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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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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<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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			For problem $1.5.17$
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			<p>
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			<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>
		−	$$ x = H \tan \alpha\;(1) $$
		+	$$ x = H \tan \alpha\quad(1) $$
			</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>
			$$ \omega = \frac{vH}{x^2+H^2} $$
			</p>
			<p>Substitute $(1)$</p>
			<p>
		@@ -90,7 +90,7 @@
			$$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$
			</p>
			<p>
		−	$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$
		+	$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$
			</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>
			$$ H=R\sin \alpha $$
		@@ -126,13 +126,13 @@
			</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>
		−	$$\omega = \frac{v \sin^2 \alpha}{H}$$
		+	$$\omega = \frac{v \sin^2 \alpha}{H}$$
			</p>


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	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>
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	<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>
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	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>
	$$ x = H \tan \alpha\;(1) $$		$$ x = H \tan \alpha\quad(1) $$
	</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>
	$$ \omega = \frac{vH}{x^2+H^2} $$		$$ \omega = \frac{vH}{x^2+H^2} $$
	</p>		</p>
	<p>Substitute $(1)$</p>		<p>Substitute $(1)$</p>
	<p>		<p>
	@@ -90,7 +90,7 @@
	$$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$		$$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$
	</p>		</p>
	<p>		<p>
	$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$		$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$
	</p>		</p>


	<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>
	$$ \Delta l=\upsilon \Delta t $$		$$ \Delta l=\upsilon \Delta t $$
	</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>
	</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>
	$$ \omega=\frac{\Delta \alpha }{\Delta t }=\frac{\Delta x }{R\Delta t } $$		$$ \omega=\frac{\Delta \alpha }{\Delta t }=\frac{\Delta x }{R\Delta t } $$
	</p>		</p>
	<p>Но $\Delta x =\upsilon \Delta t\sin \alpha$.</p>		<p>Но $\Delta x =\upsilon \Delta t\sin \alpha$.</p>
	<p>		<p>
	$$ \omega=\frac{\upsilon \Delta t\sin \alpha }{R\Delta t }=\frac{\upsilon\sin \alpha }{R} $$		$$ \omega=\frac{\upsilon \Delta t\sin \alpha }{R\Delta t }=\frac{\upsilon\sin \alpha }{R} $$
	</p>		</p>
	<p>We can also write that</p>		<p>We can also write that</p>
	<p>		<p>
	$$ H=R\sin \alpha $$		$$ H=R\sin \alpha $$
	@@ -126,13 +126,13 @@
	</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>


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