Question 10.12: The communications satellite is in a circular earth orbit of...
The communications satellite is in a circular earth orbit of period T. The body z axis always points towards the earth, so the angular velocity about the body y axis is 2π/T. The angular velocities about the body x and z axes are zero. The attitude control system consists of three momentum wheels 1, 2 and 3 aligned with the principal x, y and z axes of the satellite. Variable torque is applied to each wheel by its own electric motor. At time t = 0 the angular velocities of the three wheels relative to the spacecraft are all zero. A small, constant environmental torque M _{0} acts on the spacecraft. Determine the axial torques C^{(1)}, C^{(2)} \text { and } C^{(3)} that the three motors must exert on their wheels so that the angular velocity ω of the satellite will remain constant. The moment of inertia of each reaction wheel about its spin axis is I.
The absolute angular velocity of the xyz frame is given by
\omega=\omega_{0} \hat{ j } (a)
where \omega_{0}=2 \pi / T, a constant. At any instant, the absolute angular velocities of the three reaction wheels are, accordingly,
\begin{aligned}\omega^{(1)} &=\omega^{(1)} \hat{ i }+\omega_{0} \hat{ j } \\\omega^{(2)} &=\left[\omega^{(2)}+\omega_{0}\right] \hat{ j } \\\omega^{(3)} &=\omega_{0} \hat{ j }+\omega^{(3)} \hat{ k }\end{aligned} (b)
From (a) it is clear that \omega_{x}=\omega_{z}=\dot{\omega}_{x}=\dot{\omega}_{y}=\dot{\omega}_{z}=0. Therefore, Equations (g) of Example 10.11 become, for the case at hand,
\dot{\omega}^{(1)}=\frac{M_{G x}}{I}+\frac{B-C}{I} \omega_{0}(0)-\left\lgroup1+\frac{A}{I}+2 \frac{J}{I}\right\rgroup (0)+\omega^{(2)}(0)-\omega^{(3)} \omega_{o}
\dot{\omega}^{(2)}=\frac{M_{G y}}{I}+\frac{C-A}{I}(0)(0)-\left\lgroup 1+\frac{B}{I}+2 \frac{J}{I} \right\rgroup (0)+\omega^{(3)}(0)-\omega^{(1)}(0)
\dot{\omega}^{(3)}=\frac{M_{G z}}{I}+\frac{A-B}{I}(0) \omega_{0}-\left\lgroup 1+\frac{C}{I}+2 \frac{J}{I} \right\rgroup (0)+\omega^{(1)} \omega_{o}-\omega^{(2)}(0)
which reduce to the following set of three first-order differential equations,
\begin{aligned}\dot{\omega}^{(1)}+\omega_{0} \omega^{(3)} &=\frac{M_{G x}}{I} \\\dot{\omega}^{(2)} &=\frac{M_{G y}}{I} \\\dot{\omega}^{(3)}-\omega_{0} \omega^{(1)} &=\frac{M_{G z}}{I}\end{aligned} (c)
Equation ( c )_{2} implies that \omega^{(2)}=M_{G y} t / I+ constant, and since \omega^{(2)}=0 at t = 0, this means that for time thereafter,
\omega^{(2)}=\frac{M_{G y}}{I} t (d)
Differentiating ( c )_{3} with respect to t and solving for \dot{\omega}^{(1)} \text { yields } \dot{\omega}^{(1)}=\ddot{\omega}^{(3)} / \omega_{o}. Substituting this result into ( c )_{1} we get
\ddot{\omega}^{(3)}+\omega_{0}^{2} \omega^{(3)}=\frac{\omega_{0} M_{G x}}{I}The well-known solution of this differential equation is
\omega^{(3)}=a \cos \omega_{0} t+b \sin \omega_{0} t+\frac{M_{G x}}{I \omega_{0}}where a and b are constants of integration. According to the problem statement, \omega^{(3)}=0 when t = 0. This initial condition requires a=-M_{G x} / \omega_{0} I, so that
\omega^{(3)}=b \sin \omega_{0} t+\frac{M_{G x}}{I \omega_{0}}\left(1-\cos \omega_{0} t\right) (e)
From this we obtain \dot{\omega}^{(3)}=b \omega_{0} \cos \omega_{0} t+\left(M_{G x} / I\right) \sin \omega_{0} t which, when substituted into ( c )_{3} yields
\omega^{(1)}=b \cos \omega_{0} t+\frac{M_{G x}}{I \omega_{0}} \sin \omega_{o} t-\frac{M_{G z}}{I \omega_{0}} (f)
Since \omega^{(1)}=0 at t = 0, this implies b=M_{G z} / \omega_{0} I. In summary, therefore, the angular velocities of wheels 1, 2 and 3 relative to the satellite are
\omega^{(1)}=\frac{M_{G x}}{I \omega_{0}} \sin \omega_{0} t+\frac{M_{G z}}{I \omega_{0}}\left(\cos \omega_{0} t-1\right) ( g )_{1}
\omega^{(2)}=\frac{M_{G y}}{I} t ( g )_{2}
\omega^{(3)}=\frac{M_{G z}}{I \omega_{0}} \sin \omega_{0} t+\frac{M_{G x}}{I \omega_{0}}\left(1-\cos \omega_{0} t\right) ( g )_{3}
The angular momenta of the reaction wheels are
\begin{aligned}& H _{G_{1}}^{(1)}=I_{x}^{(1)} \omega_{x}^{(1)} \hat{ i }+I_{y}^{(1)} \omega_{y}^{(1)} \hat{ j }+I_{z}^{(1)} \omega_{z}^{(1)} \hat{ k } \\& H _{G_{2}}^{(2)}=I_{x}^{(2)} \omega_{x}^{(2)} \hat{ i }+I_{y}^{(2)} \omega_{y}^{(2)} \hat{ j }+I_{z}^{(2)} \omega_{z}^{(2)} \hat{ k } \\& H _{G_{3}}^{(3)}=I_{x}^{(3)} \omega_{x}^{(3)} \hat{ i }+I_{y}^{(3)} \omega_{y}^{(3)} \hat{ j }+I_{z}^{(3)} \omega_{z}^{(3)} \hat{ k }\end{aligned} (h)
According to (b), the components of the flywheels’ angular velocities are
\begin{array}{lll}\omega_{x}^{(1)}=\omega^{(1)} & \omega_{y}^{(1)}=\omega_{0} & \omega_{z}^{(1)}=0 \\\omega_{x}^{(2)}=0 & \omega_{y}^{(2)}=\omega^{(2)}+\omega_{0} & \omega_{z}^{(2)}=0 \\\omega_{x}^{(3)}=0 & \omega_{y}^{(3)}=\omega_{0} & \omega_{z}^{(3)}=\omega^{(3)}\end{array}Furthermore, I_{x}^{(1)}=I_{y}^{(2)}=I_{z}^{(3)}=I, so that (h) becomes
\begin{aligned}& H _{G_{1}}^{(1)}=I \omega^{(1)} \hat{ i }+I_{y}^{(1)} \omega_{0} \hat{ j } \\& H _{G_{2}}^{(2)}=I\left(\omega^{(2)}+\omega_{0}\right) \hat{ j } \\& H _{G_{3}}^{(3)}=I_{y}^{(3)} \omega_{0} \hat{ j }+I \omega^{(3)} \hat{ k }\end{aligned} (i)
Substituting (g) into these expressions yields the angular momenta of the wheels as a function of time,
\begin{aligned}H _{G_{1}}^{(1)} &=\left[\frac{M_{G_{x}}}{\omega_{0}} \sin \omega_{0} t+\frac{M_{G_{z}}}{\omega_{0}}\left(\cos \omega_{0} t-1\right)\right] \hat{ i }+I_{y}^{(1)} \omega_{0} \hat{ j } \\H _{G_{2}}^{(2)} &=\left(M_{G_{y}} t+I \omega_{0}\right) \hat{ j } \\H _{G_{3}}^{(3)} &=I_{y}^{(3)} \omega_{0} \hat{ j }+\left[\frac{M_{G_{z}}}{\omega_{0}} \sin \omega_{0} t+\frac{M_{G_{x}}}{\omega_{0}}\left(1-\cos \omega_{0} t\right)\right] \hat{ k }\end{aligned} (j)
The torque on the reaction wheels is found by applying Euler’s equation to each one. Thus, for wheel 1
\left. M _{G_{1} \text { net }}=\frac{d H _{G_{1}}^{(1)}}{d t}\right\rgroup _{ rel }+ \omega \times H _{G_{1}}^{(1)}
=\left(M_{G x} \cos \omega_{0} t-M_{G z} \sin \omega_{0} t\right) \hat{ i }+\left[M_{G z}\left(1-\cos \omega_{0} t\right)-M_{G x} \sin \omega_{0} t\right] \hat{ k }
Since the axis of wheel 1 is in the x direction, the torque is the x component of this moment (the z component being a gyroscopic bending moment),
C^{(1)}=M_{G x} \cos \omega_{0} t-M_{G z} \sin \omega_{0} tFor wheel 2,
M _{G_{2} \text { net }}=\left.\frac{d H _{G_{2}}^{(2)}}{d t}\right\rgroup _{ rel }+\omega \times H _{G_{2}}^{(2)}=M_{G y} \hat{ j }Thus,
C^{(2)}=M_{G y}Finally, for wheel 3
\left. M _{G_{3} \text { net }}=\frac{d H _{G_{3}}^{(3)}}{d t}\right\rgroup _{ rel }+\omega \times H _{G_{3}}^{(3)}
=\left[M_{G_{x}}\left(1-\cos \omega_{0} t\right)+M_{G_{z}} \sin \omega_{0} t\right] \hat{ i }+\left(M_{G_{x}} \sin \omega_{0} t+M_{G_{z}} \cos \omega_{0} t\right) \hat{ k }
For this wheel, the torque direction is the z axis, so
C^{(3)}=M_{G_{x}} \sin \omega_{0} t+M_{G_{z}} \cos \omega_{0} t