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I've been thinking about polyhedrons, when placed on a table on a certain face, will tip over and keep tipping over infinitely. I'm trying to prove mathematically that such a polyhedron doesn't exist.

Rules: the polyhedron does not need to be uniform nor convex, but at any point the mass there is non-negative. this means you can have holes and caves and points and fun stuff.

Right now I'm stuck because I'm not sure what determines if the polyhedron tips over. Either it's 1. The COM of the polyhedron, when projected onto the table, lies outside of the polygon face touching the table, and 2. The gravitational potential energy, mgh, is lowered by tipping over.

I understand with some basic geometry that 1 implies 2, but condition 2 does not necessarily imply that it will tip over.

How do I proceed to prove that such a perpetual polyhedron doesn't exist?

Edit: Thought about it some more and solved my question. 1 and 2 are equivalent in the real world because if the COM was above the base face, then tipping over would require that mgh first increase before finally adopting its final value after tipping over. If the COM is not above the face then there is no necessary activation energy. Therefore, the polyhedron tips over if and only if the energy strictly decreases, which happens if and only if the COM lies outside the face. After this process the potential energy decreases.

Now assume that the polyhedron spontaneously remains in perpetual motion. Then the energy is strictly decreasing; but the potential energy clearly has a minimum value, so it can't go on forever.

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  • $\begingroup$ As far as when the polyhedron tips over: Of your two options, 1 implies 2. I believe the correct one is 1. The states in which 2 implies it should tip, but it won't actually tip, are those where an intermediate state has a higher mgh than either the initial or final states. John Bentin uses the fact that 1 implies 2 in his answer, because 2 is a more useful thing to say about the tipping state in this context. $\endgroup$ – Aaron Dufour Jan 16 '15 at 21:52
  • $\begingroup$ I think the question might be missing a qualifier: "The COM of the polyhedron, when projected onto the table, lies outside of the polygon face touching the table" should be "The COM of the polyhedron, when projected onto the table, lies outside of the convex hull of polygon faces touching the table": example: a chair's center of mass is not over any part touching the floor. (Or I may have misunderstood you) $\endgroup$ – Mooing Duck Jan 17 '15 at 1:21
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The convex hull of the polyhedron will roll on a horizontal plane spontaneously, from being seated on one face to being seated on another, only if (not necessarily if) the centre of mass of the polyhedron is lowered in doing so. Since the convex hull of the polyhedron has only a finite number of faces, there can be only a finite number of such lowerings, and so the process must terminate in a finite number of steps.

Edit: The above answer considers only the static aspects of the problem. If we allow dynamics into the picture (which is realistic), while (unrealistically) assuming zero friction or other resistance, then a polyhedron can indeed rock backwards and forwards, or roll, forever. If we are fully realistic, perpetual motion cannot occur, for reasons of friction alone.

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  • $\begingroup$ Friction would be the major component for reasonable time standards. But even if we had zero friction, eventually all the particles in the polyhedron would decay. $\endgroup$ – Hayden Jan 17 '15 at 0:11
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It will tip over when the COM is outside the convex hull of the support area. For a polyhedron, the convex hull will be a convex polygon. The COM can be lowered by pivoting around the nearest edge of the convex hull.

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