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.
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 $a$.