$3.2.17.$ 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.
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 $$ \boxed{T=\frac{2\pi}{\omega}\sqrt{\frac{l}{R+l}}} $$
$$T=\frac{2\pi}{\omega}\sqrt{\frac{l}{R+l}}$$