A uniform disc of mass `M` and radius R is pivoted about the horizontal axis through its centre `C` A point mass m is glued to the disc at its rim, as shown in figure. If the system is released from rest, find the angular velocity of the disc when `m` reaches the bottom point B.

Explanation

Th conservation of energy yields
`/_\KE=/_\PE`
`1/2Iomega^(2)=mg(2r)+mgr`
`implies 1/2 m(2r)^(2)omega^(2)+1/2{((mr^(2))/2)+mr^(2)}omega^(2)=3mgr`
`implies 11/4 mr^(2)omega^(2)implies 3mgrimpliesomega=sqrt((12g)/(11r))`