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.

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Solutions of Savchenko Problems in Physics
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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.
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    <h3>Solution</h3>
    <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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    <h4>Answer</h4>
    <p>
        $$T=\frac{2\pi}{\omega}\sqrt{\frac{l}{R+l}}$$
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