# Derivative of angular momentum

1. May 22, 2015

### DavideGenoa

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. May 22, 2015

### theodoros.mihos

A motion with rotation on more than one main direction must be very complex. Rotation on one axis make forces on the others:
$$\textbf{τ} = \frac{\partial}{\partial{t}}\mathbf{L} = I\frac{\partial}{\partial{t}}\textbf{ω}$$