Nitpicky details of solid body motion

It should for nitpicky convenience's sake, be pointed out that unless the point we choose to calculate the moments around is the C.M of the rigid body, or that the the point moves parallell (including being at rest) to the motion of the C.M, we get an additional term on the RHS.

 

This is NOT the term I am speeking about; that is how the equation reads in body-fixed (rotating) coordinates . That is critical in order to decompose the angular motion relative to the body's principal axes.
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Let [itex]m_{i},\vec{r}_{i}, \vec{F}_{i}[/itex] be position vector, mass of the i'th particle relative to an inertial frame, while the F's is the net sum of forces acting at the i'th particle, [itex]\vec{r}_{0}[/itex] be the position vector of our moment point, so that [itex]\vec{r}_{i,0}=\vec{r}_{i}-\vec{r}_{0}[/itex]
We now sum over all particles in the body, noting that internal forces come in couples, acting along the connecting vector of the two particles, the moments from internal forces disappear, and we get:
[tex]\vec{r}_{i,0}\times\vec{F}_{i}^{(ext)}=\vec{r}_{i,0}\times{m}_{i}\vec{a}_{i}[/tex]
where superscript (ext) stands for the set of external forces.
Now, we get:
[tex]\vec{r}_{i,0}\times \vec{F}_{i}^{(ext)} =
\frac{d}{dt} (\vec{r}_{i,0}) \times {m}_{i} \vec{v}_{i})-(\vec{v}_{i} -\vec{v}_{0})\times\vec{v}_{i}{m}_{i}[/tex]
In the second expression, the cross product between the v_i's disappear, and summing over all particles, we gain, C.M subscript for Center of Mass:
[tex]\vec{r}_{i,0}\times\vec{F}_{i}^{(ext)}=\frac{d}{dt}(\vec{r}_{i,0}\times{m}_{i}\vec{v}_{i})+\vec{v}_{0}\times{M}\vec{v}_{C.M}[/tex]

The latter term disappears only when the moment point is moving parallell to the Center of Mass. The first is the rate of change of angular momentum with respect to the moment point.


The function of that last term can be appreciated considering the following scenario:
Let stuff move with constant, strictly horizontal velocity, while moment point moves strictly vertically.
No external forces acts upon stuff, but there sure is calculated rated of change of angular momentum around moment point! The last term brings the RHS side to a strict 0, as it should.

 

Again, not the term I speak of.

 

And, to emphazise:
For full 3-D rotational motion, any other representation than with rigid body coordinates is, typically, outright silly (i.e, the Eulerian representation is optimal, in which the twisting of the principal axes relative to the inertial frame is neatly represented). As long as it IS a rigid body then, which not necessarily has much to do with reality..