You might want to specify about exactly what your background is and what you want to learn in order to get more effective answers.
But given what you've asked for, I'd say the best place to learn about chaos and dynamical systems is absolutely: Steven Strogatz' Nonlinear Dynamics And Chaos: With Applications To Physics, Biology, Chemistry, And Engineering
If you already know what's in this book, you need to re-phrase the question; if you don't, this is what you need to know and where to learn it.
Enjoy - I think this book is wonderful!
There is another stable solution of total length $3\pi + h + \epsilon$ for any $\epsilon > 0$. We take a circle of constant latitude $\delta < 0$ (sufficiently small) and then connect this circle to the north pole via two diametrically opposed strings. This is then clearly stable. Furthermore this is equivalent to your solution (plus Scott Carnahan's epsilon-modification).
However, there is in fact a stable solution of total string-length $2\pi+h+\epsilon$. We simply take the almost-equatorial circle in the last solution and drag it to the south pole, so that we have a small circle there along with two diametrically opposed support strings connecting it to the north pole. This solution is (barely) stable, although physically it is not exactly easy to implement (a very small but non-negligible disturbance will suffice to remove the ball).
The reason for the two answers: this was my thought process in action.
EDIT: Please disregard the above; both of these solutions are unstable.
In fact Scott's example is part of a general class of solutions of total length $3\pi+h+\epsilon.$ Take any two diametrically opposite points, and draw a small spherical triangle containing one of the points. Connect all six possible edges between the four points by geodesics, and finally rotate the arrangement so that one of the strings passes through the north pole.
Here is a short proof that any stable configuration must have at least four points where three or more strings meet. If there are three or less points, there is a hemisphere $H$ which contains all of the points. Take the complement $H^C$ of this hemisphere; we can remove any strings in $H^C$ because they cannot be geodesics. As $H^C$ doesn't contain the north pole, the sphere can fall out.
Note that if any strings are not geodesics between junctions, we can ignore the strings.
Here is a short proof (that is not quite rigorous) that could provide a lower bound. Suppose there is not a loop (that is, a set of points connected by geodesics) that is completely contained in the southern hemisphere. Then we may drag any points in the southern hemisphere into the northern hemisphere. (This statement is the part I can't make completely rigorous.) So the sphere must be unstable. Now if we have a loop in the southern hemisphere, there must be at least two strings meeting at the north pole, as otherwise we could simply slide all of the string off one side of the sphere.
So we must have a loop in the southern hemisphere and (not necessarily direct) connections from at least two of these points to the north pole. I can't figure out how to work a good lower bound from here.
Best Answer
At least one integrable case is known, the wedge billiard with an angle of $45^\circ$, see the first paper on this system:
But, as usual in dynamical systems, completely regular cases are very special, so other examples are probably few/rare.
A link to the previous question: Billiard dynamics under gravity: a further beautiful example is given by Robert Israel.