Question 12.5: An axisymmetric oblate satellite has principal moments of in...

a) First, let us compute the angular momentum vector using Eq. (12.18)

\mathbf{H}=\mathbf{I} \boldsymbol{\omega}_{0}=\left[\begin{array}{ccc} 700 & 0 & 0 \\ 0 & 700 & 0 \\ 0 & 0 & 1,100 \end{array}\right]\left[\begin{array}{c} -0.2 \\ 0.1 \\ 8.3 \end{array}\right]=-140 \mathbf{u}_{1}+70 \mathbf{u}_{2}+9,130 \mathbf{u}_{3} \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}

Remember that the angular momentum vector \mathbf{H} computed above is expressed in the body frame at time t=0, and that the body axes are not inertially fixed due to the coning motion (however, \mathbf{H} is a fixed vector relative to an inertial frame).

Equation (12.57) shows that the nutation angle is

\theta=\tan ^{-1}\left(\frac{H_{12}}{H_{3}}\right)

where the component of \mathbf{H} in the 1-2 plane of the body frame is

\mathbf{H}_{12}=\mathbf{H}_{1}+\mathbf{H}_{2}=-140 \mathbf{u}_{1}+70 \mathbf{u}_{2}

The magnitude of \mathbf{H}_{12} is H_{12}=\sqrt{(-140)^{2}+70^{2}}=156.5248 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}. Therefore, the initial nutation angle is

\theta=\tan ^{-1}\left(\frac{H_{12}}{H_{3}}\right)=\tan ^{-1}\left(\frac{156.5248}{9,130}\right)=0.0171  \mathrm{rad}\left(=0.9822^{\circ}\right)

The oblate satellite is spinning with a very slight wobble at time t=0.

b) We can use Eq. (12.33) to compute the initial rotational kinetic energy

T_{\text {rot }, 0}=\frac{1}{2} \boldsymbol{\omega}_{0} \cdot \mathbf{I} \boldsymbol{\omega}_{0}=\frac{1}{2}\left[\begin{array}{lll} -0.2 & 0.1 & 8.3 \end{array}\right]\left[\begin{array}{c} -140 \\ 70 \\ 9,130 \end{array}\right]=37,907 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2}

where the 3 \times 1 column vector \mathbf{H}=\mathbf{I} \boldsymbol{\omega}_{0} was computed in (a). Because we are using principal axes, the rotational kinetic energy may be computed using

T_{\text {rot }, 0}=\frac{1}{2} I_{1} \omega_{1}^{2}+\frac{1}{2} I_{2} \omega_{2}^{2}+\frac{1}{2} I_{3} \omega_{3}^{2}=37,907 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2} \text { (same result) }

c) The nutation damper will eventually remove the wobble and realign the major axis (the 3 axis) with the angular momentum vector. Because the nutation angle eventually goes to zero, the final angular velocity vector (and \mathbf{H} ) will only contain a component along the 3 axis (remember that the 3 axis is a coordinate of the rotating body frame and not a coordinate of an inertial frame). Because \mathbf{H} is a constant vector, we can easily compute the final spin rate using

\omega_{f}=\frac{H}{I_{3}}=8.3012  \mathrm{rad} / \mathrm{s}

where the angular momentum magnitude is H=\|\mathbf{H}\|=9,131.34 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}. The final angular velocity vector expressed in body-fixed coordinates is

\omega_{f}=8.3012 \mathbf{u}_{3}  \mathrm{rad} / \mathrm{s}

Note that the angular momentum is \mathbf{H}=\mathbf{I} \boldsymbol{\omega}_{f}=9,131.34 \mathbf{u}_{3} \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}, which has the same magnitude as \mathbf{H}=\mathbf{I} \boldsymbol{\omega}_{0} as computed in (a).

d) The final rotational kinetic energy is easy to compute because the satellite is in a pure spin about its major axis:

T_{\text {rot }, f}=\frac{1}{2} I_{3} \omega_{f}^{2}=37,900.64 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}^{2}

Comparing this result to (b), we see that the nutation damper has dissipated a very small percentage of the initial rotational kinetic energy.