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- $mg\cos(x) - N= (mv^2)/R$

The particle leaves the surface when $N=0$ This gives us $v=\sqrt{Rg\cos(x)}$ and as $R\cos(x) =h\Rightarrow v=\sqrt{gh}$, which isn't an option.

Maybe I'm missing some concepts about circular motion and I can't understand what it is? Please help.

Answer 1

Actually, both of them are true! The one with $$v=\sqrt{2g(R-h)}$$ and $$v=\sqrt{gh}$$ But How? See the leaving condition

$$mg\cos\theta =\frac{mv^2}{R}\Rightarrow \cos \theta =\frac{v^2}{Rg}$$ and from the energy conservation at two points $$v^2=2g(R-h)$$

$$\Rightarrow \cos\theta =\frac{2(R-h)}{R}$$

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. $$v=\sqrt{\frac{2}{3}Rg}$$