Solution

Statement

$2.1.56.$ A body of mass $M$ is connected by a thread of length $l$ to an axis around which it revolves with an angular velocity $\omega$. Find the tension force of the thread. The dimensions of the body are small; neglect gravity. Replace the thread with a uniform rope of mass $m$ and find its tension force at a distance $x$ from the axis of rotation.

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 $m_x$:

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