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+ <meta name="description" content="On the inner surface of a fixed cylinder of radius 2r, a wheel of radius r rolls without slipping. Find the path of the wheel rim point.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <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>
+ $1.5.10.$ On the inner surface of a fixed cylinder of radius $2r$, a wheel of radius $r$ rolls without slipping. Find the path of the wheel rim point.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.5.10/statement.png"
+ loading="lazy" width="150" />
+ <figcaption>
+ For problem $1.5.10$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.5.10/01.gif"
+loading="lazy" width="250" />
+<figcaption>
+Construction of a hypocycloid at $r= \frac{R}{2}$
+</figcaption>
+</figure>
+</center>
+<p>
+This trajectory is the so-called <a href="https://en.wikipedia.org/wiki/Hypocycloid" target="_blank">Hypocycloid</a><br><br>
+Given by the parametric equation:
+</p>
+<p class="exp">
+$$ \begin{gathered}\begin{cases}x=r(k-1)\left(\cos t+\frac{\cos((k-1)t)}{k-1}\right)\\y=r(k-1)\left(\sin t-\frac{\sin((k-1)t)}{k-1}\right)&\end{cases}\end{gathered} $$
+</p>
+<p>
+For $r= R/2$, this trajectory lies on the diameter
+</p>
+<h3> Proof: </h3>
+<p>Let the circles initially touch at point $A$, lying on the $Ox$ axis, where point $O$ is the center of the large circle. The coordinates of point $A$ are $ (kr, 0) $, where $ k = \frac{R}{r} $. Let us consider how the coordinates of point $A$, tied to the rolling circle ($A$ goes to $ A' $), change. Let the small circle roll so that its center moves from point $ C' $ to point $ C'' $ and rotates relative to point $ O $ by angle $ t $.</p>
+<p>First, it can be shown that the rotation of the small circle relative to its center (i.e. the angle between $ CA $ and $ C' A' $) is equal to $ t - kt = -(k-1)t $. Second, the coordinates of point $ C' $ will be as follows: $ (cos(t)(k-1)r, sin(t)(k-1)r) $. Then, knowing where the center of the rolling circle will go and at what angle it has turned relative to this center, we can write down the coordinates of point $ A' $:</p>
+<p class="exp">
+$ \begin{cases}
+x = cos(t)(k-1)r + cos((k-1)t)r \\
+y = sin(t)(k-1)r - sin((k-1)t)r
+\end{cases} $
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ The trajectory of the wheel rim point passes along the cylinder diameter.
+ </p>
+
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