When we find angular momentum of an object undergoing translational and rotational motion about a point O, we use the equation $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} + {\rm I}\, \boldsymbol{\omega}$. Here is $\rm I$ the moment of inertia about centre of mass, or is it the moment of inertia about the point O?

  • $\begingroup$ This is a valid question since there is often confusion on what point quantities are summed upon in dynamics. This question is the corollary of why Newton's 2nd law applies to the center of mass only. $\endgroup$ Commented Jul 29, 2023 at 20:20
  • $\begingroup$ Related: physics.stackexchange.com/q/765483/195949 $\endgroup$ Commented Jul 29, 2023 at 21:01

2 Answers 2


In the formula

$$\boldsymbol{L}_O = \boldsymbol{r}_C \times \boldsymbol{p} + {\rm I}_C\, \boldsymbol{\omega}$$

the MMOI ${\rm I}_C$ tensor is defined about the center of mass, if $\boldsymbol{r}_C$ is the location of the center of mass relative to O.

The is the combination of the angular momentum about the center of mass $$\boldsymbol{L}_C = {\rm I}_C \boldsymbol{\omega}$$ and the law of transformation of angular momentum

$$\boldsymbol{L}_O = \boldsymbol{L}_C + \boldsymbol{r}_C \times \boldsymbol{p}$$

If the object is purely rotating about O then the above can be re-formulated as

$$ \boldsymbol{L}_O = {\rm I}_O \boldsymbol{\omega} $$

where ${\rm I}_O$ is the MMOI tensor about O.


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take the sum of the torques about point O, you obtain

$$\vec\tau_O=\vec r\times\vec F+\vec \tau_C$$

from here the angular momentum $~\vec L_O~$

$$\vec L_O=\frac{d}{dt}\left(\vec r\times\vec F+\vec \tau_C\right)= \vec r\times \frac{d}{dt}\vec F+\frac{d}{dt}\vec\tau_C$$ with $$ \frac{d}{dt}\vec F=\vec p\quad ,\frac{d}{dt}\vec\tau_c=I_C\,\vec\omega_C\quad\Rightarrow\\ \vec L_O=\vec r\times\vec p+I_C\,\vec\omega_C\quad,\vec p=m\,\vec v_C$$


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