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Transforming coordinates for a vertical hoop

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A bead of mass m slips without friction an a vertically oriented hoop of radius R.
    The hoop rotates around the z axis at a constant rate w.

    (question 4 if anyone wants to see the figure)
    http://phstudy.technion.ac.il/~wn114101/hw/wn2010_hw05.pdf" [Broken]


    2. Relevant equations

    Write the Lagrangian for the system

    3. The attempt at a solution

    I need to transform coordinates.
    I thought about transforming to spherical ones.

    constraints : r = R, [tex] \phi = wt [/tex]

    [tex] x = R sin(\theta) cos(wt), y = R sin(\theta) sin(wt) , z = R cos(\theta) [/tex]

    does this seem right? because when I try to write the velocity I get a very ugly equation

    [tex] \dot{x}^2 + \dot{y}^2 + \dot{z}^2 = \dot{\theta}^2R^2 + w^2R^2sin^2(\theta) - 2\dot{\theta}wR^2 sin(\theta)sin(wt)cos(wt)cos(\theta) +2\dot{\theta}wR^2cos(\theta)sin(wt)sin(\theta)cos(wt) + \dot{\theta}R^2sin^2(\theta) [/tex]

    which simply doesnt look right..
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 24, 2009 #2
    never mind.. i just noticed it..
     
  4. Nov 24, 2009 #3
    That doesn't look ugly! There's two terms that cancel and reduces the problem.
     
  5. Nov 24, 2009 #4
    yea.. thanks.. i noticed it..

    a question though

    is
    [tex] \frac{m}{2}\left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta)\right) + mgRcos(\theta)[/tex]

    the energy in the system?
     
  6. Nov 24, 2009 #5
    This would be the Lagrangian, but not the energy. Your potential should be negative due to the use of [itex]\cos\theta[/itex].
     
  7. Nov 24, 2009 #6
    the Lagrangian is

    [tex] L = \frac{m}{2} \left(\dot{\theta}^2R^2 + w^2R^2sin^2(\theta) + \dot{\theta}R^2sin^2(\theta)\right) - mgRcos(\theta) [/tex]

    from which i've derived the hamiltonian according to

    [tex] H = \frac{\partial L}{\partial \dot{\theta}}\dot{\theta} - L [/tex]

    which is what i wrote below. what i'm wondering is, is this the energy?
     
  8. Nov 24, 2009 #7

    You seem to be adding something here. Where did [tex]\dot{\theta}R^2sin^2(\theta)[/tex] come from? Your kinetic energy should be:

    [tex]
    T=\frac{1}{2}m\left(\dot{R}^2+R^2\dot{\theta}^2+R^2\sin^2\theta\dot{\phi}^2\right)=\frac{1}{2}m\left(R^2\dot{\theta}^2+R^2\sin^2\theta\dot{\phi}^2\right)
    [/tex]

    Also, your kinetic energy should be negative, so in the Lagrangian it should be positive: [itex]L=T-V=T+mgR\cos\theta[/itex].
     
  9. Nov 24, 2009 #8
    [tex] T = \frac{m}{2} (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) [/tex]
    and according to my first post, you can see where all the terms came from.
    also, [tex] \dot{R} = 0 [/tex]
    a small mistake, the last term should be [tex] \dot{\theta}^2R^2sin^2(\theta) [/tex]
     
  10. Nov 24, 2009 #9
    I know that R-dot is zero, I did cancel it in my equation. But I think you are confusing [tex]\dot{\mathbf{r}}\cdot\dot{\mathbf{r}}[/tex] with [tex]\dot{r}^2[/itex]. Remember that both your position and velocities are vectors, so when you see the square term, you must think to take the dot product. Your kinetic energy should be as I wrote it above.

    I probably should have caught it in the first post, but I did just wake up about an hour and a half ago...
     
  11. Nov 24, 2009 #10
    I'm trying to see what i did wrong and i can't find it. Can you see it?
     
  12. Nov 24, 2009 #11
    I get:

    [tex]
    \dot{x}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\sin^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\theta\cos\phi
    [/tex]

    [tex]
    \dot{y}^2=\dot{\theta}^2R^2\cos^2\theta\sin^2\phi+\dot{\phi}^2R^2\sin^2\theta\cos^2\phi+2\dot{\theta}\dot{\phi}R^2\sin\theta\sin\phi\cos\theta\cos\phi
    [/tex]

    [tex]
    \dot{z}^2=\dot{\theta}^2R^2\sin^2\theta
    [/tex]

    So adding them together

    [tex]
    \dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\sin^2\phi-2\dot{\phi}\dot{\theta}R^2\sin\theta\sin\phi\cos\theta\cos\phi+\dot{\theta}^2R^2\cos^2\theta\sin^2\phi
    [/tex]

    [tex]
    +\dot{\phi}^2R^2\sin^2\theta\cos^2\phi+2\dot{\theta}\dot{\phi}R^2\sin\theta\sin\phi\cos\theta\cos\phi+\dot{\theta}^2R^2\sin^2\theta
    [/tex]

    Clearly those two cross terms cancel,


    [tex]
    \dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\cos^2\theta\cos^2\phi+\dot{\phi}^2R^2\sin^2\theta\sin^2\phi+\dot{\theta}^2R^2\cos^2\theta\sin^2\phi+\dot{\phi}^2R^2\sin^2\theta\cos^2\phi+\dot{\theta}^2R^2\sin^2\theta
    [/tex]

    Some trig identities ensue:


    [tex]
    \dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2\cos^2\theta+\dot{\phi}^2R^2\sin^2\theta+\dot{\theta}^2R^2\sin^2\theta
    [/tex]

    A few more trig identities:

    [tex]
    \dot{x}^2+\dot{y}^2+\dot{z}^2=\dot{\theta}^2R^2+\dot{\phi}^2R^2\sin^2\theta
    [/tex]
     
  13. Nov 24, 2009 #12
    indeed.. i saw where i went wrong. thanks alot!!

    so the hamiltonian here is the energy of the system?
     
  14. Nov 24, 2009 #13
    If there is no explicit time dependence in the Lagrangian (as is the case here), the Hamiltonian is also the total energy of the system.
     
  15. Nov 24, 2009 #14
    thanks :smile:
     
  16. Nov 24, 2009 #15
    Not a problem, glad I could help
     
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