We will show that the shape of the rotating rope is the Bessel function.
Let us adopt a rotating coordinate system in which the rope is at rest. We shall use cylindrical coordinates, $r$ and $z$ are the radial and the vertical coordinates, where $(r,z)=(0,0)$ denotes the point of suspension and $z$ increases downwards. The rope runs along the points $(r(l),z(l))$, where $l$ is the length along the rope measured from the free end, i.e. $l=L$ at the point of suspension.
The forces acting on an infinitesimal rope element of mass $dm$ are the gravitational and the centrifugal forces ($dm\, g$ and $dm\, \omega^2 r$) and the constraint forces that arise due to the tension in the rope that make sure that the line density remains $\eta = m/L$ everywhere along the rope, where $m$ and $L$ are respectively the total mass and length. The tension in the rope at $l$ exerts a force $\vec{F}_T(l)$ which is parallel with the tangent vector of the rope. For a rope element between $l-dl$ and $l+dl$, forces arise at the bounding points i.e. $\vec{F}_T(l+dl)$ and at $\vec{F}_T(l-dl)$.
The radial and vertical equations of motion are
\begin{align}
dm \frac{d^2r}{dt^2} &= dm \,\omega^2 r - F_T(l+dl) \sin\theta(l+dl) + F_T(l-dl) \sin\theta(l-dl) = 0 ,\\
dm \frac{d^2z}{dt^2} &= dm\, g - F_T(l+dl) \cos\theta(l+dl) + F_T(l-dl) \cos\theta(l-dl) = 0 .\\
\end{align}
Here $\theta$ denotes the angle subtended by the tangent of the rope with the vertical, i.e. $\tan \theta = -dr/dz$. This generally changes with the position along the rope which we write as $\theta(l)$. The right hand side is zero given that the rope is at rest in the rotating frame. Note that the rope must be planar at rest, since no forces act in the azimuthal direction and so the rope tension cannot have an azimuthal component, the rope must have a zero derivative along the azimuthal direction.
Divide the equations with $2dl$:
\begin{align}
\frac{d}{dl} (F_T \sin\theta) &= \eta \omega^2 r, \\
\frac{d}{dl} (F_T \cos\theta) &= \eta g.
\end{align}
The vertical equation may be integrated as
$$F_T \cos\theta = \eta g l. $$
This satisfies the boundary conditions since at the lower end $l=0$ so $F_T=0$, and at $l=L$ we have $\cos \theta=1$ implying that $F_T=\eta g L = m g$. Substituting in the radial equation and using that $\sin \theta = \tan \theta \cos \theta = -(dr/dz)\cos \theta$, gives
$$\frac{d}{dl} \left(\eta g l \frac{dr}{dz}\right) = -\eta \omega^2 r. $$
In the limit that $\theta\ll 1$, we can approximate $l\approx L-z$. To simplify the equations, redefine the coordinate system with the substitution $z\rightarrow L-z$, so that $z=0$ is the lower free end of the rope and $z=L$ is the upper fixed end. This yields
$$\frac{d}{dz} \left(\eta g z \frac{dr}{dz}\right) \approx -\eta \omega^2 r. $$
This approximation allows us to obtain an exact analytic solution in a closed form. In practice we find numerically that the exact solution is very well described by this approximation.
We now manipulate this equation to arrive at the Bessel equation. Introduce a new variable $x = 2 \omega \sqrt{z/g}$ so that $z = x^2 g/(4\omega^2)$. The chain rule implies that the derivative with respect to $z$ may be written as
$$\frac{d}{dz} = \frac{dx}{dz}\frac{d}{dx} = \frac{1}{dz/dx}\frac{d}{dx} = \frac{2\omega^2}{g x} \frac{d}{dx}.$$
The constants drop out after substitution and we get
$$\frac{1}{x}\frac{d}{dx} \left(x\frac{dr}{dx}\right) = r $$
or
$$x^2 \frac{d^2 r}{dx^2} + x\frac{dr}{dx} + x^2 r = 0. $$
This is Bessel's differential equation defining the Bessel functions $J_0(x)$ and $Y_0(x)$, the general solution is a linear combination of these functions. Since $Y_0(x)$ diverges at $x=0$ and $J_0(0)=1$, the physical solution is proportional to the Bessel function of the first kind $J_0(x)$. Thus, the shape of the rotating rope is
$$r(z) = r_{\mathrm{end}}\, J_0(x(z)) = r_{\mathrm{end}} \,J_0\left( 2 \omega \sqrt{\frac{z}{g}} \right).$$
Here $r_{\mathrm{end}}$ is the radial coordinate at the loose end.
The boundary condition at the suspension is that at $z = L$ the radial coordinate satisfies $r=0$, so we must require that
$J_0\left( 2 \omega \sqrt{L/g} \right) = 0$.
The zeros of the Bessel functions are well known, they are tabulated. Let us denote the $n$th zero with $x_n = \{2.4048, 5.5201, 8.6537, 11.7915, \dots\}$. The boundary condition is satisfied when $\omega = \frac{x_n}{2} \sqrt{g/L}$ for any $n$. Thus
$$r(z) = r_{\mathrm{end}}\, J_0(x(z)) = r_{\mathrm{end}}\, J_0\left( x_n \sqrt{\frac{z}{L}} \right).$$
For larger frequencies the rope has multiple nodes, i.e. places where it crosses $r=0$. The rope has $k$ nodes if the angular frequency is $\omega = \frac{x_{k+1}}{2} \sqrt{g/L}$, one node for $\omega = \frac{x_2}{2} \sqrt{g/L} = 2.26\sqrt{g/L}$, two nodes if $4.33\sqrt{g/L}$, and three nodes if $5.896 \sqrt{g/L}$.
In the limit, $\omega \gg \sqrt{g/L}$ we can use the asymptotic formula for the zeros of Bessel functions $x_{k+1} \approx (k- \frac{3}{4}) \pi $, so it has $k$ number of nodes when
$$\omega = \frac{\pi}{2}\left(k + \frac{3}{4}\right) \sqrt{\frac{g}{L}}.$$
Solving for $k$ shows that the number of nodes for $\omega \gg \sqrt{g/L}$ is
$$k=\frac{2\omega}{\pi}\sqrt{\frac{L}{g}}-\frac{3}{4}.$$
In this case, the asymptotic form of the Bessel function
$$ J_0(x) = \sqrt{\frac{2}{\pi x}} \cos\left(x - \frac{\pi}{4}\right) + \mathcal{O}(x^{-3/2})
$$ implies that for large $\omega$ and large $z$
$$r(z) \approx r_{\mathrm{end}}\, \sqrt{\frac{1}{\pi \omega} \sqrt{\frac{g}{z}}} \cos\left( 2 \omega \sqrt{\frac{z}{g}} - \frac{\pi}{4}\right).$$
Note that we assumed that the top end fixed at $z=L$ satisfies $r(L)=0$. In the more general case, the boundary conditions are $r(L)=r_0$ and $r(0)=r_{\mathrm{end}}$. This implies that $J_0(2\omega\sqrt{L/g}) = r_0/r_{\mathrm{end}}$, and we get a similar solution for large $\omega$ if $r_0/r_{\mathrm{end}}\ll 1$.
