Let $\alpha:(0,L) \to \mathbb{R}^2$ be a $C^1$ curve satisfying $|\dot \alpha|=1$, and assume that $\alpha(t) \times \dot \alpha(t)$ is constant.
Does one of the following hold?
$(1)$ $\alpha$ is affine.
$(2)$ $\alpha$ is a circular arc.
$(3)$ $\alpha$ starts as a circular arc, then at some point becomes affine (i.e. it coincides with the tangent to the circle.)
$(4)$ $\alpha$ starts affine, then at some point it swtiches to a circular arc, and after some time, it can also switch one more time to being affine (a tangent).
Comments:
a. I proved below that the restriction of $\alpha$ to the open set $\{t\, |\,\alpha(t) \not\perp \dot \alpha(t)\}$ is $C^{\infty}$. Since a $C^2$ curve having constant momentum must be either affine or a circular arc, and since a circular arc satisfies $\alpha(t)\perp \dot \alpha(t)$, it follows that $\alpha$ is affine on each connected component of $\{t\, |\,\alpha(t) \not\perp \dot \alpha(t)\}$.
Is there a way to proceed from here? The difficulty seems to be to prove that $\{t\, |\,\alpha(t) \perp \dot \alpha(t)\}$ is a closed interval. Suppose for instance that $\alpha$ starts affine, and that it stays affine up to a certain point $t_0$ satisfying $\alpha(t_0) \perp \dot \alpha(t_0)$. Suppose that $\alpha$ is not affine after $t_0$, i.e. there is no $\epsilon>0$ such that $\alpha|_{(t_0-\epsilon,t_0+\epsilon)}$ is affine. How can we prove that $\alpha(t) \perp \dot \alpha(t)$ still holds for sufficiently small $t>t_0$?
Assuming this is not the case, there exists a decreasing sequence $t_n \to t_0$ satisfying $\alpha(t_n) \not \perp \dot \alpha(t_n)$. Thus there exists $\epsilon_n>0$, such that $\alpha|_{(t_n-\epsilon_n,t_n+\epsilon_n)}$ is affine. However, this fact alone does not imply affinity at $t_0$. Of course, in our case, all these affine parts around $t_n$ must be tangents to a given circle. Can we use this somehow?
b. There is an asymmetry between options $(3)$ and $(4)$. In option $(4)$ there are two "switching points", while in $(3)$ there is only one. Here is why there cannot be more than one switch in the scenario described in $(3)$:
On a region where $\alpha$ is part of a circle of radius $R$ centered at the origin, it satisfies $|\alpha(t) \times \dot \alpha(t)|=R$. Switching from a line to a circle is only possible when the line is tangent to that circle. (since we are $C^1$). Thus, if we start from a circle, and switch to a tangent, we cannot switch again at some point from the tangent to a tangent circle centered at the origin. (the angle at the contact from the origin is not $\pi/2$). Alternatively, after we leave the point of contact of the tangent and circle, the distance from the origin increases, so we cannot switch to a circle centered at the origin, since such a circle would have a different radius, thus violating the constancy of $|\alpha(t) \times \dot \alpha(t)|$.
c. I proved here that if $\alpha$ is assumed to be $C^2$, then the only options $(1)$ or $(2)$ are possible, i.e. $\alpha$ must be either a circular arc or affine. Clearly, a $C^2$ solution cannot alternate between circular and affine.
I don't know how to adapt the proof to this lower regularity setting, since I used there second order derivatives of $\alpha$. I guess that one possibility would be to use some distributional/weak derivatives instead. I think that a good starting point might be to show $\alpha$ is $C^2$ almost everywhere.
Proof that that $\alpha$ is $C^2$ on $\{t\, |\,\alpha(t) \not\perp \dot \alpha(t)\}$ is $C^2$:
Write $\alpha(t)=(X(t),Y(t))$. The conditions $|\dot \alpha|=1$ and $|\alpha(t) \times \dot \alpha(t)|=c$, translate into $$ \dot X^2+\dot Y^2=1, \\ X\dot Y-Y\dot X=c. $$ Thus $$ \dot Y=\frac{c+Y\dot X}{X}, $$ which together with $\dot Y^2=1-\dot X^2$ implies that $$ 0=\dot X^2(Y^2+X^2)+(2cY)\dot X+(c^2-X^2). $$ Since the discriminant $$ (2cY)^2-4(Y^2+X^2)(c^2-X^2)=4X^2(|\alpha|^2-c^2), $$ we find that whenever $|\alpha| > c$, $\dot X$ is a smooth function of $X,Y,c$.
Now, we note that $c=|\alpha(t) \times \dot \alpha(t)| \le |\alpha(t)||\dot \alpha(t)|=|\alpha(t)|$, with equality if and only if $\alpha(t) \perp \dot \alpha(t)$. Thus $$ \{t\, |\,|\alpha(t)|>c\}=\{t\, |\,\alpha(t) \not\perp \dot \alpha(t)\}, $$ and since $\dot X$ (and similarly $\dot Y$) are smooth functions of $X,Y,c$, we conclude that $\alpha$ is $C^{\infty}$ on this set.
The essential ingredient in the analysis is the time-evolution of the determinant $$ D(t) = {\rm det} \pmatrix{ \dot{x} & \dot{y} \\ -y & x}= x \dot{x} + y \dot{y} = \frac12\frac{d}{dt} (x^2+y^2).$$
We have the following properties:
$D(t)=0 \Rightarrow (x,y)\perp (\dot{x},\dot{y})$
$D(t)\neq 0 \Rightarrow$ $(x(t),y(t))$ are smooth functions at $t$ (use e.g. your proof of smoothness). From the conditions on the trajectory we see that $(\ddot{x},\ddot{y})$ is perpendicular to $(-y,x)$ and $(\dot{x},\dot{y})$. These two vectors are independent (since $D(t)\neq 0$) so the acceleration must be the zero vector. We thus have: $$ \ddot{x}(t)=\ddot{y}(t)=0 \ \ \ {\rm and} \ \ D'(t)=+1 .$$
In other words when $D(t)\neq 0$, the trajectory has constant velocity. Since $D'(t)=+1$ whenever $D(t)\neq 0$, the zero-set $\{t\in (0,L): D(t)=0\}$ is either empty, a point or an interval $I$ in $(0,L)$. The first two cases are trivial. In the case of an interval $I$ we have for $t\in I$: $x(t)^2+y(t)^2=r^2$ (constant) so the trajectory describes a circular arc. If $I$ has e.g. a right boundary point $t=b$ in $(0,L)$ then $D(t)=t-b$ for $b<t<L$ and the trajectory here follows the tangent (at constant velocity) of the circle at $(x(b),y(b))$. Similarly for a possible left end-point. So the possibilities are precisely as stated in the list.