Spinning squares and the balls that bounce in them

Suppose you have a unit square that’s rotating about its center at a constant angular speed, and there’s a moving ball inside. The ball has a constant linear speed, and there’s no friction or gravity. When the ball hits an edge of the square, it simply reflects as though the square is momentarily stationary during the briefest of moments they’re in contact. Also, the ball is not allowed to hit a corner of the square—it would get jammed in that corner, a situation we prefer to avoid.

Suppose the ball travels on a periodic (i.e., repeating) path, and that it only ever makes contact with a single point on the unit square. What is the shortest distance the ball could travel in one loop of this path?

Take a unit square that is rotating at a constant angular speed, $\omega,$ where, without loss of generality, we assume that the sides of square are perpendicular to the $x$- and $y$-axes at time zero. Further assume that that for any $n \geq 2,$ there is a ball traveling at some fixed velocity $v_n$ in a periodic path with $n$ bounces against the side of the spinning square. Since the point of contact is tracing out a circle of radius $\frac{1}{2}$ and the path of the ball is some $n$-gon inscribed in that half-unit circle. Since the ball and the point of contact are each traveling at constant speeds, the length of time between collisions and hence length of the sides of the $n$-gon must all be the same. Thus, a ball tracing a periodic path must trace out a regular $n$-gon inscribed in the half-unit circle centered at the center of the unit square.

The side length of a regular $n$-gon inscribed in a half-unit circle can be given by the law of cosines \begin{align*}s_n^2 &= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 2 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \cos \frac{2\pi}{n}\\ &=\frac{1 - \cos \frac{2\pi}{n}}{2}\\ &= \sin^2 \frac{\pi}{n}.\end{align*} Therefore the length of this periodic path is $$\ell_n = n s_n = n \sin \left( \frac{\pi}{n} \right).$$ We see that since $\sin x = x - \frac{x^3}{6} + O(x^5)$ that $$\ell_n = \pi - \frac{\pi^3}{6n^2} + O(n^{-4}),$$ so $\ell_n \uparrow \pi$ as $n \to \infty.$ So we can get that the smallest such path is $\ell_2 = 2,$ which represents the path that traces the line segment from the midpoints of the top and bottom sides of the square twice (down and back).

If we were to look for the next longest path, it would be the equilateral triangle of side length $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}.$ The total travel distance of this equilateral triangular path is thus $$\ell_3 = \frac{3\sqrt{3}}{2} \equiv 2.598076211\dots.$$