Conservation of mechanical energy for skier on sphere

[NathanLeduc1]

Homework Statement

A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. At what angle θ will the skier leave the sphere?

Homework Equations

KE= 0.5mv^2
PE = mgh
Fc = (mv^2)/r

The Attempt at a Solution

I am really quite confused and don't even know how to begin. I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.
I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh

 

[Doc Al]

I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.

That's the key point. To make use of it, set up a force equation with Newton's 2nd law.

I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh

Good. You'll need that too. (Express h in terms of θ.)

 

[NathanLeduc1]

So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)

If I plug that into the centripetal force equation, I get
F = (mv^2)/r
F = (Nv^2)/((r)(a+g))

Is this right? If so, where do I go from here? Sorry to ask such dumb questions, I'm just very confused on this problem.

 

[Doc Al]

So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).

What you want is Δh, the drop from the original position at the top. Δh ≠ r sinθ.

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)

Careful! Forces are vectors.

Hint: Consider force components perpendicular to the surface at any point.