Question 10.10: Use Equation 10.142 to obtain the equations of motion of a t...
Use Equation 10.142 to obtain the equations of motion of a torque-free, axisymmetric dual-spin satellite, such as the one shown in Figure 10.22.
\left\{ H _{G}\right\}=\left\lgroup \left[ I _{G}^{(\text {body })}\right]+\sum\limits_{i=1}^{n}\left[ I _{G}^{(i)}\right] \right\rgroup \{ \omega \}+\sum\limits_{i=1}^{n}\left[ I _{G_{i}}^{(i)}\right]\left\{ \omega _{ rel }^{(i)}\right\} (10.142)
In this case we have only one “reaction wheel,” namely, the platform p . The “body” is the rotor r. In Equation 10.142 we make the following substitutions (← means “is replaced by ”):
\omega \longleftarrow \omega^{(r)}
\omega_{ rel }^{(i)} \leftarrow \omega_{ rel }^{(p)}
\left[ I _{G}^{(b o d y)}\right] \leftarrow\left[ I _{G}^{(r)}\right]
\sum\limits_{i=1}^{n}\left[ I _{G}^{(i)}\right] \longleftarrow\left[ I _{G}^{(p)}\right]
\sum\limits_{i=1}^{n}\left[ I _{G_{i}}^{(i)}\right]\left\{ \omega _{ rel }^{(i)}\right\} \longleftarrow\left[ I _{G_{p}}^{(p)}\right]\left\{ \omega _{ rel }^{(p)}\right\}
so that Equation 10.142 becomes
\left\{ H _{G}\right\}=\left(\left[ I _{G}^{(r)}\right]+\left[ I _{G}^{(p)}\right]\right)\left\{\omega^{(r)}\right\}+\left[ I _{G_{p}}^{(p)}\right]\left\{\omega_{ rel }^{(p)}\right\} (a)
Since \left. M _{G_{\text {net }}}\right)_{\text {external }}= 0, Equation 10.143 yields
\left.\left. M _{G_{\text {net }}}\right)_{\text {external }}=\frac{d H _{G}}{d t}\right)_{\text {rel }}+\omega \times H _{G} (10.143)
\left(\left[ I _{G}^{(r)}\right]+\left[ I _{G}^{(p)}\right]\right)\left\{\dot{\omega}^{(r)}\right\}+\left[ I _{G_{p}}^{(p)}\right]\left\{\dot{\omega}_{ rel }^{(p)}\right\}+\left\{\omega^{(r)}\right\} \times\left(\left(\left[ I _{G}^{(r)}\right]+\left[ I _{G}^{(p)}\right]\right)\left\{\omega^{(r)}\right\}+\left[ I _{G_{p}}^{(p)}\right]\left\{\omega_{ rel }^{(p)}\right\}\right)=\{ 0 \} (b)
The components of the matrices and vectors in (b) relative to the principal xyz body frame axes attached to the rotor are
\left[ I _{G}^{(r)}\right]=\left[\begin{array}{ccc}A_{r} & 0 & 0 \\0 & A_{r} & 0 \\0 & 0 & C_{r}\end{array}\right] \quad\left[ I _{G}^{(p)}\right]=\left[\begin{array}{ccc}A_{p} & 0 & 0 \\0 & A_{p} & 0 \\0 & 0 & C_{p}\end{array}\right] \quad\left[ I _{G_{p}}^{(p)}\right]=\left[\begin{array}{ccc}\bar{A}_{p} & 0 & 0 \\0 & \bar{A}_{p} & 0 \\0 & 0 & C_{p}\end{array}\right] (c)
and
\left\{\omega^{(r)}\right\}=\left\{\begin{array}{c}\omega_{x}^{(r)} \\\omega_{y}^{(r)} \\\omega_{z}^{(r)}\end{array}\right\} \quad\left\{\omega_{ rel }^{(p)}\right\}=\left\{\begin{array}{c}0 \\0 \\\omega_{p}\end{array}\right\} (d)
A_{r}, C_{r}, A_{p} \text { and } C_{p} are the rotor and platform principal moments of inertia about the vehicle center of mass G , whereas \bar{A}_{p} is the moment of inertia of the platform about its own center of mass. We also used the fact that \bar{C}_{p}=C_{p}, which of course is due to the fact that G and G_{p} both lie on the z axis. This notation is nearly identical to that employed in our consideration of the stability of dual-spin satellites in Section 10.4 (wherein \omega_{r}=\omega_{z}^{(r)} \text { and } \omega_{\perp}=\omega_{x}^{(r)} \hat{ i }+\omega_{y}^{(r)} \hat{ j } ). Substituting (c) and (d) into each of the four terms in (b), we get
\left(\left[ I _{G}^{(r)}\right]+\left[ I _{G}^{(p)}\right]\right)\left\{\dot{\omega}^{(r)}\right\}=\left[\begin{array}{ccc}A_{r}+A_{p} & 0 & 0 \\0 & A_{r}+A_{p} & 0 \\0 & 0 & C_{r}+C_{p}\end{array}\right]\left\{\begin{array}{c}\dot{\omega}_{x}^{(r)} \\\dot{\omega}_{y}^{(r)} \\\dot{\omega}_{z}^{(r)}\end{array}\right\}=\left\{\begin{array}{l}\left(A_{r}+A_{p}\right) \dot{\omega}_{x}^{(r)} \\\left(A_{r}+A_{p}\right) \dot{\omega}_{y}^{(r)} \\\left(C_{r}+C_{p}\right) \dot{\omega}_{z}^{(r)}\end{array}\right\} (e)
\left\{\omega^{(r)}\right\} \times\left(\left[ I _{G}^{(r)}\right]+\left[ I _{G}^{(p)}\right]\right)\left\{\omega^{(r)}\right\}=\left\{\begin{array}{c}\omega_{x}^{(r)} \\\omega_{y}^{(r)} \\\omega_{z}^{(r)}\end{array}\right\} \times\left\{\begin{array}{c}\left(A_{r}+A_{p}\right) \omega_{x}^{(r)} \\\left(A_{r}+A_{p}\right) \omega_{y}^{(r)} \\\left(C_{r}+C_{p}\right) \omega_{z}^{(r)}\end{array}\right\}=\left\{\begin{array}{c}{\left[\left(C_{p}-A_{p}\right)+\left(C_{r}-A_{r}\right)\right] \omega_{y}^{(r)} \omega_{z}^{(r)}} \\{\left[\left(A_{p}-C_{p}\right)+\left(A_{r}-C_{r}\right)\right] \omega_{x}^{(r)} \omega_{z}^{(r)}} \\0\end{array}\right\} (f)
\left[ I _{G_{p}}^{(p)}\right]\left\{\dot{\omega}_{ rel }^{(p)}\right\}=\left[\begin{array}{ccc}\bar{A}_{p} & 0 & 0 \\0 & \bar{A}_{p} & 0 \\0 & 0 & C_{p}\end{array}\right]\left\{\begin{array}{c}0 \\0 \\\dot{\omega}_{p}\end{array}\right\}=\left\{\begin{array}{c}0 \\0 \\C_{p} \dot{\omega}_{p}\end{array}\right\} (g)
\left\{\omega^{(r)}\right\} \times\left[ I _{G_{p}}^{(p)}\right]\left\{\omega_{\text {rel }}^{(p)}\right\}=\left\{\begin{array}{c}\omega_{x}^{(r)} \\\omega_{y}^{(r)} \\\omega_{z}^{(r)}\end{array}\right\} \times\left[\begin{array}{ccc}\bar{A}_{p} & 0 & 0 \\0 & \bar{A}_{p} & 0 \\0 & 0 & C_{p}\end{array}\right]\left\{\begin{array}{c}0 \\0 \\\omega_{p}\end{array}\right\}=\left\{\begin{array}{c}C_{p} \omega_{y}^{(r)} \omega_{p} \\-C_{p} \omega_{x}^{(r)} \omega_{p} \\0\end{array}\right\} (h)
With these four expressions, (b) becomes
\left\{\begin{array}{c}\left(A_{r}+A_{p}\right) \dot{\omega}_{x}^{(r)} \\\left(A_{r}+A_{p}\right) \dot{\omega}_{y}^{(r)} \\\left(C_{r}+C_{p}\right) \dot{\omega}_{z}^{(r)}\end{array}\right\}+\left\{\begin{array}{c}{\left[\left(C_{p}-A_{p}\right)+\left(C_{r}-A_{r}\right)\right] \omega_{y}^{(r)} \omega_{z}^{(r)}} \\{\left[\left(A_{p}-C_{p}\right)+\left(A_{r}-C_{r}\right)\right] \omega_{x}^{(r)} \omega_{z}^{(r)}} \\0\end{array}\right\}+\left\{\begin{array}{c}0 \\0 \\C_{p} \dot{\omega}_{p}\end{array}\right\}+\left\{\begin{array}{c}C_{p} \omega_{y}^{(r)} \omega_{p} \\-C_{p} \omega_{x}^{(r)} \omega_{p} \\0\end{array}\right\}=\left\{\begin{array}{l}0 \\0 \\0\end{array}\right\} (i)
Combining the four vectors on the left-hand side, and then extracting the three components of the vector equation finally yields the three equations of motion of the dual-spin satellite in the body frame,
\begin{array}{r}A \dot{\omega}_{x}^{(r)}+(C-A) \omega_{y}^{(r)} \omega_{z}^{(r)}+C_{p} \omega_{y}^{(r)} \omega_{p}=0 \\A \dot{\omega}_{y}^{(r)}+(A-C) \omega_{x}^{(r)} \omega_{z}^{(r)}-C_{p} \omega_{x}^{(r)} \omega_{p}=0 \\C \dot{\omega}_{z}^{(r)}+C_{p} \dot{\omega}_{p}=0\end{array} (j)
where A and C are the combined transverse and axial moments of inertia of the dual-spin vehicle about its center of mass,
A=A_{r}+A_{p} \quad C=C_{r}+C_{p} (k)
The three equations (j) involve four unknowns, \omega_{x}^{(r)}, \omega_{y}^{(r)}, \omega_{z}^{(r)} \text { and } \omega_{p}. A fourth equation is required to account for the means of providing the relative velocity \omega_{p} between the platform and the rotor. Friction in the axle bearing between the platform and the rotor would eventually cause \omega_{p} to go to zero, as pointed out in Section 10.4. We may assume that the electric motor in the bearing acts to keep \omega_{p} constant at a specified value, so that \dot{\omega}_{p}=0. Then Equation ( j )_{3} implies that \omega_{z}^{(r)}= constant as well. Thus, \omega_{p} \text { and } \omega_{z}^{(r)} and removed from our list of unknowns, leaving \omega_{x}^{(r)} \text { and } \omega_{y}^{(r)} to be governed by the first two equations in (j). Note that we actually employed ( j )_{3} in the solution of Example 10.9 above.