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Mass moving on hoop (with hoop itself rotating)

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    vUM7fhv.png


    2. Relevant equations

    [itex] \frac{\partial\mathcal{L} }{\partial q} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}}[/itex]
    [itex] \cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)[/itex]

    3. The attempt at a solution

    The position of the center is
    [itex]
    \frac{D}{2}\left(cos(\omega t)\hat{x} + sin (\omega t)\hat{y}\right)
    [/itex]
    The vector from the center to the mass is
    [itex]
    \frac{D}{2}\left(cos(\omega t + \phi)\hat{x} + sin (\omega t +\phi)\hat{y}\right)
    [/itex]
    The position of the mass is the sume of these two vectors
    [itex]
    \frac{D}{2}\left[\left(cos(\omega t) + cos(\omega t + \phi)\right)\hat{x} +\left(sin(\omega t) + sin(\omega t + \phi)\right)\hat{y}\right]
    [/itex]
    The velocity is
    [itex]
    \frac{D}{2}\left[-\left(sin(\omega t)\omega + sin(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{x} +\left(cos(\omega t)\omega + cos(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{y}\right]
    [/itex]
    The velocity squared is
    [itex]
    \frac{D^2}{4}\left[\omega^2sin^2(\omega t) +2\omega sin(\omega t)sin(\omega t + \phi)\left(\omega + \dot{\phi}\right) +sin^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right][/itex]
    [itex]
    +\frac{D^2}{4}\left[\omega^2cos^2(\omega t) +2\omega cos(\omega t)cos(\omega t + \phi)\left(\omega + \dot{\phi}\right) +cos^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right][/itex]
    [itex]
    =\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)\left[ sin(\omega t)sin(\omega t + \phi) + cos(\omega t)cos(\omega t + \phi) \right] +\left(\omega + \dot{\phi}\right)^2\right][/itex]
    [itex]
    \cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)
    [/itex]
    [itex]
    v^2 =\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right][/itex]
    If I set [itex]\phi = 0 [/itex] and [itex]\dot{\phi} = 0 [/itex], I get [itex]v^2 = D^2\omega^2 [/itex], which is what I expect, since the diameter should just be rotating with angular frequency [itex] \omega[/itex] in this case.


    [itex]
    U = mgy = mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)
    [/itex]

    [itex]
    \mathcal{L} =\frac{mD^2}{8}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right] -mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)
    [/itex]

    [itex] \frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/itex]
    [itex] \frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{8}2\omega\left(\omega + \dot{\phi}\right)sin(\phi) -mg\frac{D}{2}\cos(\omega t + \phi) [/itex]
    [itex] \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{8}\left[ 2\omega cos(\phi) + 2\dot{\phi} + 2\omega \right] = \frac{mD^2}{4}\left[ \omega cos(\phi) + \dot{\phi} + \omega \right] [/itex]
    [itex] \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right] [/itex]
    [itex]
    \frac{mD^2}{4}\ddot{\phi} = - \frac{mD^2}{4}\omega\dot{\phi}sin(\phi) - mg\frac{D}{2}\cos(\omega t + \phi)
    [/itex]
    [itex]
    \ddot{\phi} = -\omega\dot{\phi}sin(\phi) - \frac{2g}{D}\cos(\omega t + \phi)
    [/itex]
    That doesn't look exactly like a simple pendulum to me. I'd appreciate some help.
     
  2. jcsd
  3. Apr 8, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Nicely written!

    The problem says that the system rotates about a vertical axis. So, does that imply that the system rotates in a horizontal plane?

    Check to see if you might have made a careless error when combining terms to get
    [itex] \frac{mD^2}{4}\ddot{\phi} = - \frac{mD^2}{4}\omega\dot{\phi}sin(\phi) - mg\frac{D}{2}\cos(\omega t + \phi)
    [/itex]

    I'm not sure the first term on the right should be there and I think you overlooked a term that should be there.
     
  4. Apr 8, 2013 #3
    Thanks for noticing this. I had interpreted the rotation as being in a vertical plane. So with a horizontal plane I may have U = 0. I also left out a term when connecting both sides of the Euler-Lagrange Equation. Part of the term I omitted cancels the "first term on the right" that you weren't sure should be there.

    [itex]
    \mathcal{L} =\frac{mD^2}{8}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right]
    [/itex]

    [itex] \frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/itex]
    [itex] \frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{8}2\omega\left(\omega + \dot{\phi}\right)sin(\phi) = - \frac{mD^2}{4}\omega\left(\omega + \dot{\phi}\right)sin(\phi)[/itex]


    [itex] \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{8}\left[ 2\omega cos(\phi) + 2\dot{\phi} + 2\omega \right] = \frac{mD^2}{4}\left[ \omega cos(\phi) + \dot{\phi} + \omega \right] [/itex]
    [itex] \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right] [/itex]

    [itex] \frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{4}\omega\left(\omega + \dot{\phi}\right)sin(\phi) = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right] = \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} [/itex]

    [itex] - \omega\left(\omega + \dot{\phi}\right)sin(\phi) = - \omega sin(\phi)\dot{\phi} + \ddot{\phi} [/itex]
    [itex] - \omega^2sin(\phi) -\omega\dot{\phi} sin(\phi) = - \omega sin(\phi)\dot{\phi} + \ddot{\phi} [/itex]
    The [itex] - \omega sin(\phi)\dot{\phi} [/itex] may be removed from both sides.
    [itex] \ddot{\phi} + \omega^2sin(\phi) = 0 [/itex]
     
    Last edited: Apr 9, 2013
  5. Apr 9, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Looks good to me.
     
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