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			### Statement

			$2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released?

			![ For problem $2.7.27^*$ \|394x195, 34%](../../img/2.7.27/2.7.27.png)

			### Solution

			![ Forces acting on the system \|995x447, 51%](../../img/2.7.27/2.7.27_1.png)

			Newton's second law for a rotational movement

			$$
			M = I\varepsilon
			$$

			$$
			m_1gR - m_2 g R = I\varepsilon\tag{1}
			$$

			Since there is no slippage:

			$$
			\varepsilon = \frac{a}{R}
			$$

			Substituting into $(1)$

			$$
			\boxed{(m_1 - m_2) g R^2 = Ia}\tag{2}
			$$

			Conservation of angular momentum

			$$
			I\omega = m_1vR + m_2vR\tag{3}
			$$

			Let's express velocity through angular velocity

			$$
			v=\omega R\tag{4}
			$$

			After substituting $(4)$ into $(3)$

			$$
			\boxed{I = (m_1+m_2)R^2} \tag{5}
			$$

			Let's equate the expressions $(2)$ and $(5)$

			$$
			(m_1 - m_2) g R^2 = (m_1+m_2)aR^2
			$$

			After mathematical transformations

			$$
			a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6}
			$$

			Describe the forces acting on the vertical axis

			$$
			N = (m_1+m_2)g - a (m_1-m_2)\tag{7}
			$$

		@@ -69,7 +69,7 @@Solution
			Substituting the acceleration $(6)$ into expression $(7)$

			$$
		−	N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)
		+	N = g\cdot\left((m_1+m_2)- \frac{(m_1-m_2)^2}{m_1+m_2}\right)
			$$

			From here we find the support reaction force:

			$$
			\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}
			$$

			#### Answer

			$$
			N = 4m_1m_2g/(m_1 + m_2)
			$$

			Mustafa Bakhodirov
		unchanged lines 12

	### Statement		### Statement

	$2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released?		$2.7.27^*.$ A light rod with weights of mass $m_1$ and $m_2$ fixed to its ends rests with its middle on a rigid support. At the initial moment, the rod is held horizontally and then released. With what force does it press on the support immediately after it is released?

	![ For problem $2.7.27^*$ \|394x195, 34%](../../img/2.7.27/2.7.27.png)		![ For problem $2.7.27^*$ \|394x195, 34%](../../img/2.7.27/2.7.27.png)

	### Solution		### Solution

	![ Forces acting on the system \|995x447, 51%](../../img/2.7.27/2.7.27_1.png)		![ Forces acting on the system \|995x447, 51%](../../img/2.7.27/2.7.27_1.png)

	Newton's second law for a rotational movement		Newton's second law for a rotational movement

	$$		$$
	M = I\varepsilon		M = I\varepsilon
	$$		$$

	$$		$$
	m_1gR - m_2 g R = I\varepsilon\tag{1}		m_1gR - m_2 g R = I\varepsilon\tag{1}
	$$		$$

	Since there is no slippage:		Since there is no slippage:

	$$		$$
	\varepsilon = \frac{a}{R}		\varepsilon = \frac{a}{R}
	$$		$$

	Substituting into $(1)$		Substituting into $(1)$

	$$		$$
	\boxed{(m_1 - m_2) g R^2 = Ia}\tag{2}		\boxed{(m_1 - m_2) g R^2 = Ia}\tag{2}
	$$		$$

	Conservation of angular momentum		Conservation of angular momentum

	$$		$$
	I\omega = m_1vR + m_2vR\tag{3}		I\omega = m_1vR + m_2vR\tag{3}
	$$		$$

	Let's express velocity through angular velocity		Let's express velocity through angular velocity

	$$		$$
	v=\omega R\tag{4}		v=\omega R\tag{4}
	$$		$$

	After substituting $(4)$ into $(3)$		After substituting $(4)$ into $(3)$

	$$		$$
	\boxed{I = (m_1+m_2)R^2} \tag{5}		\boxed{I = (m_1+m_2)R^2} \tag{5}
	$$		$$

	Let's equate the expressions $(2)$ and $(5)$		Let's equate the expressions $(2)$ and $(5)$

	$$		$$
	(m_1 - m_2) g R^2 = (m_1+m_2)aR^2		(m_1 - m_2) g R^2 = (m_1+m_2)aR^2
	$$		$$

	After mathematical transformations		After mathematical transformations

	$$		$$
	a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6}		a = g \frac{m_1 - m_2}{m_1 + m_2}\tag{6}
	$$		$$

	Describe the forces acting on the vertical axis		Describe the forces acting on the vertical axis

	$$		$$
	N = (m_1+m_2)g - a (m_1-m_2)\tag{7}		N = (m_1+m_2)g - a (m_1-m_2)\tag{7}
	$$		$$

	@@ -69,7 +69,7 @@Solution
	Substituting the acceleration $(6)$ into expression $(7)$		Substituting the acceleration $(6)$ into expression $(7)$

	$$		$$
	N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)		N = g\cdot\left((m_1+m_2)- \frac{(m_1-m_2)^2}{m_1+m_2}\right)
	$$		$$

	From here we find the support reaction force:		From here we find the support reaction force:

	$$		$$
	\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}		\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}
	$$		$$

	#### Answer		#### Answer

	$$		$$
	N = 4m_1m_2g/(m_1 + m_2)		N = 4m_1m_2g/(m_1 + m_2)
	$$		$$

	Mustafa Bakhodirov		Mustafa Bakhodirov
	unchanged lines 12