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.
Best Answer
In a coordinate system rotating at constant angular rate $\omega$, neither energy nor angular momentum are conserved and one has coriolis and centrifugal forces. The bead is forced outwards by the centrifugal force: the energy increases by the work done on the bead by the rotating system.
In fact, since the Hamiltonian $$H=\frac{p^2}{2m} - \frac{1}{2}\omega^2r^2$$ satisfies $$\frac{\partial H}{\partial t}=0$$ it is conserved (recall that $dH/dt=\partial H/\partial t$). It is related to energy and angular momentum via $$H=E-\omega J,$$ also known as Jacobi energy. This is still conserved if there is a conservative force that in the rotating frame is time-independent, e.g. if the bead on the string were attached to the origin by a spring.