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Conservation of Energy and Centripetal Motion Question

  • Thread starter cblack
  • Start date
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
[tex]\Sigma[/tex]F=mv[tex]^{2}[/tex][tex]/[/tex]r
mg[tex]\Delta[/tex]y=[tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]



3. 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.

[tex]\Sigma[/tex]F=Fn+mg
0=[tex]\Sigma[/tex]F-mg
0=[tex]\frac{mv^{2}}{R}[/tex]-mg
v=[tex]\sqrt{gR}[/tex]

Substituting this into the conservation of energy equation you get:

mg[tex]\Delta[/tex]y=[tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]
mg[tex]\Delta[/tex]y=[tex]\frac{1}{2}[/tex]mgR
[tex]\Delta[/tex]y=[tex]\frac{mgR}{2mg}[/tex]
[tex]\Delta[/tex]y=[tex]\frac{R}{2}[/tex]


2. Relevant 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 [tex]\frac{R}{2}[/tex] because at the point the angle between the skier and the vertical will be 90 degrees. If you use the equation [tex]\Sigma[/tex]F=mgcos[tex]\Theta[/tex]+mg where mgcos[tex]\Theta[/tex], 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?
 
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hello cblack, your physical concept is correct. you almost got it. you just left out [tex]\theta[/tex] somewhere it should be.
try to find v in terms of [tex]\theta[/tex] using conservation of E.
then construct equation of motion. Be careful the v here also relates to [tex]\theta[/tex]. then you can solve for [tex]\theta[/tex] :smile:
 

Doc Al

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[tex]\Sigma[/tex]F=Fn+mg
0=[tex]\Sigma[/tex]F-mg
0=[tex]\frac{mv^{2}}{R}[/tex]-mg
v=[tex]\sqrt{gR}[/tex]
Realize that the weight acts vertically, not centripetally. (Find the radial components of the forces.)
 

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