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Given a system of particles of masses $\{m_i\}$ whose positions relative to some origin $o$ are described by $\{\vec{r}_i(t)\}$ and whose velocities are described by $\{\vec{v}_i(t)\}$, we can formally define the following quantities: $$\vec{\tau} = \sum_i \vec{r}_i \times \vec{F}_i$$ $$\vec{L} = \sum_i m_i(\vec{r}_i \times \vec{v}_i)$$ $$\widetilde{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{bmatrix}$$

which are all mathematically defined and are not ambiguous whatsoever. We can then check that $\vec{\tau} = \frac{d}{dt}\vec{L}$, which mirrors $\vec{F} = \frac{d}{dt}\vec{p}$ for linear motion. Taking this further, perhaps we should expect that $\vec{L} = \widetilde{I}\vec{w}$ to parallel $\vec{p} = m\vec{v}$. Is there a way to formally define $\vec{w}$ for an arbitrary system of particles such that this identity holds? My textbook only defines $\vec{w}$ for the most basic of systems, and never gives a concrete formal definition for a general system of particles, so I am led to believe that it is either impossible or not worth pursuing, and I am not sure which is the case.

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    $\begingroup$ Does this link help answer your question? physics.stackexchange.com/q/79294 $\endgroup$ – Farcher Apr 14 '17 at 6:50
  • $\begingroup$ @Farcher Not particularly, it seems as if the discussion is still centered around $w$ defined for rigid bodies about their centre of mass; which is a very specific situation (treated in my textbook) when compared to the general system of particles $\endgroup$ – Joshua Lin Apr 14 '17 at 6:55
  • $\begingroup$ -1. Unclear. You seem to be saying that the velocities have no relation to positions (ie this is not a rigid body - see comment to Farcher) but your use of $I$ suggests that you are describing a rigid body. If the body is rigid then $\vec L = \widetilde{I} \vec \omega$ applies and $\vec L$ and $\vec \omega$ are not necessarily parallel. If the body is not rigid then both $\widetilde{I}$ and $\vec \omega$ will vary with time. $\endgroup$ – sammy gerbil Apr 14 '17 at 14:08
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The angular velocity vector of a system of particles is only well-defined for rigid bodies. This is because (a) the instantaneous angular velocity of a single particle relative to an origin is not well-defined, but rather has multiple possibilities; and (b) a collection of $n$ particles may not share any of these possibilities among themselves, which means that any notion of a "combined angular velocity" does not exist.

As an example of (a): Let's suppose I have a particle at position $\vec{r}$ and velocity $\vec{v}$. What is its angular velocity? Well, a particle with an angular velocity $\vec{\omega}$ should have its velocity and position vectors related by $$ \vec{v} = \vec{\omega} \times \vec{r}. $$ Right off the bat, we're going to have trouble with this definition for a general $r$; what if our particle's velocity has a radial component? The cross product will always yield a vector orthogonal to $\vec{r}$, so we can't straightforwardly define the angular velocity of a single particle if it's moving in this way.

You could try to get around this by redefining things so that the component of $\vec{v}$ perpendicular to $\vec{r}$ appears on the right-hand side. But even then, there's a more subtle problem. If you give me an $\vec{\omega}$ that satisfies the above equation, I can give you an infinite number of other $\vec{\omega}$'s that do as well: let $\vec{\omega}' = \vec{\omega} + \alpha \vec{r}$ for any $\alpha$. This new $\vec{\omega}'$ is perfectly consistent with the particle's motion, since we still have $$ \vec{\omega}' \times \vec{r} = (\vec{\omega} + \alpha \vec{r})\times \vec{r} = \vec{\omega} \times \vec{r} = \vec{v}. $$ (This can be understood as following from the fact that any particles lying along the rotation axis have zero velocity.) In other words, if you want the angular velocity of a particle to obey $\vec{v} = \vec{r} \times \vec{\omega}$, there are either no vectors $\vec{\omega}$ that satisfy this equation, or an uncountably infinite number.

You might thing that perhaps a system of many particles will get around this, since we'll get more information about which of the many $\vec{\omega}$'s is the "right" one. But consider the following system of two particles: Particle $m_1$ is at $\vec{r}_1 = \hat{x}$ and is moving with $\vec{v}_1 = \hat{y}$. Particle $m_2$ is at $\vec{r}_2 = \hat{y}$ and is moving with $\vec{v}_2 = \hat{x}$. It's not too hard to see that the possible angular velocities for each one are $$ \vec{\omega}_1 = \hat{z} + \alpha \hat{x}, \qquad \vec{\omega}_2 = -\hat{z} + \beta \hat{y}. $$ There are no possible values of $\vec{\omega}$ that these two particles share in common, and therefore there is no straightforward way to define an angular velocity for this system.

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