Question 10.10: A spacecraft in torque-free motion has three identical momen...
A spacecraft in torque-free motion has three identical momentum wheels with their spin axes aligned with the vehicle’s principal body axes. The spin axes of momentum wheels 1, 2 and 3 are aligned with the x, y and z axes, respectively. The inertia tensors of the rotationally symmetric momentum wheels about their centers of mass are, therefore,
The spacecraft moment of inertia tensor about the vehicle center of mass is
[I_{G}^{(v)}]= \left[\begin{matrix}A & 0 & 0 \\0 &B &0 \\0 & 0 & C \end{matrix} \right] (b)
Calculate the spin accelerations of the momentum wheels in the presence of external torque.
The absolute angular velocity ω of the spacecraft and the angular velocities \omega ^{(1)}_{rel}, \omega ^{(2)}_{rel}, \omega ^{(3)}_{rel} of the three flywheels relative to the spacecraft are
Therefore, the angular momentum of the spacecraft and momentum wheels is
\left\{H_{G}\right\}=[I_{G}^{(v)}]\left\{\omega \right\}+[I_{G_{1}}^{(1)}](\left\{\omega \right\}+\left\{\omega^{(1)} \right\}_{rel})+[I_{G_{2}}^{(2)}](\left\{\omega \right\}+\left\{\omega^{(2)} \right\}_{rel})
+[I_{G_{3}}^{(3)}](\left\{\omega \right\}+\left\{\omega^{(3)} \right\}_{rel}) (d)
Substituting Equations (a), (b) and (c) into this expression yields
\left\{H_{G}\right\}=\left[\begin{matrix}I & 0 & 0 \\0 &I &0 \\0 & 0 & I \end{matrix} \right]\left\{\begin{matrix} \omega ^{(1)} \\ \omega ^{(2)} \\ \omega ^{(3)} \end{matrix} \right\} +\left[\begin{matrix}A+I+2J& 0 & 0 \\0 &B+I+2J &0 \\0 & 0 &C+ I+2J \end{matrix} \right] \left\{\begin{matrix} \omega_{x} \\ \omega_{y} \\ \omega_{z} \end{matrix} \right\} (e)
In this case, Euler’s equations are
\left\{\dot{H}_{G}\right\}_{rel}+\left\{\omega \right\}\times \left\{H_{G}\right\}=\left\{M_{G}\right\} (f)
Substituting (e), we get
\left[\begin{matrix}I & 0 & 0 \\0 &I &0 \\0 & 0 & I \end{matrix} \right]\left\{\begin{matrix} \dot{\omega}^{(1)} \\ \dot{\omega}^{(2)} \\ \dot{\omega}^{(3)} \end{matrix} \right\} +\left[\begin{matrix}A+I+2J & 0 & 0 \\0 &B+I+2J &0 \\0 & 0 &C+ I+2J \end{matrix} \right] \left\{\begin{matrix} \dot{\omega}_{x} \\ \dot{\omega}_{y} \\ \dot{\omega}_{z} \end{matrix} \right\}+ \left\{\begin{matrix} \omega_{x} \\ \omega_{y} \\ \omega_{z} \end{matrix} \right\}
=\left\{\begin{matrix}M_{G_{x}} \\M_{G_{y}} \\M_{G_{z}} \end{matrix} \right\} (g)
Expanding and collecting terms yields the time rates of change of the flywheel spins (relative to the spacecraft) in terms of those of the spacecraft’s absolute angular velocity components,
\dot{\omega}^{(1)}=\frac{M_{G_{x}}}{I}+\frac{B-C}{I}\omega_{y}\omega_{z}-(1+\frac{A}{I}+2\frac{J}{I})\dot{\omega}_{x}+\omega ^{(2)}\omega_{z}-\omega^{(3)}\omega_{y}
\dot{\omega}^{(2)}=\frac{M_{G_{y}}}{I}+\frac{C-A}{I}\omega_{x}\omega_{z}-(1+\frac{B}{I}+2\frac{J}{I})\dot{\omega}_{y}+\omega ^{(3)}\omega_{x}-\omega^{(1)}\omega_{z} (h)
\dot{\omega}^{(3)}=\frac{M_{G_{z}}}{I}+\frac{A-B}{I}\omega_{x}\omega_{y}-(1+\frac{C}{I}+2\frac{J}{I})\dot{\omega}_{z}+\omega ^{(1)}\omega_{y}-\omega^{(2)}\omega_{x}