Conservation of Energy and Centripetal Motion Question

[cblack]

Homework Statement

A skier starts at rest on the top of Mt Circular, a strange, smooth, icy hill shaped like a hemisphere. The hill has a constant radius of R. Neglecting friction (it is icy!), show that the skier will leave the surface of the hill and become air-borne at a vertical distance of h=R/3, measured from the top of the hill.

Homework Equations

\SigmaF=mv^{2}/r
mg\Deltay=\frac{1}{2}mv^{2}

The Attempt at a Solution

I have found one question exactly like this one on the forum and I understand that when the skier leaves the hill, the normal force will be 0, but I keep getting R/2 as an answer.

\SigmaF=Fn+mg
0=\SigmaF-mg
0=\frac{mv^{2}}{R}-mg
v=\sqrt{gR}

Substituting this into the conservation of energy equation you get:

mg\Deltay=\frac{1}{2}mv^{2}
mg\Deltay=\frac{1}{2}mgR
\Deltay=\frac{mgR}{2mg}
\Deltay=\frac{R}{2}

Homework Equations

Is it reasonable to assume that the skier is in centripetal motion?
The only thig i don't understand is how the skier can have a normal force of zero at any other point than at \frac{R}{2} because at the point the angle between the skier and the vertical will be 90 degrees. If you use the equation \SigmaF=mgcos\Theta+mg where mgcos\Theta, which is the noraml force, can only equal 0 at 90 degrees. If the skier is leaving somewhere between 0 and 90 degrees how can this be calculated?

 

[luben]

hello cblack, your physical concept is correct. you almost got it. you just left out \theta somewhere it should be.
try to find v in terms of \theta using conservation of E.
then construct equation of motion. Be careful the v here also relates to \theta. then you can solve for \theta

 

[Doc Al]

\SigmaF=Fn+mg
0=\SigmaF-mg
0=\frac{mv^{2}}{R}-mg
v=\sqrt{gR}

Realize that the weight acts vertically, not centripetally. (Find the radial components of the forces.)