Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the earth relative to the sun's axis is the same.

Another example is the spinning bike wheel/person holding it in a chair. It has angular momentum about its axis which is equal to the angular momentum of the center of the chair say if I were holding it, but not moving.

I understand the math and the i think i understand the parallel axis theorem.

I am trying to see a concrete/more physical understanding of how this is so because the earth isnt rotation about the suns axis, so how can it have momentum about it?

Like an object moving in a straight line relative to a point still 'has some rotation'. I understand that. Trying to see how an object spinning about its axis has rotation to an axis in parallel that is not moving around in a circle.

here is a breakdown example. As you can see the spin angular momentum about Q is the same as it is for the ball center of mass. https://scripts.mit.edu/~srayyan/PERwiki/index.php?title=Module_3_--_Angular_Momentum_of_a_Rigid_Body_both_Rotating_and_Translating

  • $\begingroup$ "I am trying to see a concrete, physical understanding": could an answer to your question directly derive that the two angular momenta in your first (or second) paragraph are the same? (Without using the formula $L = \mathbf{R}_{cm} \times M \mathbf{V}_{cm} + I \omega$) $\endgroup$
    – najkim
    Commented Feb 1, 2021 at 19:58

2 Answers 2


Angular momentum isn't a quantity that stands on it own. Angular momentum does not describe the rotation axis or anything specific with motion in general. Angular momentum only describes where in space translational momentum acts. In a way, angular momentum is just the moment of momentum. Analogous to how torque is the moment of a force.

The two have the same defintion

$$ \begin{aligned} \boldsymbol{L} & = \boldsymbol{r} \times \boldsymbol{p} & \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} \end{aligned}$$

where the $\boldsymbol{r}\times$ part literatly just considers the perpendicular distance to the axis of momentum (percussion axis) or the axis of force (line of action).

But wait, isn't translational momentum always through the center of mass since $\boldsymbol{p} = m \, \boldsymbol{v}_{\rm CM}$.

One way to define momentum is where and how much to hit something in order to completely stop its motion. The impulse required has to act along the axis of momentum and is of equal and opposite magnitude to momentum. The impulse would thus instantaneously remove both translational and rotational momentum.

So a flying bullet indeed needs an impulse of $-\boldsymbol{p}$ through the center of mass and it will stop instantly.

Just like a force $\boldsymbol{F}$ through the center of mass would have an equivalent torque of $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$, the momentum $\boldsymbol{p}$ of the body has an equivalent rotational momentum of $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$.

In reverse, the presence of a torque $\boldsymbol{\tau}$ indicates there is an offset force applied, just as the presence of of rotational momentum indicates that momentum axis is offset from the origin.

The location of the line of action, or the percussion axis is also found by the same expression

$$ \begin{aligned} \boldsymbol{r} & = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2 } & \boldsymbol{r} & = \frac{ \boldsymbol{F} \times \boldsymbol{\tau}} {\| \boldsymbol{F} \|^2} \end{aligned} $$

So now let us look at a solid with mass moment of inertia. Look at the stone used for curling as it translates and rotates on a horizontal frictionless (almost) surface.

If you were to stop the stone from moving, and you just kick it along its center of mass, you are going to stop the translational momentum, but it is going to keep rotating maintaining its rotational momentum.

You would have to kick it, offset from the center of mass in order to stop all motion. By how much offset?

Here is where rotational momentum shines. The stone has intrinstic rotational momentum about the center of mass of $$ \boldsymbol{L}_{\rm CM} = \mathbf{I}_{\rm CM} \boldsymbol{\omega} $$ and the location of the offset is found by the equations above

$$ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L}_{\rm CM}} { \| \boldsymbol{p} \|^2 } $$

Now this can be simplified to the following $\| \boldsymbol{r} \| =r = \frac{I \omega}{m v} = \frac{I}{m c} $ where $I$ is the mass moment of inertia about the spin axis, $m$ is the mass and $c$ is the distance between the instant center of rotation (since the stone is rotating and translating at the same time) and the center of mass.

This is similar to pool where you hit a cue ball at an offset $r$ in order to acquire "english" or "spin". The resulting center of rotation is at $c = \frac{I}{m r}$.

As seen from the origin the rotational momentum has two parts, one because the center of mass is traveling at an offset to the origin, and the second due to its own spin.

$$ \boldsymbol{L} = \mathbf{I}_{\rm CM} \boldsymbol{\omega} + \boldsymbol{r}_{\rm CM} \times \boldsymbol{p} $$

This is the same angular momentum calculation that is done for the earth as seen from the sun, with both orbiting and spinning components.

In summary, the rotational momentum $\boldsymbol{L}$ is used to find where the axis of momentum (percussion axis) is in space using the relationship

$$ \boxed{ \boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L}} { \| \boldsymbol{p} \|^2 }} $$

This axis is significant since it describes the momentum state of a rigid body. An impulse applied along the axis, with equal and opposite magnitude to translational momentum would completely and instantaneously stop a rigid body.

There is a special case for a purely spinning object, where a single application of impulse would stop the rotation but also impart translational motion. So in this case you need an impulse couple (two equal and opposite impulses offset from each other), just as a force couple produces a torque.


Even if an object is accelerating you can take the torques about the center of mass (CM) considering only the forces in the inertial frame; using any point in the object other than the CM for evaluating torques, you must consider the non-inertial forces in the accelerating frame of reference.

If you select a fixed point (the point Q in your reference) about which to evaluate the angular momentum of a rigid body, it can be shown that the angular momentum is the sum of the angular momentum of the CM about that point plus the angular momentum of the object relative to the CM as claimed in your reference.

The text Mechanics, by Symon derives the specific relationships for the angular momentum of a system of particles, a rigid body being a special case. This falls out from manipulating the basic definition of angular momentum applied to a system of particles; specifically defined as the sum of the angular momenta of each particle. Angular momentum is somewhat tricky in that it specifically depends on the point chosen about which to calculate the quantity.

  • $\begingroup$ maybe i am not understanding, but that wasn't my question. If you see the link i attached, the ang momentum of the balls c of M is the same as the angular momentum of the ball with respect to Q(ignoring the orbital part). My question is why, since the ball is not rotating around Q $\endgroup$ Commented Feb 1, 2021 at 17:29
  • $\begingroup$ OP stated that he already understands the math, that he does not need to see another derivation, and that he would rather have a "physical, concrete" explanation $\endgroup$
    – najkim
    Commented Feb 1, 2021 at 17:30
  • $\begingroup$ I edited my response. Hope this helps. $\endgroup$
    – John Darby
    Commented Feb 1, 2021 at 17:37
  • $\begingroup$ Thank you, but yes I know the ang. Momentum can be shown to be equal, but what physical reason. If we look at the spin Ang momentum, There is no rotations about Q so why does that work. $\endgroup$ Commented Feb 1, 2021 at 17:59
  • $\begingroup$ Is it just it is what it is kinda deal lol $\endgroup$ Commented Feb 1, 2021 at 17:59

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