Question

A gramophone record of mass $M$ and radius $R$ is rotating at an angular velocity $\omega$. A coin of mass $m$ is gently placed on the record at a distance $r=R/2$ from its centre. The new angular velocity of the system is

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

Answer 1

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

The initial angular momentum of the rotating record is
$L=I\omega$
Where $I=\frac{1}{2}M{R}^2$
Let ${\omega }^{\prime }$ be the angular velocity of the record when the coin of mass $m$ is placed on it at a distance $r$ from its centre. The angular momentum of the system becomes
$L^{\prime }=(I+m{r}^2){\omega }^{\prime }$
Since no external torque acts on the system, the angular momentum is conserved. i.e.
$L^{\prime }=L$ or $(I+m{r}^2){\omega }^{\prime }=I\omega$
or ${\omega }^{\prime }=\frac{I\omega }{I+m{r}^2}=\frac{\frac{1}{2}M{R}^2\omega }{\frac{1}{2}M{R}^2+m{r}^2}$
or ${\omega }^{\prime }=\frac{\omega }{(1+\frac{2m{r}^2}{M{R}^2})}$
Putting $r=R/2$, we find that the correct choice is (a).