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nhalford
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Homework Statement
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.
Homework Equations
L = T - V
\frac{\partial L}{\partial x} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)
The Attempt at a Solution
This is related to a previous problem in which the wire is not rotating and its 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. For that problem, the resulting equation of motion is \ddot{\psi} = -\frac{gk}{a^2 + k^2}
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
<br /> 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)<br />
and the potential energy is V = mgz = mgk\psi. Then the Lagrangian becomes:
<br /> 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.<br />
This gives
<br /> \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}<br />
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. Where am I going wrong here? My instinct is that there might be a problem with my choice of coordinates; in particular, \psi is rotating, but I'm not sure if there is a better choice of coordinates.
