Правка разделов «Statement», «Solution», «Answer»

Valter правка от 14 авг. 2026 г., 18:35

правка #20029 предыдущая #20028 ← раньше

en/2.1.56.md +11 −21

		@@ -1,47 +1,41 @@
			### 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)
		+	![\|915x698, 47%](../../img/2.1.56/sol.jpg)

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

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

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

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

			$$
		−	m_x = m\\frac{l-x}{l}
		+	m_x = m\\frac{L-x}{L}
			$$

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

			$$
		−	r_x = l - \\frac{l-x}{2}
		+	r_x = L - \\frac{L-x}{2}
			$$

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

		−	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 $m_x$:
		+	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 $m_x$:

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

			#### Answer

			$$
			T = M\\omega^2l
			$$

		@@ -53,7 +47,3 @@Answer
			$$
			T_x = M\\omega^2l + m\\omega^2(l^2 - x^2)/(2l)
			$$
		−
		−	#### Answer
		−
		−	[Insert a concise answer or boxed result]

	@@ -1,47 +1,41 @@
	### Statement		### 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.		$2.1.56.$ 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		### Solution

	![~~Forces acting on the system~~\|915x698, 47%](../../img/2.1.56/sol.jpg)		![\|915x698, 47%](../../img/2.1.56/sol.jpg)

	1\\. In the case of a weightless and inextensible thread, its tension is determined as:		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}		T = F_i = \\frac{Mv^2}{L} = \\frac{M\\omega^2L^2}{L}
	$$		$$

	$$		$$
	\\boxed{T = M\\omega^2l}		\\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 $(l-x)$:		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}		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:		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 = L - \\frac{L-x}{2}
	$$		$$

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

	#### Answer		#### Answer

	$$		$$
	T = M\\omega^2l		T = M\\omega^2l
	$$		$$

	@@ -53,7 +47,3 @@Answer
	$$		$$
	T_x = M\\omega^2l + m\\omega^2(l^2 - x^2)/(2l)		T_x = M\\omega^2l + m\\omega^2(l^2 - x^2)/(2l)
	$$		$$

	#### Answer

	[Insert a concise answer or boxed result]