Lets consider a wire in the x-y plane which rotates with constant angular velocity $\omega$. The coordinates of a bead, which is forced to stay on this wire, can then be expressed as $$x=r \cos(\phi) \\ y=r \sin(\phi)$$ with the rheonomous constraint $\phi=\omega t$. The Lagrangian $L$ is then simply given as $$L=\frac{m}{2}(\dot{r}^2+r^2\omega^2)$$ and the e.o.m are just $$\ddot{r}-\omega^2r=0$$ with the solution $$r(t)=r_0\cosh(\omega t) +\frac{v_0}{\omega} \sinh(\omega t).$$ Now it follows immediately that neither the energy nor the angular momentum are conserved. I think the reason is that we are dealing with rheonomous constraints. Is that correct? But intuitively I would have guessed that a constraint like $\phi=\omega t$ does not necessarily lead to an increasing $r$ (if $v_0\geq0$). So what drives the bead outwards? The constraint only demands that $\phi$ should increase linearly. This would also be fulfilled by a simple circular motion of the bead with constant $r$. Of course, this is not a solution for the differential equation given above. But why not? Is there something more included in the Lagrangian which makes the bead go outwards (something like centrifugal force)?
The maths are absolutely clear, I am just wondering why and how the bead is driven outwards.