m)}$
• C. $\frac{\omega M}{M-2m}$
• D. $\frac{\omega (M+3m)}{M}$

Answer 1

The correct option is B $\frac{\omega M}{(M+2m)}$

Moments of inertia (MOI) of the ring before attaching the masses $I$= $M{R}^2$,

MOI of the ring after attaching the masses $I^{{}^{\prime }}=(M+2m)R^2$

Let angular momentum after the attaching the masses ${\omega }^{{}^{\prime }}$

Since there is no external torque, so we use conservation of angular momentum.

$I\times \omega =I^{{}^{\prime }}\times {\omega }^{{}^{\prime }}$

$\Rightarrow M{R}^2\times \omega =\frac{M{R}^2\times {\omega }^{{}^{\prime }}}{M+2m}$,

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