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cblack · 2008-04-15 04:02:48

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.

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

The attempt at a solution
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= $\frac{R}{2}$

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?

donaldos · 2008-04-25 23:16:43

You got the right equations but you're not using them properly :

$\frac{mv^2}{R}=F_{n}+mg\underline{\cos \theta}$ (radial component!)

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

$F_n=0 \quad \Rightarrow \quad mgh=\frac 1 2 m v^2=\frac 1 2 mg(R-h)$

hence $h=\frac R 3$

Announcement

#1 2008-04-15 04:02:48

Conservation of Energy and Centripetal Motion?

#2 2008-04-25 23:16:43

Re: Conservation of Energy and Centripetal Motion?

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