Question 10.14: A satellite is in torque-free motion. A control moment gyro,...
A satellite is in torque-free motion. A control moment gyro, spinning at the constant rate \omega_{s}, is gimbaled about the spacecraft y and z axes, with φ = 0 and φ = 90° (cf. Figure 10.26 ). The spacecraft angular velocity is \omega=\omega_{z} \hat{ k }. If the spin axis of the gyro, initially along the x direction, is rotated around the y axis at the rate \dot{\theta}, what is the resulting angular acceleration of the spacecraft?
Substituting \omega_{x}=\omega_{y}=\dot{\omega}_{s}=\phi=0 and θ = 90° into Equations 10.152 gives
A \dot{\omega}_{x}+C^{(w)} \omega_{s} \dot{\theta} \cos \phi \cos \theta-C^{(w)} \omega_{s} \dot{\phi} \sin \phi \sin \theta+C^{(w)} \dot{\omega}_{s} \cos \phi \sin \theta +\left(C^{(w)} \omega_{s} \cos \theta+C \omega_{z}\right) \omega_{y}-\left(C^{(w)} \omega_{s} \sin \phi \sin \theta+B \omega_{y}\right) \omega_{z}=M_{G_{x}} (10.152a)
B \dot{\omega}_{y}+C^{(w)} \omega_{s} \dot{\theta} \sin \phi \cos \theta+C^{(w)} \omega_{s} \dot{\phi} \cos \phi \sin \theta+C^{(w)} \dot{\omega}_{s} \sin \phi \sin \theta -\left(C^{(w)} \omega_{s} \cos \theta+C \omega_{z}\right) \omega_{x}+\left(C^{(w)} \omega_{s} \cos \phi \sin \theta+A \omega_{x}\right) \omega_{z}=M_{G_{y}} (10.152b)
C \dot{\omega}_{z}-C^{(w)} \omega_{s} \dot{\theta} \sin \theta+C^{(w)} \dot{\omega}_{s} \cos \theta -\left(C^{(w)} \omega_{s} \cos \phi \sin \theta+A \omega_{x}\right) \omega_{y}+\left(C^{(w)} \omega_{s} \sin \phi \sin \theta+B \omega_{y}\right) \omega_{x}=M_{G_{z}} (10.152c)
A \dot{\omega}_{x}=0
B \dot{\omega}_{y}+C^{(w)} \omega_{s}\left(\omega_{z}+\dot{\phi}\right)=0
C \dot{\omega}_{z}-C^{(w)} \omega_{S} \dot{\theta}=0
Thus, the components of vehicle angular acceleration are
\dot{\omega}_{x}=0 \quad \dot{\omega}_{y}=-\frac{C^{(w)}}{B} \omega_{s}\left(\omega_{z}+\dot{\phi}\right) \quad \dot{\omega}_{z}=\frac{C^{(w)}}{C} \omega_{s} \dot{\theta}We see that pitching the gyro at the rate \dot{\theta} around the vehicle y axis alters only \omega_{z}, leaving \omega_{x} unchanged. However, to keep \omega_{y}=0 clearly requires \dot{\phi}=-\omega_{z}. In other words, for the control moment gyro to control the angular velocity about only one vehicle axis, it must therefore be able to precess around that axis (the z axis in this case). That is why the control moment gyro must have two gimbals.