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 new angular velocity of ring is

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

Answer 1

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

$$\begin{array}{c}Accorrdingtoquestion.......................\\ Here,\\ I_1{\omega }_1=I_2{\omega }_2...............\left(1\right)\\ I_1=m{R}^2..................\left(2\right)\\ and{I}_2=m{R}^2+2M{R}^2--------\left(3\right)\\ Now,fromequation\left(1\right),\left(2\right)and\left(3\right)\\ then,weget....\\ {\omega }_2=\frac{I_1}{I_2}\omega \\ {\omega }_2=\frac{m{R}^2}{m{R}^2+2M{R}^2}\times \omega \\ \therefore {\omega }_2=\frac{m\omega }{m+2M}\\ SothecorrectoptionisB.\end{array}$$