Rolling body kinematics

Discussion in 'Physics & Math' started by arfa brane, Nov 26, 2011.

  1. prometheus viva voce! Moderator

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    2,045
    As I said, it's simply a definition. I can choose zero to be anywhere I like.

    Sometime. Believe it or not my life does not revolve around this forum.

    That's their prerogative. As I said before, I got my inspiration from a book that I linked to previously. Why don't you file an errata with the publisher and see how that goes?

    See what a stupid argument that is now?
     
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  3. Tach Banned Banned

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    I did not dispute that, I am just pointing out that you get the arc length to be a negative entity when \(-\pi <\alpha<0\) with your approach. How do you plan to solve this?


    Actually, this is not such a stupid idea. Did you file yours?
     
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  5. prometheus viva voce! Moderator

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    2,045
    Quite obviously you say that the distance from \(-\pi\) to 0 is the same as minus the distance from 0 to \(-\pi\), because I have defined zero to be in the middle of the trajectory. It's not complicated at all.

    Yep. Go for it, and let us know how it goes.
     
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  7. Tach Banned Banned

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    5,265
    Nope, the change (reversal) in traversing parameter space (\([-\pi,0]\) vs \([0,-\pi]\)) does not warrant changing (reversing) the sign for s since s is an absolute quantity.


    Arc length is a positive entity, by definition. Your solution to your error is to simply stick a minus sign in front of its expression doesn't cut it.


    I asked you if you filed your error report, why the deflection? Did you file?
     
    Last edited: Dec 9, 2011
  8. prometheus viva voce! Moderator

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    2,045
    What exactly is your problem? I showed in post 59 that
    \(ds = 2 R \cos \frac{\alpha}{2} d \alpha\) and hence \(s = 2 R \int_{-\pi}^{\pi} \cos \frac{\alpha}{2} d \alpha = 8R\) in exact agreement with AN's post. There is no issue with negative arc length.

     
  9. Tach Banned Banned

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    5,265
    The problem is that you arrived to \(s=4R sin(\frac{\alpha}{2})\) .
     
  10. prometheus viva voce! Moderator

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    2,045
    why is that a problem?
     
  11. Tach Banned Banned

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    5,265
    We've been over this, for \(-\pi <\alpha<0\) your \(s\) is negative. Your attempt at reversing the parametrization doesn't fix the problem.
     
  12. prometheus viva voce! Moderator

    Messages:
    2,045
    Why is this a problem? For your parameterisation s is going to be something like \(\cos \frac{\alpha}{2}\) which is negative for \(\pi < \alpha <2 \pi\).
     
  13. Tach Banned Banned

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    5,265
    Nope, \(s=4R sin( \theta/2) \) , positive for all \(0<\theta< 2 \pi\)
     
    Last edited: Dec 9, 2011
  14. prometheus viva voce! Moderator

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    2,045
    \(x = R(\alpha- \sin \alpha), \quad y = R(1+\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}\)

    Equation of motion: \(\frac{d^2 s}{dt^2} + \frac{g}{4R} s=0\)

    Solution subject to the boundary conditions: \(s(0) = s_0, \quad \frac{ds}{dt}(0) = 0\) is \(s = s_0 \cos(\omega t)\) where \(\omega^2 = \frac{g}{4R} \)

    hence the velocity is \(\frac{ds}{dt} = - s_0 \omega \sin (\omega t)\) and the kinetic energy is \(KE = \frac{1}{2} m \omega^2 s_0^2 \sin^2 (\omega t)\)

    The potential energy is \(PE = mgy\). Using \(y = 2 R \sin^2 \frac{\alpha}{2}\) and \(s = 4 R \sin\frac{\alpha}{2}\) we can show \(y = \frac{s^2}{8R}\). Hence \(PE = \frac{mgs^2}{8R} = \frac{mg s_0^2 \cos^2(\omega t)}{8R} \)

    Therefore the total energy having subbed for \(\omega^2\) is \(E = KE + PE = \frac{mgs_0^2}{8R}\left(\sin^2 (\omega t) + \cos^2 (\omega t) \right) = \frac{mgs_0^2}{8R} \) and \( \frac{\partial E}{\partial t} = 0\)
     
  15. prometheus viva voce! Moderator

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    2,045
    By my reckoning for \(x = R(\alpha - \sin \alpha), \quad y = R(1- \cos \alpha)\) we have \(ds = 2 R \sin \frac{\alpha}{2} d\alpha\) and hence \( s = -4 R \cos\frac{\alpha}{2}\) which is negative for \(\alpha\) between 0 and \(\pi\)
     
  16. Tach Banned Banned

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    5,265
    Nope, the starting point is \(ds^2=2R^2 (1-cos (\theta))\) .
    What does this tell you?
     
  17. Tach Banned Banned

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    5,265
    Err, you surely meant \(y = R(1+\cos \alpha)=2 R \cos^2 \frac{\alpha}{2}\), right?


    Kind of oK, so far (modulo the trigonometry flub).

    How can that be? At \(\alpha=-\pi\) your \(y=0\) and \(PE=0\) which is clearly not the case.
    At the bottom of the trajectory , i.e. at \(\alpha=0\) your \(PE=2mgR\) , which is clearly not the case.
     
    Last edited: Dec 9, 2011
  18. prometheus viva voce! Moderator

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    2,045
    Yikes! That was a rotten typo. What I actually meant was \(x = R (\alpha + \sin \alpha), \quad y = R(1-\cos \alpha)=2 R \sin^2 \frac{\alpha}{2}\) in line with my original solution. That also clears up the potential energy issue.
     
  19. prometheus viva voce! Moderator

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    2,045
    You mean \(ds^2=2R^2 (1-cos (\theta))d \theta^2 \) which means \(ds^2=4R^2 \sin^2 \frac{\theta}{2} d \theta^2 \Rightarrow ds=2R \sin \frac{\theta}{2} d \theta\) and s is as I wrote above.
     
  20. Tach Banned Banned

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    5,265
    Yes.


    Not exactly.
     
  21. prometheus viva voce! Moderator

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    2,045
    Go on...
     
  22. Tach Banned Banned

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    5,265
    \(ds^2=4R^2 sin^2(\theta/2) d \theta^2 => ds=2R d \theta |sin(\theta/2)|\)

    ..and this should also fix the problems with your derivations.
     
  23. prometheus viva voce! Moderator

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    2,045
    No...

    \(ds^2=4R^2 sin^2(\theta/2) d \theta^2 \Rightarrow ds= \pm 2R d \theta sin(\theta/2)\) which means there is always a region where s < 0.

    I just did a big lol at this. Right, I think we've established that s can be positive or negative as a function of \(\alpha\)...
     

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