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 (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 $\pi (193E^{-12})^2 = 1.17 \times 10^{-19}$ , meaning $F_{yield} = 5.85 \times 10^{-11}$
Compared to the force of air resistance partially calculated above, a tungsten line is barely strong enough to take the load.
You can have a hypothetical material ten times the strength of tungsten (5 GPa). That would allow a velocity of around 3 ${m} \over {s}$
Performance—
In order to proceed, we need to figure out the counterweight : if the monoline is stable at 1 ${m} \over {s}$
Since we’ve chosen a velocity and radius of 1, the mass of our counterweight in kilograms is conveniently equal to the air resistance $2.6 \times 10^{-11}$ kilograms. Or, about 30 nanograms. Or about the mass of 30 thousand bacteria stacked together. The weight of the line is orders of magnitude lighter than the counterweight. I’m ignoring the line weight for what comes next.
The energy in the line is $E = {{1} \over {2}} m v^2$. Which is equal to $1.3 \times 10^{-10}$ joules.
For low speeds (and 3 meters per second qualifies) you can very roughly estimate cutting power as $E = F_{yield, target} d$, where d is the depth of the cut and $F_{yield, target} = T_{yield, target} \times 2 \times r_{line} \times 1$
Plugging in values for human bone (50 MPa) : 0.0193 joules of energy are required per meter of cut.
This isn’t going to do much injury.
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 our “sci-fi ten times tungsten” line can handle is 585 Newtons.
Air resistance has gone up with the thicker line to $2.54 \times 10^{-5} \times v^2$ Newtons.
The top speed of the whole assembly, then, can be up to 4,800 meters per second.
The flyweight might need to be recalculated, but this post is getting long. At 4,800 meters per second, the faux monofilament line has 0.000299 joules behind it. Each cutting hit with the line will cut away 0.015 meters, or 1.5 centimeters of bone-equivalent material.
Like a weed eater, as long as the line is not broken, it can loop back around again (4,800 times per second) biting away more material. Expressed as a cutting rate (provided a motor could keep up) you could cut about 74 meters per second.
Solidness—
The last category is solidness. This is expressed by the formula ${{rpm} \times {{l_1 + l_2} \over {v}}}$. L1 is the length of the incoming object. L2 is the length of the line (negligible unless the incoming object is also very small). And v is the velocity of the incoming object. What you end up with is a number that, if greater than 1, indicates the number of times the line will strike the incoming object (the integer part). If the number is less than one, the value represents the percentage chance that the line will hit the incoming object at all.
