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?

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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?

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

    <h4>Answer</h4>
    <p>
        $$N = 4m_1m_2g/(m_1 + m_2)$$
    </p>

    <p style="text-align: right; font-style: italic; font-size: 14;">
        Mustafa Bakhodirov<br>
    </p>


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