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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>
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			<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.3.15.$ A ball is launched along the inner surface of a smooth vertical cylinder of radius $R$ at an angle $\alpha$ to the vertical. What initial velocity must be given to it so that it returns to the starting point?
			</p>

			<h3>Solution</h3>
			<p>
			<p>We will divide the movement into vertical and in the plane of the cross section of the cylinder</p>
			<p class="exp">
			$x = vt \sin \alpha$
			</p>
			<p class="exp">
			$$ y = vt \cos \alpha - \frac{gt^2}{2} $$
			</p>
			<p>We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0:</p>

			<p class="exp">
			$$ vt \cos \alpha - \frac{gt^2}{2} = 0 $$
			</p>
		@@ -70,7 +70,7 @@
			$$vt \sin \alpha = 2\pi n R,\,n\in \mathbb{N}$$
			We find time $t$ as
			$$ t = \frac{2\pi n R}{v \sin \alpha} $$
		−	$$ 2v \cos \alpha = g\frac{2\pi n R}{v \sin \alpha} $$
		+	$$ 2v \cos \alpha = g\frac{2\pi n R}{v \sin \alpha} $$
			<p>From here we express $v$:
			<p class="exp">
			$$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$
			</p>
			</p>

			<h4>Answer</h4>
			<p>
			$v=\sqrt{2\pi Rgn/\sin2\alpha}$, where $n$ is any natural number; when $\alpha = 0$, the speed can be any in absolute value.

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	<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>
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	<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.3.15.$ A ball is launched along the inner surface of a smooth vertical cylinder of radius $R$ at an angle $\alpha$ to the vertical. What initial velocity must be given to it so that it returns to the starting point?		$1.3.15.$ A ball is launched along the inner surface of a smooth vertical cylinder of radius $R$ at an angle $\alpha$ to the vertical. What initial velocity must be given to it so that it returns to the starting point?
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<p>We will divide the movement into vertical and in the plane of the cross section of the cylinder</p>		<p>We will divide the movement into vertical and in the plane of the cross section of the cylinder</p>
	<p class="exp">		<p class="exp">
	$x = vt \sin \alpha$		$x = vt \sin \alpha$
	</p>		</p>
	<p class="exp">		<p class="exp">
	$$ y = vt \cos \alpha - \frac{gt^2}{2} $$		$$ y = vt \cos \alpha - \frac{gt^2}{2} $$
	</p>		</p>
	<p>We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0:</p>		<p>We take into account that the displacement along $Ox$ is equal to the circumference, and along $Oy$ — 0:</p>

	<p class="exp">		<p class="exp">
	$$ vt \cos \alpha - \frac{gt^2}{2} = 0 $$		$$ vt \cos \alpha - \frac{gt^2}{2} = 0 $$
	</p>		</p>
	@@ -70,7 +70,7 @@
	$$vt \sin \alpha = 2\pi n R,\,n\in \mathbb{N}$$		$$vt \sin \alpha = 2\pi n R,\,n\in \mathbb{N}$$
	We find time $t$ as		We find time $t$ as
	$$ t = \frac{2\pi n R}{v \sin \alpha} $$		$$ t = \frac{2\pi n R}{v \sin \alpha} $$
	$$ 2v \cos \alpha = g\frac{2\pi n R}{v \sin \alpha} $$		$$ 2v \cos \alpha = g\frac{2\pi n R}{v \sin \alpha} $$
	<p>From here we express $v$:		<p>From here we express $v$:
	<p class="exp">		<p class="exp">
	$$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$		$$\fbox{$v=\sqrt{2\pi Rgn/\sin2\alpha}$}$$
	</p>		</p>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$v=\sqrt{2\pi Rgn/\sin2\alpha}$, where $n$ is any natural number; when $\alpha = 0$, the speed can be any in absolute value.		$v=\sqrt{2\pi Rgn/\sin2\alpha}$, where $n$ is any natural number; when $\alpha = 0$, the speed can be any in absolute value.

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