EulersMomentEquations 2005-03-10 09:43:27 2005-03-21 19:45:15 Definition \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} Euler's Moment Equations in terms of the principle axis 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 angular momentum of a rigid body $$ \vec{H_B} = I \omega = \left [ \begin{array}{c c c} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -I_{zx} & -I_{zy} & I_{zz} \end{array} \right ] \left [\begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right ] $$ Since the vector is in the body frame and we want the Moment in an inertial frame we need to use the transport theorem since our body is in a non-inertial reference frame to express the derivative of the angular momentum vector in this frame. So the Moment is given by $$ \vec{M} = \dot{\vec{H_I}} = \dot{\vec{H_B}} + \vec{\omega} \times \vec{H_B} $$ Since we are assuming the Inertia Tensor is expressed using the principal axis 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 Moments of Inertia are constant, when we take the derivative of the Inertia Tenser it is zero, so $$ \dot{\vec{H_B}} = \left [ \begin{array}{c c c} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \end{array} \right ] \left [\begin{array}{c} \dot{\omega_x} \\ \dot{\omega_y} \\ \dot{\omega_z} \end{array} \right ] + \left [\begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right ] \times \left [ \begin{array}{c c c} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \end{array} \right ] \left [\begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right ] $$ Carrying out the matrix multiplication $$ \dot{\vec{H_B}} = \left [ \begin{array}{c} I_x \dot{\omega_x} \\ I_y \dot{\omega_y} \\ I_z \dot{\omega_z} \end{array} \right ] + \left [\begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \end{array} \right ] \times \left [ \begin{array}{c} I_x \omega_x \\ I_y \omega_y \\ I_z \omega_z \end{array} \right ] $$ after evaluating the cross product, we are left with adding the vectors $$ \dot{\vec{H_B}} = \left [ \begin{array}{c} I_x \dot{\omega_x} \\ I_y \dot{\omega_y} \\ I_z \dot{\omega_z} \end{array} \right ] + \left [\begin{array}{c} \omega_y \omega_z (I_z - I_y) \\ \omega_x \omega_z (I_x - I_z) \\ \omega_x \omega_y (I_y - I_x) \end{array} \right ] $$ Once we add these vectors we are left with Euler's Moment Equations $$ M_x = I_x \dot{\omega_x} + (I_z - I_y) \omega_y \omega_z $$ $$ M_y = I_y \dot{\omega_y} + (I_x - I_z) \omega_x \omega_z $$ $$ M_z = I_z \dot{\omega_z} + (I_y - I_x) \omega_x \omega_y $$