70G45-EulersEquationForRigidBodies.tex

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\pmtitle{Euler's equation for rigid bodies}
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\begin{document}
Let $1$ be an inertial frame body (a rigid body) and $2$ a rigid body in motion respect to an observer located at $1$. Let $Q$ be an arbitrary point (fixed or in motion) and $C$ the center of mass of $2$. Then,
\begin{align}
\mathbf{M}_Q=\boldsymbol{\mathbb{I}}^Q\boldsymbol{\alpha}_{21}+
\boldsymbol{\omega}_{21}\times(\boldsymbol{\mathbb{I}}^Q\boldsymbol{\omega}_{21})+
m\mathbf{\overline{QC}}\times \mathbf{a}^{Q2}_1,
\end{align}
where $m$ is the mass of the rigid body, $\mathbf{\overline{QC}}$ the position vector of $C$ respect to $Q$, $\mathbf{M}_Q$ is the moment of forces system respect to $Q$, $\boldsymbol{\mathbb{I}}^Q$ the tensor of inertia respect to orthogonal axes embedded in $2$ and origin at $Q2$ {\footnote{That is possible because the kinematical concept of frame extension.}}, and $\mathbf{a}^{Q2}_1$, $\boldsymbol{\omega}_{21}$, $\boldsymbol{\alpha}_{21}$, are the acceleration of $Q2$, the angular velocity and acceleration vectors respectively, all of them measured by an observer located at $1$. \\
This equation was got by Euler by using a  fixed system of principal axes with origin at $C2$. In that case we have $Q=C$, and therefore
\begin{align}
\mathbf{M}_C=\boldsymbol{\mathbb{I}}^C\boldsymbol{\alpha}_{21}+ \boldsymbol{\omega}_{21}\times(\boldsymbol{\mathbb{I}}^C\boldsymbol{\omega}_{21}).
\end{align}
Euler used three independent scalar equations to represent (2). It is well known that the number of degrees of freedom associate to a rigid body in free motion in $\mathbb{R}^3$ are six, just equal the number of independent scalar equations necessary to solve such a motion. (Newton's law contributing with three) \\
Its is clear if $2$ is at rest or in uniform and rectilinear translation, then $\mathbf{M}_Q=\mathbf{0}$, one  of the necessary and sufficient conditions for the equilibrium of the system of forces applied to a rigid body. (The other one is  the force resultant $\mathbf{F}=\mathbf{0}$)




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