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

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Homework Help Overview

The problem involves a skier descending a hemispherical hill and determining the height at which the skier becomes airborne. The context is centered around concepts of forces acting on the skier, particularly the normal force and gravitational force, as well as energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the normal force and the skier's angular position, suggesting the use of free body diagrams and equations of motion. There are inquiries about how to express the normal force in terms of the skier's position and height.

Discussion Status

Some participants have offered hints regarding the use of energy conservation to relate the height and the angle of departure. There is ongoing exploration of the necessary equations and relationships without reaching a consensus on the specific approach to proving the height at which the skier becomes airborne.

Contextual Notes

Participants are working under the assumption that friction is negligible and are focused on deriving relationships based on the geometry of the hill and the forces acting on the skier. There is a correction noted regarding the height at which the skier goes airborne, emphasizing the need for clarity in the problem statement.

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 :
[tex]mg \cdot \cos \theta - N = m \cdot \frac{V^2}{R} \rightarrow \cos \theta = \frac{ \frac{V^2}{R} + N}{g}[/tex]
 
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.
 
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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