Euler’s equation for rigid bodies

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,

\\displaystyle\\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},

(1)

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 $Q2$ ¹¹That 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

\\displaystyle\\mathbf{M}_{C}=\\boldsymbol{\\mathbb{I}}^{C}\\boldsymbol{\\alpha}_{21%}+\\boldsymbol{\\omega}_{21}\\times(\\boldsymbol{\\mathbb{I}}^{C}\\boldsymbol{\\omega%}_{21}).

(2)

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 $\\mathbf{F}=\\mathbf{0}$ )