Question 10.11: A spacecraft has three identical momentum wheels with their ...
A spacecraft 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,
\left[ I _{G_{1}}^{(1)}\right]=\left[\begin{array}{ccc}I & 0 & 0 \\0 & J & 0 \\0 & 0 & J\end{array}\right] \quad\left[ I _{G_{2}}^{(2)}\right]=\left[\begin{array}{ccc}J & 0 & 0 \\0 & I & 0 \\0 & 0 & J\end{array}\right] \quad\left[ I _{G_{3}}^{(3)}\right]=\left[\begin{array}{lll}J & 0 & 0 \\0 & J & 0 \\0 & 0 & I\end{array}\right] (a)
The spacecraft moment of inertia tensor about the vehicle center of mass is
\left[ I _{G}^{(v)}\right]=\left[\begin{array}{ccc}A & 0 & 0 \\0 & B & 0 \\0 & 0 & C\end{array}\right] (b)
Calculate the spin accelerations of the momentum wheels in the presence of external torque.
For n = 3, Equation 10.140 becomes
\left\{ H _{G}\right\}=\left\lgroup \left[ I _{G}^{(v)}\right]+\sum\limits_{i=1}^{n}\left[ I _{G_{i}}^{(i)}\right] \right\rgroup \{ \omega \}+\sum\limits_{i=1}^{n}\left[ I _{G_{i}}^{(i)}\right]\left\{ \omega _{ rel }^{(i)}\right\} (10.140)
\left\{ H _{G}\right\}=\left(\left[ I _{G}^{(v)}\right]+\left[ I _{G_{1}}^{(1)}\right]+\left[ I _{G_{2}}^{(2)}\right]+\left[ I _{G_{3}}^{(3)}\right]\right)\{ \omega \}+\left[ I _{G_{1}}^{(1)}\right]\left\{ \omega _{ rel }^{(1)}\right\}+\left[ I _{G_{2}}^{(2)}\right]\left\{ \omega _{ rel }^{(2)}\right\}+\left[ I _{G_{3}}^{(3)}\right]\left\{ \omega _{ rel }^{(3)}\right\} (c)
The absolute angular velocity ω of the spacecraft and the angular velocities \omega_{ rel }^{(1)}, w _{ rel }^{(2)}, w _{ rel }^{(3)} of the three flywheels relative to the spacecraft are
\{\omega\}=\left\{\begin{array}{l}\omega_{x} \\\omega_{y} \\\omega_{z}\end{array}\right\} \quad\left\{\omega^{(1)}\right\}_{ rel }=\left\{\begin{array}{c}\omega^{(1)} \\0 \\0\end{array}\right\} \quad\left\{\omega^{(2)}\right\}_{ rel }=\left\{\begin{array}{c}0 \\\omega^{(2)} \\0\end{array}\right\} \quad\left\{\omega^{(3)}\right\}_{ rel }=\left\{\begin{array}{c}0 \\0 \\\omega^{(3)}\end{array}\right\} (d)
Substituting (a), (b) and (d) into (c) yields
\left\{ H _{G}\right\}=\left(\left[\begin{array}{lll}A & 0 & 0 \\0 & B & 0 \\0 & 0 & C\end{array}\right]+\left[\begin{array}{lll}I & 0 & 0 \\0 & J & 0 \\0 & 0 & J\end{array}\right]+\left[\begin{array}{lll}J & 0 & 0 \\0 & J & 0 \\0 & 0 & I\end{array}\right]+\left[\begin{array}{lll}J & 0 & 0 \\0 & J & 0 \\0 & 0 & I\end{array}\right]\right) \left\{\begin{array}{c}\omega_{x} \\\omega_{y} \\\omega_{z}\end{array}\right\}+\left[\begin{array}{lll}I & 0 & 0 \\0 & J & 0 \\0 & 0 & J\end{array}\right]\left\{\begin{array}{c}\omega^{(1)} \\0 \\0\end{array}\right\}+ \left[\begin{array}{lll}J & 0 & 0 \\0 & I & 0 \\0 & 0 & J\end{array}\right]\left\{\begin{array}{c}0 \\\omega^{(2)} \\0\end{array}\right\}+\left[\begin{array}{lll}J & 0 & 0 \\0 & J & 0 \\0 & 0 & I\end{array}\right]\left\{\begin{array}{c}0 \\0 \\\omega^{(3)}\end{array}\right\}or
\left\{ H _{G}\right\}=\left[\begin{array}{ccc}A+I+2 J & 0 & 0 \\0 & B+I+2 J & 0 \\0 & 0 & C+I+2 J\end{array}\right]\left\{\begin{array}{c}\omega_{x} \\\omega_{y} \\\omega_{z}\end{array}\right\}+\left[\begin{array}{ccc}I & 0 & 0 \\0 & I & 0 \\0 & 0 & I\end{array}\right]\left[\begin{array}{c}\omega^{(1)} \\\omega^{(2)} \\\omega^{(3)}\end{array}\right\} (e)
Substituting this expression for \left\{ H _{G}\right\} into Equation 10.143, we get
\left.\left. M _{G_{\text {net }}}\right)_{\text {external }}=\frac{d H _{G}}{d t}\right\rgroup _{\text {rel }}+\omega \times H _{G} (10.143)
\left[\begin{array}{lll}I & 0 & 0 \\0 & I & 0 \\0 & 0 & I\end{array}\right]\left\{\begin{array}{l}\dot{\omega}^{(1)} \\\dot{\omega}^{(2)} \\\dot{\omega}^{(3)}\end{array}\right\}+\left[\begin{array}{ccc}A+I+2 J & 0 & 0 \\0 & B+I+2 J & 0 \\0 & 0 & C+I+2 J\end{array}\right]\left\{\begin{array}{c}\dot{\omega}_{x} \\\dot{\omega}_{y} \\\dot{\omega}_{z}\end{array}\right\} +\left\{\begin{array}{l}\omega_{x} \\\omega_{y} \\\omega_{z}\end{array}\right\} \times\left(\left[\begin{array}{lll}I & 0 & 0 \\0 & I & 0 \\0 & 0 & I\end{array}\right]\left\{\begin{array}{l}\omega^{(1)} \\\omega^{(2)} \\\omega^{(3)}\end{array}\right\}+\left[\begin{array}{ccc}A+I+2 J & 0 & 0 \\0 & B+I+2 J & 0 \\0 & 0 & C+I+2 J\end{array}\right]\left\{\begin{array}{l}\omega_{x} \\\omega_{y} \\\omega_{z}\end{array}\right\}\right)=\left\{\begin{array}{l}M_{G_{x}} \\M_{G_{y}} \\M_{G_{z}}\end{array}\right\} (f)
Expanding and collecting terms yields the time rates of change of the flywheel spins (relative to the spacecraft) in terms of the spacecraft’s absolute angular velocity components,
\begin{aligned}&\dot{\omega}^{(1)}=\frac{M_{G_{x}}}{I}+\frac{B-C}{I} \omega_{y} \omega_{z}-\left\lgroup 1+\frac{A}{I}+2 \frac{J}{I} \right\rgroup \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}-\left\lgroup 1+\frac{B}{I}+2 \frac{J}{I} \right\rgroup \dot{\omega}_{y}+\omega^{(3)} \omega_{x}-\omega^{(1)} \omega_{z} \\&\dot{\omega}^{(3)}=\frac{M_{G_{z}}}{I}+\frac{A-B}{I} \omega_{x} \omega_{y}-\left\lgroup 1+\frac{C}{I}+2 \frac{J}{I} \right\rgroup \dot{\omega}_{z}+\omega^{(1)} \omega_{y}-\omega^{(2)} \omega_{x}\end{aligned} (g)