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+ <meta name="description" content="The rod, with one end pivotally fixed on a horizontal plane, lies on the cylinder. Angular velocity of the rod \omega. There is no slippage between the cylinder and the plane. Find the dependence of the angular velocity of the cylinder on the angle \alpha between the rod and the plane.">
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+ <title>The rod, with one end pivotally fixed on a horizontal plane, lies on the cylinder. Angular velocity of the rod \omega. There is no slippage between the cylinder and the plane. Find the dependence of the angular velocity of the cylinder on the angle \alpha between the rod and the plane.</title>
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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>
+ $1.5.18.$ The rod, with one end pivotally fixed on a horizontal plane, lies on the cylinder. Angular velocity of the rod $\omega$. There is no slippage between the cylinder and the plane. Find the dependence of the angular velocity of the cylinder on the angle $\alpha$ between the rod and the plane.
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.5.18/statement.png"
+ loading="lazy" width="250" />
+ <figcaption>
+ For problem $1.5.18$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.5.18/draw.png"
+loading="lazy" width="250" />
+<figcaption>
+Cylinder inscribed in angle $\alpha$
+</figcaption>
+</figure>
+</center>
+
+<p>NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution <a href="../1.5.17">1.5.17</a></p>
+<p>By the time $t$ the rod will form an angle $\alpha$</p>
+<p>Consider the change in coordinate $x = AC$ over a short period of time $dt$</p>
+<p>
+$$ dx = d\left(\frac{R}{\tan (\frac{\alpha}{2})}\right) $$
+</p>
+ <p>I share both honors with $dt$</p>
+ <p>
+$$ \frac{dx}{dt} = \frac{d\left(\frac{R}{\tan \alpha/2}\right)}{dt} $$
+</p>
+<p>Considering that $\omega = \frac{d \alpha}{dt}$ and $v = \frac{dx}{dt}$</p>
+<p>
+$$ v = \frac{\omega R}{2 \sin ^2 \left(\frac{\alpha}{2}\right)} $$
+</p>
+<p>Let's write down the condition of no slippage</p>
+<p>
+$$ \omega ' R = v $$
+</p>
+<p>From here</p>
+<p>
+$$ \fbox{$\omega ' = \frac{\omega}{2 \sin ^2 \left(\frac{\alpha}{2}\right)}$} $$
+</p>
+ </p>
+
+ <h4>Answer: $ω_0 = ω/2 \sin^2(α/2)$</h4>
+
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