A uniform circular disc of mass `50 kg` and radius `0.4 m` is rotating wit an angular velocity of `10 rad//s` about its own axis, which is vertical. Two uniform circular rings, each of mass `6.25 kg` radius `0.2 m` are gently place symmetrically on the disc in such a manner that tey are touching each other along the axis of the disc and are horizontal. assume that the friction is large enough such that the rings are at rest to the disc and the system rotates about the original axis. the new angular velocity (in rad `s`) of the system is

Explanation

Moment of inertia of the disc about the axis.
`I_(1) = 1/2MR^(2) =1/2 xx 50 kg xx (0.4m)^(2) = 4 kg m^(2)`
`therefore` Initial angular momentum of the disc is `L_(1) = I_(1)omega_(1)`
When two uniform circular rings are placed symmetrically on the disc such that they are touching each other along the axis of the disc and are horizontal as shown. Then the moment of inertia of the system about the given axis.
`I_(2) = 1/2MR^(2) + 2mr^(2) + 2mr^(2) =1/2MR^(2) + 4mr^(2)`

`=1/2 xx 50 kg xx (0.4 m)^(2) + 4 xx 6.25 kg xx (0.2 m)^(2)`
`=4 kg m^(2) + 1 kg m^(2) = 5 kg m^(2)`
Let `omega_(2)` be the final angular speed of the system.
Final angular momentum of the system is `L_(2) = I_(2)omega_(2)`
According to law of conservation of angular momentum, we get
`L_(1) = L_(2)` or `I_(1)omega_(1) = I_(2)omega_(2)`, `omega_(2)=I_(1)/I_(2)omega_(1) = (4 kg m^(2))/(5 kgm^(2)) xx 10 "rad"s^(-1) = 8 "rad" s^(-1)`