Balancing sphere A cube is balanced on the very top of a sphere, and is given a small push in some direction, at which point it begins sliding down that side of the cube (due to gravity).
Now, I want to keep the cube on top of the sphere without touching the cube. How can I move the sphere around in a horizontal plane so that the cube will not lose contact with the sphere at any point?
I imagine there might be an infinite number of ways to do it; for example, oscillating the sphere back and forth so that it is exactly 180 degrees out of phase with the cube's oscillation might work; however, I can't imagine how I would actually prove this.
 A: The easiest way, though it may not be what you want: make the sphere and cube out of rubber (so it doesn't slide) and choose a large enough sphere.
The constraint of rolling without slipping can get rather complicated in 3D, but we can simplify things to the 2D case. 
First consider a circle radius $1$ fixed at the origin. A square side length $2a$ moves without slipping OR ROLLING (changing angle) on top of it. Then points on the circle trace out $(\cos(\theta),\sin(\theta))$ while the center of the cube is at $(-\theta,1+a)$.
Rotate the whole system by $\theta$ clockwise, and you find that the circle is now fixed, and the cube is now at position:
$(-\theta\cos(\theta)+(1+a)\sin(\theta),\theta \sin(\theta)+(1+a)\cos(\theta))$. Expand the $y$ position in a power series:
$$\begin{align} y& =\theta (\theta+\cdots)+(1+a)(1-\frac{1}{2}\theta^2+\cdots) \\
&=(1+a)+\theta^2\left(1-\frac{1+a}{2}\right)+\cdots \\
&=C+\theta^2\left(\frac{1-a}{2}\right)+\cdots
\end{align}$$
if $1>a$, the square's center of mass will have to lift up for it to roll. Since gravity pushes the center of mass down, that means that if $1>a$ it will be a stable equilibrium.
Other than that, you might want to look at the theory of small oscillations: like Parametric resonance. You can make some unstable systems stable by varying parameters in time.
