Added 2.7.27

astrosander правка от 19 окт. 2024 г., 18:35

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		+	<title>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?</title>
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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<i><b>knowledge must be free</b></i>
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		+
		+	<h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3>
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		+	<h3> Statement </h3>
		+	<p>
		+	$2.7.27^*.$ 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?
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="2.7.27.png"
		+	loading="lazy" width="230" />
		+	<figcaption>
		+	For problem $2.7.27^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	Newton's second law for a rotational movement
		+	$$M = I\varepsilon$$
		+	$$m_1gR - m_2 g R = I\varepsilon\quad(1)$$
		+	Since there is no slippage:
		+	$$\varepsilon = \frac{a}{R}$$
		+	Substituting into $(1)$
		+	$$\boxed{(m_1 - m_2) g R^2 = Ia}\quad(2)$$
		+	Conservation of angular momentum
		+	$$I\omega = m_1vR + m_2vR\quad(3)$$
		+	Let's express velocity through angular velocity
		+	$$v=\omega R\quad(4)$$
		+	After substituting $(4)$ into $(3)$
		+	$$\boxed{I = (m_1+m_2)R^2} \quad (5)$$
		+	Let's equate the expressions $(2)$ and $(5)$
		+	$$(m_1 - m_2) g R^2 = (m_1+m_2)aR^2$$
		+	After mathematical transformations
		+	$$a = g \frac{m_1 - m_2}{m_1 + m_2}\quad(6)$$
		+	Describe the forces acting on the vertical axis
		+	$$N = (m_1+m_2)g - a (m_1-m_2)\quad(7)$$
		+	Substituting the acceleration $(6)$ into expression $(7)$
		+	$$N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)$$
		+	From here we find the support reaction force:
		+	$$\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}$$
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$N = 4m_1m_2g/(m_1 + m_2)$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Bakhodirov Mustafa<br>
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			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="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?">
			<meta property="og:image" content="img/logo.png">
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			<title>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?</title>
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			</p>
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			<h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$2.7.27^*.$ 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?
			</p>
			<center>
			<figure>
			<img src="2.7.27.png"
			loading="lazy" width="230" />
			<figcaption>
			For problem $2.7.27^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			Newton's second law for a rotational movement
			$$M = I\varepsilon$$
			$$m_1gR - m_2 g R = I\varepsilon\quad(1)$$
			Since there is no slippage:
			$$\varepsilon = \frac{a}{R}$$
			Substituting into $(1)$
			$$\boxed{(m_1 - m_2) g R^2 = Ia}\quad(2)$$
			Conservation of angular momentum
			$$I\omega = m_1vR + m_2vR\quad(3)$$
			Let's express velocity through angular velocity
			$$v=\omega R\quad(4)$$
			After substituting $(4)$ into $(3)$
			$$\boxed{I = (m_1+m_2)R^2} \quad (5)$$
			Let's equate the expressions $(2)$ and $(5)$
			$$(m_1 - m_2) g R^2 = (m_1+m_2)aR^2$$
			After mathematical transformations
			$$a = g \frac{m_1 - m_2}{m_1 + m_2}\quad(6)$$
			Describe the forces acting on the vertical axis
			$$N = (m_1+m_2)g - a (m_1-m_2)\quad(7)$$
			Substituting the acceleration $(6)$ into expression $(7)$
			$$N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)$$
			From here we find the support reaction force:
			$$\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}$$
			</p>

			<h4>Answer</h4>
			<p>
			$$N = 4m_1m_2g/(m_1 + m_2)$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Bakhodirov Mustafa<br>
			</p>


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