Body falling off a smooth hemisphere - simple but I'm missing something. What is it?

Question

So here's the question that I'm stuck on.

A particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface the speed of the particle when it leaves the surface is:

I can simply apply Conservation of Mechanical Energy(or Work-Energy Theorem, if you think like that) and get the first option as my answer =$ sqrt{2g(R-h)}$
But, if I make a free body diagram at any arbitrary angle from the vertical I make the following equation-   $mgcos(x) - N= (mv^2)/R$
The particle leaves the surface when $N=0$
This gives us $v=sqrt{Rgcos(x)}$ and as $Rcos(x) =hRightarrow v=sqrt{gh}$, which isn't an option.
Maybe I'm missing some concepts about circular motion and I can't understand what it is? Please help.

Young Kindaichi · Answer

Actually, both of them are true! The one with
$$v=sqrt{2g(R-h)}$$
and $$v=sqrt{gh}$$
But How? See the leaving condition
$$mgcostheta =frac{mv^2}{R}Rightarrow cos theta =frac{v^2}{Rg}$$
and from the energy conservation at two points
$$v^2=2g(R-h)$$
$$Rightarrow costheta =frac{2(R-h)}{R}$$
If you look at the diagram, You will find from simple geometry that
$$cos theta =frac{h}{R}=frac{2(R-h)}{R}$$
$$Rightarrow h=frac{2R}{3}$$
If you put this into both of the expressions for $v$, You find the same result.
$$v=sqrt{frac{2}{3}Rg}$$

Body falling off a smooth hemisphere - simple but I'm missing something. What is it?

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