Not sensitive to initial conditions.
As originally stated, it was. The point of my diagram was to prove that. If the initial conditions were as shown then the system would have stayed in that state.
The question is about the steady stage conditions of the spuinning ball in a flat horizontal plane above the ground.
Ah -- you didn't specify the flat horizontal plane was above the ground. I exhibited the below-the-ground solution.
Size of the ball? Ignore size and treat it as a point. Assume the wire has zero mass.
That changes the system to this:
Code:
Motor
SSS
SSS
SSS
SSS
w
w
w
w
w
w
gggggggggBggggggg
ggggggggggggggggg
Both the wire and the ball have zero moment of inertia, so they spin up instantly. The steady state condition is the ball rotates on its axis, (if that is even meaningful for a point ball -- perhaps it has quantum spin like a real point mass: an electron), in a flat horizontal plane exactly at ground level. So why would the steady state condition have the plane of rotation above the ground -- why would the ball ever lift off? The forces on it are balanced.
For extra credit if the wire breaks how long does it take for the ball to hit the ground?
You're introducing self-contradictory conditions: if you let the wire break, the breaking event is evidently not a steady state, which means there is no steady state and the answer will necessarily depend on initial conditions and transients.
Assuming you set the initial conditions in some way that causes the ball to fly in a helix above ground level, accelerating until the wire breaks, how long until the ball hits the ground -- if it ever does -- will depend on the breaking strength of the wire. Weak wire: a little under half a second. Strong wire: a bit longer. (Assuming a spherical Earth. If you do the experiment in a valley it will hit the ground sooner.) Sufficiently* strong wire, and assuming you're raising the angular velocity far above 100*2*pi radian/second in your futile attempt to get the rotation plane up to 1 meter high: the ball reaches orbital velocity and never hits the ground. (There's no atmospheric drag on a point mass.)
(* Back of the envelope, the ball would need to be pulling about 3,000,000 "g"s.)
