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 fllen a height h.

Explanation

Consider the two shown positions of the cylinder. As it does not slip, its total mechanical energy will be conserved.
Energy at position 1 is `E_(1)` = mgh
Energy at position 2 is
`E_(2) = (1)/(2) mv_(c.m.)^(2) + (1)/(2) I_(c.m.) omega^(2)`
`because" "(V_(c.m.))/(r) = omega and I_(c.m.) = (mr^(2))/(2)`
`rArr" "E_(2) = (3)/(4) mv_(c.m.)^(2)`
From law of conservation of energy, `E_(1) = E_(2)`
`rArr" "V_(c.m.) = sqrt((4)/(3)gh)`
Note : In the previous example we used conservation principle, while in the one above we used Newton.s laws. Either one leads to the correct result, it is only a matter of convenience as to which method we choose. Conservation of angular momentum also helps in tackling problems concerning collisions of rolling bodies. Applying conservation of angular momentum about the point of collision helps to eliminate the external torques due to large impulsive forces.