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According to Wikipedia, the general formula for the angular velocity of a particle in three dimensions is $$\boldsymbol \omega = \frac{\mathbf r \times \mathbf v}{\left |\mathbf r\right|^2}.$$

But if this were true in general, wouldn't it follow that the angular momentum, $$L = \mathbf r \times (m\mathbf v),$$ is always parallel to $\boldsymbol \omega $?

Or is it simply that we can use different reference points to measure $\mathbf r$ in each formula (as appropriate), and thus we end up obliged to have tensors in our equations?

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    $\begingroup$ I think they are parallel: $\vec L=I\vec\omega$. $\endgroup$
    – Ryan Unger
    Commented Aug 31, 2015 at 13:15
  • $\begingroup$ @0celo7 Really? Kindly see my edit too. $\endgroup$ Commented Aug 31, 2015 at 13:19
  • $\begingroup$ @0celo7 Watch out for with the claim that the angular momentum is parallel to angular velocity. In general $I$ is a second rank tensor and they are not guaranteed parallel. $\endgroup$ Commented Aug 31, 2015 at 17:12
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    $\begingroup$ @dmckee OP was referring to a particle, not a collection. For a particle they are parallel. $\endgroup$
    – Ryan Unger
    Commented Aug 31, 2015 at 20:12

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For a single particle, yes they're parallel.

For a system of particles, $$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$

(you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted differently. "$\alpha$" is some constant of proportionality; another phrasing is that the two sides don't have to be linearly dependent.)

You write something about tensors/different reference points, but that isn't relevant. The components of each ${\bf r}_i$ are just regular functions of time with respect to a fixed origin in an inertial frame, as usual.

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    $\begingroup$ Key point in this answer: "for a single particle". And worth pointing out that the Wikipedia equation is also "for a particle", not an object of finite size. $\endgroup$
    – Floris
    Commented Aug 31, 2015 at 13:29
  • $\begingroup$ How is the angular velocity sum-able this way? Given a system composed of a 1-kg object and only one atom, both moving in the same circular path of the same radius with the same speed but with opposite directions, would the angular velocity vanish? $\endgroup$ Commented Aug 31, 2015 at 13:33
  • $\begingroup$ @kalkanistovinko Yes I suppose they would! That's a topic for another question. It's explained poorly in the wiki article en.wikipedia.org/wiki/… . But in practice I've never seen that definition of angular velocity used -- you usually specify a frame with Euler angles, construct a base change matrix $R(\theta,\varphi,\psi)$ in which all particles of the rigid body are fixed, and go from there. $\endgroup$
    – user12029
    Commented Aug 31, 2015 at 13:43
  • $\begingroup$ A common first counter-example is a massless bar with two point masses on the ends rotating about a axis through its center that is neither co-linear with nor perpendicular to the bar. $\endgroup$ Commented Aug 31, 2015 at 19:12
  • $\begingroup$ @NeuroFuzzy But if it's zero in this case regardless of how high the speed is then the angular momentum is zero too if $L_i = I_{ji}\omega_j}$ isn't it? $\endgroup$ Commented Sep 22, 2015 at 3:41

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