A spacecraft in torque-free motion has three identical momentum wheels with their spin axes aligned with the vehicle’s principal body axes. The spin axes of momentum wheels 1, 2 and 3 are aligned with the x, y and z axes, respectively. The inertia tensors of the rotationally symmetric momentum

Question

A spacecraft in torque-free motion has three identical momentum wheels with their spin axes aligned with the vehicle’s principal body axes. The spin axes of momentum wheels 1, 2 and 3 are aligned with the x, y and z axes, respectively. The inertia tensors of the rotationally symmetric momentum wheels about their centers of mass are, therefore,

[latex]\begin{matrix}[I_{G_{1}}^{(1)}]= \left[\begin{matrix} I & 0 & 0 \0 &J &0 \0 & 0 & J \end{matrix} ight] &[I_{G_{2}}^{(2)}]= \left[\begin{matrix} J & 0 & 0 \0 &I &0 \0 & 0 & J \end{matrix} ight] & [I_{G_{3}}^{(3)}]= \left[\begin{matrix} J & 0 & 0 \0 &J &0 \0 & 0 & I\end{matrix} ight]\end{matrix}[/latex] (a)

The spacecraft moment of inertia tensor about the vehicle center of mass is
[latex][I_{G}^{(v)}]= \left[\begin{matrix}A & 0 & 0 \0 &B &0 \0 & 0 & C \end{matrix} ight][/latex] (b)

Calculate the spin accelerations of the momentum wheels in the presence of external torque.

Accepted Answer

The absolute angular velocity ω of the spacecraft and the angular velocities [latex]\omega ^{(1)}_{rel}, \omega ^{(2)}_{rel}, \omega ^{(3)}_{rel} [/latex] of the three flywheels relative to the spacecraft are

[latex]\begin{matrix} \left\{\omega ight\}=\left\{\begin{matrix} \omega_{x} \\omega_{y}\\omega _{z}\end{matrix} ight\} &\left\{\omega ^{(1)} ight\}_{rel}=\left\{\begin{matrix} \omega ^{(1)} \0\0 \end{matrix} ight\} &\left\{\omega ^{(2)} ight\}_{rel}=\left\{\begin{matrix} 0 \\omega ^{(2)}\0 \end{matrix} ight\}&\left\{\omega ^{(3)} ight\}_{rel}=\left\{\begin{matrix} 0 \0\\omega ^{(3)}\end{matrix} ight\} \end{matrix}[/latex] (c)

Therefore, the angular momentum of the spacecraft and momentum wheels is
[latex]\left\{H_{G} ight\}=[I_{G}^{(v)}]\left\{\omega ight\}+[I_{G_{1}}^{(1)}](\left\{\omega ight\}+\left\{\omega^{(1)} ight\}_{rel})+[I_{G_{2}}^{(2)}](\left\{\omega ight\}+\left\{\omega^{(2)} ight\}_{rel})[/latex]
[latex]+[I_{G_{3}}^{(3)}](\left\{\omega ight\}+\left\{\omega^{(3)} ight\}_{rel})[/latex] (d)
Substituting Equations (a), (b) and (c) into this expression yields
[latex]\left\{H_{G} ight\}=\left[\begin{matrix}I & 0 & 0 \0 &I &0 \0 & 0 & I \end{matrix} ight]\left\{\begin{matrix} \omega ^{(1)} \ \omega ^{(2)} \ \omega ^{(3)} \end{matrix} ight\} +\left[\begin{matrix}A+I+2J& 0 & 0 \0 &B+I+2J &0 \0 & 0 &C+ I+2J \end{matrix} ight] \left\{\begin{matrix} \omega_{x} \ \omega_{y} \ \omega_{z} \end{matrix} ight\}[/latex] (e)

In this case, Euler’s equations are
[latex]\left\{\dot{H}_{G} ight\}_{rel}+\left\{\omega ight\} imes \left\{H_{G} ight\}=\left\{M_{G} ight\}[/latex] (f)

Substituting (e), we get
[latex]\left[\begin{matrix}I & 0 & 0 \0 &I &0 \0 & 0 & I \end{matrix} ight]\left\{\begin{matrix} \dot{\omega}^{(1)} \ \dot{\omega}^{(2)} \ \dot{\omega}^{(3)} \end{matrix} ight\} +\left[\begin{matrix}A+I+2J & 0 & 0 \0 &B+I+2J &0 \0 & 0 &C+ I+2J \end{matrix} ight] \left\{\begin{matrix} \dot{\omega}_{x} \ \dot{\omega}_{y} \ \dot{\omega}_{z} \end{matrix} ight\}+ \left\{\begin{matrix} \omega_{x} \ \omega_{y} \ \omega_{z} \end{matrix} ight\}[/latex]

[latex] imes\left( \left[\begin{matrix}I & 0 & 0 \0 &I &0 \0 & 0 & I \end{matrix} ight]\left\{\begin{matrix} {\omega}^{(1)} \ {\omega}^{(2)} \ {\omega}^{(3)} \end{matrix} ight\} +\left[\begin{matrix}A+I+2J & 0 & 0 \0 &B+I+2J &0 \0 & 0 &C+ I+2J \end{matrix} ight] \left\{\begin{matrix} {\omega}_{x} \ {\omega}_{y} \ {\omega}_{z} \end{matrix} ight\} ight)[/latex]

[latex]=\left\{\begin{matrix}M_{G_{x}} \M_{G_{y}} \M_{G_{z}} \end{matrix} ight\}[/latex] (g)

Expanding and collecting terms yields the time rates of change of the flywheel spins (relative to the spacecraft) in terms of those of the spacecraft’s absolute angular velocity components,
[latex]\dot{\omega}^{(1)}=\frac{M_{G_{x}}}{I}+\frac{B-C}{I}\omega_{y}\omega_{z}-(1+\frac{A}{I}+2\frac{J}{I})\dot{\omega}_{x}+\omega ^{(2)}\omega_{z}-\omega^{(3)}\omega_{y}[/latex]
[latex]\dot{\omega}^{(2)}=\frac{M_{G_{y}}}{I}+\frac{C-A}{I}\omega_{x}\omega_{z}-(1+\frac{B}{I}+2\frac{J}{I})\dot{\omega}_{y}+\omega ^{(3)}\omega_{x}-\omega^{(1)}\omega_{z}[/latex] (h)
[latex]\dot{\omega}^{(3)}=\frac{M_{G_{z}}}{I}+\frac{A-B}{I}\omega_{x}\omega_{y}-(1+\frac{C}{I}+2\frac{J}{I})\dot{\omega}_{z}+\omega ^{(1)}\omega_{y}-\omega^{(2)}\omega_{x}[/latex]

Question 10.10: A spacecraft in torque-free motion has three identical momen...

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