Calculating Airborne Distance on a Hemispherical Hill

AI Thread Summary
To determine when a skier leaves a hemispherical hill, the key is analyzing the forces acting on the skier, including the normal force and gravitational force. The skier will become airborne when the normal force equals zero, which can be derived using conservation of energy principles. The calculations involve setting up equations of motion and understanding the skier's acceleration as they descend the hill. The critical height at which the skier loses contact with the surface is found to be h = R/3, measured from the top. This approach combines dynamics and energy conservation to solve the problem effectively.
SABander
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Hey, I am new and i am hoping someone can give me a hand. I desperately need assitance on this question and any advice would be great. Thanks
A skier starts at rest on the top of 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.
 
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So far i have thought that it could be a circular motion question and find the maximum velocity that the object could go around. But given no numbers i am having trouble. This is what i have so far
E=Mgd
E=MGr
FD=MGr

And that got me no where. Then i tried this approach
(MV^2/r)D=MGR
V^2D=GR^2

PLEASE HELP!
Thanks
 
At an arbitrory point on the path of the skier, write down the equation of motion. The forces are the normal force, weight, and the cetripetal force.

Conservation of energy should give a second equation.

Remember that the skier will leave the surface when the reaction force (normal force) = 0.

USe this condition to find h.
 
I think the motion along the hemispherical surface is doing you harm. You write the equation of acceleration by referring to a point on the hemisphere and take the angle made by the radius drawn to that point with the radius drawn to topmost point. Now integrate this acceleration with respect to theta. I think this will provide you the result. Was the explanation out of your head-I don't think so. Reread it once more if you don't undertstand.
 
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