question_answer

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\[\omega '\]is equal to     AIEEE  Solved  Paper-2006

A) \[\frac{\omega (m+2M)}{m}\]  

B)        \[\frac{\omega (m-2M)}{(m-2M)}\]

C) \[\frac{\omega m}{(m+M)}\]    

D)        \[\frac{\omega m}{(m+2M)}\]

Correct Answer: D

Solution :

As external torque is acting about the axis, angular momentum of system remains conserved. \[\therefore \]  \[{{l}_{1}}\omega ={{l}_{2}}\omega '\] \[\Rightarrow \]\[m{{R}^{2}}\omega =(m{{R}^{2}}+2M{{R}^{2}})\omega '\] \[\Rightarrow \]\[\omega '=\left( \frac{m}{m+2M} \right)\omega \]