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.

Thanks in advance.

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.

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

A particle slides from 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 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 = [ 2g(R-h)]^(1/2)

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=[Rgcos(x)]^(1/2) and as Rcos(x) is "h", v=(gh)^(1/2) ( which isn't an option) I am missing some concept about circular motion and i can't understand what it is.Please help.

Thanks in advance.

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.

Thanks in advance.