A particle is released on a vertical smooth semicircular track from point X so that, O X makes angle θ from the vertical (see figure). The normal reaction of the track on the particle vanishes at the point Y where O Y makes an angle ϕ with the horizontal. Then

Step-by-step solution

m g R cos θ - R sin ϕ = 1 2 m V 2     . . . . . . . (1) On losing contact N =0⇒ m g sinϕ = m V 2 R     . . . . . . (2) ⇒ m g R cosθ - sinϕ = 1 2 m g R sinϕ ⇒ 2 cos θ = 3 sinϕ