(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A sphere is on the ground, with the sphere having a radius R. A person (skier) is standing on top of the sphere. neglecting friction, find the angle at which the person will slide off the sphere if he moves from rest to a constant velocity of v.

http://img63.imageshack.us/img63/930/untitledai7.jpg [Broken]

2. Relevant equations

F = ma

F = mv^2/r (centripetal force)

U + K = U + K (conservation of energy)

3. The attempt at a solution

1/2mv^2 + mgh = 1/2mv^2 + mgh

0J + m(9.8m/s^2)(2R) = 1/2mv^2 + m(9.8m/s^2)(R + Rcos (angle))

So that simplifies to v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle))

Drawing a free body diagram, I know that the centripetal force is equal to the Normal foce*cos(angle):

mv^2/r = mgcos(angle)

v^2/r = 9.8m/s (cos (angle)

v^2 = 9.8 m/s (Rcos(angle))

Substitute that into v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle)) and get:

9.8m/s Rcos (angle)) = 19.6 m/s(R) - 19.6m/s (Rcos (angle))

That reduces to:

cos (angle) = 2/3

so the angle must be 48.2 degrees

My teacher said it was wrong, can anyone assist me?

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# Homework Help: Finding the angle of a falling person on a sphere

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