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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#2.7">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.7">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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. | | $2.7.42.$ 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. |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="160" /> | | loading="lazy" width="160" /> |
| <figcaption> | | <figcaption> |
| For problem $2.7.42$ | | For problem $2.7.42$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| Geometry of the problem | | Geometry of the problem |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Angular momentum | | Angular momentum |
| $$\left\{\begin{matrix} | | $$\left\{\begin{matrix} |
| L=m_2v\frac{l}{2} \\ | | L=m_2v\frac{l}{2} \\ |
| L=I\omega | | L=I\omega |
| \end{matrix}\right.$$ | | \end{matrix}\right.$$ |
| Parallel axis theorem | | Parallel axis theorem |
| $$I=I_0+m_2d^2=\frac{1}{3}m_1l^2+m_2\left(\frac{l}{2}\right)^2$$ | | $$I=I_0+m_2d^2=\frac{1}{3}m_1l^2+m_2\left(\frac{l}{2}\right)^2$$ |
| $$m_2v\frac{l}{2}=\omega\left(\frac{1}{3}m_1l^2+\frac{1}{4}m_2l^2\right)$$ | | $$m_2v\frac{l}{2}=\omega\left(\frac{1}{3}m_1l^2+\frac{1}{4}m_2l^2\right)$$ |
| From where | | From where |
| $$\omega = \frac{6m_2}{4m_1+3m_2}\frac{v}{l}$$ | | $$\omega = \frac{6m_2}{4m_1+3m_2}\frac{v}{l}$$ |
| From the figure | | From the figure |
| $$h=\frac{l}{2}(1-\cos\alpha)$$ | | $$h=\frac{l}{2}(1-\cos\alpha)$$ |
| Law of conservation of energy in rotational motion | | Law of conservation of energy in rotational motion |
| $$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha)$$ | | $$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha)$$ |
| Expressing $\cos\alpha$ | | Expressing $\cos\alpha$ |
| $$1-\cos\alpha=\frac{I\omega^2}{(m_1+m_2)l}$$ | | $$1-\cos\alpha=\frac{I\omega^2}{(m_1+m_2)l}$$ |
| $$\boxed{\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$ | | $$\boxed{\cos\alpha=1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
