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suppose we need to calculate $\overrightarrow{L}$ about an axis, but the rigid body is not rotating about this axis. Can we define the $\overrightarrow{L}_{axis}$ still? I think we should be able to since $\overrightarrow{L}$ (for an infinitesimal point on the rigid body) is just $dm(\overrightarrow{r}\times\overrightarrow{v})$

If it is possible then is there a way to determine angular momentum in such cases more systematic approach (with proof) for

  1. Special case where we have to only deal with axis parallel to axis of rotation

  2. The general case i.e. the axis of rotation is tilted/skewed w.r.t to axis of rotation.

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  • $\begingroup$ Do you mean to say that you are observing an object undergoing rotation about an axis from some external axis? $\endgroup$ – Triatticus Jan 8 at 20:58
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Yes. If the angular momentum of a body about the origin is $\vec{L}=\vec{r} \times \vec{p}$, the angular momentum about an axis in the direction $\hat{n}$ is given by- $$L_{axis}=|\hat{n}.\vec{L}|$$ You can see that for yourself pretty easily. This is simply the projection of the angular momentum of the body about that axis. The perpendicular component of the angular momentum will vanish in the axis frame.

This answers all the remaining questions as well.

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