We know angular momentum $L = mvr$, where $v$ is the velocity in the direction perpendicular to the distance from the source to the object whose $L$ we are trying to measure. My question is why $L$ was defined as $mvr$ where power of $r$ is 1 and not other integer greater than 1? And what experiments were conducted to come to this conclusion or was it only intuition?

And also when a dancer rotates with respect to her body it is said $L$ will be conserved if she pulls her hand closer to her body. As a result her angular velocity will increase. Where does this extra kinetic energy come from?

  • $\begingroup$ Because angular momentum is the moment of momentum ,and $r$ is the moment arm. It is literally how far away is momentum from the origin. $\endgroup$
    – JAlex
    Mar 23, 2021 at 18:04

3 Answers 3


Using Newton's law one can easily check that $\vec L= \vec r \times \vec p$ is a conserved vector for a motion of a particle in a spherically symmetric potential (while axial symmetry implies conservation of one component of $\vec L$). If you try to modify this definition, you will not obtain a conserved quantity.

If you are familiar with analytical mechanics, the form of angular momentum can also be derived from the Lagrangian using Noether's procedure.

Summarizing, I would say that the reason for this definition of angular momentum is not necessarily related to any particular experiment (although I don't know how it was developed historically). That's because motion of bodies is determined already by Newton's law and and there is no need to give separate laws for circular motions -- they can be derived from Newton's law (or other formulations of classical mechanics). It happens to be true that angular momentum is an extremely useful quantity in solving the Newton's equations.


Historically, this was mostly derived from observation (for example, conservation of angular momentum appears straightforwardly in Kepler's 2nd Law for the motion of planets).

Nowadays, however, conservation laws (of momentum, angular momentum, energy…) can be found directly from symmetries of the problem, through Noether's wonderful Theorem.

Simply put, Noether's Theorem states that if a system is invariant through a continuous transformation (for example, any rotation around a given point), then some quantity is a constant of motion (for this example, this turns out to be the usual angular momentum).

More specifically, if the Lagrangian $L$ describing a system does not depend on some quantity $q_k$, then the associated momentum $p_k=\frac{\partial L}{\partial \dot q_k}$ is a constant of motion. Writing the expression of $L$ for a given system gives the expression of the conserved quantity.

In our example, if a system is not changed by a rotation $\theta$ around origin, then L is independent of $\theta$ ; Noether's theorem states then that $\frac{\partial L}{\partial \dot \theta}$ is a constant of motion. Replacing L with the classical Lagrangian of a point mass, leads to the usual angular momentum.

All this works with other symmetries as well, leading to conservation of energy, of linear momentum ; and the same formalism applies to fields as well as particles, leading to modern gauge theories.

Regarding conservation of energy

To pull her arms towards her body, the dancer has to do work (against the centrifugal force) in the rotating frame. In the non-rotating frame, the corresponding force is towards the axis of rotation, and if the arm moves inward, it does work all the same.

That's unrelated to the conservation of momentum though! Momentum can be conserved (for rotational invariant systems) even if energy is not (for non time-invariant systems)

  • $\begingroup$ This answer should be expanded to include a comment re the dancer's conservation of energy $\endgroup$ Mar 23, 2021 at 10:17
  • $\begingroup$ I'm not quite sure I understand what you mean… If you're speaking about the cancer accelerating its rotation by bringing its arms close to the axis of rotation, that's about angular momentum conservation. The fact that energy is conserved (or not) depends on the details of the process and is an unrelated matter (conservation of energy is related to time-shift invariance through Noether's theorem, not rotation). $\endgroup$
    – Nicolas
    Mar 23, 2021 at 17:24
  • $\begingroup$ Oh, OK I get the point… $\endgroup$
    – Nicolas
    Mar 23, 2021 at 17:27

I was in the middle of writing an answer when I saw 2 perfectly good answers, but I think at least one important point remains unanswered. Also, just in case you don't know about vector products and/or Lagrangians, here goes my explanation.

Angular momentum is not just any combination of coordinates and velocities. It's the one that doesn't change when the system is invariant under rotations.

Angular momentum is proportional to the following particular combination of coordinates and velocities. The body is assumed to rotate in the $x$, $y$ plane, and the axis of rotation is along $z$, in the perpendicular direction: $$ xv_{y}-yv_{x} $$ It's not too difficult to see that, if you rotate your system infinitesimally, with angle $\epsilon$, the quantities involved change as, $$ x\mapsto x-\epsilon y $$ $$ v_{x}\mapsto v_{x}-\epsilon v_{y} $$ $$ y\mapsto y+\epsilon x $$ $$ v_{y}\mapsto v_{y}+\epsilon v_{x} $$ And our proposed conserved quantity changes as, $$ xv_{y}-yv_{x}\mapsto\left(x-\epsilon y\right)\left(v_{y}+\epsilon v_{x}\right)-\left(y+\epsilon x\right)\left(v_{x}-\epsilon v_{y}\right)= $$ $$ =xv_{y}+\epsilon xv_{x}-\epsilon yv_{y}-\epsilon^{2}yv_{x}-yv_{x}-\epsilon xv_{x}+\epsilon yv_{y}+\epsilon^{2}xv_{y}= $$ $$ =xv_{y}-yv_{x}+\epsilon^{2}\left(xv_{y}-yv_{x}\right) $$ Now you make this infinitesimal $\epsilon$ go to zero and you realise immmediately why this particular combination is special when there is rotation symmetry. How do we recover your special case, when the system is rotating itself at a constant rate around a fixed axis? In Cartesian coordinates: $$ v_{x}=-\omega y $$ $$ v_{y}=\omega x $$ If we restore the mass: $$ L=m\left(xv_{y}-yv_{x}\right)=m\left(\omega x^{2}+\omega y^{2}\right)=mr^{2}\omega=mrv $$

Kinetic energy (rotational)

Energy is conserved (approximately) in your dancer case. So there's no extra kinetic energy. The reason why the dancer (I prefer to think of an ice-skater; she illustrates better the point) spins faster when she pulls her arms closer is because this rotational energy is, $$ \frac{1}{2}mv^{2}=\frac{1}{2}mr^{2}\omega^{2}=\textrm{const.} $$ So if energy is to be conserved, $\omega$, the angular velocity, must increase when the average distance to the axis of rotation decreases.

  • $\begingroup$ +1 Excellent answer. $\endgroup$
    – user290607
    Mar 23, 2021 at 9:49
  • $\begingroup$ Thanks a lot, Feynstein. But if you think about it, it's not that good an answer. @Blazej's and Nicolas' answers are more rigorous --I'm voting them up to highlight their answers better. I was deliberately ambiguous about the words "doesn't change" for the sake of pedagogy. Noether's theorem is very simple, but far more subtle. I just wanted to show that the particular combination $xp_y-yp_x$ is special when you rotate the whole thing. $\endgroup$
    – joigus
    Mar 23, 2021 at 12:19
  • $\begingroup$ Your equation doesn't coincide with conservation of angular momentum. $K=\frac{L^2}{2mr^2).$ $\endgroup$
    – Bill N
    Mar 23, 2021 at 12:43
  • $\begingroup$ Doesn't it? $$K=\frac{L^{2}}{2mr^{2}}=\frac{m^{2}r^{4}\omega^{2}}{2mr^{2}}=\frac{1}{2}mr^{2}\omega^{2}$$ The dancer isn't a rigid body either, but still... $\endgroup$
    – joigus
    Mar 23, 2021 at 13:09
  • $\begingroup$ Oh, I see your point. But the OP said "extra energy." So energy was the point there. $\endgroup$
    – joigus
    Mar 23, 2021 at 13:15

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