A coin is sliding down on a smooth hemispherical surface of radius R. The height from the bottom, where it looses contact with the surface is

In the given  problem, a coin is moving smoothly along a hemispherical surface with radius R. Here, the smooth surface makes it very evident that there is no surface friction. Find the height from the bottom where the coin loses contact with hemispherical surfaces using the information provided.

We know that the change in potential energy of the coin as it slides from a height R to a height h is converted to kinetic energy by the law of conservation of energy. In terms of mathematics,

$$\begin{gathered} \frac{1}{2}m{v}^2=mgR-mgh \\ h=R\cos \theta \\ \frac{1}{2}m{v}^2=mgR-mgR\cos \theta \\ {v}^2=2gR(1-\cos \theta ) \end{gathered}$$

Additionally, the coin will experience zero normal force as it separates from the surface. Let's examine the free body illustration.

Let's now balance the forces acting in the normal direction,

$N=mg\cos \theta -\frac{m{v}^2}{R}$

But, the normal force is known to be zero when the coin separates from the surface. So,

$$\begin{gathered} mg\cos \theta =\frac{m{v}^2}{R} \\ \Rightarrow g\cos \theta =\frac{v^2}{R} \\ \Rightarrow g\cos \theta =\frac{2gR(1-\cos \theta )}{R} \\ \Rightarrow \cos \theta =2-2\cos \theta \\ \Rightarrow \cos \theta =\frac{2}{3}\_\_\_\_\_\_\left(1\right) \end{gathered}$$

$from the figure,$

$\cos \theta =\frac{h}{R}\_\_\_\_\_\_\left(2\right)$

$$\begin{gathered} by equation 1 and 2 \\ \frac{h}{R}=\frac{2}{3} \\ \Rightarrow h=\frac{2}{3}R \end{gathered}$$

Hence the correct answer is $\frac{2}{3}R.$