A particle is released on a vertical smooth semicircular track from point X so that, OX makes an angle $$\theta$$ from the vertical (see the figure). The normal reaction of the track on the particle vanishes at point Y, where OY makes an angle $$\phi$$ with the horizontal. Then:

Let the potential energy be zero at the center $$O$$. The initial height of the particle at point $$X$$ relative to $$O$$ is $$h_X = R \cos \theta$$

At point $$Y$$, where $$OY$$ makes an angle $$\phi$$ with the horizontal, the height relative to $$O$$ is $$h_Y = R \sin \phi$$

By the Law of Conservation of Energy between points $$X$$ and $$Y$$:

$$mgR \cos \theta = mgR \sin \phi + \frac{1}{2}mv^2$$

$$v^2 = 2gR(\cos \theta - \sin \phi) \quad \dots (1)$$

At point $$Y$$, the forces acting on the particle along the radial direction are the normal reaction $$N$$ and the radial component of weight $$mg \sin \phi$$. The centripetal force equation is:  $$mg \sin \phi - N = \frac{mv^2}{R}$$

The normal reaction vanishes at point $$Y$$ ($$N = 0$$). Therefore, $$mg \sin \phi = \frac{mv^2}{R}$$

$$v^2 = gR \sin \phi \quad \dots (2)$$

Equating $$(1)$$ and $$(2)$$: $$gR \sin \phi = 2gR(\cos \theta - \sin \phi)$$

$$\sin \phi = 2 \cos \theta - 2 \sin \phi$$

$$3 \sin \phi = 2 \cos \theta$$

$$\sin \phi = \frac{2}{3} \cos \theta$$