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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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    Angular momentum is $\sum_i \vec{r}_i \times (m \vec{v}_i)$ which only equals your equation if the origin is the center of rotation. – John Alexiou Nov 16 '18 at 15:22
  • thank you, that solved the problem. My definitions were all wrong –  Nov 19 '18 at 11:44
  • When in doubt always go back to fundamentals. It all starts from momentum in mechanics. – John Alexiou Nov 19 '18 at 13:46
  • I'm voting to close this question as off-topic because the problem was resolved by the OP. – John Alexiou Nov 19 '18 at 15:35
  • @ja72 if we closed every question because it was resolved, then we'd have no open questions... – Kyle Kanos Nov 24 '18 at 12:15
  • @KyleKanos - I would expect the OP to write an answer with how he resolved the issue and then the question to be closed. That retains a record of what happened and triages open questions to unresolved issues. – John Alexiou Nov 24 '18 at 17:24
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    @ja72 that isn't at all how things operate here. Just because a question was answered does not mean we can/should close it. That would literally be a waste of everyone's time to do it that way & never let anyone else answer another question. – Kyle Kanos Nov 24 '18 at 17:34
  • @KyleKanos - point taken. I'll retract my vote. – John Alexiou Nov 24 '18 at 17:38
  • The factoring of the particle-basis expression for a rigid body into an angular momentum of the object as a whole (calculated using the motion of the CoM) plus a angular momentum of the object about it’s own CoM is covered in all the mechanics texts I’m familiar with. – dmckee --- ex-moderator kitten Dec 15 '19 at 06:27
  • "this axis is parallel to the axis it orbits about". That's not correct. The Moon's mean obliquity (the angle between its rotational axis and its orbital axis) is 6.687°, according to https://en.wikipedia.org/wiki/Orbit_of_the_Moon – PM 2Ring Apr 13 '20 at 08:07

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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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  • You said 'which would be the same as the moment of inertia of a point mass in place of the moon's COM' but that neglects the fact that the moon is spherical? Are you assuming this because it is small? The parallel axis theorem should tell us what L is for a moon that is not spinning! How is the rest justified? –  Nov 14 '18 at 11:52
  • @lucky-guess "Are you assuming this because it is small?" I am assuming that it is a point mass to show, that, if we neglect the fact that it is a spinning sphere (with its own moment of inertia), we'll get an incorrect answer. – V.F. Nov 14 '18 at 13:53
  • @lucky-guess "The parallel axis theorem should tell us what L is for a moon that is not spinning!" $L=I\omega$ is an angular momentum: it depends on the angular velocity, i.e., it would be zero for an object which is not spinning. The parallel axis theorem is about moments of inertia, I, which is what I am calculating before finding angular momentum, which is what the question is about. – V.F. Nov 14 '18 at 13:58
  • i meant not spinning about its own axis, but still orbiting around earth –  Nov 14 '18 at 14:44
  • for example, what would the answer be if the moon was not orbiting but just spinning about it's own axis? –  Nov 14 '18 at 15:06
  • @lucky-guess The moon is spinning around the earth and around its own axis at the same time and at the same angular velocity. The moment of inertia due to both rotations is greater than moment of inertia due to just orbiting (point mass case). Therefore, angular momentum is greater, which is what we show and that's the answer to your question. – V.F. Nov 14 '18 at 15:07
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    @lucky-guess "for example, what would the answer be if the moon was not orbiting but just spinning about it's own axis?" $L=I\omega$. You should be able to calculate I (moment of inertia) for a sphere of a known mass and radius. – V.F. Nov 14 '18 at 15:11
  • Thank you, but it doesn't seem trivial that the angular momentum stays the same regardless of origin –  Nov 14 '18 at 15:15
  • Moving to chat... – V.F. Nov 14 '18 at 15:51
  • Let us continue this discussion in chat. – V.F. Nov 14 '18 at 15:51
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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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