What's the angular momentum of a two point system of unit mass, velocity and separation when the points are orthogonal and after time t? Question
Let the universe comprise two points travelling orthogonally with separation distance $1$, speed $1$.  In the absence of gravity, what is the angular momentum of one around the other at time $0$ and after time $t$?
Discussion
As time goes on, the line between them moves away from being orthogonal to the motion, thereby reducing the angular velocity.  But counter to this effect, their separation increases which increases the radius of rotation, thereby increasing the moment of intertia.  Does either effect dominate or are they balanced?  As I understand it, the claim in the comments here requires that they be perfectly balanced.  Is this correct?
Attempt
About 35 years since I did this but I think angular momentum as they travel orthogonally is $I\omega=\dfrac1{2\pi}$
Then after time $t$ I have $I\omega=mr^2\dfrac{d\theta}{dt}$
$\dfrac{d\theta}{dt}=\dfrac{v\sin\theta}{2\pi}$
and $\sin\theta=\dfrac1{\sqrt{1+t^2}}$
So that gives $I\omega=\dfrac{\sqrt{1+t^2}}{2\pi}$
And that all appears to check out nicely because it generalises the original expression for $t=0$. (Boom I've still got it).
Is that all correct?
 A: I'm a little confused by your notation in the comments, but I'll put up an answer showing how it's supposed to work, and I can try to edit it if it's not quite the situation you're considering.
Suppose at $t = 0$ we have particle 1 at rest at $(0,0)$ and particle 2 at $(1,0)$ moving with $\vec{v} = (0,1)$.  This means that at a later time $t$, the vector pointing from 1 to 2 is $\vec{r} = (1,t)$.
The angle between $\vec{v}$ and $\vec{r}$ is given by $\tan \theta = t$.  To see this, draw a right triangle with vertices at particle 1, particle 2, and $(1,0)$;  the legs have lengths $1$ and $t$ and the hypotenuse has length $r = \sqrt{1 + t^2}$.  If we differentiate both sides of our equation for $\theta$ with respect to $t$, we have
$$
\frac{1}{\cos^2 \theta} \frac{d\theta}{dt} = 1.
$$
But $\cos \theta = 1/r$ from that same triangle, and so we have
$$
\omega = \frac{d\theta}{dt} = \cos^2 \theta = \frac{1}{r^2}.
$$
This means that the angular momentum is $L = m r^2 \omega = m r^2 (1/r^2) = m$, which is constant.
A: In the absence of gravity place a coordinate system in the middle of the line that separates the two paths. Each line (path of object) is a fixed distance $h$ from the coordinate system

At least I think this is the setup you are describing.
Now angular momentum for one particle summed at the origin is
$$ \boldsymbol{L}_0 = \boldsymbol{r} \times m \boldsymbol{v} $$
which results in the magnitude
$$ L_0 = h\,m v $$
regardless of where along the path $\boldsymbol{r} = \boldsymbol{h} + \boldsymbol{v}\,t$ the particle is, since any part of $\boldsymbol{r}$ that is parallel to the velocity $\boldsymbol{v}$ is ignored by the cross product.
Now the same angular momentum can be used in $L_0 = I_0 \omega$ to find the apparent rotational speed that each particle has. Here $I_0 = m r^2 = m ( h^2 +(v\,t)^2 )$ and so you have
$$ h\,m\,v = m ( h^2 +(v\,t)^2 ) \omega $$
which is solved for $$ \omega = \frac{h^2}{h^2 + (v\,t)^2}$$
