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Precession: Deriation of formula

  1. Dec 15, 2007 #1
    Please see this link first: http://en.wikipedia.org/wiki/Precession#Classical_.28Newtonian.29

    How do they get to the formula
    [tex]\textbf{wp}[/tex] = [tex]\frac{Q}{I*w}[/tex]

    I also note that I*w = L (angular momentum).

    How do they get to that equation? I've thought and though, I just don't know how. I really hope that someone can give me a link to a deriation of it, or perhaps help me themselves.

    Thanks in advance.
     
  2. jcsd
  3. Dec 15, 2007 #2

    Doc Al

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  4. Dec 15, 2007 #3
    Wow! Thank you so much.

    Just one thing though: In equation (4), the d0, what is that? Change in angle? And where does that equation (4) come from? Never seen it before? Why is that true? :s
     
  5. Dec 15, 2007 #4
    I think I get it now actually. It is because the change as well as angle are inifitsmally small... :) Right? :D
     
  6. Dec 15, 2007 #5

    Doc Al

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    Yes, that's it. And because the torque is perpendicular to the angular momentum.

    Now that I looked it over more carefully, I'm not happy with that derivation that I linked. It seems a bit sloppy. I'll post my own version in a bit.
     
  7. Dec 15, 2007 #6
    Oh thank you so much! :eek: It's deeply appreciated.
     
  8. Dec 15, 2007 #7

    Doc Al

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    I was just about to post my own derivation, when I found this on hyperphysics (one of my favorite educational sites--I highly recommend it): Precession of Spinning Top

    This is almost exactly what I would have written, so it saves me the trouble! If you have questions about this derivation, let me know.
     
  9. Dec 15, 2007 #8
    Thank you once more. What I need is acutally a combination of the two. The first is good because it fits the precession of a bicycle wheel (which is what I need to explain), and the next is good because of the smart derivation.

    I've tried to write it down, how I think I need it. But I'm not sure, if I did any errors... I would very much like if you could just quickly look it through?
    Here's the link: http://peecee.dk/?id=85095 (danish upload site: Click "Download fil").
    At the moment it hasn't got an illustration, I will make one however.

    Thanks in advance, once again. :smile:
     
  10. Dec 15, 2007 #9
    Oh, and a last question: Is this good argumentation:
    If I call the angle v. Can I say this:

    "It is obvious that
    [tex]sin(v) = \frac{\Delta L}{L}[/tex]​
    For infinitsimal small angles sin(v)=v, so that:
    [tex]dv = \frac{dL}{L}[/tex]​
    since the angle is small when delta L is small."

    Thanks in advance.
     
    Last edited: Dec 16, 2007
  11. Dec 16, 2007 #10

    Doc Al

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    It looks OK to me. Note that your "r" is the moment arm--I assume your bicycle wheel is oriented perpendicular to the vertical?

    It is certainly true that for small angles [itex]\sin\theta \approx \theta[/itex], as long as the angle is in radians. But you can also argue directly that as [itex]\Delta L[/itex] becomes small it more closely approximates the arc length of a circle, thus [itex]\Delta L/L[/itex] becomes the radian measure of the angle.
     
  12. Dec 16, 2007 #11
    Thank you once more! :) I think I get it now. :D
     
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