A uniform spherical shell of mass `M` and radius `R` rotates, about a vertical axis on frictionless bearing. A massless cord passes around the equator of the shell., over a pulley of rotational inertia `I` and radius `r` and is attached to small object of mass `m` that is otherwise free to fall under the influence of gravity. There is no friction of pulley's axle, the cord does not slip on the pulley. What is the speed of the object after it has fallen a distance `h`. from rest? Use work-energy considerations.

Explanation

Using conservation of energy principle, we have fall in `PE` of te block `=` Gain in `KE` of the block `+` Rotational `KE` of the pulley `+` Rotational `KE` of the shell.
or `mgh=1/2mv^(2)+1/2Iomega^(2)+1/2((2MR^(2))/3)omega^('2)`
where `omega=v/r` and `omega'=v/R`
After substituting these values in the above equation and solving we get
`v^(2)=(mgh)/((m/2+I/(2r^(2))+M/3))`