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?

Okay. Now I'm probably over/underthinking this but I'm just going to try to see when the parabola the particle would take if the circle disappeared travels under the circle... bleh my brain no work good tonight

I'm not really sure where you're defining theta in this problem, but should find where the force normal to the track becomes vanishes, which will occur for the top half of the circle if the velocity is zero (so, the starting condition) or it leaves the track.

Ohhhh, I just remembered about curvature. That's what I have been trying to do...

So just force the curvature of the circle to be greater than that of the ball if it were free

Theta is the angle the ball makes with respect to the center of the circular track, measuring from the horizontal. So, taking normal convention, if it's on the right hand side of the track the angle is 0, if it's at the top its angle is pi/2...

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

>>6
We got the same answer. What I used was:

v(theta)=sqrt(2*g*R*(1-sin(theta)))
vx(theta)=-v(theta)*sin(theta)
vy(theta)=v(theta)*cos(theta)
curvature_free_ball=-vx(theta)*g/(vx(theta)^2+vy(theta)^2)^(3/2)
cuvrature_loop=1/R
solve(curvature_free_ball(theta)=curvature_loop(theta),theta)
wow i shouldn't have put theta after all the functions, but yeah. it also works out to
theta=.73 rads


thanks :D

Basics?

Lol, i don't understand a thing.....