A disc of mass $M$ and radius $R$ is rotating with angular velocity $\omega$. A second identical disc is placed gently on top of the first. The final angular velocity of the system is:

Solution

1. The initial angular momentum of the system is $L_i=I\omega$, where $I=\frac{1}{2}M{R}^2$ is the moment of inertia of the first disc.
2. When the second identical disc is placed on top, the total moment of inertia becomes $$\begin{array}{rl}{I}_{total}& =I+I\\ & =2I\end{array}$$.
3. By the conservation of angular momentum, $I\omega =(2I){\omega }_f$.
4. Solving for the final angular velocity gives ${\omega }_f=\frac{\omega }{2}$.
5. Hence the answer is (A).