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Transforming to a Rotating Reference Frame

  1. Feb 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ## \mathbf{r} ## be the position of a point in a rigid body relative to some origin ##O##. Let ##\mathbf{R}## be the position of the centre of mass from that origin. ##\mathbf{r^{*}} = (\mathbf{r}-\mathbf{R})##. ## d\boldsymbol{\phi} ## is the infitesimal vector directed along the axis of rotation, with magnitude of the rotation angle ##d\phi##. Then, show:

    $$ \mathbf{v} = \mathbf{V}+\boldsymbol{\Omega} \times \mathbf{r^{*}} $$,

    Where $$ \mathbf{v} = \frac{d\mathbf{r}}{dt} \quad \mathbf{V}=\frac{d\mathbf{R}}{dt} \quad \boldsymbol{\Omega} = \frac{d\boldsymbol{\phi}}{dt} $$

    Further, assume ## \frac{d\boldsymbol{\Omega}}{dt} = \vec{0}## and prove:

    $$ \frac{d\mathbf{v}}{dt} =\frac{d\mathbf{V}}{dt}+ 2 \boldsymbol{\Omega} \times \mathbf{V} + \boldsymbol{\Omega} \times \big( \boldsymbol{\Omega} \times \boldsymbol{r^{*}} \big) $$


    2. Relevant equations


    3. The attempt at a solution

    I think it is possible I have got my origins muddled, here. The result for the velocity I can get to, after Landau [See Landau, Course of Theoretical Physics Vol 1, pg 96-97]

    We realize that any displacement of the body (I think this is due to Euler's rotation theorem) can be thought of as a combination of a fixed axis rotation about the centre of mass, and a displacement of the centre of mass itself. So:

    $$ d\mathbf{r} =d \mathbf{R} + d\boldsymbol{\phi} \times \mathbf{r^{*}} $$
    $$ \mathbf{v} = \mathbf{V} + \boldsymbol{\Omega} \times \mathbf{r^{*}} $$

    As required. However, I then run into trouble when I want to look at ##d\mathbf{v}##. Can I write the following:

    $$ d\mathbf{v} = d\mathbf{V} + d\boldsymbol{\Omega} \times \mathbf{r^{*}} + \boldsymbol{\Omega} \times d(\mathbf{r}-\mathbf{R}) $$

    Then:
    $$ \frac{d\mathbf{v}}{dt} = \frac{d\mathbf{V}}{dt} + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{r^{*}} + \boldsymbol{\Omega} \times \bigg( \mathbf{V} + \boldsymbol{\Omega} \times \mathbf{r^{*}}\bigg) - \boldsymbol{\Omega} \times \frac{d\mathbf{R}}{dt}$$

    Which using the assumption about ##\boldsymbol{\dot{\Omega}}## and the definition ##\mathbf{V} = \frac{d\mathbf{R}}{dt}## simplifies to:

    $$ \frac{d\mathbf{v}}{dt} = \frac{d\mathbf{V}}{dt} + \boldsymbol{\Omega}\times \big( \boldsymbol{\Omega} \times \mathbf{r^{*}} \big) $$

    Which is completely missing the Coriolis term!! Where has it gone?

    Thanks!
     
  2. jcsd
  3. Feb 28, 2017 #2

    Charles Link

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    Homework Helper

    I'm not sure I agree with what this problem is attempting to prove. (I'd really need to look it over much more carefully, but a preliminary evaluation seems to suggest it may be incorrect.) The way I know it, any vector ## \vec{A} ## in a rotating frame has ## [\frac{d \vec{A}}{dt}]_o=\frac{d \vec{A}}{dt}+\Omega \times \vec{A} ##.(The subscript zero implies the true rate of change in the vector A, relative to the stationary frame. The unscripted rate of change of the vector A is that measured in the rotating frame.) In this case, the velocity of the particle is part of a rigid body, so that the velocity in the rotating frame of the particle is equal to zero. The Coriolis term, as I know it, involves the velocity in the rotating frame, and the velocity in this term is not the velocity ## V ## of the origin of the rotating frame relative to the origin of the fixed frame. I need to look this over further, but I'm not so sure the original problem is correctly stated. ## \\ ## Editing: The two dimensional case of a moving particle referenced to polar coordinates is worth considering. In this case the coordinate system rotates at ## \omega=\dot{\theta} ## and the particle moves along the rotating frame axis x'. ## \vec{v}=\dot{r} \hat{a}_r+r \dot{\theta} \hat{a}_{\theta} ##. The acceleration is computed and is ## \vec{a}=(\ddot{r} -r \dot{\theta}^2) \hat{a}_r+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{a}_{\theta} ##. (Notice the ## 2 \dot{r} \dot{\theta} ## Coriolis term.) In the problem at hand, ## \dot{r}=0 ## because the particle is part of a rigid body. If ## V ## is responsible for some ## \Omega ##, then you might get a Coriolis term out of it. If there was some ## v ## in the rotating frame, e.g. a non-zero ## \dot{r} ##, then a Coriolis term would appear... ## \\ ## Additional editing: As a counterexample as to why the expression can not be correct as given (with a ## 2 \Omega \times V ## term), consider a finite and large ## V ## and a large ## \Omega ## with the particle of interest close to the origin of the rotating frame=(close to center of mass, etc.). It then has a small velocity in the rotating frame, but the computed Coriolis term in the formula as given in the OP formula will be huge.
     
    Last edited: Feb 28, 2017
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