What is the minimum height you would have to place a solid ball of mass M and radius R (I = 2MR^2 / 5) on a hill in order for the ball to make it completely around a circular loop at the bottom of the hill? The diameter of the loop is D. Assume the ball rolls without slipping.

Answer 1

Height = H
Ke at bottom = m g H =.5 m v^2 + .5 (2mR^2 / 5) (v/R)^2 = total energy

Now to make it around the loop the velocity, u at the top of the loop must be high enough so centripetal acceleration = g
height = D and radius = D/2
(D/2) omega^2 = g
u = R omega
(D/2)u^2/R^2 = g so u^2 = 2 g R^2/D
total energy = m g D + (1/2)mu^2 + (1/2)(2mR^2 / 5)(u^2/R^2)
= m g D + (1/2)m u^2 + (1/5)m u^2
m g H= m g D + (7/10) m u^2
but we know that to just make it u^2 = 2 g R^2/D
h (H-D) = (7/5) (R^2/D)g
check my arithmetic !!!!!