Translated 3.1.1-3.1.17

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		+	<div id = "logo">
		+	<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
		+	</div>
		+	<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>
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		+	<h3> Statement </h3>
		+	<p>
		+	$3.1.9.$ A ball of mass $m$ and radius $r$ slides on the surface of a well whose radius of curvature is $R$. Find the dependence of the potential energy of the ball on a small displacement $x$ from the equilibrium position.
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/3/3.1.9/statement.png"
		+	loading="lazy" width="180" />
		+	<figcaption>
		+	For problem $3.1.9$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	As in <a href="../3.1.5">3.1.5</a>, we will propose two possible solutions: through small transformations and through integration</p><p>
		+	When the body deviates horizontally by a small value $x$, its height changes by the value
		+	$$ h=R'-\sqrt{{R'}^2-x^2}=R'\left(1-\sqrt{1-\frac{x^2}{{R'}^2}}\right)\quad(1) $$
		+	Where $R'$ is the distance from the center of curvature to the center of mass of the ball.
		+	$$ R'=R-r $$
		+	Considering the smallness of $x$
		+	$$ x\ll R'\Leftrightarrow \frac{x}{R'}\ll 1 $$
		+	So we can use approximation $(1+\xi)^a\approx 1+\xi a$, where $\xi \ll 1$
		+	$$ \left(1-\frac{x^2}{{R'}^2}\right)^{1/2}=1-\frac{x^2}{2{R'}^2} $$
		+	Substitute into $(1)$
		+	$$ h=R'\cdot \frac{x^2}{2{R'}^2}=\frac{x^2}{2R'} $$
		+	Substitute into the formula for potential energy
		+	$$ U=mgh=mg\frac{x^2}{2R'} $$
		+	Substituting the value for $R'$, we obtain the final expression
		+	$$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Aliaksandr Melnichenka<br>
		+	</p>
		+	<h3>Alternative solution</h3>
		+	<p>
		+	The change in energy with a small change in coordinate can be found by definition
		+	$$ dU=Fdx $$
		+	In this case, we will find the horizontal force that will return the ball to stable equilibrium taking into account the smallness of the angle $\varphi$
		+	$$ F=mg\sin\varphi=mg\frac{x}{R-r} $$
		+	Where we find the potential energy through integration $dU$
		+	$$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{R-r}dx=\frac{mg}{R-r}\int_{0}^{x}xdx $$
		+	This integral is solved as
		+	$$ \int_{0}^{x}xdx=\frac{mg}{R-r}\cdot\frac{x^2}{2} $$
		+	Where does the potential energy of the ball come from?
		+	$$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Dzikan Mikita<br>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$U=\frac{mgx^2}{2(R-r)}$$
		+	</p>
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			<meta name="description" content="A ball of mass m and radius r slides on the surface of a well whose radius of curvature is R. Find the dependence of the potential energy of the ball on a small displacement x from the equilibrium position.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
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			<div id = "logo">
			<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
			</div>
			<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>
			$3.1.9.$ A ball of mass $m$ and radius $r$ slides on the surface of a well whose radius of curvature is $R$. Find the dependence of the potential energy of the ball on a small displacement $x$ from the equilibrium position.
			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/3/3.1.9/statement.png"
			loading="lazy" width="180" />
			<figcaption>
			For problem $3.1.9$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			As in <a href="../3.1.5">3.1.5</a>, we will propose two possible solutions: through small transformations and through integration</p><p>
			When the body deviates horizontally by a small value $x$, its height changes by the value
			$$ h=R'-\sqrt{{R'}^2-x^2}=R'\left(1-\sqrt{1-\frac{x^2}{{R'}^2}}\right)\quad(1) $$
			Where $R'$ is the distance from the center of curvature to the center of mass of the ball.
			$$ R'=R-r $$
			Considering the smallness of $x$
			$$ x\ll R'\Leftrightarrow \frac{x}{R'}\ll 1 $$
			So we can use approximation $(1+\xi)^a\approx 1+\xi a$, where $\xi \ll 1$
			$$ \left(1-\frac{x^2}{{R'}^2}\right)^{1/2}=1-\frac{x^2}{2{R'}^2} $$
			Substitute into $(1)$
			$$ h=R'\cdot \frac{x^2}{2{R'}^2}=\frac{x^2}{2R'} $$
			Substitute into the formula for potential energy
			$$ U=mgh=mg\frac{x^2}{2R'} $$
			Substituting the value for $R'$, we obtain the final expression
			$$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Aliaksandr Melnichenka<br>
			</p>
			<h3>Alternative solution</h3>
			<p>
			The change in energy with a small change in coordinate can be found by definition
			$$ dU=Fdx $$
			In this case, we will find the horizontal force that will return the ball to stable equilibrium taking into account the smallness of the angle $\varphi$
			$$ F=mg\sin\varphi=mg\frac{x}{R-r} $$
			Where we find the potential energy through integration $dU$
			$$ U=\int_{0}^{U}dU=\int_{0}^{x}mg\frac{x}{R-r}dx=\frac{mg}{R-r}\int_{0}^{x}xdx $$
			This integral is solved as
			$$ \int_{0}^{x}xdx=\frac{mg}{R-r}\cdot\frac{x^2}{2} $$
			Where does the potential energy of the ball come from?
			$$ \boxed{U=\frac{mgx^2}{2(R-r)}} $$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Dzikan Mikita<br>
			</p>

			<h4>Answer</h4>
			<p>
			$$U=\frac{mgx^2}{2(R-r)}$$
			</p>


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