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Normal Force Going to Zero

  1. Apr 22, 2005 #1
    I need help on a physics problem I've been working on...

    A skier starts at rest at the top of a large hemispherical hill with radius (height) = R. Neglecting friction, show that the skier will leave the hill and become airborne at the distance of h = R/3 below the top of the hill.

    I understand that at the point the skier goes airborne, the normal force is zero, but how do I conceptually show it? When it's at the crest of the hill, there are obviously two vertical forces in play: the weight of the skier (downward) and the normal force (upward).

    So the question, once again, is how do I show specifically (to prove) that the skier goes airborne at height = 3/R.

    Thanks in advance.
     
  2. jcsd
  3. Apr 22, 2005 #2
    Physics,

    Can you find an equation for the normal force as a function of the skier's angular position, where angle is measured from the vertical. In other words he starts at theta=0 and flies off somewhere between theta=0 and theta=90degs? You need an equation for the normal force in terms of theta.
     
  4. Apr 23, 2005 #3
    like it was said by jdavel, you need to know the angle. if you do a "free body diagram", you will notice that :
    [tex] mg \cdot \cos \theta - N = m \cdot \frac{V^2}{R} \rightarrow \cos \theta = \frac{ \frac{V^2}{R} + N}{g} [/tex]
     
  5. Apr 23, 2005 #4

    Doc Al

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    Staff: Mentor

    more hints

    Since you are asked to find the point of departure in terms of height h below the hilltop, rewrite [itex]\cos\theta[/itex] in terms of h and R. Hint: You'll need to use conservation of energy.
     
  6. Apr 27, 2005 #5
    *Correction**

    The skier goes airborne at height h= R/3
     
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