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

$$\Sigma$$F=mv$$^{2}$$$$/$$r
mg$$\Delta$$y=$$\frac{1}{2}$$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.

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

Substituting this into the conservation of energy equation you get:

mg$$\Delta$$y=$$\frac{1}{2}$$mv$$^{2}$$
mg$$\Delta$$y=$$\frac{1}{2}$$mgR
$$\Delta$$y=$$\frac{mgR}{2mg}$$
$$\Delta$$y=$$\frac{R}{2}$$

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 $$\Sigma$$F=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]

$$\Sigma$$F=Fn+mg
0=$$\Sigma$$F-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.)