question_answer

A circular disc of mass M and radius R is rotating with an angular velocity to about an axis passing through its centre and perpendicular to the plane of the disc. A small point like part of mass m detaches from the rim of the disc and continues to move with same angular speed. The angular velocity of remaining disc just after detaching will become

A) \[\left( \frac{M-2m}{M+m} \right)\omega \]     

B)        \[\left( \frac{M+2m}{M+m} \right)\omega \]    

C)        \[\left( \frac{M-2m}{M-m} \right)\omega \]      

D)        \[\left( \frac{M+2m}{M-m} \right)\omega \]

Correct Answer: C

Solution :

Key Idea: Angular momentum remains conserved. From law of conservation of angular momentum if no external torque is acting upon a body rotating about an axis, then    the    angular momentum of the body remains constant. \[J=I\omega =\text{constant}\]  (\[\omega =\] angular velocity) Moment of inertia of disc \[=\frac{1}{2}M{{R}^{2}}\] \[\therefore \]  \[\frac{1}{2}M{{R}^{2}}\omega =\frac{1}{2}(M-m){{R}^{2}}\omega +m{{R}^{2}}\omega \] \[\Rightarrow \]               \[(M-m)\omega =M\omega -2m\omega \] \[\Rightarrow \]               \[\omega =\frac{(M-2m)}{(M-m)}\omega \]