- #1

DavideGenoa

- 155

- 5

##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}##

##=\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!