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

Answer: (C)

$\frac{{{I}_{1}}\omega }{{{I}_{1}}+{{I}_{2}}}$

Answer 2

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., $L_{1} =L_{2}$ $I_{1} \omega =\left(I_{1}+I_{2}\right) \omega'$ $\Rightarrow \omega' =\frac{I_{1} \omega}{I_{1}+I_{2}}$