basic physics

all right.. i've been wasting my mind all summer so this problem is giving me a hard time.

If you release a ball from the height of a circular track (h0=2R) and allow it to start traveling around the loop, at what angle (with respect to the horizontal through the center of the loop) will it leave the loop?

So with a couple lines of work, using conservation of energy and circular acceleration = v^2/R it's easy to get |a|=2g(1-sin(theta)), the vertical component of which is 2g*sin(theta)*(1-sin(theta))
so.. now that I have that, I was thinking that I should set the vertical component equal to gravity, since if gravity was pulling down harder on the particle than what the normal acceleration would be without gravity, then that would mean the particle would leave the track.

That obviously isn't right, though... because at theta=0, there's no vertical component so the ball would fall no matter what... and the same with low angles.


So what am I missing?

For the forces in the radial direction,
F = ma
N + mg sin(theta) = m v^2/R

When ball leaves track, N = 0 ->
g sin(theta) = v^2/R <- A

Convservation of energy yields
mgR = mgR sin(theta) + m v^2/2
gR(1-sin(theta)) = v^2/2
2g(1-sin(theta)) = v^2/R <- B

Combining A and B, one gets
g sin(theta) = 2g(1-sin(theta))
sin(theta) = 2-2sin(theta)
sin(theta) = 2/3