At time t = 0 , a disk of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of α = 2 3 rad s - 2 . A small stone is stuck to the disk. At t = 0 , it is at the contact point of the disk and the plane. Later, at time t = π s , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m ) reached by the stone measured from the plane i

Step-by-step solution

Using second equation of motion, The angle rotated by disc in t = π   s is θ = ω 0 t + 1 2 α t 2 ⇒ θ = 1 2 × 2 3 π 2 = π 3   rad Now using first equation of motion, the angular velocity of disc is ω = ω 0 + α t ⇒ ω = 0 + 2 π 3 = 2 π 3   rad   s - 1 And, velocity of disk about centre of mass is v cm = ω R = 2 π 3 × 1 = 2 π 3   m   s - 1 At the moment the stone detaches, the situation is