A solid cylinder of mass m and radius r starts rolling down an inclined plane of inclination `theta`. Friction is enough to prevent slipping. Find the speed of its centre of mass when its centre of mass has fallen a height `h`.

Explanation

Considering the two shown positions of the cylinder. As it does not slip hence total mechanical energy will be conserved.

Energy at position 1 is, `E_(1) = mgh`
Enery at position 2 is
`E_(2)=(1)/(2)mv_(CM)^(2)+(1)/(2)I_(CM)omega^(2)`
`because (v_(CM))/(r)=omega and I_(CM)=(mr^(2))/(2)rArrE_(2)=(3)/(4)mv_(CM)^(2)`
From law of conservation of energy, `E_(1)=E_(2)`
`rArr v_(CM)=sqrt((4)/(3)gh)`.