Answer:
2ω/3
Explanation:
The moments of inertia of the 1st disk is:
[tex]I_1 = (2MR^2)/2 = MR^2[/tex]
The moments of inertia of the 2nd disk is:
[tex]I_2 = MR^2/2[/tex]
So the total moments of inertia of the system of 2 disks after the drop is:
[tex]I = I_1 + I_2 = MR^2 + MR^2/2 = 1.5MR^2[/tex]
Using law of angular momentum conservation we have the following equation for before and after the drop
[tex]I_1\omega = I\omega_2[/tex]
[tex]\omega_2 = \omega\frac{I_1}{I} = \omega\frac{MR^2}{1.5MR^2} = 2\omega/3[/tex]
