New solution

Valter edited
View solution History
revision #20028 newer →
en/2.1.56.md +59 −0
@@ -0,0 +1,59 @@
+### Statement
+
+$2.1.56.$ [Insert the problem statement]
+
+### 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
+
+![Forces acting on the system|915x698, 47%](../../img/2.1.56/sol.jpg)
+
+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 $m_x$:
+
+$$
+\\boxed{T_x = M\\omega^2l + \\frac{m\\omega^2(l^2-x^2)}{2l}}
+$$
+
+#### Answer
+
+$$
+T = M\\omega^2l
+$$
+
+$$
+T_x = M\\omega^2l + m\\omega^2(l^2 - x^2)/(2l)
+$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]