In pure rotation, angular momentum is measured about the rotation axis. This angular momentum is measured with respect to that axis. Why do we consider it universal? It should change with respect to other points, e.g. if the reference is taken on the body away from the axis of rotation doesn't the angular momentum change?
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Angular momentum is not universal. You are correct in stating that angular momentum should always be considered relative to an axis. We never consider an axis as being universal.
answered Jan 23 at 11:36
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The earth that spins and orbits has angular momentum as measured about the sun, and also as measured about its axis. It is not universal which one describes a certain situation.
When considering angular momentum about the sun we are describing the geometry or the orbit in terms of its major radius and eccentricity.
But when considering angular momentum about the earth's axis then we are looking at the properties of the earth's spin and its stability/wobbles.
Consider a different example. Say you have a bullet that is traveling in a straight line with momentum $\vec{p}$. You are using a high-speed camera and at time instant you measure the location of the bullet as $\vec{r}$.
The angular momentum of the bullet with respect to the camera is $$ \vec{L} = \vec{r} \times \vec{p}$$
Notice that nothing is rotating here, and there is not axis to measure angular momentum about. You are taking a valid measurement of angular momentum for this situation.
Furthermore, you can use the measured angular momentum to find the location in space of the path of the bullet (let us assume you didn't know $\vec{r}$, but only $\vec{L}$) reversing the calculation above.
The closest point of the path to the observer is $$\vec{r}_\perp = \frac{ \vec{p} \times \vec{L} }{ \| \vec{p} \|^2}$$ and the direction of motion is parallel to $\vec{p}$.
In the above scenario $\vec{L}$ encodes the geometry of the problem as it provides the moment-arm of of the path of linear momentum.
You can provide the above with the help of some vector algebra
$$\require{cancel} \small \begin{aligned}\vec{L} & =\vec{r}_{\perp}\times\vec{p}\\ & =\left(\frac{\vec{p}\times\vec{L}}{\|\vec{p}\|^{2}}\right)\times\vec{p}\\ & =\frac{\left(\vec{p}\times\vec{L}\right)\times\vec{p}}{\vec{p}\cdot\vec{p}}\\ & =\frac{\vec{L}\left(\vec{p}\cdot\vec{p}\right)-\vec{p}\left(\cancel{\vec{L}\cdot\vec{p}}\right)}{\vec{p}\cdot\vec{p}}\\ & =\vec{L}\frac{\vec{p}\cdot\vec{p}}{\vec{p}\cdot\vec{p}}=\vec{L} \end{aligned}$$
answered Jan 23 at 14:19
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Indeed as explained by Níckolas Alves, the angular momentum is always associated to an axis of reference. You might confuse "universal" with conserved. The reason why angular momentum is so interesting is that it is conserved in some conditions (for central forces and all situations symmetric under rotations). If the system is symmetric around a given axis, then the angular momentum will be conserved (in the sense of Lagrangian mechanics, the Lagrangian should be invariant under rotations around this axis). If the system is spherical, then no matter what axis you choose, it will be conserved.
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$\begingroup$ Regardless if symmetry that applies if a torque is present then angular momentum is not conserved. $\endgroup$ Commented Jan 23 at 13:57
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$\begingroup$ Well if a torque is present the system is not symmetric under rotation anymore right? $\endgroup$ Commented Jan 23 at 14:00
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$\begingroup$ Symmetry has to do with geometry. What do you mean by "symmetric around an axis" if you are not describing the geometry of the system? $\endgroup$ Commented Jan 23 at 14:04
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$\begingroup$ You are right, I am talking here at the lagrangian level not at the geometry in space level. I should have make my point more clear. Angular momentum around an axis is conserved if the Lagrangian after and before rotation gives the same equations of motions. $\endgroup$ Commented Jan 23 at 14:23
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$\begingroup$ "As stated above" isn't a set thing on this site. Answer orders change based on votes and the choice of answer sorting by each user. Best to mention / link the reference directly $\endgroup$ Commented Jan 23 at 14:47
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Choosing an axis is our wish, and the angular momentum changes about the axis for different positions.
