Translated 2.1.8 & 3.1.15
en/3.1.15.md
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| + | <title>The amplitude of small oscillations of a mathematical pendulum standing on a cart is x_0, and the amplitude of oscillations of the cart is y_0. The length of the pendulum thread is l.Determine the maximum speed of the pendulum and cart. Neglect friction.</title> | ||
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| + | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
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| + | <h3 id="back-link"><a href="../../#3.1">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $3.1.15.$ The amplitude of small oscillations of a mathematical pendulum standing on a cart is $x_0$, and the amplitude of oscillations of the cart is $y_0$. The length of the pendulum thread is $l$. | ||
| + | Determine the maximum speed of the pendulum and cart. Neglect friction. | ||
| + | </p> | ||
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| + | <figure> | ||
| + | <img src="https://savchenkosolutions.com/3/3.1.15/3.1.15.png" | ||
| + | loading="lazy" width="130" /> | ||
| + | <figcaption> | ||
| + | For problem $3.1.15$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | Law of conservation of energy | ||
| + | $$ \frac{mg(x_0+y_0)^2}{2l}=\frac{m\upsilon^2}{2}+\frac{MU^2}{2} $$ | ||
| + | Law of conservation of momentum | ||
| + | $$ m\upsilon =MU $$ | ||
| + | Because the center of mass does not change position | ||
| + | $$ mx_0=My_0 $$ | ||
| + | Let's solve the upper equation for $\upsilon$:<br><br> | ||
| + | $$ \upsilon =\sqrt{\frac{Mg}{(M+m)l}(x_0+y_0)^2}=\sqrt{\frac{x_0g}{(x_0+y_0)l}(x_0+y_0)^2} $$ | ||
| + | Maximum speed of the pendulum | ||
| + | $$ \boxed{v=x_0\sqrt{\frac{g}{l}(1+\frac{y_0}{x_0})}} $$ | ||
| + | Let's solve the upper equation for $U$:<br><br> | ||
| + | $$ U=\sqrt{\frac{mg}{(M+m)l}(x_0+y_0)^2}=\sqrt{\frac{y_0g}{(x_0+y_0)l}(x_0+y_0)^2} $$ | ||
| + | Maximum speed of the trolley | ||
| + | $$ \boxed{U=y_0\sqrt{\frac{g}{l}(1+\frac{x_0}{y_0})}} $$ | ||
| + | |||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$\upsilon =x_0\sqrt{\frac{g}{l}(1+\frac{y_0}{x_0}});\quad U=y_0\sqrt{\frac{g}{l}(1+\frac{x_0}{y_0})}$$ | ||
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| <meta name="description" content="The amplitude of small oscillations of a mathematical pendulum standing on a cart is x_0, and the amplitude of oscillations of the cart is y_0. The length of the pendulum thread is l.Determine the maximum speed of the pendulum and cart. Neglect friction."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="The amplitude of small oscillations of a mathematical pendulum standing on a cart is x_0, and the amplitude of oscillations of the cart is y_0. The length of the pendulum thread is l.Determine the maximum speed of the pendulum and cart. Neglect friction."> | |||
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| <div id="logo"> | |||
| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
| </div> | |||
| </a> | |||
| <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="../../#3.1">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $3.1.15.$ The amplitude of small oscillations of a mathematical pendulum standing on a cart is $x_0$, and the amplitude of oscillations of the cart is $y_0$. The length of the pendulum thread is $l$. | |||
| Determine the maximum speed of the pendulum and cart. Neglect friction. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/3/3.1.15/3.1.15.png" | |||
| loading="lazy" width="130" /> | |||
| <figcaption> | |||
| For problem $3.1.15$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| Law of conservation of energy | |||
| $$ \frac{mg(x_0+y_0)^2}{2l}=\frac{m\upsilon^2}{2}+\frac{MU^2}{2} $$ | |||
| Law of conservation of momentum | |||
| $$ m\upsilon =MU $$ | |||
| Because the center of mass does not change position | |||
| $$ mx_0=My_0 $$ | |||
| Let's solve the upper equation for $\upsilon$:<br><br> | |||
| $$ \upsilon =\sqrt{\frac{Mg}{(M+m)l}(x_0+y_0)^2}=\sqrt{\frac{x_0g}{(x_0+y_0)l}(x_0+y_0)^2} $$ | |||
| Maximum speed of the pendulum | |||
| $$ \boxed{v=x_0\sqrt{\frac{g}{l}(1+\frac{y_0}{x_0})}} $$ | |||
| Let's solve the upper equation for $U$:<br><br> | |||
| $$ U=\sqrt{\frac{mg}{(M+m)l}(x_0+y_0)^2}=\sqrt{\frac{y_0g}{(x_0+y_0)l}(x_0+y_0)^2} $$ | |||
| Maximum speed of the trolley | |||
| $$ \boxed{U=y_0\sqrt{\frac{g}{l}(1+\frac{x_0}{y_0})}} $$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\upsilon =x_0\sqrt{\frac{g}{l}(1+\frac{y_0}{x_0}});\quad U=y_0\sqrt{\frac{g}{l}(1+\frac{x_0}{y_0})}$$ | |||
| </p> | |||
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| Dzikan Mikita<br> | |||
| </p> | |||
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| <br> | |||
| <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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