Is it possible to determine maximum frequency of a disc using a relative circumference? On Wikipedia there is an article that shows how to calculate the relativistic effects on a rigid spinning disc, I am referencing the formula under the "Ehrenfest's argument" section in the link.
The formula, as I interpret it, shows how the circumference of the disc changes as speed of rotation changes. Higher the speed, bigger the circumference.
If you derive a new circumference and you have a velocity you can determine frequencies. $$Velocity/Relative Circumference=Frequency$$  
With significant relativistic effects needing a good fraction of the speed of light you can spin a disc up really fast while increasing its frequency. At one point though the Circumference "grows" faster than the velocity of its spin and the frequency slows down.
Playing around with the formulas maximum frequency appears at velocity equal to $1/sqrt(2)*c$  
Let me know if this formula is doing what I am interpreting it to be doing? Can it be used like this?
You can see the plot with frequency on the y axis and velocity on the x axis in the link below.
Wolfram
 A: I believe that the frequency for the rotating disc of radius $r$ can be anything below $1/r$ (or $c/r$ in SI units).
I think the picture you consider is flawed because you do not take into account the measurement of the speed by the observer on the disc. He would have to time the passing by 2 measuring roads at the previously known laboratory distance ($v = L / \Delta t$) — but the measured time will appear dilated with respect to laboratory frame! So, the Lorentz factors will cancel out in your formula.
While researching the Ehrenfest's paradox, I've found an interesting analogy between it and Bell's spaceship paradox on the internet: http://www.physicspages.com/2015/02/26/lorentz-contraction-in-a-rotating-disk-ehrenfests-and-bells-spaceship-paradoxes/. I believe, this might provide you some ideas.
Now please, bear with me for a minute, I have a story to tell. This is how I reached those conclusions, I find this quite entertaining.
From the point of view of an observer riding the disc, disc's geometry is unchanged although a very strong force points towards the rim of the disc. Basically, he feels gravitational potential $U(r) = -\frac12 r^2 \omega^2$ where $\omega$ is the laboratory frame rotation frequency (see https://physics.stackexchange.com/a/53771/119172 and https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html for details):
$$ ds^2 = -(1 - r^2 \omega^2) dt^2 + dr^2 + r^2 d\phi^2 + dz^2 $$
A cool feature of this situation is that here, like in the case of the Schwarzschild black hole, we have an apparent singularity at the distance $r = 1 / \omega$. And if you compute the scalar curvature in this metric, you will get
$$ R = -\frac{2 \omega ^2 \left(r^2 \omega^2 -2\right)}{\left(r^2 \omega ^2-1\right)^2} $$
We see that at the curvature is singular at $r=1 / \omega$ providing us a physical singularity unlike the one in Schwarzschild's case.
What does it mean? As the observer does not known a thing about him being rotating, he can interpret the pull towards the rim as if something massive was surrounding the disc (although symmetric mass distribution should have zero net force). And at some distance this force goes through the roof as if there was a gorgeous cylindrical black hole!
On the other hand, lets go back to the laboratory frame. Say, we have this disc of the radius $r$ and we want to spin it up to frequency $\omega = 1 / r$. What linear speed will correspond to that? Well, $v = \omega r = 1$. So, the rim of the disc would have to move with the speed of light!
