While studying solids of revolution at college, I came across a problem related to physics that seems to have an answer difficult to prove mathematically, which I have not been able to obtain.


Consider the surface $z=|x|^n+|y|^n$ for $0 \le z \le h$ and $n \ge 1$. If this surface were a physical object standing in a table, it is very clear that for values of $n \approx 1$ it will be in an unstable equilibrium, while at great values, such as $n=10$ it will be at an stable equilibrium, that is, the object may or may not tumble/fall under any force $F>0$ depending on the value of $n$. The question is then at which value of $n$ does it transition from unstable to stable?

enter image description hereenter image description here $n=1.5$ and $n=10$

Problem statement

From the motivating problem I was able to obtain the equation below. Solving for $n$ as $a \rightarrow 0$ gives the answer, for any constant $c>0$

in the physical problem, $c$ is the z-coordinate of the center of mass of the object and $a$ is associated with at which point it will tumble. Taking $a \rightarrow 0$ provides a situation in which it may fall if moved by any infinitesimal amount, which is an unstable equilibrium

$$\left (c-a^n \right )na^{n-2}=1$$ However, there seems to be no feasible way to isolate $n$ in order to apply the limit $a \rightarrow 0$. Using Wolfram Alpha, it seems the value of $n$ approaches $2$ by using $a \approx 0$, but I see no way to prove it.

  • $\begingroup$ I think it should not be $2$ in general, since let's just assume that $n=2$, then what will happen if we set $n=2$ and then take the limit as $a\rightarrow 0$. In this case, you will end up with something like $2c=1$. Thus, this implies that the $z$-coordinate of the center of mass is $\frac{1}{2}$ which I doubt is true since you are cutting off the $z$ value at an arbitrary height $h$. $\endgroup$ Aug 31, 2023 at 16:32
  • $\begingroup$ I did not understand the exact meaning of $a$. For example, if $n=1$, then your equation has the solution $a = c/2$. What does that mean? $\endgroup$ Aug 31, 2023 at 17:57
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    $\begingroup$ Please edit your question to provide a more meaningful and informative title. Thank you. $\endgroup$
    – D.W.
    Aug 31, 2023 at 18:30
  • $\begingroup$ Here is a solution that someone can formalise. Let $c\mapsto2c$ and rewrite as $a^n=c\pm\sqrt{c^2-a^2/n}$. Taking the positive root leads to $n\to0$ as $a\to0$. Taking the negative root, a series expansion gives $a^n=a^2/(2cn)+O(a^4)$ whence $n\to2$ as $a\to0$. There are important pieces of rigour needed between each of these steps, including justifying whether the limits come from above or below. $\endgroup$
    – TheSimpliFire
    Aug 31, 2023 at 19:18
  • $\begingroup$ I have used ContourPlot from Mathematica, and it looks like if $c < 1/2$, then $n$ converges to $2$ from the left as $a\to 0^+$. $\endgroup$ Aug 31, 2023 at 20:46

1 Answer 1


To avoid problems with noninteger powers of negative numbers, we assume $a>0$. Let $\varepsilon\in (0,1]$ be any number. If $n\ge 2+\varepsilon$, $a<\min\left\{e^{-1/2}, (c(2+\varepsilon))^{-1/\varepsilon}\right\}$ then $\left (c-a^n \right )na^{n-2}<cna^{n-2}\le c(2+\varepsilon)a^{\varepsilon}\le 1.$ The previous to the last inequality holds because the function $cna^{n-2}$ has the derivative $ca^{n-2}(1+n\ln a)$, and so is decreasing when $a$ is a fixed numbers less then $e^{-1/2}$. On the other hand, if $n\le 2-\varepsilon$, $a<\min\left\{1, \left(\frac c2\right)^{1/\varepsilon}\right\}$ then $\left (c-a^n \right )na^{n-2}\ge \frac{c}{2}a^{-\varepsilon}>1$. It follows $n\to 2$ when $a\to 0$.


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