Question 10.9: Use Equation 10.136 to obtain the equations of motion of a t...

M_{G_{net}}=\left[\frac{dH^{(v)}_{G}}{dt})_{rel}+\omega \times H^{(v)}_{G} \right] +\left[\frac{dH^{(w)}}{dt})_{rel}+\omega \times H^{(w)} \right]                  (10.136)
In the dual-spin satellite, we may arbitrarily choose the rotor as the body of the vehicle, to which the body frame is attached. The coaxial platform will play the role of the single reaction wheel. The center of mass G of the satellite lies on the axis of rotational symmetry (the z axis), between the center of mass of the rotor (G_{r}) and  that of the platform (G_{p}). For this torque-free system, Equation 10.136 becomes
\left.\frac{dH^{(v)}_{G}}{dt}\right)_{rel}+\omega^{(r)} \times H^{(v)}_{G} +\left.\frac{dH^{(p)}_{G_{p}}}{dt}\right)_{rel}+\omega^{(r)}\times H^{(p)}_{G_{p}}=0                                                         (a)
in which r signifies the rotor and p the platform.
The vehicle angular momentum about G is that of the rotor plus that of the platform center of mass,
\left\{H^{(v)}_{G}\right\} =\left[I^{(r)}_{G}\right]\left\{\omega ^{(r)}\right\}+\left[I^{(p)}_{m_{G}}\right]\left\{\omega ^{(r)}\right\}=([I^{(r)}_{G}]+[I^{(p)}_{m_{G}}]) \left\{\omega ^{(r)}\right\}                    (b)
[I^{(p)}_{m_{G}}]is the moment of inertia tensor of the concentrated mass of the platform about the system center of mass, and it is calculated by means of Equation 9.44. The components of and are constants, so from (b) we obtain
[I_{m}]=\left[\begin{matrix} m(y^{2} + z^{2})& −mxy& −mxz\\−mxy& m(x^{2} + z^{2}) &−myz\\−mxz& −myz &m(x^{2} + y^{2}) \end{matrix} \right]                                            (9.44)

\left.\frac{d\left\{H_{G}^{(v)}\right\}}{dt}\right) _{rel}=([I_{G}^{(r)}]+[I_{m_{G}}^{(p)}])\left\{\dot{\omega }^{(r)} \right\}                                                         (c)
The angular momentum of the platform about its own center of mass is
\left\{H_{G_{p}}^{(p)}\right\}=([I_{G_{p}}^{(p)}]\left\{\omega ^{(p)}\right\}                               (d)
For both the platform and the rotor, the z axis is an axis of rotational symmetry. Thus, even though the platform is not stationary in xyz, the moment of inertia matrix is not time dependent. It follows that
\left.\frac{d\left\{H^{(v)}_{G_{p}}\right\} }{dt}\right) _{rel}=[I_{G_{p}}^{(p)}]\left\{\dot{\omega}^{(p)}\right\}               (e)
Using (b) through (e), we can write the equation of motion (a) as
([I_{G}^{(p)}]+[I_{m_{G}}^{(p)}])\left\{\dot{\omega}^{(r)} \right\}+\left\{\omega^{(r)} \right\}\times ([I_{G}^{(r)}]+[I_{m_{G}}^{(p)}])\left\{\omega^{(r)} \right\}
+\left[I_{G_{p}}^{(p)}\right]+\left\{\dot{\omega}^{(p)} \right\}+\left\{\omega^{(r)} \right\}\times \left[I_{G_{p}}^{(p)}\right]\left\{\omega^{(p)} \right\}= \left\{0\right\}                               (f)
The angular velocity \omega^{(p)} of the platform is that of the rotor, \omega^{(r)}, plus the angular velocity of the platform relative to the rotor, \omega ^{(p)}_{rel} . Hence, we may replace\left\{\omega ^{(p)}\right\}with\left\{\omega ^{(r)}\right\} +\left\{\omega ^{(r)}_{rel}\right\}, so that, after a little rearrangement, (f) becomes
(\left[I_{G}^{(r)}\right]+\left[I_{G}^{(p)}\right])\left\{\dot{\omega}^{(r)} \right\}+\left\{\omega ^{(r)}\right\} \times ([I_{G}^{(r)}]+[I_{G}^{(p)}])\left\{\omega ^{(r)}\right\}
+[I_{G_{p}}^{(p)}]\left\{\dot{\omega }^{(p)}_{rel} \right\}+\left\{\omega ^{(r)}\right\}\times [I_{G_{p}}^{(P)}]\left\{\omega _{rel}^{(p)}\right\}=0                        (g)
in which
[I_{G}^{(p)}]=[I_{m_{G}}^{(p)}]+[I_{G_{p}}^{(p)}](Parallel axis formula.)                     (h)
The components of the matrices and vectors in (g) relative to the principal xyz body frame axes are

and

A_{r}, C_{r}, A_{p}  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 tothe 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^{(r)}_{z}  and  \omega_{\bot}= \omega^{(r)}_{x}\hat{i}+\omega^{(r)}_{y}\hat{j}). Substituting (i) and (j) into each of thefour terms in (g), we get
([I^{(r)}_{G}]+[I^{(p)}_{G}])\left\{\dot{\omega} ^{(r)}\right\}=\left[\begin{matrix} A_{r}+A_{p} & 0 & 0 \\ 0 & A_{r}+A_{p} & 0 \\ 0 & 0 &C_{r}+C_{p}\end{matrix} \right]\left\{\begin{matrix} \dot{\omega}^{(r)}_{x} \\\dot{\omega}^{(r)}_{y}\\\dot{\omega}^{(r)}_{z} \end{matrix} \right\}

=\left\{\begin{matrix} (A_{r}+A_{p})\dot{\omega}^{(r)}_{x} \\(A_{r}+A_{p})\dot{\omega}^{(r)}_{y}\\(C_{r}+C_{p})\dot{\omega}^{(r)}_{z} \end{matrix} \right\}                                    (k)

=\left\{\begin{matrix}[(C_{p}-A_{p})+(C_{r}-A_{r})]{\omega}^{(r)}_{y} {\omega}^{(r)}_{z} \\ [(A_{p}-C_{p})+(A_{r}-C_{r})]{\omega}^{(r)}_{x}{\omega}^{(r)}_{z}\\0 \end{matrix} \right\}            (l)

[I^{(p)}_{G_{p}}]\left\{\dot{\omega}^{(p)}_{rel} \right\}=\left[\begin{matrix} \bar{A}_{p} & 0 & 0 \\ 0 & \bar{A}_{p} &0 \\ 0 & 0 &C_{p} \end{matrix} \right]\left\{\begin{matrix} 0 \\ 0 \\ \dot{\omega}_{p} \end{matrix} \right\}= \left\{\begin{matrix} 0 \\ 0 \\ C_{p}\dot{\omega}_{p} \end{matrix} \right\}(m)

\left\{\omega ^{(r)}\right\} \times [I^{(p)}_{G_{p}}]\left\{{\omega}^{(p)}_{rel} \right\}= \left\{\begin{matrix}\omega _{x}^{(r)} \\ \omega _{y}^{(r)} \\ \omega _{z}^{(r)} \end{matrix} \right\}\times \left[\begin{matrix} \bar{A}_{p} & 0 & 0 \\ 0 & \bar{A}_{p} &0 \\ 0 & 0 &C_{p} \end{matrix} \right]\left\{\begin{matrix} 0 \\ 0 \\ {\omega}_{p} \end{matrix} \right\}= \left\{\begin{matrix}C_{p}\omega _{y}^{(r)}\omega _{p} \\ -C_{p}\omega _{x}^{(r)}\omega _{p} \\0 \end{matrix} \right\}                 (n)

With these four expressions, (g) becomes
\left\{\begin{matrix}(A_{r}+A_{p})\dot{\omega }_{x}^{(r)} \\ (A_{r}+A_{p})\dot{\omega }_{y}^{(r)}\\(C_{r}+C_{p})\dot{\omega }_{z}^{(r)} \end{matrix} \right\} +\left\{\begin{matrix}[(C_{p}-A_{p})+(C_{r}-A_{r})]{\omega}^{(r)}_{y} {\omega}^{(r)}_{z} \\ [(A_{p}-C_{p})+(A_{r}-C_{r})]{\omega}^{(r)}_{x}{\omega}^{(r)}_{z}\\0 \end{matrix} \right\}

+\left\{\begin{matrix} 0 \\ 0 \\C_{p}\dot{\omega }_{p} \end{matrix} \right\} +\left\{\begin{matrix}C_{p}\omega _{y}^{(r)}\omega _{p} \\ -C_{p}\omega _{x}^{(r)}\omega _{p} \\0 \end{matrix} \right\}=\left\{\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right\}                       (o)

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,
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                    (p)
C\dot{\omega}_{z}^{(r)}+C_{p}\dot{\omega}_{p}=0
where A and C are the combined transverse and axial moments of inertia of the dual-spin vehicle about its center of mass,
\begin{matrix}A = A_{r} + A_{p}& C = C_{r} + C_{p}\end{matrix}                (q)
The three equations (p) involve four unknowns , \omega^{(r)}_{x}, \omega^{(r)}_{y}, \omega^{(r)}_{z} and ωp. A fourthequation is required to account for the means of providing the relative velocity  ω_{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 ωp constant at a specified value, so that \dot{\omega}_{p}= 0. Then Equation  (p)_{3} implies \omega^{(r)}_{z} = constant as well. Thus,  \omega_{p}  and  \omega^{(r)}_{z} are removed from our list of unknowns, leaving  \omega^{(r)}_{x}  and  \omega^{(r)}_{y}to be governed by the first two equations in (p).