Question 10.9: A disk is attached to a plate at their common center of mass...
A disk is attached to a plate at their common center of mass. Between the two is a motor mounted on the plate which drives the disk into rotation relative to the plate. The system rotates freely in the xy plane in gravity-free space. The moments of inertia of the plate and the disk about the z axis through G are I_{p} \text { and } I_{w}, respectively. Determine the change in the relative angular velocity \omega_{\text {rel }} of the disk required to cause a given change in the inertial angular velocity ω of the plate.
The plate plays the role of the body of a spacecraft and the disk is a momentum wheel. At any given time, the angular momentum of the system about G is given by Equation 10.142,
\left\{ H _{G}\right\}=\left\lgroup \left[ I _{G}^{(\text {body })}\right]+\sum\limits_{i=1}^{n}\left[ I _{G}^{(i)}\right] \right\rgroup \{ \omega \}+\sum\limits_{i=1}^{n}\left[ I _{G_{i}}^{(i)}\right]\left\{ \omega _{ rel }^{(i)}\right\} (10.142)
H_{G}=\left(I_{p}+I_{w}\right) \omega+I_{w} \omega_{\text {rel }}At a later time (denoted by primes), after the torquing motor is activated, the angular momentum is
H_{G}^{\prime}=\left(I_{p}+I_{w}\right) \omega^{\prime}+I_{w} \omega_{ rel }^{\prime}Since the torque is internal to the system, we have conservation of angular momentum, H_{G}^{\prime}=H_{G}, which means
\left(I_{p}+I_{w}\right) \omega^{\prime}+I_{w} \omega_{ rel }^{\prime}=\left(I_{p}+I_{w}\right) \omega+I_{w} \omega_{ rel }Rearranging terms we get
I_{w}\left(\omega_{\text {rel }}^{\prime}-\omega_{\text {rel }}\right)=-\left(I_{p}+I_{w}\right)\left(\omega^{\prime}-\omega\right)Letting Δω = ω’ – ω, this can be written
\Delta \omega_{ rel }=-\left\lgroup 1+\frac{I_{p}}{I_{w}} \right\rgroup \Delta \omegaThe change \Delta \omega_{\text {rel }} in the relative rotational velocity of the disk is due to the torque applied to the disk at G by the motor mounted on the plate. An equal torque in the opposite direction is applied to the plate, producing the angular velocity change Δω opposite in direction to \Delta \omega_{\text {rel }}.
Notice that if I_{p} \gg I_{w}, which is true in an actual spacecraft, then the change in angular velocity of the momentum wheel must be very much larger than the required change in angular velocity of the body of the spacecraft.