Solution

1. In the case of a weightless and inextensible thread, its tension is determined as:

$$T = F_i = \frac{Mv^2}{L} = \frac{M\omega^2L^2}{L}$$

$$\boxed{T = M\omega^2L}$$

2. Let us isolate a specified cross-section of the rope at distance $x$ and determine the mass of its outer part of length $(L-x)$:

$$m_x = m\frac{L-x}{L}$$

3. Let us determine the distance from the rotation axis $Oz$ to the center of mass of this rope segment:

$$r_x = L - \frac{L-x}{2}$$

$$r_x = \frac{L+x}{2}$$

4. The tension of the rope at section $x$ will be caused by the inertial (centrifugal) force of the rotating mass $M$ and the mass of the outer rope segment $m_x$:

$$\boxed{T_x = M\omega^2L + \frac{m\omega^2(L^2-x^2)}{2L}}$$