A frictionless pulley has the shape of a uniform solid disc of mass `M` and radius `R`. A stone of mass `m` is attached to a light wire that is wrapped around the rim of the pulley and a system is released from rest. If a stone falls by a distance `h`, what is the angular speed of pulley at that moment?

Explanation

Loss in the potential energy of stone
`=` gain in the kinetic energy of (stone `+` pulley)
`mgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`v=R omega , I=(1)/(2)MR^(2)`
`mgh=(1)/(2)m(Romega^(2))+(1)/(2)xx(1)/(2)MR^(2)omega^(2)`
`=R^(2)omega^(2)((2m+M)/(4))`
`omega^(2)=(4mgh)/((M+2m)R^(2))`
` omega=sqrt((4mgh)/((M+2m)R^(2)))`