A $C^2$ curve with constant angular momentum is either a straight line or a circular arc This is a self-answered question. I post it here since it wasn't trivial for me. Alternative solutions are welcomed, of course.
Let $\alpha:(0,L) \to \mathbb{R}^2$ be a $C^2$ curve  satisfying $|\dot \alpha|=1$, and assume that $\alpha(t) \times \dot \alpha(t)$ is constant.
Must $\alpha$ be either a circular arc, or affine?
This is not true if we only assume $\alpha \in C^1$; in that case $\alpha$ can alternate between a circular arc and a tangent polygon to a given circle.
 A: I prove below that $
\{t \, | \, \ddot \alpha(t) \neq 0 \} \subseteq \{ t\,|\,\alpha(t) \perp \dot \alpha(t)\}$.
In particular, on any open interval where $\ddot \alpha \neq 0$, $\alpha$ has a constant norm. Since $\{t \, | \, \ddot \alpha(t) \neq 0 \}$ is open, it is a union of open intervals, so $\alpha$ has a constant norm on it. Thus,
$$
 (|\alpha|^2)''=2(1+\langle \alpha, \ddot \alpha\rangle).
$$
Set $f:=\langle \alpha, \ddot \alpha\rangle$. We showed that $\ddot \alpha(t) \neq 0 \Rightarrow |\alpha|=\text{const} \Rightarrow  f(t)=-1$. Clearly
$ \ddot \alpha(t) = 0 \Rightarrow  f(t)=0$. So
$$
\{ \ddot \alpha \neq 0 \}=f^{-1}(\{-1\}),\,\,\,\{ \ddot \alpha = 0 \}=f^{-1}(\{0\}),
$$
Since by assumption $f$ is continuous, its level sets are closed. In particular, the set $\{ \ddot \alpha \neq 0 \}$ is clopen, hence it must be empty or everything.
If it's empty, then $\alpha$ is affine; If it's everything, then $\alpha$ has a constant norm, i.e. it lies on a circle,  and since it's connected, it must be a circular arc.

Proof that $
\{t \, | \, \ddot \alpha(t) \neq 0 \} \subseteq \{ t\,|\,\alpha(t) \perp \dot \alpha(t)\}$:
Differentiating $\alpha \times \dot \alpha$, we get $
\alpha \times \ddot \alpha=0$, so if $\ddot \alpha(t) \neq 0$, then
$$\alpha(t) \parallel  \ddot \alpha(t). \tag{1}$$
Differentiating $|\dot \alpha|=1$, we get $$\dot \alpha \perp \ddot \alpha. \tag{2}$$
From equations $(1),(2)$ we deduce that
$$
\ddot \alpha(t) \neq 0 \Rightarrow  \alpha(t) \perp \dot \alpha(t).
$$
