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My assumed definition of angular momentum is the sum over $i$ of $L_i =r_i\times{\omega_i}\times{r_i}$ for each particle about some origin.

We have two spheres rotating about the centre of orbit. For simplicity's sake, let's assume it is the centre of the earth. The moon has an orbit rate of $\omega_{m1}$ and a spin rate of $\omega_{m2}$. It is spinning around its own axis which passes through the centre of the moon, this axis is parallel to the axis it orbits about. The earth spins about an axis that is tilted at some angle to the axis of the moon's orbit.

what is the angular momentum of the moon, in terms of its mass, radius, distance to earth's centre, and its angular velocities?

This question answers my conceptual query of how rotation about a parallel axis affects angular momentum. I have derived the relation for when the rotation acts about an axis passing through the origin. $L_i = \omega r_i\cdot{r_i}-r_i\omega\cdot{r_i}$ which is obviously $I\omega$ since $\omega$ is constant. This tells us that the A.M of the earth is simply given by the formula of the momentum of inertia of a sphere times its angular velocity, given the initial assumption. However, it is not trivial to say that this stays the same when the system is shifted to the right for the moon.

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The equation $$\boldsymbol{L} = \sum_i m_i \boldsymbol{r}_i \times ( \boldsymbol{\omega} \times \boldsymbol{r}_i ) $$

only applies if the velocity of each body is $\boldsymbol{v}_i = \boldsymbol{\omega} \times \boldsymbol{r}_i$, and this only happens if the origin is at the center of rotation.

The easiest way to tackle this is to define the center of mass and decompose the motion of each object as the motion of the center of mass, and a rotation about it.

$$\boldsymbol{v}_i = \boldsymbol{v}_C + \boldsymbol{\omega} \times (\boldsymbol{r}_i-\boldsymbol{r}_C) $$

Then define the angular momentum about the center of mass as

$$\boldsymbol{L}_C = \sum_i (\boldsymbol{r}_i-\boldsymbol{r}_C) \times m_i \boldsymbol{v}_i $$

I think you will be able to take it from here. You can also read this answer to see how to proceed from momentum to the equations of motion for rigid bodies.

John Alexiou
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The parallel axis theorem tells us how to determine the moment of inertia of the moon relative to the earth's axis, $I_e$, if we know the moment of inertia of the moon relative to its COM, $I_{COM}$.

So, first we have to determine $I_{COM}$, which should be easy, if we assume that the moon is a uniform sphere.

Then, applying the parallel axis theorem, we can determine $I_e$.

The next step would be to assume that the moon is not spinning around its COM and calculate its moment of inertia relative to the earth's axis, $I_e'$, which would be the same as the moment of inertia of a point mass in place of the moon's COM.

Having calculated $I_e$ and $I_e'$, we can calculate the angular momentums, $L_e$ and $L_e'$, keeping in mind that the angular velocity of the moon relative to its COM is equal to the orbital angular velocity of the moon relative to the earth.

The comparison will show that $L_e$ is greater than $L_e'$, which means that the spin of the moon does change (increase) the angular momentum of the moon relative to the earth's axis.

V.F.
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If you wish to be more accurate, you should calculate the angular momentum of the system relative to the common center of mass, which is just below the surface of the Earth. Calculate Iw for each sphere. Then calculate r x P for each, treating each as a point mass a distance r from the common center of mass. Then add these four vectors. ( The tilt of the Earth requires a vector sum.) The big problem is getting an accurate value for I, since the density of each sphere varies with depth (and may contain some lumps.)

R.W. Bird
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