Question

A gramophone record of mass $M$ and radius $R$ is rotating with angular speed $\omega$. If two pieces of wax each of mass $m$ are kept on it at a distance $\frac{R}{2}$ from the centre on opposite sides, then the new angular velocity will be

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

Answer 1

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

Initial angular momentum will be same as final angular momentum as there is no external torque acting on it.
Therefore,
$I\omega =I\text{'}\omega \text{'}$
$\Rightarrow \frac{M{R}^2}{2}\omega =\left(\frac{M{R}^2}{2}+2m\left(\frac{R^2}{4}\right)\right]\omega \text{'}$
On solving we get

$\Rightarrow M\omega =\left(M+m\right)\omega \text{'}$

$\Rightarrow \omega \text{'}=\frac{M\omega }{M+m}$