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Derivative of angular momentum

  1. May 22, 2015 #1
    Hi, friends! Let the quantity ##I\boldsymbol{\omega}## be given, where ##I## is an inertia matrix and ##\boldsymbol{\omega}## a column vector representing angular velocity; ##I\boldsymbol{\omega}## can be the angular momentum of a rigid body rotating around a static point or around its -even moving- centre of mass, with ##I## calculated with respect to such a point. I have tried to derive, without any difficulty, a key formula from which Euler 's equation derive in the following way. By using a moving Cartesian frame, dextrogyre as is usual for angular quantities, having the basis ##(\mathbf{i},\mathbf{j},\mathbf{k})=(\mathbf{i}(t),\mathbf{j}(t),\mathbf{k}(t))##, solidal with the rigid body, whose vectors I write as the columns of a matrix ##E=\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)##, we see that

    ##I\boldsymbol{\omega}=EI_m E^{-1} \boldsymbol{\omega}=EI_m \boldsymbol{\omega}_{m}##​

    where the index ##m## means that the respective quantities are expressed with respect to basis ##\{\mathbf{i},\mathbf{j},\mathbf{k}\}##. By differentiating we get

    ##\frac{d(I\boldsymbol{\omega})}{dt}=\frac{dE}{dt}I_m\boldsymbol{\omega}_{m}+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}##
    and, since Poisson formulae give us ##\frac{d}{dt}\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)=\left( \begin{array}{ccc}\boldsymbol{\omega}\times\mathbf{i}&\boldsymbol{\omega}\times\mathbf{j}&\boldsymbol{\omega}\times\mathbf{k} \end{array} \right)##, if we call the ##i##-th row of matrix ##I_m## as ##I_m^{(i)}##, we see that

    ##\frac{d(I\boldsymbol{\omega})}{dt}=\frac{dE}{dt}I_m\boldsymbol{\omega}_{m}+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}=\boldsymbol{\omega}\times( I_m^{(1)}\cdot\boldsymbol{\omega}_m\mathbf{i})+\boldsymbol{\omega}\times( I_m^{(2)}\cdot\boldsymbol{\omega}_m\mathbf{j})+\boldsymbol{\omega}\times( I_m^{(3)}\cdot\boldsymbol{\omega}_m\mathbf{k})++E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}##
    ##=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+E\frac{d(I_m\boldsymbol{\omega}_{m})}{dt}##​

    and, since ##I_m## does not depend upon ##t## because it is calculated according to the moving basis,


    ##\frac{d(I\boldsymbol{\omega})}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)I_m\frac{d\boldsymbol{\omega}_{m}}{dt}##​

    or, equivalently,​

    ##\frac{d(I\boldsymbol{\omega})}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I \left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right) \frac{d\big(\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)^{-1}\boldsymbol{\omega}\big)}{dt}=\boldsymbol{\omega}\times (I\boldsymbol{\omega})+I\Big(\frac{d(\boldsymbol{\omega}\cdot\mathbf{i})}{dt}\mathbf{i}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{j})}{dt}\mathbf{j}+\frac{d(\boldsymbol{\omega}\cdot\mathbf{k})}{dt}\mathbf{k}\Big)##​

    I suppose that such a formula is used to simplify some calculations, but I am not sure where the simplification is. Is it used because it is easier to differentiate ##\boldsymbol{\omega}_m=(\boldsymbol{\omega}\cdot\mathbf{i},\boldsymbol{\omega}\cdot\mathbf{j},\boldsymbol{\omega}\cdot\mathbf{k})## (which does not seem so easy to compute to my eyes...) and left-multiply by ##\left( \begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \end{array} \right)I_m##, where ##I_m## can be easy to compute, than acting upon ##I##? I would like to collect ideas on how such formula is used. Thank you very much for any answer!
     
  2. jcsd
  3. May 22, 2015 #2
    A motion with rotation on more than one main direction must be very complex. Rotation on one axis make forces on the others:
    $$ \textbf{τ} = \frac{\partial}{\partial{t}}\mathbf{L} = I\frac{\partial}{\partial{t}}\textbf{ω} $$
     
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