Updated spacing between @ latex expressions
en/1.5.17.md
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| <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> | |||
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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | |||
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| <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="../../#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 $v$. | |||
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| <figcaption> | |||
| For problem $1.5.17$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| @@ -69,7 +69,7 @@ | |||
| <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> | |||
| − | $$ x = H \tan | ||
| + | $$ x = H \tan\alpha\quad(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} $$ | |||
| @@ -87,7 +87,7 @@ | |||
| </p> | |||
| <p>Substitute $(1)$</p> | |||
| <p> | |||
| − | $$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$ | ||
| + | $$ \omega = \frac{v}{H(1 + \tan^2 \alpha )} $$ | ||
| </p> | |||
| <p> | |||
| $$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$ | |||
| </p> | |||
| @@ -97,7 +97,7 @@ | |||
| <h3> Alternative solution: </h3> | |||
| <p>In time $\Delta t$ the truck will travel a distance</p> | |||
| <p> | |||
| − | $$ \Delta l=\upsilon | ||
| + | $$ \Delta l=\upsilon\Delta t $$ | ||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="https://savchenkosolutions.com/1/1.5.17/draw.png" | |||
| loading="lazy" width="280" /> | |||
| <figcaption> | |||
| Log and truck | |||
| </figcaption> | |||
| </figure> | |||
| @@ -110,23 +110,23 @@ | |||
| </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> | |||
| − | $$ \Delta \alpha=\frac{\Delta x }{R} $$ | ||
| + | $$ \Delta \alpha =\frac{\Delta x }{R} $$ | ||
| </p> | |||
| <p>Angular velocity is defined as</p> | |||
| <p> | |||
| − | $$ \omega=\frac{\Delta \alpha | ||
| + | $$ \omega =\frac{\Delta \alpha}{\Delta t }=\frac{\Delta x }{R\Delta t } $$ | ||
| </p> | |||
| − | <p>Но $\Delta x =\upsilon | ||
| + | <p>Но $\Delta x =\upsilon\Delta t\sin\alpha$.</p> | ||
| <p> | |||
| − | $$ \omega=\frac{\upsilon | ||
| + | $$ \omega =\frac{\upsilon\Delta t\sin\alpha}{R\Delta t }=\frac{\upsilon\sin\alpha}{R} $$ | ||
| </p> | |||
| <p>We can also write that</p> | |||
| <p> | |||
| − | $$ H=R\sin | ||
| + | $$ H=R\sin\alpha$$ | ||
| </p> | |||
| <p>With this in mind</p> | |||
| <p> | |||
| − | $$ \omega=\frac{\upsilon\sin^2 \alpha}{H} $$ | ||
| + | $$ \omega =\frac{\upsilon\sin^2 \alpha}{H} $$ | ||
| </p> | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\omega = \frac{v \sin^2 \alpha}{H}$$ | |||
| </p> | |||
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| <p> | |||
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| <meta name="date" content="2023-10" scheme="YYYY-MM"> | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| <div id="logo"> | <div id="logo"> | ||
| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | ||
| </div> | </div> | ||
| </a> | </a> | ||
| <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> | ||
| </header> | </header> | ||
| <h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3> | <h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3> | ||
| <h3> Statement </h3> | <h3> Statement </h3> | ||
| <p> | <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$. | $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> | </p> | ||
| <center> | <center> | ||
| <figure> | <figure> | ||
| <img src="https://savchenkosolutions.com/1/1.5.17/statement.png" | <img src="https://savchenkosolutions.com/1/1.5.17/statement.png" | ||
| loading="lazy" width="250" /> | loading="lazy" width="250" /> | ||
| <figcaption> | <figcaption> | ||
| For problem $1.5.17$ | For problem $1.5.17$ | ||
| </figcaption> | </figcaption> | ||
| </figure> | </figure> | ||
| </center> | </center> | ||
| <p> | <p> | ||
| </p> | </p> | ||
| <h3>Solution</h3> | <h3>Solution</h3> | ||
| <p> | <p> | ||
| @@ -69,7 +69,7 @@ | |||
| <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>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>At time $t$ the car will travel the distance $vt$ and the bottom will have horizontal coordinate $x$</p> | ||
| <p> | <p> | ||
| $$ x = H \tan |
$$ x = H \tan\alpha\quad(1) $$ | ||
| </p> | </p> | ||
| <p>Let's consider the change in angle $\alpha = \arctan \frac{H}{x}$ over a small period of time $dt$</p> | <p>Let's consider the change in angle $\alpha = \arctan \frac{H}{x}$ over a small period of time $dt$</p> | ||
| <p> | <p> | ||
| $$ d\alpha = d\left(\arctan \frac{H}{x}\right) $$ | $$ d\alpha = d\left(\arctan \frac{H}{x}\right) $$ | ||
| </p> | </p> | ||
| <p> | <p> | ||
| $$ d\alpha = \frac{1}{1+\frac{H^2}{x^2}} \frac{H}{x^2} $$ | $$ d\alpha = \frac{1}{1+\frac{H^2}{x^2}} \frac{H}{x^2} $$ | ||
| </p> | </p> | ||
| <p> | <p> | ||
| $$ d\alpha = dx\frac{H}{x^2+H^2} $$ | $$ d\alpha = dx\frac{H}{x^2+H^2} $$ | ||
| </p> | </p> | ||
| <p>Considering that $\omega = \frac{d \alpha}{dt}$ and $v = \frac{dx}{dt}$</p> | <p>Considering that $\omega = \frac{d \alpha}{dt}$ and $v = \frac{dx}{dt}$</p> | ||
| <p> | <p> | ||
| $$ \omega = \frac{vH}{x^2+H^2} $$ | $$ \omega = \frac{vH}{x^2+H^2} $$ | ||
| @@ -87,7 +87,7 @@ | |||
| </p> | </p> | ||
| <p>Substitute $(1)$</p> | <p>Substitute $(1)$</p> | ||
| <p> | <p> | ||
| $$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$ | $$ \omega = \frac{v}{H(1 + \tan^2 \alpha )} $$ | ||
| </p> | </p> | ||
| <p> | <p> | ||
| $$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$ | $$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$ | ||
| </p> | </p> | ||
| @@ -97,7 +97,7 @@ | |||
| <h3> Alternative solution: </h3> | <h3> Alternative solution: </h3> | ||
| <p>In time $\Delta t$ the truck will travel a distance</p> | <p>In time $\Delta t$ the truck will travel a distance</p> | ||
| <p> | <p> | ||
| $$ \Delta l=\upsilon |
$$ \Delta l=\upsilon\Delta t $$ | ||
| </p> | </p> | ||
| <center> | <center> | ||
| <figure> | <figure> | ||
| <img src="https://savchenkosolutions.com/1/1.5.17/draw.png" | <img src="https://savchenkosolutions.com/1/1.5.17/draw.png" | ||
| loading="lazy" width="280" /> | loading="lazy" width="280" /> | ||
| <figcaption> | <figcaption> | ||
| Log and truck | Log and truck | ||
| </figcaption> | </figcaption> | ||
| </figure> | </figure> | ||
| @@ -110,23 +110,23 @@ | |||
| </center> | </center> | ||
| <p>The log will drop slightly, turning at a small angle $\Delta \alpha$. If the length of the arc of rotation is $\Delta x$, then by definition of the radian measure of angle</p> | <p>The log will drop slightly, turning at a small angle $\Delta \alpha$. If the length of the arc of rotation is $\Delta x$, then by definition of the radian measure of angle</p> | ||
| <p> | <p> | ||
| $$ \Delta \alpha=\frac{\Delta x }{R} $$ | $$ \Delta \alpha =\frac{\Delta x }{R} $$ | ||
| </p> | </p> | ||
| <p>Angular velocity is defined as</p> | <p>Angular velocity is defined as</p> | ||
| <p> | <p> | ||
| $$ \omega=\frac{\Delta \alpha |
$$ \omega =\frac{\Delta \alpha}{\Delta t }=\frac{\Delta x }{R\Delta t } $$ | ||
| </p> | </p> | ||
| <p>Но $\Delta x =\upsilon |
<p>Но $\Delta x =\upsilon\Delta t\sin\alpha$.</p> | ||
| <p> | <p> | ||
| $$ \omega=\frac{\upsilon |
$$ \omega =\frac{\upsilon\Delta t\sin\alpha}{R\Delta t }=\frac{\upsilon\sin\alpha}{R} $$ | ||
| </p> | </p> | ||
| <p>We can also write that</p> | <p>We can also write that</p> | ||
| <p> | <p> | ||
| $$ H=R\sin |
$$ H=R\sin\alpha$$ | ||
| </p> | </p> | ||
| <p>With this in mind</p> | <p>With this in mind</p> | ||
| <p> | <p> | ||
| $$ \omega=\frac{\upsilon\sin^2 \alpha}{H} $$ | $$ \omega =\frac{\upsilon\sin^2 \alpha}{H} $$ | ||
| </p> | </p> | ||
| </p> | </p> | ||
| <h4>Answer</h4> | <h4>Answer</h4> | ||
| <p> | <p> | ||
| $$\omega = \frac{v \sin^2 \alpha}{H}$$ | $$\omega = \frac{v \sin^2 \alpha}{H}$$ | ||
| </p> | </p> | ||
| <footer class="row container"> | <footer class="row container"> | ||
| <br> | <br> | ||
| <p> | <p> | ||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | ||
| </p> | </p> | ||
| <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> | <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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