Translated 3.2.1-3.2.17
en/3.2.14.md
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| + | <title>A board of mass m lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is L, the coefficient of friction when the board slides on the roller is \mu. Find the frequency of longitudinal oscillations of the board.</title> | ||
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| + | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | ||
| + | </div> | ||
| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $3.2.14.$ A board of mass $m$ lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is $L$, the coefficient of friction when the board slides on the roller is $\mu$. Find the frequency of longitudinal oscillations of the board. | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/3/3.2.14/3.2.14.png" | ||
| + | loading="lazy" width="230" /> | ||
| + | <figcaption> | ||
| + | For problem $3.2.14$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | |||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/3/3.2.14/3.2.14_1.png" | ||
| + | loading="lazy" width="230" /> | ||
| + | <figcaption> | ||
| + | Forces acting on the board | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | Newton's second law for the horizontal axis | ||
| + | $$ ma=F_{fr1}-F_{fr2} $$ | ||
| + | Equilibrium condition for the vertical axis | ||
| + | $$ mg=N_1+N_2\quad(1) $$ | ||
| + | In the equilibrium condition, the sum of the moment of external forces must be equal to zero. | ||
| + | $$ N_1\left(\frac{l}{2}-x\right)-N_2\left(\frac{l}{2}+x\right)=0 $$ | ||
| + | Where does the reaction force on the second support come from? | ||
| + | $$ N_2=N_1\left(\frac{L/2-x}{L/2+x}\right) $$ | ||
| + | Substitute into $(1)$ | ||
| + | $$ mg=N_1\left(1+\frac{L/2-x}{L/2+x}\right) $$ | ||
| + | |||
| + | $$ N_1=mg\left(\frac{L/2+x}{L}\right);\quad N_2=mg\left(\frac{L/2-x}{L}\right) $$ | ||
| + | Newton's second law for the horizontal axis | ||
| + | $$ m\ddot{x}=\mu N_2-\mu N_1= - \mu mg \frac{2x}{L}\quad(2) $$ | ||
| + | We transform the obtained expression and obtain the equation of harmonic oscillations | ||
| + | $$ \ddot{x}(t)+\frac{2\mu g}{l}x(t)=0 $$ | ||
| + | Where does the angular frequency of oscillations come from? | ||
| + | $$ \boxed{\omega=\sqrt{\frac{2\mu g}{l}}} $$ | ||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | Dzikan Mikita<br> | ||
| + | Aliaksandr Kanashenka<br> | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$\omega=\sqrt{\frac{2\mu g}{l}}$$ | ||
| + | </p> | ||
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| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
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| <meta name="description" content="A board of mass m lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is L, the coefficient of friction when the board slides on the roller is \mu. Find the frequency of longitudinal oscillations of the board."> | |||
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| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <div id = "logo"> | |||
| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | |||
| </div> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $3.2.14.$ A board of mass $m$ lies on two rollers rotating with high speed towards each other. The distance between the axes of the rollers is $L$, the coefficient of friction when the board slides on the roller is $\mu$. Find the frequency of longitudinal oscillations of the board. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/3/3.2.14/3.2.14.png" | |||
| loading="lazy" width="230" /> | |||
| <figcaption> | |||
| For problem $3.2.14$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/3/3.2.14/3.2.14_1.png" | |||
| loading="lazy" width="230" /> | |||
| <figcaption> | |||
| Forces acting on the board | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| Newton's second law for the horizontal axis | |||
| $$ ma=F_{fr1}-F_{fr2} $$ | |||
| Equilibrium condition for the vertical axis | |||
| $$ mg=N_1+N_2\quad(1) $$ | |||
| In the equilibrium condition, the sum of the moment of external forces must be equal to zero. | |||
| $$ N_1\left(\frac{l}{2}-x\right)-N_2\left(\frac{l}{2}+x\right)=0 $$ | |||
| Where does the reaction force on the second support come from? | |||
| $$ N_2=N_1\left(\frac{L/2-x}{L/2+x}\right) $$ | |||
| Substitute into $(1)$ | |||
| $$ mg=N_1\left(1+\frac{L/2-x}{L/2+x}\right) $$ | |||
| $$ N_1=mg\left(\frac{L/2+x}{L}\right);\quad N_2=mg\left(\frac{L/2-x}{L}\right) $$ | |||
| Newton's second law for the horizontal axis | |||
| $$ m\ddot{x}=\mu N_2-\mu N_1= - \mu mg \frac{2x}{L}\quad(2) $$ | |||
| We transform the obtained expression and obtain the equation of harmonic oscillations | |||
| $$ \ddot{x}(t)+\frac{2\mu g}{l}x(t)=0 $$ | |||
| Where does the angular frequency of oscillations come from? | |||
| $$ \boxed{\omega=\sqrt{\frac{2\mu g}{l}}} $$ | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| Dzikan Mikita<br> | |||
| Aliaksandr Kanashenka<br> | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\omega=\sqrt{\frac{2\mu g}{l}}$$ | |||
| </p> | |||
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| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
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