Edit: This is not a duplicate question.
The other question asked how angular momentum remained constant if the distance varied.
This question asks why you can select any point and calculate angular momentum from there, instead of intuitively choosing to calculate the angular momentum with reference to the centre of mass/pivot.

A box is moving with constant velocity in a straight line. (The box is not rotating about its centre of mass)

But apparently, you can set the axis of rotation at any point, and the box will have an angular momentum of r x p (r is perpendicular distance from axis of rotation, p is momentum)

But why can you select the axis of rotation at any point instead of only at a pivot/centre of mass?enter image description here

  • $\begingroup$ its not just angular momentum. linear momentum depends on the frame of reference as well. you can get different momentum for different observers (moving observers) but the conservation laws still hold. $\endgroup$ Feb 2, 2019 at 8:50
  • $\begingroup$ "why can you select the axis of rotation at any point?" This suggests that you think some point has a special claim to be selected. Which point and why? $\endgroup$ Feb 2, 2019 at 9:24
  • $\begingroup$ The pivot/centre of mass. I've always thought that rotation has to be about a pivot, because you'd describe a wheel rotating about its centre of mass as "rotating", but a box moving in a straight line as "not rotating". $\endgroup$
    – helpme
    Feb 2, 2019 at 14:38
  • $\begingroup$ Possible duplicate of Can angular momentum not be conserved in a straight line motion? $\endgroup$
    – anna v
    Feb 2, 2019 at 14:48
  • 1
    $\begingroup$ I voted it as a duplicate because the mathematical answer is given in the link physics.stackexchange.com/questions/250448/… . $\endgroup$
    – anna v
    Feb 2, 2019 at 14:49

2 Answers 2


Short answer: you can calculate the angular momentum from anywhere you want, as long as the vectors $r$ and $p$ are defined. If what you calculate is useful of easy is another issue, but nothing prevents you from calculating a vector product of two things.

  • $\begingroup$ Interesting. Is there a long answer for this? I still find it weird to define vector r with reference to a randomly selected point. $\endgroup$
    – helpme
    Feb 3, 2019 at 10:39
  • 1
    $\begingroup$ There might be a long answer, but there's not much more to say. If you have a ruler, yo can measure positions, from where you choose! Not only you can measure distances, you acn also choose where to set $x=0$. This is kind of the same, you can choose any origin for your angular momenta. Of course there is one reasonable choice, because it is easier, but you can choose any. $\endgroup$
    – FGSUZ
    Feb 3, 2019 at 11:00

You can choose the point about which you calculate angular momentum, just as you can choose the point about which you take moments in a statics set-up (such as an unevenly loaded plank resting on trestles).

Indeed for a particle displaced by $\vec r$ from a point O, the rate of change of angular momentum, $\vec J,$ about that point is$$\frac{d \vec J}{dt}=\frac{d[\vec r \times (m\vec v)]}{dt}=m\left(\frac{d\vec r}{dt}\times \vec v\ + \vec r \times \frac{d \vec v}{dt}\right)= \vec r \times \frac{d(m\vec v)}{dt}=\vec r \times \vec F$$

[The first term in the big brackets is the cross product of a vector by itself and is therefore zero.] So we have established the important result that

Rate of change of particle's angular momentum about O = Moment of force on particle about point O.

This applies for any point O that we choose, making it a powerful principle because of its generality!

Although we can choose any point O, there may be strategic reasons for choosing a particular point. For example, if we choose the centre of the Sun as O, then a planet has a fixed angular momentum about that point, because there is no moment of the Sun's force on the planet about that point, as the force is radially inwards towards that point (to a good approximation). The constancy of angular momentum about O 'explains' Kepler's observed equal area law.

  • $\begingroup$ This seems arbitrary, is there any use of selecting a random point to calculate angular momentum in a straight line? Could angular momentum be used instead of momentum to analyse collisions? $\endgroup$
    – helpme
    Feb 3, 2019 at 10:37
  • $\begingroup$ (a) The choice of the point,O, about which to calculate angular momentum is indeed arbitrary. As I've tried to show, the connection to moment of a force works for any choice of O. (b) A similar arbitrary choice can be made in statics, where the Principle of Moments states that for a body to be in equilibrium the sum of moments (taking account of their directions) about 𝑎𝑛𝑦 point is zero. (c) Angular momentum is defined for any instant, so it doesn't really matter whether a body is moving in a straight line or in a curved path. $\endgroup$ Feb 3, 2019 at 16:57
  • $\begingroup$ One person's arbitrariness is another's generality, and we're all seeking generality, are we not? $\endgroup$ Feb 3, 2019 at 17:15

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