While I have studied some resources as to these terms, in fact, I cannot yet get what could be considered as their difference. Or, when could they be interchangeable, when not?

Any comment would greatly be appreciated 


Question: Shape of rotating rope (lasso problem?)
Answer: This is not a complete solution but I'm making it a community wiki so that others can supplement and correct it.

Frédéric Grosshans is almost correct in the above, but misses a crucial part of the problem: the rope may coil around the axis. The general method of assuming a stable equilibrium shape and minimising the (effective) potential energy in a rotating frame is a good one, but perhaps one should be wary of assuming that such a stable equilibrium even exists.

Pressing on, we parametrise the rope as $(z(l), r(l), \phi(l))$, and the potential energy is: $$U = \int_0^L dl \; \rho \left(\frac{1}{2} \omega^2 r^2 - g z\right)$$ where $\rho$ is the density per length (apparently irrelevant in the end, but we will keep it to make sure that the units are sane) and we take the convention that $z$ increases downwards.

The constraint is that each small element of the rope must be at a fixed length, so we introduce a Lagrange multiplier field (this is a field theory in 1D!): $$L = \rho \left(\frac{1}{2} \omega^2 r^2 - g z\right) + \frac{1}{2}T\left(r&#39;^2 + z&#39;^2 + r^2 \phi&#39;^2 - 1\right)$$ where $T$ is the Lagrange multiplier field (the name will become obvious, as will the random factor of 1/2).

Pushing through the Euler-Lagrange equations yields:
$$\begin{align*}
\rho g + (Tz&#39;)&#39; &amp;= 0 \\
\rho \omega^2 r + T r \phi&#39;^2 - (Tr&#39;)&#39; &amp;= 0 \\
(T r^2 \phi &#39;)&#39; &amp;= 0 \\
r&#39;^2 + z&#39;^2 + r^2 \phi&#39;^2 &amp;= 1
\end{align*}$$

The first equation encodes the meaning of $T$ as the tension in the rope. We may integrate it and obtain the quite obviously sensible $$Tz&#39; = \rho g (L - l)$$ (using the boundary condition that $T = 0$ at $l = L$) which simply encodes that the vertical component of the tension must balance the weight of the rope below. 

The remaining equations are 1st order in $z$ and 2nd order in $\phi$ and $r$. We have the coordinate setting equations that $z(0) = 0$ and $\phi(0) = 0$. There is a genuine extra parameter $r(0)$ which dictates the radius of the driving force at the top. However, one still needs another boundary condition for $r$ and $\phi$ (or possibly their derivatives) for the problem to be well posed. In addition, the condition for $T$ leads to a tricky integral equation, since the natural boundary condition is that $T(L) = 0$, and one would then need to integrate this to obtain the boundary at $l = 0$. 

Finally, I should mention that this partial solution as it stands has one serious deficiency which may or may not be connected to the boundary condition problem above: the equations are symmetric for $\omega \rightarrow - \omega$, so apparently an available solution would have the rope pointing forwards of the rotating force. But perhaps this is a maximum of the energy rather than the minimum.

In any case, the problem seems highly non-linear, and I would be surprised (and delighted) if an analytic solution is possible. One might hope however that there are easy to understand regimes (apart from the trivial ones at $\omega = 0$ or $\infty$).