Control Moment Gyroscope Gimbal Motor Torque

by Viroos
Tags: control, gimbal, gyroscope, moment, motor, torque
 P: 4 Hi ! My name is Max and I'm a Mechanical Engineer. I'm working on Control Moment Gyroscope project, but because for a long time I didn't deal with theory, I'm stuck at some very basic dynamics problem. In this article http://www.spacecraftresearch.com/fi...nPeck_JGCD2009 the author writes that the total CMG torque is ${\tau _{cmg}} = \dot \phi \hat g \times {h_r}$ (eq. 8) and the gimbal motor torque is ${\tau _g} = {I_{cmg}}\ddot \phi + {I_{cmg}}{\mathop \omega \limits^B ^{B/N}} \cdot \hat g + ({\omega ^{B/N}} \times {h_r}) \cdot \hat g$ (eq. 9) But what happens practically if I mount the CMG at the top of a static table at my room, connect both motors (flywheel and gimbal) to power and wait for some steady state ? In the first equation we get some constant torque that CMG applies to the table, because $\dot \phi$ is some constant value, but on the other side, the gimbal motor torque, ${\tau _g}$ has to be zero, because there is no angular acceleration: $\ddot \phi = 0$, and the table is static, that is, ${\mathop \omega \limits^B ^{B/N}} = 0$ and ${\mathop \omega \limits^{B/N}} = 0$. That is, we need torque only to accelerate the gimbal. My question is how the CMG can to apply torque to the controlled body (table, satellite etc), actually without torque at a gimbal motor ? Many thanks in advance, Max
 P: 1 Its torque is applied through conservation of angular momentum. Let h_1 equal the angular momentum stored in the flywheel of the CMG. As the CMG rotates there is a new momentum, h_2 of the same magnitude but a different direction. The vector difference in h_2 and h_1 represents the integral of a torque applied to the CMG, therefore an equal and oposite torque had to have been transfered to the test stand, satellite, etc. That's why "torque amplification" exists. Small torques induce gimbal rates, which in turn create huge torques in the form of conservation of angular momentum.

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