A solid cylinder is released from rest from the top of an inclined plane of inclination 30 ° and length 60   cm . If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ m   s - 1 . (Given g = 10   m   s - 2 )

Step-by-step solution

Let speed of cylinder upon reaching the bottom of the inclined plane be v . Here, gain in kinetic energy = loss in potential energy So, 1 2 I ω 2 + 1 2 m v 2 = m g l sin θ ⇒ 1 2 m r 2 2 ω 2 + 1 2 m v 2 = m g l sin 30 ° For pure rolling, v = r ω ⇒ 1 2 m r 2 2 v r 2 + 1 2 m v 2 = m g l sin 30 ° ⇒ 1 4 m v 2 + 1 2 m v 2 = m g l sin 30 ° ⇒ 3 4 m v 2 = m g l 2 ⇒ v = 4 g l 6 Putting the values, we have ⇒ v = 4 × 10 × 0 . 6 6 = 2   m &#16