A bead of mass $m$ is threaded around a smooth spiral wire and slides downwards without friction due to gravity. The $z$-axis points upwards vertically. Suppose the spiral wire is rotated about the $z$-axis with a fixed angular velocity $\Omega$. Determine the Lagrangian and the equation of motion.
This is related to a previous problem in which the wire shape is given as $$ z = k\psi, \hspace{3mm} x = a\cos\psi, \hspace{3mm} y = a\sin\psi $$ where $a$ and $k$ are both positive.
My attempt at a solution: We still have $z = k\psi$, but now $x = a\cos(\psi + \Omega t)$ and $y = a\sin(\psi + \Omega t)$. This gives $\dot{z} = k\dot{\psi}$, $\dot{x} = -a(\dot{\psi} + \Omega)\sin(\psi + \Omega t)$ and $\dot{y} = a(\dot{\psi} + \Omega)\cos(\psi + \Omega t)$. Then the kinetic energy is $$ T = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) = \frac{1}{2}m\left(a^2\left(\dot{\psi}^2 + 2\dot{\psi}\Omega + \Omega^2\right) + k^2\dot{\psi}^2\right) $$ and the potential energy is $V = mgz = mgk\psi$. Then the Lagrangian becomes: $$ L = T - V = \frac{1}{2}m\left(a^2\left(\dot{\psi}^2 + 2\dot{\psi}\Omega + \Omega^2\right) + k^2\dot{\psi}^2\right) - mgk\psi. $$ This gives $$ \frac{\partial L}{\partial \psi} = -mgk, \hspace{3mm} \frac{\partial L}{\partial \dot{\psi}} = m\left(a^2\dot{\psi} + a^2\Omega + k^2\dot{\psi}\right), \hspace{3mm} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\psi}}\right) = m\left(a^2 + k^2\right)\ddot{\psi} $$ so plugging into Lagrange's Equation gives $-mgk = m\left(a^2 + k^2\right)\ddot{\psi}$ or $\ddot{\psi} = -\frac{gk}{a^2 + k^2}$, which is the exact same equation of motion as in the case with the coil not rotating. Obviously this isn't correct. I think the problem might be related to my choice of coordinates (in particular, having $\psi$ rotating), but I'm not sure what a better choice of coordinates would be.
