This tweet claims to give an explanation for why one should expect the perimeter of a circle with a rational radius to be irrational. It doesn't strike me as that convincing (although feel free to disagree).
A simple refutation would be to exhibit a smooth curve which is never straight, and yet has a rational arc length. Furthermore it needs to clearly not be smuggling any irrationality into its parameterization, so it should at least start and end at rational coordinates. (This second requirement is definitely a bit soft).
So, what's the simplest curve one can come up with which fits the criteria?
Here's a simple family of curves such that the arc length over any interval $[a, b]$ with rational endpoints (and, say, $0 < a < b$) is rational.
Fix $n \in \{2, 3, 4, \ldots\}$ and $c \in \Bbb Q \setminus \{0\}$ and define $$f(x) = c x^{n + 1} + \frac{1}{4 (n^2 - 1) c x^{n - 1}} .$$ In particular, the endpoints $(a, f(a))$ and $(b, f(b))$ are rational. The arc length element for the graph of $f$ is $$ds = \sqrt{1 + f'(x)^2} \,dx = \left((n + 1) c x^n + \frac{1}{4 (n + 1) c x^n}\right) dx,$$ and so the arc length of $f$ over $[a, b]$ is $$\int ds = \left. c x^{n + 1} - \frac{1}{4 (n^2 - 1) c x^{n - 1}}\right\vert_a^b = c (b^{n + 1} - a^{n + 1}) - \frac{1}{4(n^2 - 1)c}\left(\frac{1}{b^{n + 1}} - \frac{1}{a^{n + 1}}\right),$$ which in particular is rational.