Why Does a Skier Go Airborne at h=R/3 on a Hemispherical Hill?

AI Thread Summary
A skier on a hemispherical hill with radius R will become airborne at a height of h = R/3 due to the balance of forces acting on them. At this point, the normal force acting on the skier becomes zero, indicating the transition from contact with the hill to free flight. The analysis involves using conservation of energy to relate the skier's potential energy at the top to their kinetic energy and the forces at play as they descend. By applying a free body diagram and deriving the normal force in terms of the skier's angular position, it can be shown that the skier departs the hill at the specified height. This understanding is crucial for solving similar physics problems involving motion on curved surfaces.
PhysicsinCalifornia
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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.
 
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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.
 
like it was said by jdavel, you need to know the angle. if you do a "free body diagram", you will notice that :
mg \cdot \cos \theta - N = m \cdot \frac{V^2}{R} \rightarrow \cos \theta = \frac{ \frac{V^2}{R} + N}{g}
 
more hints

Since you are asked to find the point of departure in terms of height h below the hilltop, rewrite \cos\theta in terms of h and R. Hint: You'll need to use conservation of energy.
 
PhysicsinCalifornia said:
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.

*Correction**

The skier goes airborne at height h= R/3
 
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