The equations you are looking for are :

* Centripedal Force -- ${m v^2}\over{r}$

  Where $m$ is the mass of the line, equal to ($\rho_{line} \pi r_{line}^2 l$), and $r$ is the radius of the rotation, from the center of rotation (which is not necessarily the same thing as the line length, $l$.

  For the sake of an example, let's assume the line is tungsten ($\rho$ = 19,300 ${kg}\over{m^3}$ ), and this is true monoatomic line with a [radius][1] (for tungsten) of $193 \times 10^{-12} m$

* Air Resistance -- ${1\over2}\rho A C_d v^2$, 
  
  where $\rho$ is the density of air (1.225 ${kg} \over {m^3}$ ), $A$ is the area of the line relative to the wind ($d_{line} \times l$), $C_d$ is the drag coefficient of the line (let's assume its designed to have a low drag profile; I see drag coefficients as low as 0.22 for some cars)

For the line to ever be taught, the derivative of Centripedal Force (CF) with respect to velocity ($v$) must be greater than the derivative of Air Resistance (AR) with respect to $v$.

* ${{\delta CF} \over {\delta v}} = {{2mv} \over r} $
* ${{\delta AR} \over {\delta v}} = {\rho A C_d v}$

Plugging in the values we know --

* ${{\delta CF} \over {\delta v}} = {{2 \times 19,300 \times \pi \times (193E^{-12})^2 \times 1 v} \over 1} = 1.44 \times 10^{-15} v$ 
   
   (using an arbitrary 1 meter radius and line for this example)

* ${{\delta AR} \over {\delta v}} = {1.225 \times 193E^{-12} \times 1 \times  0.22 \times v} = 5.2 \times 10^{-11}$

**So, no. Not exactly as written.** 

The monomolecular line doesn't have enough mass to overcome air resistance. However, you might have a weight built-in on the far end to give the line tension and overcome air resistance.

In that case, the next trouble will be the strength of the line. I'm having trouble finding good strengths for tungsten, but see enough references to 500 MPa, to use that.

The equation in charge here is :

* $F_{yield} = T_{yield} \times A$

  Where $A$ in this case is $\pi (193E^{-12})^2 = 1.17 \times 10^{-19}$ 

**Compared to the force of air resistance partially calculated above, a tungsten line is not strong enough to take the load.** 

Here's where science fiction comes in.

* You could make your line of a hypothetical new material that is strong enough to carry the load. How much stronger? It'll need to be at least 100 million times ($10^8$) times stronger than tungsten to overcome to air resistance of a rather leisurely 1 ${m} \over {s}$ spin.

   For some context, you could build submarines that could travel to the Earth's core with materials only a hundred times or thousand times stronger than tungsten. You would be looking for a material almost a million times stronger than that.

**Alternatively, it might not be "true" monomolecular line.**

If you want to imagine that "monomolecular" is just sales puffery from whoever is manufacturing these things, we have some more options :

* If the line is 193 micrometers ($10^{-6}$) in radius, the max tension a normal tungsten line can handle is $1.17 \times 10^{-7}$. Enough to handle a spin of ... well, the mass of the line also increases with diameter, increasing then tensile strength past break.

You need another material entirely — something almost as light as air and almost as strong as tungsten. [Graphene aerogel][2] is close to fitting the requirements.

Can we go back to true monowire? The exceptional tensile strength required was to creating enough tension to overcome air resistance. That value is independent of mass. However, it has gone up with the thicker line to $2.86 \times 10^{-7}$ Newton’s.

Assuming a sci-fi graphene aerogel with ten times the strength of tungsten (5 GPa) of strength, counterweighted to help it overcome air resistance, your mono  line can handle velocities out to around  2 ${m} \over {s}$

Performance—  

In order to proceed, we need to figure out the counterweight : if the monoline is stable at 1 ${m} \over {s}$

Work in Progress 

  [1]: http://www.astronoo.com/en/articles/size-of-atoms.html
  [2]: https://www.science.org/doi/10.1126/sciadv.aat6951