A straight wire rotates with constant angular speed $\omega$ about one of its end points (the origin $O$) in a horizontal plane containing $e_1$ and $e_2$. A bead of mass $m$ is free to slide along the wire, but is connected to $O$ by a spring coiled around the wire, as illustrated below. The spring has natural length $a$ and spring constant $k$. The magnitude of the frictional force is proportional to the radial speed of the bead with a frictional constant $K$.
Assuming $\omega\neq\omega_s$, find the equilibrium length of the spring, say $R$.
I have worked through previous parts of the question and have derived the equation of motion of the bead which is $$\ddot{r}+\alpha\dot{r}=(\omega^2-\omega_s^2)r+a\omega_s^2,$$ where $\alpha=\frac{K}{m}$ and $\omega_s^2=\frac{k}{m}$.
But I struggle on how to derive the equilibrium length of the spring, I don't know how to start.
Thank you.
$\endgroup$ 31 Answer
$\begingroup$The bead is free to slide on the wire and we’re not given any information about static friction, so if the bead is whirling around at a constant radius $R$, the only thing providing the centripetal force for this is the spring. Equate the two forces $$-mR\omega^2=-k(R-a)$$ and solve for $R$. As a check on the calculation, make sure that the result is greater than $a$.
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