Translated 3.2.1-3.2.17

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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3>
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+ <h3> Statement </h3>
+ <p>
+ $3.2.17.$ A spacecraft rotates around its axis with angular velocity $\omega$. How does the period of oscillation of a pendulum of length $l$ depend on the distance $R$ of the suspension point to the axis of rotation? The plane of oscillation passes through the axis of rotation.
+ </p>
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+ <h3>Solution</h3>
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+ <figure>
+ <img src="https://savchenkosolutions.com/3/3.2.17/3.2.17_1.png"
+ loading="lazy" width="60" />
+ <figcaption>
+ The pendulum deflected a small distance $x$
+ </figcaption>
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+<p>
+When rotating along an arc of a circle of radius $l$, the ball is subject to centripetal acceleration
+$$ a = \omega^2(R+l) $$
+Newton's second law for a ball:
+$$ m\ddot{x}(t)=-m\omega^2(R+l)\sin\varphi $$
+Neglecting gravitational interaction with the Earth, we write Newton's second law
+$$ m\ddot{x}(t)+m\omega^2(R+l)\sin\varphi=0\quad(1) $$
+Let's use the approximation for small angles $(\varphi \ll 1)$:
+$$ \sin\varphi\approx\varphi=\frac{x}{l} $$
+Using the approximation for $(1)$, we obtain the equation of harmonic oscillations
+$$ \ddot{x}(t)+\frac{\omega^2(R+l)}{l}x(t)=0 $$
+We solve the equation of harmonic oscillations of the form $\ddot{x}+\omega ^2x(t)=0$ using the standard method and obtain the desired oscillation period
+$$ \boxed{T=\frac{2\pi}{\omega}\sqrt{\frac{l}{R+l}}} $$
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+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Dzikan Mikita<br>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$T=\frac{2\pi}{\omega}\sqrt{\frac{l}{R+l}}$$
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