A disc of radius $R$ and mass $M$ is rotating with angular velocity $\omega$. A second identical disc is dropped coaxially onto the first. What is the final angular velocity?

Solution

1. The initial angular momentum of the system is $L_i=I_1\omega$, where the moment of inertia of the first disc is $I_1=\frac{1}{2}M{R}^2$.
2. When the second identical disc is dropped, the total moment of inertia becomes $$\begin{array}{rl}{I}_f& =I_1+I_2\\ & =\frac{1}{2}M{R}^2+\frac{1}{2}M{R}^2\\ & =M{R}^2\end{array}$$.
3. By the conservation of angular momentum, $I_1\omega =I_f{\omega }_f$, which simplifies to $(\frac{1}{2}M{R}^2)\omega =(M{R}^2){\omega }_f$.
4. Solving for the final angular velocity gives ${\omega }_f=\frac{\omega }{2}$.
5. Hence the answer is (B).