Q110 / 125 · BITSAT 2021 · Physics
A round disc of moment of inertia $\mathrm{I}_{2}$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $\mathrm{I}_{1}$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is
- A$\frac{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega}{\mathrm{I}_{1}}$
- B$\frac{\mathrm{I}_{2} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$
- C$\omega$
- Correct answerD$\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$
Solution
Total angular momentum of the system initially \(\mathrm{L}_{\mathrm{i}}=\mathrm{I}_{1} \mathrm{w}+\mathrm{I}_{2}(0)=\mathrm{I}_{1} \mathrm{w}\)
Total angular momentum of the system finally \(L_{f}=\left(I_{1}+I_{2}\right) w_{2}\)
According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)
\(\begin{aligned}
&\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2} \\
&\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}
\end{aligned}\)
Hence, the correct answer is option: 'D'.