A solid cylinder of radius R is at rest at a height h on an inclined plane. If it rolls down then its velocity on reaching the ground is

Step-by-step solution

The given situation is shown below Let v be the velocity of solid cylinder on reaching the ground. According to conservation of energy, Total energy at point A= Total energy at point B i. e m g h+0= 1 2 I ^2+ 1 2 m v^2+0 m g h= 1 2 ( 1 2 M R^2 ) ( v R )^2+ 1 2 m v^2 (Where, I= 1 2 M R^2 and = v R ) aligned & g h= v^2 4 + v^2 2 g h= 3 v^2 4 & v^2= 4 g h 3 v= 4 g h 3 aligned