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How Is This Torque Equation Derived?

  1. Jul 8, 2004 #1
    I have this book, which provides the following torque equation for a rigid body:

    [tex]\sum\tau_{cg} = \frac{dL_{cg}}{dt} = I\frac{d\omega}{dt} + (\omega \times (I\omega))[/tex]

    Where [tex]L_{cg}[/tex] is the angular momentum around the CG. The moments, inertia tensor, and angular velocity are all expressed in local (body) coordinates.

    This is all fine and dandy, but what is not provided is a derivation. I'm wondering if anyone knows its derivation, because I don't fully understand its meaning. :uhh:

    Any help would be greatly appreciated.

    Thanks!
     
  2. jcsd
  3. Jul 8, 2004 #2
    I believe the answer has something to do with the fact you are dealing with motion in a noninertial reference frame. You may be dealing with "fixed" axes and "rotating" axes. The derivation may lie in the section pertaining to rotating coordinate systems under the noninertial reference frames chapter. For example, in my dynamics text I found a derivation for an equation matching the one you show, except that it was generalized for an arbitrary vector. In other words, just replace your vector Lcg and I(omega), which are equivalent, with a general vector and you get the equation in my book. This equation was later used with torque when dealing with Euler's equations for a rigid body. Hope this helps.

    BTW I would have given you the equation in my book directly, however, I do not yet know how to input formulas in this forum yet!
     
  4. Jul 9, 2004 #3
    I know it, but it's long and boring. Beleive me, you're not missing anything crucial. If you want the proof I have it complete in my lecture notes. They are not in English, but I think you would have no problem understanding the proof. If you are interesting I can email you the doc.
     
  5. Jul 9, 2004 #4
    Thanks to both of you for the feedback.

    quarkman, if you want to learn about using latex, check out:

    Latex


    tomkeus, even if it's boring, I'd love to see it. :biggrin:

    I'm not sure if this forum provides an email attachment capability... But you can send it to jwalway@hotmail.com

    Thanks again!
     
  6. Jul 9, 2004 #5
    I have sent it.
     
  7. Jul 9, 2004 #6
  8. Jul 9, 2004 #7

    arildno

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    Given a set of body-fixed unit coordinate vectors
    [tex]\vec{i}_{x,b},\vec{i}_{y,b},\vec{i}_{z,b}[/tex]
    for a body that rotates absolutely with [tex]\vec{\omega}[/tex],
    the time change of these absolutely is given by:
    [tex]\frac{d\vec{i}_{x,b}}{dt}_{abs}=\vec{\omega}\times\vec{i}_{x,b}[/tex]
    and similarly for the two others.
    (Clearly, this is a statement, not a proof, but ought to be obvious)

    Let a vector be given in body coordinates:
    [tex]\vec{L}=L_{x}\vec{i}_{x,b}+L_{y}\vec{i}_{y,b}+L_{z}\vec{i}_{z,b}[/tex]
    Hence, the absolute time derivative is given by:
    [tex]\frac{d\vec{L}}{dt}_{abs}=\frac{d\vec{L}}{dt}_{rel}+\vec{\omega}\times\vec{L}[/tex]
    Here, we have the relative time derivative (i.e, in which the axes is considered fixed):
    [tex]\frac{d\vec{L}}{dt}_{rel}=\frac{dL_{x}}{dt}\vec{i}_{x,b}+\frac{dL_{y}}{dt}\vec{i}_{y,b}+\frac{dL_{z}}{dt}\vec{i}_{z,b}[/tex]
     
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