Question 10.13: A satellite is in torque-free motion. A control moment gyro,...

Substituting \omega_{x} ={\omega}_{y}=\dot{H^{(w)}} =\phi=\theta  and  \theta= 90° into Equations 10.146 gives
A\dot{\omega}_{x}+H^{(W)}\dot{\theta}\cos \phi \cos \theta -H^{(W)}\dot{\phi}\sin \phi \sin \theta +\dot{H}^{(W)}\cos \phi \sin \theta
+(H^{(W)}\cos \theta +C\omega_{z} )\omega _{y}-(H^{(W)}\sin \phi \sin \theta +B\omega _{y})\omega_{z}=M_{G_{net_{x}}}           (10.146a)
B\dot{\omega}_{y}+H^{(W)}\dot{\theta}\sin \phi \cos \theta+ H^{(W)}\dot{\phi}\cos \phi \sin \theta +\dot{H}^{(W)}\sin \phi \sin \theta
-(H^{(W)}\cos \theta +C\omega_{z} )\omega _{x}+(H^{(W)}\cos \phi \sin \theta +A\omega _{x})\omega_{z}=M_{G_{net_{y}}}           (10.146b)
C\dot{\omega}_{z}-H^{(W)}\dot{\theta}\sin\theta+\dot {H^{(W)}}\cos \theta -(H^{(W)}\cos \phi \sin \theta +A\omega _{x})\omega _{y}
+(H^{(W)}\sin \phi \sin \theta +B\omega_{y} )\omega _{x}=M_{G_{net_{z}}}                                                           (10.146c)
A\dot{\omega}_{x}=0
B\dot{\omega}_{y}=H^{(w)}({\omega}_{z}+\dot{\phi})=0
C\dot{\omega}_{z}-H^{(w)}\dot{\theta}=0
where H^{(w)} = C^{(w)}\omega_{s}. Thus, the components of vehicle angular acceleration are
\underline{\begin{matrix}\dot{\omega}_{x}=0 &\dot{\omega}_{y}=-\frac{C^{(w)}}{B}\omega_{s}(\omega _{z}+\dot{\phi }) & \dot{\omega}_{z}=\frac{C^{(w)}}{C}\omega _{s}\dot{\theta}\end{matrix}}

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.