Two horizontal discs rotate freely about a vertical axis passing through their centres. The moments of inertia of the discs relative to this axis are equal to `I_1` and `I_2`, and the angular velocities to `omega_1` and `omega_2`. When the upper disc fell on the lower one, both discs began rotating, after some time, as a single whole (due to friction). Find:
(a) teh steady-state angular rotation velocity of the discs,
(b) the work performed by the friction forces in this process.

Explanation

(a) Frome the law of conservation of angular momentum of the system relative to vertical axis z, it follows that:
`I_1omega_(1z)+I_zomega_(2z)=(I_1+I_2)omega_z`
Hence `omega_z=(I_1omega_(1z)+I_2omega_(2z))//(I_1+I_2)` (1)
Not that for `omega_zgt0`, the corresponding vector `vecomega` coincides with the poitive direction to the z axis, and vice versa. As both discs rotates about the same vertical axis z, thus in vector form.
`vecomega=I_1vecomega_1+I_2vecomega_2//(I_1+I_2)`
However, the problem makes sense only if `vecomega_1uarruarrvecomega_2` or `vecomega_1uarrdarrvecomega_2`
(b) From the equation of increment of mechanical energy of a system: `A_(f r)=DeltaT`.
`=1/2(I_1+I_2)omega_z^2-1/2I_1omega_(1z)^2+1/2I_2omega_(2z)^2`
Using Eq. (1)
`A_(f r)=-(I_1I_2)/(2(I_1+I_2))(omega_(1z)-omega_(2z))^2`