What is the Lagrangian and equation of motion for a bead on a rotating coil?

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SUMMARY

The discussion focuses on determining the Lagrangian and equation of motion for a bead of mass m sliding down a rotating spiral wire with a fixed angular velocity \(\Omega\). The Lagrangian is derived as \(L = \frac{1}{2}m\left(a^2\left(\dot{\psi}^2 + 2\dot{\psi}\Omega + \Omega^2\right) + k^2\dot{\psi}^2\right) - mgk\psi\). The resulting equation of motion is \(\ddot{\psi} = -\frac{gk}{a^2 + k^2}\), which matches the non-rotating case, raising questions about the choice of coordinates. The discussion highlights the importance of considering fictitious forces in a rotating frame.

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  • Familiarity with rotational dynamics
  • Knowledge of kinetic and potential energy calculations
  • Concept of fictitious forces in non-inertial reference frames
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  • Explore the effects of centrifugal and Coriolis forces on motion
  • Learn about coordinate transformations in dynamics
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Physics students, mechanical engineers, and anyone studying dynamics involving rotating systems will benefit from this discussion.

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.
 
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Hello nhalford. Welcome to PF!

Why do you say your result is incorrect?

Suppose you went to the rotating frame. Then in this frame the wire would not be rotating but you would have centrifugal and Coriolis fictitious forces. Consider the directions of these fictitious forces and decide whether or not they would have any effect on the z equation of motion (or equivalently the ##\psi## equation of motion, since z and ##\psi## are proportional).
 
Ah, you're right, thanks. It just seemed counterintuitive to me.
 

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