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| − | <meta name="description" content="Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge Q is fixed directly under the equilibrium position of the ball at a distance h from it. Length of the thread l, mass of the ball m, charge q."> | ||
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| − | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| − | <title>Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge Q is fixed directly under the equilibrium position of the ball at a distance h from it. Length of the thread l, mass of the ball m, charge q.</title> | ||
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| − | <header style="text-align:center;"> | ||
| − | <h2>Solutions of Savchenko Problems in Physics</h2> | ||
| − | <p class="author"> | ||
| − | Aliaksandr Melnichenka <br/> | ||
| − | October 2023 | ||
| − | </p> | ||
| − | </header> | ||
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| − | <h3> Statement </h3> | ||
| − | | ||
| − | $7.1.22^*.$ | ||
| − | Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge $Q$ is fixed directly under the equilibrium position of the ball at a distance $h$ from it. Length of the thread $l$, mass of the ball $m$, charge $q$. | ||
| − | </p> | ||
| − | <center> | ||
| − | <figure> | ||
| − | <img src="statement.png" | ||
| − | loading="lazy" alt="Mountain landscape" width="150" /> | ||
| − | <figcaption> | ||
| − | For the 7.1.22 problem | ||
| − | </figcaption> | ||
| − | </figure> | ||
| − | </center> | ||
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| + | </script> | ||
| − | | ||
| − | <center> | ||
| − | <figure> | ||
| − | | ||
| − | loading="lazy" alt="Mountain landscape" width="250" /> | ||
| − | <figcaption> | ||
| − | Forces acting on pendulum. | ||
| − | </figcaption> | ||
| − | </figure> | ||
| − | </center> | ||
| − | <p> | ||
| − | Applying Newton's Second Law: | ||
| − | <div class="scroll-wrapper"> | ||
| − | $$mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} = m\frac{dv}{dt} \; (1)$$ | ||
| − | </div> | ||
| − | where, | ||
| − | $$\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$$ | ||
| − | and, it is known that | ||
| − | $$\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$$ | ||
| − | as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then | ||
| − | $$\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$$ | ||
| − | | ||
| − | $$\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$$ | ||
| − | Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ | ||
| − | <div class="scroll-wrapper"> | ||
| − | $$mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$$ | ||
| − | </div> | ||
| − | Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained | ||
| − | <div class="scroll-wrapper"> | ||
| − | $$\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$$ | ||
| − | </div> | ||
| − | Graphically it can be seen that, with cosine and sine approximations applied, | ||
| − | $$\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$$ | ||
| − | Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically | ||
| − | <div class="scroll-wrapper"> | ||
| − | $$\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$$ | ||
| − | </div> | ||
| − | Since there are small oscillations, $\theta\rightarrow0$, but terms $\theta^2$ and $\theta^3$ tend to zero faster than $\theta$, so they are negligible respect to $\theta$. | ||
| − | $$\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$$ | ||
| − | as $k = \frac{1}{4\pi\epsilon_0}$, | ||
| − | $$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$$ | ||
| − | Finally, | ||
| − | $$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$$ | ||
| − | valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ | ||
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| + | <div class="col flash"> <header class = "margin-main" style="text-align:center;"> | ||
| + | <h2>Решение задач из Савченко О.Я.</h2> | ||
| + | <p class="author"> | ||
| + | Aliaksandr Melnichenka <br/> | ||
| + | October 2023 | ||
| </p> | |||
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| + | <a href="../#1">←Назад</a> | ||
| + | </h2> | ||
| + | </div> | ||
| + | </div> | ||
| − | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| − | BSc. Luis Daniel Fernández Quintana<br> | ||
| − | Physics Department (FCNE)<br> | ||
| − | Universidad de Oriente, Cuba<br> | ||
| − | </p> | ||
| + | <div id="1" class="row odd"> | ||
| + | <div class="col col-lg-12"> | ||
| + | <h3> | ||
| + | Problem statement: | ||
| + | </h3> | ||
| + | <div class="task"> | ||
| + | <p> | ||
| + | \(7.1.22^*\) | ||
| + | Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge $Q$ is fixed directly under the equilibrium position of the ball at a distance $h$ from it. Length of the thread $l$, mass of the ball $m$, charge $q$. <br> | ||
| + | </p> | ||
| + | <div class="StatementDiv"> | ||
| + | <img class="Statement" src="statement.png" alt="9.1.14"> | ||
| + | </div> | ||
| + | </div> | ||
| + | <hr> | ||
| + | <h3> | ||
| + | Solution: | ||
| + | </h3> | ||
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| + | <img class="ScreenS" src="drawing.png" style="max-width: 250;"> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Applying Newton's Second Law: | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} $$= m\frac{dv}{dt} \; (1)$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | where, | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | and, it is known that | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Modifying $(2)$, | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Graphically it can be seen that, with cosine and sine approximations applied, | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Since there are small oscillations, $\theta\rightarrow0$, but terms $\theta^2$ and $\theta^3$ tend to zero faster than $\theta$, so they are negligible respect to $\theta$. | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | as $k = \frac{1}{4\pi\epsilon_0}$, | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | Finally, | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: center;"> | ||
| + | $\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions"> | ||
| + | valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ | ||
| + | </p> | ||
| + | <br> | ||
| + | <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | BSc. Luis Daniel Fernández Quintana<br> | ||
| + | Physics Department (FCNE)<br> | ||
| + | Universidad de Oriente, Cuba<br> | ||
| + | </p> | ||
| + | </div> | ||
| + | |||
| + | <h3> | ||
| + | Answer: $\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}$ | ||
| + | </h3> | ||
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| <h2>Solutions of Savchenko Problems in Physics</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| <script src="../../js/jquery-1.10.1.min.js"></script> | |||
| <h3> Statement </h3> | <script type="text/javascript" async | ||
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src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"> | ||
| $7.1.22^*.$ | </script> | ||
| Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge $Q$ is fixed directly under the equilibrium position of the ball at a distance $h$ from it. Length of the thread $l$, mass of the ball $m$, charge $q$. | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="statement.png" | |||
| loading="lazy" alt="Mountain landscape" width="150" /> | |||
| <figcaption> | |||
| For the 7.1.22 problem | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| |
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| <figure> | MathJax.Hub.Config({ | ||
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extensions: ['tex2jax.js'], | ||
| loading="lazy" alt="Mountain landscape" width="250" /> | jax: ['input/TeX','output/HTML-CSS'], | ||
| <figcaption> | tex2jax: { | ||
| Forces acting on pendulum. | inlineMath: [ ['$','$'], ['\\(','\\)'] ], | ||
| </figcaption> | processEscapes: true, | ||
| </figure> | processClass: 'tex2jax', | ||
| </center> | ignoreClass: 'html' | ||
| <p> | }, | ||
| Applying Newton's Second Law: | showProcessingMessages: false, | ||
| <div class="scroll-wrapper"> | messageStyle: 'none' | ||
| $$mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} = m\frac{dv}{dt} \; (1)$$ | }); | ||
| </div> | </script> | ||
| where, | </head> | ||
| $$\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$$ | |||
| and, it is known that | <body style=""> | ||
| $$\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$$ | <div class="container row"> | ||
| as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then | <div class="col flash"> <header class = "margin-main" style="text-align:center;"> | ||
| $$\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$$ | <h2>Решение задач из Савченко О.Я.</h2> | ||
| |
<p class="author"> | ||
| $$\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$$ | Aliaksandr Melnichenka <br/> | ||
| Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ | October 2023 | ||
| <div class="scroll-wrapper"> | |||
| $$mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$$ | |||
| </div> | |||
| Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained | |||
| <div class="scroll-wrapper"> | |||
| $$\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$$ | |||
| </div> | |||
| Graphically it can be seen that, with cosine and sine approximations applied, | |||
| $$\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$$ | |||
| Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically | |||
| <div class="scroll-wrapper"> | |||
| $$\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$$ | |||
| </div> | |||
| Since there are small oscillations, $\theta\rightarrow0$, but terms $\theta^2$ and $\theta^3$ tend to zero faster than $\theta$, so they are negligible respect to $\theta$. | |||
| $$\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$$ | |||
| as $k = \frac{1}{4\pi\epsilon_0}$, | |||
| $$\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$$ | |||
| Finally, | |||
| $$\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$$ | |||
| valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ | |||
| </p> | </p> | ||
| </header> | |||
| </div> | |||
| <div id="main" class="col col-lg-9"> | |||
| <div class="row"> | |||
| <div class="col caption"> | |||
| <h2> | |||
| <a href="../#1">←Назад</a> | |||
| </h2> | |||
| </div> | |||
| </div> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | <div id="1" class="row odd"> | ||
| BSc. Luis Daniel Fernández Quintana<br> | |||
| Physics Department (FCNE)<br> | |||
| Universidad de Oriente, Cuba<br> | |||
| </p> | |||
| <div class="col col-lg-12"> | |||
| <h3> | |||
| Problem statement: | |||
| </h3> | |||
| <div class="task"> | |||
| <p> | |||
| \(7.1.22^*\) | |||
| Find the frequency of small oscillations of the mathematical pendulum relative to its lower equilibrium position, if a charge $Q$ is fixed directly under the equilibrium position of the ball at a distance $h$ from it. Length of the thread $l$, mass of the ball $m$, charge $q$. <br> | |||
| </p> | |||
| <div class="StatementDiv"> | |||
| <img class="Statement" src="statement.png" alt="9.1.14"> | |||
| </div> | |||
| </div> | |||
| <hr> | |||
| <h3> | |||
| Solution: | |||
| </h3> | |||
| <div class="solutions" > | |||
| |
<img class="ScreenS" src="drawing.png" style="max-width: 250;"> | ||
| <br> | <br> | ||
| <p> | <p class="TxtSolutions"> | ||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | Applying Newton's Second Law: | ||
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| <p> | <br> | ||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small> | <p class="TxtSolutions" style="text-align: center;"> | ||
| </p> | $mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2{\theta}}\sin{(\beta+\theta)} $$= m\frac{dv}{dt} \; (1)$ | ||
| </footer> | </p> | ||
| </body> | <br> | ||
| <p class="TxtSolutions"> | |||
| where, | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\tan{\beta} = \frac{l\sin{\theta}}{h+l\left(1-\cos\theta\right)} \; (2)$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| and, it is known that | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Modifying $(2)$, | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\tan{\beta} = \frac{l\theta}{h+\frac{l\theta^2}{2}} \; (4)$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Substituting $(3)$ and $(4)$ into $(1)$ and taking account $\sin{\theta}\approx\theta$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Graphically it can be seen that, with cosine and sine approximations applied, | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\cos{\beta} = \frac{h+\frac{l\theta^2}{2}}{\sqrt{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Putting $(4)$ and $(6)$ into $(5)$ and developing algebraically | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Since there are small oscillations, $\theta\rightarrow0$, but terms $\theta^2$ and $\theta^3$ tend to zero faster than $\theta$, so they are negligible respect to $\theta$. | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\left[\frac{g}{l}-\frac{kqQ\left(l+h\right)}{mh^3l}\right]\theta = \ddot{\theta}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| as $k = \frac{1}{4\pi\epsilon_0}$, | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\left[\frac{g}{l}-\frac{qQ\left(l+h\right)}{4\pi\epsilon_0mh^3l}\right]\theta = \ddot{\theta}$ | |||
| </p> | |||
| <br> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| Finally, | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: center;"> | |||
| $\boxed{\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}}$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions"> | |||
| valid for the condition $\frac{qQ(h+l)}{4\pi\epsilon_0h^3} < mg$ | |||
| </p> | |||
| <br> | |||
| <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> | |||
| BSc. Luis Daniel Fernández Quintana<br> | |||
| Physics Department (FCNE)<br> | |||
| Universidad de Oriente, Cuba<br> | |||
| </p> | |||
| </div> | |||
| <h3> | |||
| Answer: $\omega = \sqrt{\frac{g}{l}-\frac{qQ\left(h+l\right)}{4\pi\epsilon_0mh^3l}}$ | |||
| </h3> | |||
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