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?