A horizontal disk is rotating with angular velocity ω about a vertical axis passing through its centre. A ball is placed at the centre of groove and pushed slightly. The velocity of the ball when it comes out of the groove-

Step-by-step solution

Let us consider the motion of the ball with respect to disk N e t f o r c e a l o n g g r o o v e = m ω 2 r sin ⁡ θ = m ω 2 r x r = m ω 2 x ∴ m a = m ω 2 x ⇒ v d v d x = ω 2 x ⇒ ∫ 0 v v d v = ∫ 0 R 2 ω 2 x d x ⇒ v 2 2 = ω 2 2 R 2 2 ⇒ v = ω R 2 v → B a l l , G r o u n d = v → B a l l , D i s k + v → Disk , G r o u n d ∴ v → Ball , Ground = ω R 2 + ω R 2 2 + 2 ω R ω R 2 cos ⁡ 120 ° 1 2 = 3 2 ω R