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Euler's equations

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Disregard the title of this thread.

    Say you have a fixed coordinate system and a rotating coordinate system. Say that the rotating coordinate system rotates with angular velocity [tex]\vec{\omega} [/tex]. Is it always true that the components of [tex]\omega[/tex] will be the same in both coordinate systems? If not, when is it true?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 21, 2008 #2

    pam

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    If I understand what you wrote, omega would be zero in the rotating system.
     
  4. Feb 21, 2008 #3
    No. [itex]\omega[/tex] is definitely not zero in the fixed frame. Therefore it cannot be zero in the rotating. There is no way you can transform a nonzero vector in the fixed frame into a zero vector in the rotating frame. Remember that [itex]\omega[/tex] is the angular velocity vector that describes the rotation of the rotating coordinate system w.r.t the fixed coordinate system. I am asking about how this vector transforms into the rotating coordinate system
     
  5. Feb 21, 2008 #4

    D H

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    No. It is trivial mater to construct a counter-example. What is true is that if there is no nutation or procession, (i.e., rotation about a fixed axis), the angular velocity vector will have a constant direction in both coordinate frames. If the rotation rate is constant as well, the angular velocity vector will be constant in both frames.

    Suppose there is nutation or precession (i.e., the angular velocity vector is not constant). What can you say about the derivative of the angular velocity vector as observed in the inertial and rotating frames?
     
  6. Feb 21, 2008 #5
    It is the same.

    Firstly, can you actually give me a counterexample?

    Secondly, is it true that if the fixed frame ever coincides with the rotating frame, then the components of omega are always the same in both systems (I think that follows obviously from the statement above)?
     
    Last edited: Feb 21, 2008
  7. Feb 21, 2008 #6

    D H

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    [itex]\dot \omega[/itex] is the same vector whether observed in the rotating or inertial frame. Note well: This is not the case for most vectors. In general, the derivative of a vector as observed in the inertial frame and the derivative rotating frame are different vectors. For example, consider a point fixed in the rotating frame some distance [itex]r[/itex] away from the rotation axis. The time derivative of the location of this fixed point is obviously zero in the rotating frame and equally obviously [itex]\omega r[/itex] in the inertial frame.

    While the time derivative of the angular velocity vector, [itex]\dot {\vect{\omega}}[/itex], is the same vector in both frames, that does not mean it has the same coordinates in both frames.

    Not if this is homework.
     
  8. Feb 21, 2008 #7
    Thanks. Is what I said after "secondly" true?
     
  9. Feb 21, 2008 #8

    D H

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    You still haven't said whether this is homework. I guess saying a qualified "yes" isn't offering too much help. Qualified meaning constant angular velocity. If the angular velocity is not constant, the statement is obviously not true.
     
  10. Feb 21, 2008 #9
    This is not homework.

    I think it is true even if the angular velocity is not constant. The derivative of the angular velocity vector measured in both the coordinate systems is the same, so if omega has the same components in both coordinate systems at time t, it must always have the same components in both coordinate systems even if those components are changing.
     
  11. Feb 21, 2008 #10

    D H

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    Here's a counterexample: Build a second reference frame from another frame by rotating 90 degrees about the x axis. Then [itex](\hat x_2,\hat y_2,\hat z_2) = (\hat x_1,\hat z_1,-\hat y_1)[/itex]. Now set this second frame in rotation about the y' axis. The angular velocity vector is [itex][0,0,\omega]^T[/itex] in the non-rotating frame, [itex][0,\omega,0]^T[/itex] in the rotating frame.

    Counterexample again: Consider a cylinder with uniform mass distribution but a non-spherical inertia tensor. Define a coordinate system based on the cylinder. This will be our rotating frame. Set the cylinder spinning in space (no external torques) such that the angular velocity has non-zero components along and normal to the cylinder axis. I can always define an inertial frame that is instantaneously co-aligned with the rotating frame at some given point in time. At this point in time, the angular velocity vector tautologically the same components in the inertial and rotating frame. By construction, the rotating frame is tumbling. Therefore, at some other point the angular velocity vector will not have the same components in the inertial and rotating frame.
     
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