Question 10.13: A satellite is in torque-free motion (MGnet = 0). A nongimba...
A satellite is in torque-free motion \left( M _{G_{\text {net }}}= 0 \right). A nongimbaled gyro (momentum wheel) is aligned with the vehicle’s x axis and is spinning at the rate \omega_{s_{0}}. The spacecraft angular velocity is \omega=\omega_{x} \hat{ i }. If the spin of the gyro is increased at the rate \dot{\omega}_{s}, find the angular acceleration of the spacecraft.
Using Figure 10.26 as a guide, we set φ = 0 and φ = 90° to align the spin axis with the x axis. Since there is no gimbaling, \dot{\theta}=\dot{\phi}=0. Equations 10.152 then yield
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}+C^{(w)} \dot{\omega}_{s}=0
B \dot{\omega}_{y}=0
C \dot{\omega}_{z}=0
Clearly, the angular velocities around the y and z axes remain zero, whereas
\dot{\omega}_{x}=-\frac{C^{(w)}}{A} \dot{\omega}_{s}Thus, a change in the vehicle’s roll rate around the x axis can be initiated by accelerating the momentum wheel in the opposite direction. See Example 10.9.
