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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.7">$\leftarrow$Back</a></h3>
+
+ <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.
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="160" />
+ <figcaption>
+ For problem $2.7.42$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ </p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Geometry of the problem
+ </figcaption>
+ </figure>
+</center>
+<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 $\cos\alpha$
+$$1-\cos\alpha=\frac{I\omega^2}{(m_1+m_2)l}$$
+$$\boxed{\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}$$
+ </p>
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