How would you figure out the fuel cost or delta V for spinning up a S/C to match the orbital period of 27.32 days (moon orbit)? Given that the S/C is initially not rotating about any axes.
For simplicity say the S/C is in an equatorial orbit and only needs to rotate about one axis. If the max torque is 40 N-cm, max thrust is 3N, total mass is 2kgs and ISP is 200s. Just the general equations are sufficient.
OK. So given the torque and moment of inertia we can work out the angular acceleration.
$$\dot\omega = T/I = 0.4/0.00045 = 1000 s^{-2}$$
Now the angular velocity you want is $$2\pi/(27.32*86400) = 2.6\times 10^{-6} s^{-1}$$ So full thrust for about $2.6$ nanoseconds would be needed to spin it up. Now a thruster with an $I_{sp}$ of 200s and a thrust of $3N$ consumes $3/200g kg$ of fuel per second (where $g$ is the acceleration of Earth's gravity at the surface) or about $1.5gs^{-1}$. So your spacecraft would need roughly 4 nanograms of fuel to achieve the desired rotation.
In practice, controlling a thruster of this power sufficiently accurately to achieve such a precise spin rate is likely to be a harder problem than fuelling it.
