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Say i have a bead of mass $m$ sliding on a friction-less rode (or wire) that is rotating with a permanent angular velocity $ω$. The whole system is under the influence of a uniform gravitational field $F=mg$.

How many degrees of freedom do the system have?

Assuming we use polar coordinates, it seems as if we have only one degree of freedom and that is the radial coordinate - r, that is because the polar angle theta is set to be ω*t. But on the other hand, r cant be smaller than 0, while it is obvious that in the physical system - if the angular velocity is slow and gravity is strong, the bead can slide to the other side of the rotation axis.

So will this system's Lagrangian depend on both r and theta and so the Euler–Lagrange equation should be applied once for each of them?

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  • $\begingroup$ The radial coordinate can be smaller than 0. I'd say there's only 1 DOF. You don't need to apply the Euler–Lagrange equation to theta. $\endgroup$ – Laff70 Apr 6 at 14:29
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The bead is constrained to move along a line, so, as you thought, there is only 1 degree of freedom. The problem with your polar coordinates choice is that you did not choose the right coordinates system. The coordinate is "position along the rod", not "distance from center". One way of thinking about this is to presume you want to use Cartesian coordinates and say that since both x and y can vary, you have 2 degrees of freedom. That would not make any sense. The number of degrees of freedom does not depend on the coordinates system used. The trick is to use the right coordinates system, one in which all states of the system can be represented with the number of variables equal to the number of degrees of freedom. So, polar coordinates is simply not the right coordinate system in your example. If you use instead "position along the rod" with the center of the rod being the origin and position being allowed to have both positive and negative values, you will have the right coordinates system.

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