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 φ = 90° (cf. Figure 10.26 ). The spacecraft angular 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 φ = 0 and φ = 90° (cf. Figure 10.26 ). 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{\omega}_{s}=\phi=0[/latex] and θ = 90° into Equations 10.152 gives

[latex]A \dot{\omega}_{x}+C^{(w)} \omega_{s} \dot{ heta} \cos \phi \cos heta-C^{(w)} \omega_{s} \dot{\phi} \sin \phi \sin heta+C^{(w)} \dot{\omega}_{s} \cos \phi \sin heta +\left(C^{(w)} \omega_{s} \cos heta+C \omega_{z} ight) \omega_{y}-\left(C^{(w)} \omega_{s} \sin \phi \sin heta+B \omega_{y} ight) \omega_{z}=M_{G_{x}}[/latex] (10.152a)

[latex]B \dot{\omega}_{y}+C^{(w)} \omega_{s} \dot{ heta} \sin \phi \cos heta+C^{(w)} \omega_{s} \dot{\phi} \cos \phi \sin heta+C^{(w)} \dot{\omega}_{s} \sin \phi \sin heta -\left(C^{(w)} \omega_{s} \cos heta+C \omega_{z} ight) \omega_{x}+\left(C^{(w)} \omega_{s} \cos \phi \sin heta+A \omega_{x} ight) \omega_{z}=M_{G_{y}}[/latex] (10.152b)

[latex]C \dot{\omega}_{z}-C^{(w)} \omega_{s} \dot{ heta} \sin heta+C^{(w)} \dot{\omega}_{s} \cos heta -\left(C^{(w)} \omega_{s} \cos \phi \sin heta+A \omega_{x} ight) \omega_{y}+\left(C^{(w)} \omega_{s} \sin \phi \sin heta+B \omega_{y} ight) \omega_{x}=M_{G_{z}}[/latex] (10.152c)

[latex]A \dot{\omega}_{x}=0[/latex]
[latex]B \dot{\omega}_{y}+C^{(w)} \omega_{s}\left(\omega_{z}+\dot{\phi} ight)=0[/latex]
[latex]C \dot{\omega}_{z}-C^{(w)} \omega_{S} \dot{ heta}=0[/latex]

Thus, the components of vehicle angular acceleration are

[latex]\dot{\omega}_{x}=0 \quad \dot{\omega}_{y}=-\frac{C^{(w)}}{B} \omega_{s}\left(\omega_{z}+\dot{\phi} ight) \quad \dot{\omega}_{z}=\frac{C^{(w)}}{C} \omega_{s} \dot{ heta}[/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 10.14: A satellite is in torque-free motion. A control moment gyro,...

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