A skier slides down a giant snowball ( = sphere of radius R) with negligible fri

Question

A skier slides down a giant snowball ( = sphere of radius R) with negligible friction, as shown. He starts at the top with very small velocity. Determine the angle theta f where the skier will come off the surface. Use these principles of dynamics: Energy is a constant of the motion. The normal force of contact decreases as the skier descends; and N = O at the point where the skier comes off the surface.

Explanation / Answer

As you slide down, the angle of the snowball changes. This means the decomposition of the gravitational force changes (at the top all of it is normal to the snowball. As you descend, more of it becomes parallel to the slope and less is normal). To remain on the snowball requires that there is sufficient centripetal force. If the actual centripetal force available is less than that required for circular motion at the particular velocity, then you won't remain on the snowball. Finally the velocity is a function of energy, which is dependent on the distance descended. The vertical distance descended is (1 - cos(theta)) * R So putting it all together: KE = 1/2 m * v^2 = m*g * (1 - cos(theta)) * R v^2 = 2 * g * (1-cos(theta)) * R F(c) is m * v^2 / R F(c) = m * (2gR(1-cos(theta)) / R F(c) = 2gm (1-cos(theta)) normal force is m*g*cos(theta). Contact lost when they're equal (as the normal force starts out greater than necessary F(c) but goes to zero as theta -> pi/2). F(c) = F(n) 2gm(1-cos(theta)) = m*g*cos(theta) 2gm - 2gm cos(theta) = 1gm cos(theta) 2gm = 3gm cos(theta) 2/3 = cos(theta) arccos(2/3) = theta 48 degrees = theta

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