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I have a physics question for you. Let us say a polygon shaped object is stable on a side when the center of mass "falls" inside the base. Is it possible in 2D to build an object that is unstable on all sides? Intuitevely I would think no, but I would like to see a proof. Alternatively, what is the max number of unstable sides for a polygon with N sides?

How about in 3D?

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    $\begingroup$ I kinda think this is for MathSE, at least the second part, regarding a proof, rather than experimental verification. $\endgroup$
    – user108787
    Apr 28, 2016 at 20:24
  • $\begingroup$ Yeah i think you have a point, but as long as we are not talking about N-dimensions, a physicist should be able to answer the question just fine :) $\endgroup$
    – Stefano
    Apr 28, 2016 at 20:31
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    $\begingroup$ No it is not possible, its existence would be equivalent to a perpetum mobile $\endgroup$
    – user83548
    Apr 28, 2016 at 20:41
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    $\begingroup$ As you can see, a physicist just did. @brucesmitherson is correct. No such object can exist, otherwise one could extract endless energy from it rolling around. A sphere on a plane, by the way, is only borderline stable. It takes only an infinitesimal amount of force to get it moving from any position. Now you can generalize that: how much force does it take to get an approximation of a sphere moving? As little as you like, you just have to make it smooth enough. $\endgroup$
    – CuriousOne
    Apr 28, 2016 at 21:05

2 Answers 2

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People have constructed three-dimensional convex objects of constant density with only one stable point. These are called gömböcs (if you want to be pedantic, the Hungarian plural is gömböcök). It is straightforward to approximate these by polyhedra and obtain polyhedra with only one stable side.

Polygons and polyhedra must have at least one stable side. Suppose we have a polygonal or polyhedral object that is unstable. Put it on a flat surface, one side down. It can only topple onto another side if the center of mass is lower on the second side.

Thus, if we choose the side that makes the center of mass lowest, that side is stable.

Polygons need to have at least two stable sides. Here is a proof that any planar convex object must have at least two stable balancing points. If the object is a polygon, these stable points must be sides.

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About second part of your question, I should say that I couldn't understand it because it may the polygon like a star has no side contacted with the ground.

About first part of your question, I should say "It is not possible".

“Let us say a polygon shaped object is stable on a side when the center of mass "falls" inside the base”. If you accept this phrase without any more explanation, then we have easy job.

When we talk about a polygon it is clear that we have more than 2 vertices. It is enough to prove that in any arbitrary polygon, certainly 2 vertices exist with below conditions:

"Those 2 vertices have contact with the ground at the same time, and if we draw 2 lines parallel with the gravity and pass from them; then center of mass is between those 2 lines" enter image description here

I hope you could understand what I mean. My English is terrible

OK. Assume that we have a polygon on the ground that 2 vertices of it (we name them A and B) have contact with the ground and we have drawn 2 lines as describe above. Three situations are possible.

  1. CM is between A and B. So, the problem is solved.
  2. CM is in the left side of the A.
  3. CM is in the right side of the B.

enter image description here

1 and 3 have same explanation. So, assume that CM is in the left side of the A. We will prove that at least 1 vertex of the polygon exists in the left side of the CM (relative to A). If we prove that topic, then problem will be solved clearly.

We use reductio ad absurdum. Assume there is no vertex in the left side of the CM. If we choose A as the origin, then it is clear that x component of CM position vector is negative. But this is impossible.

So, we have at least 1 vertex in the left side of the CM and proof is finished.

If you want a proof for existence of 2 vertices of an arbitrary polygon that have contact with the ground at the same time (You may say it is possible that at each moment we have 1 vertex that has contact with the ground), then I should say I cannot wrestle with dictionaries anymore :)

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  • $\begingroup$ This is not a mathematically rigorous proof. You've shown that if a polygon falls to the left, then it has a vertex to the left of the edge it's sitting on. But you haven't shown that there is a stable edge containing that vertex. In fact, there might not be—the stable edge might contain two different vertices. $\endgroup$
    – Peter Shor
    Jun 19, 2016 at 12:09
  • $\begingroup$ @PeterShor The OP has accepted this statement: _ Let us say a polygon shaped object is stable on a side when the center of mass "falls" inside the base_ and I have used that to prove the main statement. $\endgroup$
    – lucas
    Jun 19, 2016 at 13:56
  • $\begingroup$ I have linked a rigorous proof of a stronger theorem from my answer. $\endgroup$
    – Peter Shor
    Jun 19, 2016 at 19:20

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