An object is sliding from top of a ball to the ground. This object starts from rest and slides without friction. At what height will the object get apart from the ball's surface if the radius of the ball is r meters?
Potential energy will be measured with respect to
the center of the ball. Let t be the angle, measured
at the center of the ball, from the object's initial
position to its current position. If m is the object's
mass, v its velocity, and g the accelertion due to
gravity, then the object's
kinetic energy + potential energy
at its current position is
(1/2)*m*v^2 + m*g*r*cos(t)
This quantity is the same for its initial position
and its current position. Therefore,
(1/2)*m*v^2 + m*g*r*cos(t) = m*g*r (1)
The object will lose contact with the ball when the
normal force equals the centrifugal force,
m*g*cos(t) = m*v^2/r (2)
Solving (1) and (2) gives
cos(t) = 2/3
when the object loses contact with the ball. Thus,
the height is
r + r*cos(t) = (5/3)*r

Posted by Bractals
on 20060801 22:06:57 