In cylindrical coordinates, let the shape of the rope be parameterized by r = r(s), $\theta=\theta(s)$, and z = z(s), where s is distance measured along the rope. Then a unit vector along the rope is given by: $$\mathbf{i_s}=\frac{dr}{ds}\mathbf{i_r}+r\frac{d\theta}{ds}\mathbf{i_{\theta}}+\frac{dz}{ds}\mathbf{i_z}$$For an inextensible rope, we must have:$$\left(\frac{dr}{ds}\right)^2+\left(r\frac{d\theta}{ds}\right)^2+\left(\frac{dz}{ds}\right)^2=1$$Letting T(s) represent the tension in the rope at location s along the rope, the force balance on the section of rope between s and s + ds is given by:$$\frac{d(T\mathbf{i_s})}{ds}-\rho g\mathbf{i_z}=-\rho \omega^2r\mathbf{i_r}$$where $\rho=M/L$. In component form, this becomes:$$\frac{d}{ds}\left(T\frac{dz}{ds}\right)=\rho g$$$$\frac{d}{ds}\left(T\frac{dr}{ds}\right)-r\left(\frac{d\theta}{ds}\right)^2T=-\rho\omega^2r$$$$\frac{d}{ds}\left(Tr\frac{d\theta}{ds}\right)+T\frac{dr}{ds}\frac{d\theta}{ds}=0$$The z equation can be integrated once immediately to yield: $$T\frac{dz}{ds}=\left[T\frac{dz}{ds}\right]_{s=0}+\rho g s$$
Similarly, the $\theta$ equation can be integrated to yield: $$Tr^2\frac{d\theta}{ds}=\left[Tr^2\frac{d\theta}{ds}\right]_{s=0}$$
We have 4 equations in the four unknowns r, z, $\theta$, and T, but integrating these equations seems daunting.
ADDENDUM
After further consideration, I see no reason why $\theta$ needs to be anything other than zero over the entire length of the rope (i.e., no variations in $\theta$). So that's what I'm going to assume from this point on. With this assumption, the in extensibility condition now becomes: $$\left(\frac{dr}{ds}\right)^2+\left(\frac{dz}{ds}\right)^2=1$$If we let $\phi(s)$ represent the contour angle of the rope with respect to the horizontal at location s along the rope, then we can write:
$$\frac{dr}{ds}=\cos{\phi}\tag{A}$$
$$\frac{dz}{ds}=-\sin{\phi}\tag{B}$$These equations satisfy the in extensibility condition exactly. Once the function $\phi(s)$ is established, this determines the shape of the rope over its entire length.
In terms of $\phi$, the unit tangent vector along the rope is given by: $$\mathbf{i_s}=\cos{\phi}\mathbf{i_r}-\sin{\phi}\mathbf{i_z}$$ and the derivative with respect to s of the unit tangent vector (i.e., the unit normal vector times the curvature) is given by:$$\frac{d\mathbf{i_s}}{ds}=-(\sin{\phi}\mathbf{i_r}+\cos{\phi}\mathbf{i_z})\frac{d\phi}{ds}$$
If we substitute these equations into the differential force balance equation, we obtain: $$(\cos{\phi}\mathbf{i_r}-\sin{\phi}\mathbf{i_z})\frac{dT}{ds}-T(\sin{\phi}\mathbf{i_r}+\cos{\phi}\mathbf{i_z})\frac{d\phi}{ds}-\rho g\mathbf{i_z}=-\rho \omega^2r\mathbf{i_r}$$If we dot this equation with the unit tangent vector and then also with respect to the unit normal vector, we obtain:
$$\frac{dT}{ds}=-\rho g \sin{\phi}-\rho\omega^2r\cos{\phi}=\rho g\frac{dz}{ds}-\rho g \omega^2 r\frac{dr}{ds}\tag{1}$$and$$T\frac{d\phi}{ds}=-\rho g\cos{\phi}+\rho \omega^2r\sin{\phi}\tag{2}$$
Eqn. 1 can be integrated immediately to yield the tension T: $$T=T(0)+\rho g z-\frac{\rho \omega^2 (r^2-R^2)}{2}\tag{3}$$If we combined Eqns. 2 and 3, we obtain an equation for the derivative of $\phi$ with respect to s:
$$\frac{d\phi}{ds}=\frac{-\rho g\cos{\phi}+\rho \omega^2r\sin{\phi}}{T(0)+\rho g z-\frac{\rho \omega^2 (r^2-R^2)}{2}}\tag{4}$$
This equation could be integrated numerically together with equations A and B to get the rope shape if we knew that initial values for T and $\phi$. The initial tension must be such that the value of T at s = L is zero. Also, since the denominator must be equal to zero at s = L, the numerator must also be zero at this location in order for the curvature to be finite. So, at s = L, we must have $$r(L)\tan{\phi(L)}=\frac{g}{\omega^2}$$This is a pretty nasty boundary condition that would have to be satisfied. But, conceptually, we could solve the problem by using the shooting method, and adjusting the initial values of T and $\phi$ until the required conditions are satisfied at s = L.
CONTINUATION
Before continuing and presenting a method for solving the differential equations for the shape of the rope, I'm first going to follow @Hussein's recommendation, and reduce the equations to dimensionless form. This is done simply by scaling all the spatial parameters r, z, s, and L by the radius R of the drum. In terms of the new dimensionless variables, our equations now become:
$$\frac{dr}{ds}=\cos{\phi}\tag{5}$$
$$\frac{dz}{ds}=-\sin{\phi}\tag{6}$$
$$\frac{d\phi}{ds}=\frac{-\cos{\phi}+\beta r\sin{\phi}}{[z-z(L)]-\beta\frac{(r^2-r^2(L))}{2}}\tag{7}$$where $$\beta=\frac{\omega^2R}{g}\tag{8}$$and the dimensionless tension is given by $$\tau=\frac{T}{\rho g R}=[z-z(L)]-\beta\frac{(r^2-r^2(L))}{2}\tag{9}$$
and our zero-tension boundary condition at s = L now becomes $$r(L)\tan{\phi(L)}=\frac{1}{\beta}\tag{10}$$
In our subsequent development, we are going to also need to know the value of the dimensionless curvature $d\phi/ds$ at s = L. Because of the zero-tension boundary condition (Eqn. 10) at s = L, both the numerator and denominator of Eqn. 7 for $d\phi/ds$ approach zero at this location. However, we can still obtain the value for $d\phi/ds$ by applying l'Hospital's rule; this yields:$$\left[\frac{d\phi}{ds}\right]_{s=L}=-\frac{\beta^2r(L)}{2[1+(\beta r(L))^2]^{3/2}}\tag{11}$$
METHOD OF SOLUTION
The differential equation can be integrated, subject to the prescribed boundary conditions, by either stating at s = 0 and integrating forward to increasing radii, or by starting at s = L and integrating backward toward lower radii. For various reasons that I won't get into here, it is more straightforward to start at s = L and to integrate backwards.
To integrate backwards, we make a change of variable according to $$S=L-s$$ Our differental equation and initial conditions in terms of S then become:
$$\frac{dr}{dS}=-\cos{\phi}\tag{5a}$$
$$\frac{dz}{dS}=\sin{\phi}\tag{6a}$$
$$\frac{d\phi}{dS}=\frac{\cos{\phi}-\beta r\sin{\phi}}{[z-z(0)]-\beta\frac{(r^2-r^2(0))}{2}}\tag{7a}$$where the dimensionless tension is now given by $$\tau=\frac{T}{\rho g R}=[z-z(0)]-\beta\frac{(r^2-r^2(0))}{2}\tag{8a}$$Eqn. 7a applies at all values of S except S = 0, where $$\left[\frac{d\phi}{dS}\right]_{S=0}=+\frac{\beta^2r(0)}{2[1+(\beta r(0))^2]^{3/2}}\tag{11a}$$In addition, at S = 0, we have the initial condition of $\phi$ as:
$$r(0)\tan{\phi(0)}=\frac{1}{\beta}\tag{10a}$$And, with no loss of generality we can take $$z(0)=0$$
Prior to carrying out the integration of these equations as an initial value problem, we don't know the value of r(0) that will be necessary for r(L) to be unity at S = L. So we can choose various values of r(0) and perform the integration, iterating on r(0) until we obtain a solution where r(L) = 1.0. Or we can just choose different value of r(0) and generate an array of solutions for the values of L that each of them implies at S = L.
The easiest way to integrate these equations numerically as an initial value problem is to employ forward Euler with a small step size for good accuracy.
RESULTS OF SAMPLE CALCULATION
I have carried out a numerical solution of the model differential equations on an Excel Spreadsheet using the approach described above. The objective was to compare with @rob's results. The case considered was with L=10 R and $\beta=0.25$, where $\beta = 0.25$ corresponds to rob's case of $\omega= 0.5 \omega_0$.
This shows the dimensionless vertical coordinate vs the dimensionless radial coordinate for the rope. To the eye, the results are a very close match to rob's results for the same case in his figure. In particular, the dimensionless vertical drop is about 4.75 and the dimensionless radial location of the rope tail is at about 9.75. The dimensionless rope tension at the drum for this case was about 16.5
RESULTS FOR CASE REQUESTED BY Alex Trounev
Alex Trounev has requested that I perform the calculation for the following case: $\omega=2\pi$, R = 0.1 meters, L = 1 meter, and $g = 9.81/ m^2/sec$. For these parameter values, we have that the dimensionless radial acceleration $\beta$ is given by $$\beta=\frac{\omega^2R}{g}=\frac{(2\pi)^2(0.1)}{9.81}=0.4024$$and the dimensionless length of the rope is $L/R=10$. The calculated shape of the rope for this case is shown in the figure below:

The vertical drop of the rope from the drum to the free end is predicted to be about 0.3 meters, and the radial extent of the rope from the drum to the free end is predicted to run from 0.1 meters to 1.053 meters.
The predicted dimensionless tension in the rope at the drum is predicted to be $\tau=25.1$. The actual dimensional tension is related to the dimensionless tension by $$T=\rho g R \tau=\rho g L\frac{R}{L}\tau=W\frac{R}{L}\tau$$where W is the weight of the rope. So, in this case, $$T=(0.1)(25.1)W=2.51W$$That is 2.51 times the weight of the rope. Of course, the vertical component of the tension at the drum must be equal to the weight of the rope. So the remainder of the tension in the rope is the effect of the horizontal component associated with the angular acceleration.