We know angular momentum is defined as $mvr$. In the context of Lagrangians and Noether's theorem, this definition pops up as the conserved quantity due to rotational symmetry of the system. Is there anyway to justify this definition without the use of Lagrangians and Noether's theorem? As in if we try to generalise linear momentum and linear force to the rotational case, it is clear we should have $L=I\omega$ like $p=mv$, and $\tau=I\alpha$ like $F=ma$ for some inertia $I$. But then why should $I=mr^2$ (which would imply $L=mr^2\omega=mvr)$ ?

This question came to me because I was reviewing classical mechanics after a long time, but then started to wonder how rotations were treated before Lagrangian mechanics. There must have been some intuitive reason for defining these quantities in this way, right?

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    $\begingroup$ No, the reason why we even care about momenta is that they are conserved for "free" particles, i.e. their conservation, approximate (and thus slow variation) or not, is fundamental to why we even care about them. It is $L=\vec r\times m\vec v$ that is conserved, and that forces $I=mr^2$ to be the appropriate mass equivalent for rotation. $\endgroup$ Commented Mar 28 at 10:08
  • $\begingroup$ @naturallyinconsistent How do we show the quantity $r \times mv$ is conserved? I mean yeah if we differentiate we get an $r \times a$ term which is zero for free motion. But the derivative of other quantities (like simply $m\omega$) is also zero for free motion. So I guess we need to show that the total $r \times p$ must be constant despite internal interactions. How do we show this? $\endgroup$
    – ksnad
    Commented Mar 28 at 10:32
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    $\begingroup$ @ksnad $r \times mv$ is conserved thanks to Newton's third law: the internal forces come in pairs that are in a straight line and opposite. $\endgroup$ Commented Mar 28 at 10:55
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    $\begingroup$ The connection of angular momentum with rotational symmetry does not require Lagrangians. $\endgroup$ Commented Mar 28 at 11:01
  • $\begingroup$ @TomekCzajka: The requirement that the forces are "in a straight line" (i.e., directed along the line connecting the objects) is not part of the usual statement of Newton's Third Law. $\endgroup$ Commented Mar 28 at 11:33

2 Answers 2


The concept of angular momentum has a precursor: Kepler's law of areas. In retrospect: Kepler's law of areas is an instance of conservation of angular momentum.

Isaac Newton showed in his work 'Principia' that Kepler's law of areas generalizes to any central force.

There is a 2022 answer (by me) that presents the logic of Newton's derivation of the law of areas

The key concept in Newton's derivation is that if the force that is involved is a central force then there is a rotational symmetry; the reasoning is independent of the orientation.

The area law shows that there is a conserved quantity that is proportional to $\omega r^2$

Angular momentum is an entity that exists in a plane; in order to have an angular momentum at all there must be some form of circumnavigating motion, so the minimum space that is needed is a plane.

One way to motivate the concept of moment of inertia is to require consistency between linear kinetic energy and angular kinetic energy.

Let's say a car is coasting along a straight section of road, at some velocity $v$, so we attribute a kinetic energy of $\tfrac{1}{2}mv^2$
Let the car be on a banked circuit and the car enters a corner. The car is moving along a circular path now, but obviously it still has the same kinetic energy.

We can express the kinetic energy in terms of the instantaneous tangent velocity vector, or we can express the kinetic energy in terms of angular velocity

We have: $v = \omega r$

If we want to express kinetic energy in terms of angular velocity, then in order to have overall consistency:

$$ \tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2 r^2 $$

Rearranging the second expression:

$$ \tfrac{1}{2}mr^2 \omega^2 $$

That suggests making the quantity $mr^2$ a standardized item: $I=mr^2$.

Then angular momentum is $I\omega$ and angular kinetic energy is $ \tfrac{1}{2}I \omega^2 $

So we have two lines of reasoning that slot in with each other: the law of areas, and self-consistent attribution of kinetic energy

It seems to me that to the scholars who worked in the century or so after Newton the concept of moment of inertia defined as $mr^2$ was pretty much inescapable.

  • $\begingroup$ This was really helpful. Thanks! $\endgroup$
    – ksnad
    Commented Mar 29 at 16:23

Starting with Newton law

$$ \frac{d\vec p}{dt}=\vec F$$

where $~\vec p=m\dot{\vec{r}}~$ is the linear momentum. $~\vec p~$ is conserved if the force $~\vec F~$ central force; $~\vec F=f(r)\,\frac {\vec r}{|\vec r|}~$

now the angular momentum $~\vec L~$ must be also conserved , thus we defined it as

$$\vec L=\vec r\times\vec p$$

form here $$\frac{d\vec L}{dt}=\frac{d}{dt}\left(\vec r\times\vec p\right)= \underbrace{\dot{\vec{r}}\times\vec p}_{=0}+\vec r\times\dot{\vec{p}}=\vec r\times\vec F=\vec\tau$$

thus if the force $~\vec F~$ central force $~\frac{d\vec L}{dt}=0~$ die angular momentum $~\vec L~$ is constant (conserved).

the angular momentum remain conserved if $~\vec L=\mathbf I\,\vec\omega~$ and $~\vec \tau~=\vec 0$

$\Rightarrow$ your case $~\omega=\frac vr~$ ,$~r~$ constant

$$\frac{d}{dt}\left(m\,r^2\,\frac vr\right)= m\,r\,\dot v=r\,F=\tau$$

the angular mometum $~L=m\,r\,v~$ is conserved if $~\tau~$ equal zero


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