Imagine an infinite collection of nested, concentric spheres, of radius 1, $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, and so on. Suppose they are somehow suspended in space, fixed on their common center $x$. Then the outermost sphere is "released" from its center, and falls vertically under the influence of gravity, while all the other spheres remain "pinned" with their centers on $x$. Next, the top interior of the $r=1$ sphere collides with the top exterior of the $r=\frac{1}{2}$ sphere, knocking it loose from $x$ via a perfectly elastic collision, sending it downward. And so on.
Essentially my question is: What happens? It would be pleasing to understand the behavior of this system without resorting to explicit calculation of all the interactions. Assume the spheres are made of some homogenous, thin material so that their weight is proportional to their surface area (or circumference if you'd prefer to drop down to $\mathbb{R}^2$). I cannot see intuitively the sequence of collisions and overall behavior, and I have not yet tried careful calculations. Perhaps there is a line of reasoning that demystifies the apparent complexities...?
Infinite sequence of nested, falling, colliding spheres
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geometry
classical-mechanics
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0At least you could say that the system will behave symmetrically around the "gravitational axis", so I guess the only difference between the 2D and 3D-cases are the ratios of the masses of the different circles/spheres. – 2011-09-07
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0@Thijs: Yes, the only dependence on dimension is the mass ratios, just as you observe. – 2011-09-07