A satellite is in torque-free motion. A control moment gyro, spinning at the constant rate ωs, is gimbaled about the spacecraft y and z axes, with φ = 0 and θ = 90° (cf.Figure 10.25). The spacecraft angula velocity is ω = ωzk. If the spin axis of the gyro, initially along the x direction, is

Question

A satellite is in torque-free motion. A control moment gyro, spinning at the constant rate [latex] {\omega}_{s}[/latex] , is gimbaled about the spacecraft y and z axes, with [latex] \phi= 0 and θ = 90°[/latex] (cf.Figure 10.25). The spacecraft angular velocity is [latex]{\omega} = {\omega}_{z}\hat{k}[/latex] . If the spin axis of the gyro, initially along the x direction, is rotated around the y axis at the rate [latex]\dot{ heta}[/latex], what is the resulting angular acceleration of the spacecraft?

Accepted Answer

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

We see that pitching the gyro at the rate [latex]\dot{ heta}[/latex] around the vehicle y axis alters only [latex]\omega_{z}[/latex] , leaving [latex]\omega_{x}[/latex] unchanged. However, to keep [latex]\omega_{y}= 0 [/latex] clearly requires [latex] \dot{\phi} =-\omega_{z}[/latex]. 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.

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

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