PlanetPhysics/Euler's Moment Equations

Euler's Moment Equations in terms of the principle axes is given by

$M_x=I_x\dot{{\omega }_x}+(I_z-I_y){\omega }_y{\omega }_z$ $M_y=I_y\dot{{\omega }_y}+(I_x-I_z){\omega }_x{\omega }_z$ $M_z=I_z\dot{{\omega }_z}+(I_y-I_x){\omega }_x{\omega }_y$

In order to derive these equations, we start with the [[../MolecularOrbitals/|angular momentum]] of a [[../RigidBody/|rigid body]]

$\overrightarrow{H_B}=I\omega =\left[\begin{array}{ccc}{I}_{xx}& -I_{xy}& -I_{xz}\\ -I_{yx}& I_{yy}& -I_{yz}\\ -I_{zx}& -I_{zy}& I_{zz}\end{array}\right]\left[\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right]$

Since the [[../Vectors/|vector]] is in the body frame and we want the Moment in an inertial frame we need to use the transport [[../Formula/|theorem]] since our body is in a non-inertial [[../CosmologicalConstant/|reference frame]] to express the derivative of the angular momentum vector in this frame. So the Moment is given by

$\overrightarrow{M}=\dot{\overrightarrow{H_I}}=\dot{\overrightarrow{H_B}}+\overrightarrow{\omega }\times \overrightarrow{H_B}$

Since we are assuming the [[../InertiaTensor/|inertia tensor]] is expressed using the principal axes of the body the Products of Inertia are zero

$I_{yx}=I_{xy}=I_{xz}=I_{zx}=I_{zy}=I_{yz}=0$

and using the shorter notation

$I_{xx}=I_x$ $I_{yy}=I_y$ $I_{zz}=I_z$

Also since the [[../MomentOfInertia/|moments of inertia]] are constant, when we take the derivative of the Inertia Tenser it is zero, so

$\dot{\overrightarrow{H_B}}=\left[\begin{array}{ccc}{I}_x& 0& 0\\ 0& I_y& 0\\ 0& 0& I_z\end{array}\right]\left[\begin{array}{c}\dot{{\omega }_x}\\ \dot{{\omega }_y}\\ \dot{{\omega }_z}\end{array}\right]+\left[\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right]\times \left[\begin{array}{ccc}{I}_x& 0& 0\\ 0& I_y& 0\\ 0& 0& I_z\end{array}\right]\left[\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right]$

Carrying out the [[../Matrix/|matrix multiplication]] $\dot{\overrightarrow{H_B}}=\left[\begin{array}{c}{I}_x\dot{{\omega }_x}\\ I_y\dot{{\omega }_y}\\ I_z\dot{{\omega }_z}\end{array}\right]+\left[\begin{array}{c}{\omega }_x\\ {\omega }_y\\ {\omega }_z\end{array}\right]\times \left[\begin{array}{c}{I}_x{\omega }_x\\ I_y{\omega }_y\\ I_z{\omega }_z\end{array}\right]$

after evaluating the [[../VectorProduct/|cross product]], we are left with adding the vectors $\dot{\overrightarrow{H_B}}=\left[\begin{array}{c}{I}_x\dot{{\omega }_x}\\ I_y\dot{{\omega }_y}\\ I_z\dot{{\omega }_z}\end{array}\right]+\left[\begin{array}{c}{\omega }_y{\omega }_z(I_z-I_y)\\ {\omega }_x{\omega }_z(I_x-I_z)\\ {\omega }_x{\omega }_y(I_y-I_x)\end{array}\right]$

Once we add these vectors we are left with Euler's Moment Equations

$M_x=I_x\dot{{\omega }_x}+(I_z-I_y){\omega }_y{\omega }_z$ $M_y=I_y\dot{{\omega }_y}+(I_x-I_z){\omega }_x{\omega }_z$ $M_z=I_z\dot{{\omega }_z}+(I_y-I_x){\omega }_x{\omega }_y$