Question 10.12: A satellite is in torque-free motion (MGnet = 0).A non-gimba...
A satellite is in torque-free motion (M_{G_{net}} = 0). A non-gimbaled 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 10.25 as a guide, we set \phi=\theta and \theta= 90° to align the spin axis with the x axis. Since there is no gimbaling ,\dot{\theta}=\dot{\phi} =\theta. Equations 10.146 then yield
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}+\dot{H^{(w)}}=0
B\dot{\omega}_{y}=0
C\dot{\omega}_{z}=0
Clearly, the angular velocities around the y and z axes remain zero, whereas,
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.
