Doubts about the parallel-axis theorem

Question

I was studying about the parallel-axis theorem but had trouble making intuitive sense looking at it's derivation and statement. My confusion is that how is the moment of inertia about the center of mass axis affecting the moment of inertia about any other axis parallel to the C.O.M axis? Also why is the condition such that the new axis must be parallel to the C.O.M axis and not in any other direction?

My though process is along the lines that the moment of inertia about any other parallel axis is determined by decomposing it about the C.O.M axis and then accounting for the change in position of the new axis w.r.t to the C.O.M axis thus resulting in the "moment of inertia of that transformation" giving the result for the parallel-axis theorem. For the reasoning of why it would apply to only parallel axes and not some other axis it may be due to "component of moment of inertia" in some other direction(as it is not in the direction of the C.O.M axis) resulting in the moment of inertia in a direction which the C.O.M axis does not account for but then I think there could be formula like the one below with a "new factor" to account for the direction.

$I_{axis} = md^2 + \mathit factor + I_{c.o.m}$

All of this is speculation on my part and I know that these reasonings are probably very wrong so it would be helpful if someone could explain it.

Answer 1

Consider a body orbiting at some distance around a fixed point and spinning about its center of mass (such that both rotations are in the same plane).

The orbital angular momentum will be that of a point particle positioned at the center of mass: $$L_{orbit} = mr_{CM}^2\omega_{orbit}$$

The spin angular momentum will be: $$L_{spin} = I_{CM}\omega_{spin}$$

So the total angular momentum is: $$L = mr_{CM}^2\omega_{orbit} + I_{CM}\omega_{spin}$$

Now try to picture that a body rotating at angular frequency $\omega$ about a point a distance $d$ from its center of mass is nothing but the previously considered system, with that point as the center of orbit and where the orbit is in "tidal lock", i.e. $r_{CM} = d$ and $\omega_{orbit} = \omega_{spin} = \omega$. The angular momentum is thus

$$L = md^2\omega + I_{CM}\omega = (md^2+I_{CM})\omega$$

And hence we arrive at the parallel axis theorem:

$$I = md^2 + I_{CM}$$