Question 12.3: An axisymmetric oblate satellite has the principal moments o...
An axisymmetric oblate satellite has the principal moments of inertia I_{1}=I_{2}=50 \mathrm{~kg}-\mathrm{m}^{2} and I_{3}=90 \mathrm{~kg}-\mathrm{m}^{2}. Its initial angular velocity vector (in body-fixed 123 coordinates) is
\boldsymbol{\omega}(0)=\left[\begin{array}{c} 0 \\ 0.18 \\ 0.30 \end{array}\right]=0.18 \mathbf{u}_{2}+0.30 \mathbf{u}_{3} \mathrm{rad} / \mathrm{s}
The initial Euler angles are \psi_{0}=0 and \phi_{0}=0. The satellite is equipped with a camera with body-fixed position vector \mathbf{r}_{\text {cam }}=0.9 \mathbf{u}_{1} \mathrm{~m} (i.e., the camera points along the satellite’s 1 axis) as shown in Figure 12.15.
a) Determine the nutation angle \theta and angle \gamma.
b) Compute the angular momentum vector \mathbf{H} in inertial coordinates at time t=0.
c) Determine the satellite’s angular velocity vector \boldsymbol{\omega} in 123 body coordinates at time t=25 \mathrm{~s}.
d) Determine the position vector of the satellite’s camera at time t=0 and t=25 \mathrm{~s} as expressed in the inertial O X Y Z frame.
e) Recalculate the angular momentum vector \mathbf{H} in inertial coordinates at time t=25 \mathrm{~s} using the satellite’s angular velocity vector \boldsymbol{\omega} (expressed in body coordinates) and show that it remains constant.
a) Figure 12.15 shows the initial configuration of the spinning oblate satellite. Note that there is no angular velocity component along the 1 axis at t=0 and therefore \omega_{12} =\omega_{2}(0)=0.18 \mathrm{rad} / \mathrm{s}. Using Eq. (12.56), we can determine the angle \gamma
\gamma=\tan ^{-1}\left(\frac{\omega_{12}}{\omega_{3}}\right)=\tan ^{-1}\left(\frac{0.18}{0.30}\right)=30.964^{\circ}
Using Eq. (12.59), the nutation angle is
\theta=\tan ^{-1}\left(\frac{I_{1} \omega_{12}}{I_{3} \omega_{3}}\right)=\tan ^{-1}\left(\frac{9}{27}\right)=18.435^{\circ}
\theta{=}\tan^{-1}\left({\frac{I_{1}\omega_{12}}{I_{3}n}}\right) (12.59)
b) First, we compute the angular momentum vector in 123 body coordinates:
\mathbf{H}_{\text {body }}=\mathbf{I} \boldsymbol{\omega}_{0}=\left[\begin{array}{ccc} 50 & 0 & 0 \\ 0 & 50 & 0 \\ 0 & 0 & 90 \end{array}\right]\left[\begin{array}{c} 0 \\ 0.18 \\ 0.30 \end{array}\right]=\left[\begin{array}{c} 0 \\ 9 \\ 27 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}
To obtain \mathbf{H} in the inertial frame, we need the rotation matrix \widetilde{\mathbf{R}} evaluated with the appropriate Euler angles. Using Eq. (12.74) with initial Euler angles \psi_{0}=0, \phi_{0}=0, and \theta_{0}=18.435^{\circ}, we obtain
\widetilde{\mathbf{R}}_{0}=\left[\begin{array}{ccc} c_{\psi} c_{\phi}-s_{\psi} s_{\phi} c_{\theta} & -c_{\psi} s_{\phi}-s_{\psi} c_{\phi} c_{\theta} & s_{\psi} s_{\theta} \\ s_{\psi} c_{\phi}+c_{\psi} s_{\phi} c_{\theta} & -s_{\psi} s_{\phi}+c_{\psi} c_{\phi} c_{\theta} & -c_{\psi} s_{\theta} \\ s_{\phi} s_{\theta} & c_{\phi} s_{\theta} & c_{\theta} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0.9487 & -0.3162 \\ 0 & 0.3162 & 0.9487 \end{array}\right]
We see that the initial rotation matrix is a pure rotation about the inertial X axis; intuitively this makes sense if we revisit Figure 12.13 with \psi_{0}=0 and \phi_{0}=0. Hence, the angular momentum in the inertial frame O X Y Z is
\mathbf{H}_{\mathrm{fix}}=\widetilde{\mathbf{R}}_{0} \mathbf{H}_{\text {body }}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0.9487 & -0.3162 \\ 0 & 0.3162 & 0.9487 \end{array}\right]\left[\begin{array}{c} 0 \\ 9 \\ 27 \end{array}\right]=\left[\begin{array}{c} 0 \\ 0 \\ 28.4605 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}
We know that for torque-free motion the satellite’s angular momentum vector \mathbf{H}_{\mathrm{fix}} is constant; we will verify this fact in (e).
c) We will show two ways to determine the satellite’s angular velocity at a future time. Because the satellite is axisymmetric \left(I_{1}=I_{2}\right), we may use Eqs. (12.52) and (12.53)
\begin{array}{c}{{\omega_{1}(t)=\omega_{12}\cos(\lambda t+\beta)}} & (12.52)\\ {{\omega_{2}(t)=\omega_{12}\sin(\lambda t+\beta)}} & (12.53) \end{array}
to project the 1-2 spin components to a future time:
\omega_{1}(t)=\omega_{12} \cos (\lambda t+\beta), \quad \omega_{2}(t)=\omega_{12} \sin (\lambda t+\beta)
Because \omega_{12}=\omega_{2}(0)=0.18 \mathrm{rad} / \mathrm{s} (as shown in Figure 12.15), the phase angle must be \beta=90^{\circ}(=\pi / 2 \mathrm{rad}). Next, we use Eq. (12.43) to find the “coning-motion frequency”
\lambda=\frac{I_{3}-I_{2}}{I_{1}} n=\frac{90-50}{50}(0.3 \mathrm{rad} / \mathrm{s})=0.24 \mathrm{rad} / \mathrm{s}
\lambda={\frac{I_{3}-I_{2}}{I_{1}}}n={\frac{I_{3}-I_{1}}{I_{2}}}n (12.43)
where n=\omega_{3} is the constant spin rate about the 3 axis. Hence, the two spin components at t=25 \mathrm{~s} are
\begin{aligned} & \omega_{1}(25)=\omega_{12} \cos (0.24 \cdot 25+\pi / 2)=0.0503 \mathrm{rad} / \mathrm{s}(=2.88 \mathrm{deg} / \mathrm{s}) \\ & \omega_{2}(25)=\omega_{12} \sin (0.24 \cdot 25+\pi / 2)=0.1728 \mathrm{rad} / \mathrm{s}(=9.90 \mathrm{deg} / \mathrm{s}) \end{aligned}
The angular velocity vector at t=25 \mathrm{~s} is \boldsymbol{\omega}(25)=0.0503 \mathbf{u}_{1}+0.1728 \mathbf{u}_{2}+0.3 \mathbf{u}_{3} \mathrm{rad} / \mathrm{s}.
We can also determine the body-spin components from the Euler rates by using Eqs. (12.60)-(12.62):
\omega_{1}=\dot{\psi} \sin \theta \sin \phi, \quad \omega_{2}=\dot{\psi} \sin \theta \cos \phi, \quad \omega_{3}=\dot{\phi}+\dot{\psi} \cos \theta
Recall that the nutation angle \theta is constant for torque-free motion. Equation (12.67)
{\dot{\phi}}=-\lambda (12.67)
shows that the Euler angle rate \dot{\phi} is constant and opposite the coning frequency, that is, \dot{\phi}=-\lambda=-0.24 \mathrm{rad} / \mathrm{s}. The precession rate \dot{\psi} is determined using Eq. (12.69):
\dot{\psi}=\frac{I_{3} \dot{\phi}}{\left(I_{1}-I_{3}\right) \cos \theta}=0.5692 \mathrm{rad} / \mathrm{s}(=32.61 \mathrm{deg} / \mathrm{s})
The Euler angles \phi and \psi at time t=25 \mathrm{~s} are
\begin{aligned} & \phi(25)=\phi_{0}+\dot{\phi} t=0-(0.24)(25)=-6 \mathrm{rad}=-343.77^{\circ}\left(=16.23^{\circ}\right) \\ & \psi(25)=\psi_{0}+\dot{\psi} t=0+(0.5692)(25)=14.230 \mathrm{rad}=95.32^{\circ} \end{aligned}
Using the Euler angles and their rates, the body-frame spin components at t=25 \mathrm{~s} are
\begin{aligned} & \omega_{1}=\dot{\psi} \sin \theta \sin \phi=0.0503 \mathrm{rad} / \mathrm{s} \\ & \omega_{2}=\dot{\psi} \sin \theta \cos \phi=0.1728 \mathrm{rad} / \mathrm{s} \\ & \omega_{3}=\dot{\phi}+\dot{\psi} \cos \theta=0.3 \mathrm{rad} / \mathrm{s} \end{aligned}
These alternate calculations match the previous solutions.
d) We can easily determine the inertial position of the satellite’s camera at time t=0 by multiplying matrix \widetilde{\mathbf{R}}_{0} with the camera’s body-fixed coordinates \mathbf{r}_{\text {cam: }} :
\mathbf{r}_{\mathrm{fix}}=\widetilde{\mathbf{R}}_{0} \mathbf{r}_{\mathrm{cam}}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0.9487 & -0.3162 \\ 0 & 0.3162 & 0.9487 \end{array}\right]\left[\begin{array}{c} 0.9 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{c} 0.9 \\ 0 \\ 0 \end{array}\right] \mathrm{m}
Because the body 1 axis is initially aligned with the inertial X axis (recall that \psi_{0}=\phi_{0} =0 ) and the camera points along the 1 axis, the inertial and rotating coordinates of the camera are coincident at time t=0.
Determining the inertial position of the camera at 25 \mathrm{~s} requires the rotation matrix \widetilde{\mathbf{R}} evaluated with the Euler angles at t=25 \mathrm{~s}, that is, \psi(25)=95.32^{\circ}, \phi(25)=-343.77^{\circ}, and \theta=18.435^{\circ} :
\widetilde{\mathbf{R}}_{25}=\left[\begin{array}{ccc} c_{\psi} c_{\phi}-s_{\psi} s_{\phi} c_{\theta} & -c_{\psi} s_{\phi}-s_{\psi} c_{\phi} c_{\theta} & s_{\psi} s_{\theta} \\ s_{\psi} c_{\phi}+c_{\psi} s_{\phi} c_{\theta} & -s_{\psi} s_{\phi}+c_{\psi} c_{\phi} c_{\theta} & -c_{\psi} s_{\theta} \\ s_{\phi} s_{\theta} & c_{\phi} s_{\theta} & c_{\theta} \end{array}\right]=\left[\begin{array}{ccc} -0.3532 & -0.8810 & 0.3149 \\ 0.9314 & -0.3629 & 0.0294 \\ 0.0884 & 0.3036 & 0.9487 \end{array}\right]
The inertial position of the camera at t=25 \mathrm{~s} is
\mathbf{r}_{\text {fix }}(25)=\widetilde{\mathbf{R}}_{25} \mathbf{r}_{\mathrm{cam}}=\left[\begin{array}{ccc} -0.3532 & -0.8810 & 0.3149 \\ 0.9314 & -0.3629 & 0.0294 \\ 0.0884 & 0.3036 & 0.9487 \end{array}\right]\left[\begin{array}{c} 0.9 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{c} -0.318 \\ 0.838 \\ 0.080 \end{array}\right] \mathrm{m}
Figure 12.16 shows the orientation of the satellite with respect to the inertial O X Y Z frame at times t=0 and t=25 \mathrm{~s}. At t=0 (Figure 12.16a), the satellite’s camera is pointing along the inertial +X axis which is collinear with the body 1 axis because \psi_{0}=\phi_{0}=0. Figure 12.16 \mathrm{~b} shows that at t=25 \mathrm{~s} the camera is primarily pointing along the +Y axis with a negative X-axis component.
e) The angular momentum vector in 123 body coordinates at t=25 \mathrm{~s} is
\mathbf{H}_{\text {body }}=\mathbf{I} \boldsymbol{\omega}(25)=\left[\begin{array}{ccc} 50 & 0 & 0 \\ 0 & 50 & 0 \\ 0 & 0 & 90 \end{array}\right]\left[\begin{array}{c} 0.0503 \\ 0.1728 \\ 0.30 \end{array}\right]=\left[\begin{array}{c} 2.515 \\ 8.640 \\ 27 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}
The angular momentum in the inertial frame O X Y Z is determined by multiplying this result by the rotation matrix at t=25 \mathrm{~s} :
\begin{aligned} \mathbf{H}_{\text {fix }}(25)=\widetilde{\mathbf{R}}_{25} \mathbf{H}_{\text {body }} & =\left[\begin{array}{ccc} -0.3532 & -0.8810 & 0.3149 \\ 0.9314 & -0.3629 & 0.0294 \\ 0.0884 & 0.3036 & 0.9487 \end{array}\right]\left[\begin{array}{c} 2.515 \\ 8.640 \\ 27 \end{array}\right] \\ & =\left[\begin{array}{c} 0 \\ 0 \\ 28.4605 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s} \end{aligned}
This calculation verifies that the satellite’s angular momentum vector remains constant in a torque-free environment when referenced to the inertial O X Y Z frame (see Figures 12.16a and 12.16b).
