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Equations relating changes between rotating and inertial frames

  1. Apr 6, 2008 #1
    i understand the reason and steps leading to the equation that relates acceleration in the inertial frame to acceleration in the rotating frame i.e.

    a(I) = a(R) + 2(omega)Xv(R) + (omega)X(omega) X r

    a(I) = acceleration in inertial frame
    a(R) = acceleration in rotating frame
    omega = rotation vector
    X = cross product
    v(R) = velocity in rotating frame

    r = position vector

    now i understand why this (or a rearrangement thereof) can be applied to a system like focaults pendulum, becuase you have acceleration going on, so obviously you would want to relate the changes between the two frames.

    however, everyone remembers the simple problems where you simply use the coriolis term to calculate the direction and magnitude of the coriolis force on something like a moving train (i.e. 500 tonne train moving north at 100kph experiences a coriolis force of something liek 1500N eastwards)

    however it struck me, that if the train is moving with constant velocity, then surely the above equation doesnt apply? and surely the equation linking the two vectors a together is simply the very original relation between the two frames for vector that is fixed in the rotating frame namely:

    V(I) = V(R) + omega X (V(R))

    or is it more subtle than this?
  2. jcsd
  3. Apr 6, 2008 #2


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    Your original equation shows that there is a Coriolis acceleration even if v is constant wilth respect to the Earth. That just means a(R)=0.
  4. Apr 6, 2008 #3
    ah i see, that makes sense.

    but when would

    V(I) = V(R) + omega X (V(R))

    be applicable? or is it just an intermediary?
  5. Apr 6, 2008 #4

    D H

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    Staff Emeritus
    Science Advisor

    Never! Look at the units on the right-hand side. The first term has units of velocity, the second, acceleration. The equation you are looking for is

    [tex]\mathbf v_I = \mathbf v_R + \mathbf \omega \times \mathbf r[/tex]

    So when does this equation apply? Simple: When you want to know the velocity of some object as seen by an observer fixed in the inertial frame.
  6. Apr 6, 2008 #5
    ah i see now. cheers.
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