Translated 1.5.11-1.5.20

astrosander edited
View solution History
revision #6330 GitHub 020c5e6 newer →
en/1.5.17.md +146 −0
@@ -0,0 +1,146 @@
+<!DOCTYPE html>
+<html lang="en">
+
+<head>
+ <meta charset="utf-8">
+ <meta name="viewport" content="width=device-width, initial-scale=1.0">
+ <meta http-equiv="content-language" content="en">
+ <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
+ <meta name="description" content="The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height H from the ground. Find the angular velocity of the log as a function of the angle \alpha between it and the horizontal if the truck is moving away from the wall at speed v.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
+ <meta property="og:title" content="The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height H from the ground. Find the angular velocity of the log as a function of the angle \alpha between it and the horizontal if the truck is moving away from the wall at speed v.">
+ <meta property="og:image" content="img/logo.png">
+ <meta property="og:description" content="The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height H from the ground. Find the angular velocity of the log as a function of the angle \alpha between it and the horizontal if the truck is moving away from the wall at speed v.">
+ <meta name="yandex-verification" content="6cfda41f74038368">
+ <title>The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height H from the ground. Find the angular velocity of the log as a function of the angle \alpha between it and the horizontal if the truck is moving away from the wall at speed v.</title>
+ <link rel="stylesheet" href="https://savchenkosolutions.com/css/css-latex/style.css">
+ <link rel="icon" href="https://savchenkosolutions.com/img/logo.png" type="image/png">
+ <script src="https://savchenkosolutions.com/js/jquery-1.10.1.min.js"></script>
+ <script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
+ <script type="text/x-mathjax-config">
+ MathJax.Hub.Config({
+ extensions: ['tex2jax.js'],
+ jax: ['input/TeX', 'output/HTML-CSS'],
+ tex2jax: {
+ inlineMath: [['$', '$'], ['$', '$']],
+ processEscapes: true,
+ processClass: 'tex2jax',
+ ignoreClass: 'html'
+ },
+ showProcessingMessages: false,
+ messageStyle: 'none'
+ });
+ </script>
+</head>
+ <body style="">
+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.5.17.$ The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height $H$ from the ground. Find the angular velocity of the log as a function of the angle $\alpha$ between it and the horizontal if the truck is moving away from the wall at speed $v$.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.5.17/statement.png"
+ loading="lazy" width="250" />
+ <figcaption>
+ For problem $1.5.17$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution <a href="../1.5.16">1.5.16</a></p>
+<p>At time $t$ the car will travel the distance $vt$ and the bottom will have horizontal coordinate $x$</p>
+<p>
+$$ x = H \tan \alpha\;(1) $$
+</p>
+<p>Let's consider the change in angle $\alpha = \arctan \frac{H}{x}$ over a small period of time $dt$</p>
+<p>
+$$ d\alpha = d\left(\arctan \frac{H}{x}\right) $$
+</p>
+ <p>
+$$ d\alpha = \frac{1}{1+\frac{H^2}{x^2}} \frac{H}{x^2} $$
+</p>
+ <p>
+$$ d\alpha = dx\frac{H}{x^2+H^2} $$
+</p>
+<p>Considering that $\omega = \frac{d \alpha}{dt}$ and $v = \frac{dx}{dt}$</p>
+<p>
+$$ \omega = \frac{vH}{x^2+H^2} $$
+</p>
+<p>Substitute $(1)$</p>
+<p>
+$$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$
+</p>
+ <p>
+$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$
+</p>
+
+
+<h3> Alternative solution: </h3>
+<p>In time $\Delta t$ the truck will travel a distance</p>
+<p>
+$$ \Delta l=\upsilon \Delta t $$
+</p>
+<center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.5.17/draw.png"
+loading="lazy" width="280" />
+<figcaption>
+Log and truck
+</figcaption>
+</figure>
+</center>
+<p>The log will drop slightly, turning at a small angle $\Delta \alpha$. If the length of the arc of rotation is $\Delta x$, then by definition of the radian measure of angle</p>
+<p>
+$$ \Delta \alpha=\frac{\Delta x }{R} $$
+</p>
+<p>Angular velocity is defined as</p>
+<p>
+$$ \omega=\frac{\Delta \alpha }{\Delta t }=\frac{\Delta x }{R\Delta t } $$
+</p>
+<p>Но $\Delta x =\upsilon \Delta t\sin \alpha$.</p>
+<p>
+$$ \omega=\frac{\upsilon \Delta t\sin \alpha }{R\Delta t }=\frac{\upsilon\sin \alpha }{R} $$
+</p>
+<p>We can also write that</p>
+<p>
+$$ H=R\sin \alpha $$
+</p>
+<p>With this in mind</p>
+<p>
+$$ \omega=\frac{\upsilon\sin^2 \alpha}{H} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$\omega = \frac{v \sin^2 \alpha}{H}$$
+ </p>
+
+
+ <footer class="row container">
+ <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>
+ </p>
+ </footer>
+ </body>
+
+</html>