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

In summary: To prove this, you need to find an equation for the normal force as a function of the skier's angular position, where the angle is measured from the vertical. This can be done by creating a free body diagram and using conservation of energy. Once you have an equation for the normal force in terms of theta, you can rewrite it in terms of h and R to show that the normal force becomes zero at h= R/3.
  • #1
PhysicsinCalifornia
58
0
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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  • #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.
 
  • #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]
 
  • #4
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.
 
  • #5
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
 

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

1. What is normal force going to zero?

Normal force going to zero refers to a situation in which the force exerted by a surface on an object becomes zero. This can happen when the object is either not in contact with the surface or when the surface is unable to exert a force on the object due to external factors.

2. Why does normal force go to zero?

Normal force goes to zero because it is a reactive force that is exerted by a surface in response to an object's weight. If the object is not in contact with the surface, there is no weight to be countered and thus no normal force is present. Additionally, external forces such as gravity or other objects may also affect the surface's ability to exert a normal force.

3. What are the implications of normal force going to zero?

When normal force goes to zero, it means that the object is no longer being supported by the surface and may experience changes in its motion or behavior. For example, if the normal force goes to zero while an object is on a slope, the object may begin to slide down the slope due to the force of gravity.

4. Can normal force going to zero be dangerous?

In some cases, normal force going to zero can be dangerous. For example, if a person is standing on a surface and the normal force suddenly goes to zero, they may experience a loss of balance and fall. Additionally, if a structure or object is relying on normal force for support, a sudden loss of normal force can lead to structural failure or collapse.

5. How can normal force going to zero be prevented or controlled?

Normal force going to zero can be prevented or controlled by ensuring that objects are securely supported on a surface and that external forces are accounted for. Additionally, proper maintenance and structural design can also help prevent sudden loss of normal force. In situations where normal force going to zero is expected or desired, such as in certain physics experiments, special equipment and techniques can be used to precisely control and measure the forces involved.

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