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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
| </p> | | </p> |
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| <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> |
| $2.7.27^*.$ 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? | | $2.7.27^*.$ 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> | | </p> |
| $$\varepsilon = \frac{a}{R}$$ | | $$\varepsilon = \frac{a}{R}$$ |
| Substituting into $(1)$ | | Substituting into $(1)$ |
| $$\boxed{(m_1 - m_2) g R^2 = Ia}\quad(2)$$ | | $$\boxed{(m_1 - m_2) g R^2 = Ia}\quad(2)$$ |
| Conservation of angular momentum | | Conservation of angular momentum |
| $$I\omega = m_1vR + m_2vR\quad(3)$$ | | $$I\omega = m_1vR + m_2vR\quad(3)$$ |
| Let's express velocity through angular velocity | | Let's express velocity through angular velocity |
| $$v=\omega R\quad(4)$$ | | $$v=\omega R\quad(4)$$ |
| After substituting $(4)$ into $(3)$ | | After substituting $(4)$ into $(3)$ |
| $$\boxed{I = (m_1+m_2)R^2} \quad (5)$$ | | $$\boxed{I = (m_1+m_2)R^2} \quad (5)$$ |
| Let's equate the expressions $(2)$ and $(5)$ | | Let's equate the expressions $(2)$ and $(5)$ |
| $$(m_1 - m_2) g R^2 = (m_1+m_2)aR^2$$ | | $$(m_1 - m_2) g R^2 = (m_1+m_2)aR^2$$ |
| After mathematical transformations | | After mathematical transformations |
| $$a = g \frac{m_1 - m_2}{m_1 + m_2}\quad(6)$$ | | $$a = g \frac{m_1 - m_2}{m_1 + m_2}\quad(6)$$ |
| Describe the forces acting on the vertical axis | | Describe the forces acting on the vertical axis |
| $$N = (m_1+m_2)g - a (m_1-m_2)\quad(7)$$ | | $$N = (m_1+m_2)g - a (m_1-m_2)\quad(7)$$ |
| Substituting the acceleration $(6)$ into expression $(7)$ | | Substituting the acceleration $(6)$ into expression $(7)$ |
| $$N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)$$ | | $$N = g\cdot\left((m_1+m_2)+ \frac{(m_1-m_2)^2}{m_1+m_2}\right)$$ |
| From here we find the support reaction force: | | From here we find the support reaction force: |
| $$\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}$$ | | $$\boxed{N = \frac{4m_1m_2g}{m_1+m_2}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
