I did work out the shape of a rotating bar in a moving reference frame via the Lorentz transform. This is simpler than the see-saw case to analyze, but one can gain some insight of the see-saw case from the rotating bar case.
It's convenient to write the coordinate of any point on the bar as a function of r, which we set to zero at the origin of the bar (the origin being the point on the bar with zero proper acceleration). We will call the inertial frame of reference in which the origin of the bar is at rest "the rest frame of the bar", even though only the origin of the bar is at rest in this frame. r is a parameter that picks out a specific point on the bar. We also need a time coordinate, which we will take as the proper time ##\tau## of the point we just specified by specifying r. We will let the angular frequency of the rotation be ##\omega## in the bar's rest frame. Then in the bar's rest frame, we write:
$$x = r \cos \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad y = r \sin \frac{\omega \tau}{\sqrt{1-r^2\omega^2}} \quad t = \frac{\tau}{\sqrt{1-r^2\omega^2}}$$
Applying the Lorentz transform, we can transform these coordinates to a moving frame. Let the velocity be determined by the dimensionless parameter ##\beta##, so that v = \beta c. And let ##\gamma = 1/\sqrt{1-\beta^2}##. Then we can find the position of a point on the bar in a moving frame with coordinates x1, y1, t1 via the Lorentz transform.
$$x1 = \gamma \left(x + \beta\,c\,t \right) \quad y1=y \quad t1 = \gamma \left( t +\beta x/c \right) $$
We can substitute the expressions for ##x(r,\tau)##, ##y(r,\tau)##, and ##t(r,\tau)## from the first set of equations into the second to find ##x1(r, \tau), y1(r,\tau), t1(r,\tau)##. It's also convenient to set c=1 at this point, unless one really wants to keep tract of it (I did not).
We now wish to find and plot x1 and y1 as a function of t1, rather than as a function of ##\tau##. To do this, we need to solve the equation
$$
t1 = \frac{1}{\sqrt{1-\beta^2}} \left( {\frac {\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}}+\frac{\beta}{c}\,r\cos
\left( {\frac {\omega\,\tau}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right)
\right)$$
for ##\tau## as a function of r and t1. This is not something that has a closed form solution. We will call this ##\tau = f^{-1}(r, t1)##
Then we can write the coordinates x1 and y1 of the bar at time t1 as a function of r, where I have now omitted the factors of c:
$$ x1 =
\left( r\cos \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) +{\frac {\beta\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) {
\frac {1}{\sqrt {1-{\beta}^{2}}}}$$
$$ y1=r\sin \left( {\frac {\omega\,f^{-1}}{\sqrt {1-{r}^{2}{\omega}^{2}}}} \right) $$
This would be quite messy to carry out by hand - I used a computer algebra package to do the algebraic manipulations and to do some graphs for ##\beta = .9## and ##\omega = .5## with r varying from 0 to 1. While I don't have a convenient way of posting the results, I can say that the bar is generally not straight (there's an exception when it's vertical), that one can see the effects of time dilation (the bar rotates more slowly), and that the bar is length contracted as well as bent when it's not vertical.
As I mentioned in a previous post, this is more or less to be expected. While the Lorentz transformation is a linear transformation and must map straight lines into straight lines, the worldlines of points on the bar (a fixed value of r as ##\tau## varies) are not straight lines for r>0. So we don't expect the bar to be straight, and we don't expect the bar to have a constant length, and the calculations demonstrate this.
I like this approach because a) I'm used to it, and b) the coordinates r and ##\tau## have physical significance, being the proper distance of a point on the bar from the origin in the former case, and being the proper time of a point on the bar in the later case. Both the proper distance and the proper times are independent of any coordinate choices, so it's a convenient representation of the spinning bar.
