A rod of mass m_1 and length l is suspended on a hinge. A small piece of plasticine of mass m_2 sticks to the middle of the rod, moving until it collides with it horizontally at a speed of V. Find the maximum angle of deviation of the rod from the vertical. Ignore the friction in the hinge.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

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    <h3> Statement </h3>
    <p>
        $2.7.42.$ A rod of mass $m_1$ and length $l$ is suspended on a hinge. A small piece of plasticine of mass $m_2$ sticks to the middle of the rod, moving until it collides with it horizontally at a speed of $V$. Find the maximum angle of deviation of the rod from the vertical. Ignore the friction in the hinge.

    <h3>Solution</h3>
    <p>
        </p>

Angular momentum $$\left\{\begin{matrix} L=m_2v\frac{l}{2} \\ L=I\omega \end{matrix}\right.$$ Parallel axis theorem $$I=I_0+m_2d^2=\frac{1}{3}m_1l^2+m_2\left(\frac{l}{2}\right)^2$$ $$m_2v\frac{l}{2}=\omega\left(\frac{1}{3}m_1l^2+\frac{1}{4}m_2l^2\right)$$ From where $$\omega = \frac{6m_2}{4m_1+3m_2}\frac{v}{l}$$ From the figure $$h=\frac{l}{2}(1-\cos\alpha)$$ Law of conservation of energy in rotational motion $$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha)$$ Expressing $\cos\alpha$ $$1-\cos\alpha=\frac{I\omega^2}{(m_1+m_2)l}$$ $$\boxed{\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$

    <h4>Answer</h4>
    <p>
        $$\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Yuldashev Ulugbek<br>
    </p>


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