Question 13.2: Satellite operators want to accurately determine a spacecraf...
Satellite operators want to accurately determine a spacecraft’s principal moment of inertia after it has fired an onboard rocket and expended propellant mass. The spacecraft is equipped with a reaction wheel, and both the spacecraft and wheel are initially at rest (i.e., zero angular momentum). The satellite operators command the reaction wheel to spin up to a constant angular velocity vector of \boldsymbol{\omega}_{w}=3,000 \mathbf{u}_{3} rpm by using the DC motor’s speed control mode. After the wheel reaches its constant spin rate, the onboard gyroscope measures the satellite’s angular velocity vector as \boldsymbol{\omega}_{3}=-0.182 \mathbf{u}_{3} \mathrm{rpm}. If the reaction wheel’s moment of inertia is I_{w}=0.055 \mathrm{~kg}-\mathrm{m}^{2}, determine the satellite’s moment of inertia about its 3 axis without the inertia contribution from the reaction wheel.
Because the satellite and wheel are initially at rest, the angular momentum is zero. We use Eq. (13.28) to determine the angular momentum
H=I_{3} \omega_{3}+I_{w} \omega_{w}=0
where I_{3} is the total moment of inertia (satellite + wheel) about the 3 axis. Solving this expression for the total moment of inertia, we obtain
I_{3}=-I_{w} \frac{\omega_{w}}{\omega_{3}}=\left(-0.055 \mathrm{~kg}-\mathrm{m}^{2}\right)(3,000 \mathrm{rpm}) /(-0.182 \mathrm{rpm})=906.5934 \mathrm{~kg}-\mathrm{m}^{2}
Note that we set the satellite’s spin rate \omega_{3} as a negative value because it spins in the opposite direction as the reaction wheel. Finally, the satellite’s moment of inertia without the reaction wheel is
I_{\mathrm{sat}}=I_{3}-I_{w}=906.5384 \mathrm{~kg}-\mathrm{m}^{2}