Question

A round disc of moment of inertia $ {{I}_{2}} $ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $ {{I}_{1}} $ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is

• A. $ \frac{{{I}_{2}}\omega }{{{I}_{1}}+{{I}_{2}}} $
• B. $ \omega $
• C. $ \frac{{{I}_{1}}\omega }{{{I}_{1}}+{{I}_{2}}} $
• D. $ \frac{({{I}_{1}}+{{I}_{2}})\omega }{{{I}_{1}}} $

Answer 1

The Correct Option is C

Key Idea : When no external torque acts on a system of particles, then the total angular momentum of the system remains always a constant. The angular momentum of a disc of moment of inertia $I_{1}$ and rotating about its axis with angular velocity $\omega$ is $L_{1}=I_{1} \omega$ When a round disc of moment of inertia $I_{2}$ is placed on first disc, then angular momentum of the combination is $L_{2}=\left(I_{1}+I_{2}\right) \omega^{\prime}$ In the absence of any external torque, angular momentum remains conserved i.e., $ L_{1} =L_{2} $ $I_{1} \omega =\left(I_{1}+I_{2}\right) \omega' $ $\Rightarrow \omega' =\frac{I_{1} \omega}{I_{1}+I_{2}}$