A cylinder of mass 2 kg is released from rest from the top of an inclined plane of inclination 30 ° and length 1 m . If the cylinder rolls without slipping, then its speed when it reaches the bottom is [ g = 10 m s - 2 ]

Step-by-step solution

Potential energy of cylinder at top position U = m g A B = 2 × 10 1 . sin 30 °    ∵ sin 30 ° = A B B C ∴ A B = 1 sin 30 ° = 10   J If v is the linear speed of cylinder when it reaches at bottom, then at bottom, total potential energy of cylinder is converted into kinetic energy, i.e. K = U 1 2 I ω 2 + 1 2 m v 2 = 10 1 2 · m r 2 2 · v 2 r 2 + 1 2 m v 2 = 10 ∵ ω = v r ⇒    3 4 m v 2 = 10 v = 40 3 m = 40 3 × 2 = 20 3   &