Question

A round disc of moment of inertia ${I_2}$ about its axis perpendicular to its plane and passing through its center is placed over another disc of the moment of inertia ${I_1}$ rotating with an angular velocity about the same axis. The final angular velocity of the combination of disc is
$A) \,\dfrac{{{I_2}\omega}}{{{I_1} + {I_2}}} \\\ B) )\,\omega \\\ C) )\,\dfrac{{{I_1}\omega}}{{{I_1} + {I_2}}} \\\ D) \,\dfrac{{\left( {{I_1} + {I_2}} \right)\omega}}{{{I_1}}} \\\$

Answer 1

As in this system there is no external torque, so the momentum will remain conserved. Angular momentum is represented by L.
As the angular momentum is conserved. So initial angular momentum is equal to final angular momentum which can be written as
${L_i} = {L_f}$
Formula for angular momentum is
$L = I\omega$
Where I is inertia
And $\omega$ is angular velocity
Using these two relations we can easily find the final angular velocity of the combination of discs.
Step by step complete solution:
Initially we have a disc of moment of inertia ${I_1}$ rotating with an angular velocity $\omega$ about an axis
Angular momentum is given as $L = I\omega$
So initial angular momentum is
${L_i} = {I_1}\omega$
After this we place a round disc of moment of inertia ${I_2}$ on disc with inertia ${I_1}$
Let the final angular velocity be ${\omega_f}$
The moment of inertia of this system the sum of ${I_1}$ and ${I_2}$
So final angular momentum is
${L_f} = \left( {\;{I_1} + {I_2}} \right){\omega_f}$
As in this system there is no external torque. So the angular momentum will remain conserved. As the angular momentum is conserved initial angular momentum is equal to final angular momentum. So,
${L_i} = {L_f}$
${I_1}\omega = \left( {{I_1} + {I_2}} \right){\omega_f} \\\ {\omega_f} = \dfrac{{{I_1}\omega}}{{\left( {{I_1} + {I_2}} \right)}} \\\$

So option C is correct.

Note: The moment of inertia for a combination of bodies having the same axis of rotation can simply be added. For a system with no external torque, angular momentum remains conserved.