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Bead on a rotating coil

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data
    A bead of mass [itex]m[/itex] is threaded around a smooth spiral wire and slides downwards without friction due to gravity. The [itex]z[/itex]-axis points upwards vertically. Suppose the spiral wire is rotated about the [itex]z[/itex]-axis with a fixed angular velocity [itex]\Omega[/itex]. Determine the Lagrangian and the equation of motion.


    2. Relevant equations
    [tex]L = T - V[/tex]
    [tex]\frac{\partial L}{\partial x} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)[/tex]

    3. 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 [tex]z = k\psi, \hspace{3mm} x = a\cos\psi, \hspace{3mm} y = a\sin\psi[/tex] where [itex]a[/itex] and [itex]k[/itex] are both positive. For that problem, the resulting equation of motion is [itex]\ddot{\psi} = -\frac{gk}{a^2 + k^2}[/itex]

    We still have [itex]z = k\psi[/itex], but now [itex]x = a\cos(\psi + \Omega t)[/itex] and [itex]y = a\sin(\psi + \Omega t)[/itex]. This gives [itex]\dot{z} = k\dot{\psi}[/itex], [itex]\dot{x} = -a(\dot{\psi} + \Omega)\sin(\psi + \Omega t)[/itex] and [itex]\dot{y} = a(\dot{\psi} + \Omega)\cos(\psi + \Omega t)[/itex]. Then the kinetic energy is
    [tex]
    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)
    [/tex]
    and the potential energy is [itex]V = mgz = mgk\psi[/itex]. Then the Lagrangian becomes:
    [tex]
    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.
    [/tex]
    This gives
    [tex]
    \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}
    [/tex]
    so plugging into Lagrange's Equation gives [itex]-mgk = m\left(a^2 + k^2\right)\ddot{\psi}[/itex] or [itex]\ddot{\psi} = -\frac{gk}{a^2 + k^2}[/itex], 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, [itex]\psi[/itex] is rotating, but I'm not sure if there is a better choice of coordinates.
     
  2. jcsd
  3. Feb 3, 2014 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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).
     
  4. Feb 5, 2014 #3
    Ah, you're right, thanks. It just seemed counterintuitive to me.
     
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