# Angular momentum with respect to the centre of mass

I have been told [Warning: I leave this because it's what I asked and allows to understand the dialogues in the comments, but Azad, whom I thank, has pointed that the formula does not hold in general in the form it is expressed] that the angular momentum of and rigid body with respect to any point $P$ can always be expressed as $$\mathbf{L}_{P}=\mathbf{r}_{cm}\times M\mathbf{v}_{cm}+\big(\sum_im_iR_i^2\big)\boldsymbol{\omega}$$ where $\mathbf{r}_{cm}$ is the position of the centre of mass with respect to $P$, $M$ the mass of the body, $R_i$ the distance of the $i$-th point, having mass $m_i$, composing the body, and $\sum_im_iR_i^2=I$ its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity $\boldsymbol{\omega}$.

I know that the velocity $\mathbf{v}_i$ of each point $P_i$, having mass $m_i$, of a rigid body of mass $M$ can be see as the sum of a translation velocity of one of its points $C$ plus a rotation velocity around that point: $\mathbf{v}_i=\mathbf{v}_{C}+\boldsymbol{\omega}\times\overrightarrow{CP_i}$. If we chose $C$ as the centre of mass I see that $$\mathbf{L}_{cm}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{i}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{cm}+\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$$$=\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$because, if I am not wrong, $\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_C=(\sum_i m_i\overrightarrow{CP_i})\times\mathbf{v}_C=\mathbf{0}$ since $\sum_i m_i\overrightarrow{CP_i}$ is the position of the centre of mass with respect to itself, which is $\mathbf{0}$.

How can it be proved that $\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=(\sum_im_iR_i^2)\boldsymbol{\omega}$? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, $\forall t\quad I(t)= I(t_0)$, the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}=I\boldsymbol{\alpha}_{cm}$ where $\boldsymbol{\alpha}$ is the angular acceleration around the centre of mass. I heartily thank you for any answer!

Some unfruitful trials: by using the "BAC CAB identity" as suggested by Azad, whom I heartily thank, $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$, I can see that$$\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=\sum_im_i\|\overrightarrow{CP_i}\|^2\boldsymbol{\omega}-m_i(\overrightarrow{CP_i}\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$which, by decomposing every $\overrightarrow{CP_i}$ into an axial component $\mathbf{A}_i$ and a radial component $\mathbf{R}_i$, whose norms respectively are $A_i$ and $R_i$, with $R_i$ as the distance from $i$ to the axis of rotation, becomes $$\sum_im_iR_i^2\boldsymbol{\omega}+\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$but I cannot prove that $\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}=\mathbf{0}$.

• Isn't this just a consequence of the parallel axis theorem? Proofs of the theorem should be easy to Google. May 12, 2015 at 9:09
• Your first equation describes total angular momentum around origin and you're just a BAC CAB away from the proof
May 12, 2015 at 9:20
• I am not sure that I would call this a proof in any sense of the word. The calculation merely leads to the derivation of $I$, i.e. it's the definition of moments of inertia more than anything else. May 12, 2015 at 9:34
• I consider it a pretty bad habit to introduce things into physics that one "knows a-priori" just to get some sense of "proving something". Your a-priori knowledge is a trivial derivation of the very expression that you are using in your "proof". The much better way is to treat everything as a derivation. It works just as well and it doesn't clash with the definition of science, which is not the same as the definition of mathematics. May 12, 2015 at 9:47
• I recommend you read the first chapter of Goldstein's Classical Mechanics. It has a good and brief survey of these things.
May 12, 2015 at 10:22

I think you are overcomplicating this. Consider an arbitrary point P moving with linear speed $\mathbf{v}_A$.

• Linear momentum is $$\mathbf{P} = m \mathbf{v}_{cm}$$
• Angular momentum at the center of mass is $$\mathbf{L}_{cm} = I_{cm} \mathbf{\omega}$$
• Linear velocity of the center of mass is $$\mathbf{v}_{cm} = \mathbf{v}_A + \mathbf{\omega} \times \mathbf{r}_{cm}$$ where $\mathbf{r}_{cm}$ is the location of the center of mass relative to A.
• Linear momentum in terms of the motion of A is $$\mathbf{P} = m (\mathbf{v}_A + \mathbf{\omega} \times \mathbf{r}_{cm})$$ $$\boxed{ \mathbf{P} = m \mathbf{v}_A - m \mathbf{r}_{cm} \times \mathbf{\omega} }$$
• Angular momentum at A is $$\mathbf{L}_A =\mathbf{L}_{cm} +\mathbf{r}_{cm} \times \mathbf{P}$$ which is expanded as $$\mathbf{L}_A =I_{cm} \mathbf{\omega} +\mathbf{r}_{cm} \times m \mathbf{v}_{cm} = I_{cm} \mathbf{\omega} +\mathbf{r}_{cm} \times m (\mathbf{v}_A + \mathbf{\omega} \times \mathbf{r}_{cm})$$

$$\boxed{ \mathbf{L}_A = I_{cm} \mathbf{\omega}-m \mathbf{r}_{cm} \times\mathbf{r}_{cm} \times \mathbf{\omega} + m \mathbf{v}_{A}}$$

• Combined the spatial momenum at A yields the 6×6 spatial inertia matrix at A

$$\hat{\ell}_A = I_A \hat{v}_A$$ $$\begin{Bmatrix} \mathbf{P} \\ \mathbf{L}_A \end{Bmatrix} = \begin{bmatrix} m & -m [\mathbf{r}_{cm}\times] \\ m [\mathbf{r}_{cm}\times] & I_{cm}-m\,[\mathbf{r}_{cm}\times][\mathbf{r}_{cm}\times] \end{bmatrix} \begin{Bmatrix}\mathbf{v}_{A} \\ \mathbf{\omega} \end{Bmatrix}$$

NOTE: For the wierd $[\mathbf{r}\times]$ notation that seems to be missing a vector see What is the Vector/Cross Product?

• The mass momenent of inertia at A is thus defined as $$I_A = I_{cm}-m\,[\mathbf{r}_{cm}\times][\mathbf{r}_{cm}\times]$$ This is an vector representation of the parallel axis theorem.
• Finally you need to differentiate the momentum expressions to arrive at the 6 Newton-Euler equations of motion (See https://physics.stackexchange.com/a/80449/392)
• Thank you very much! Forgive me: I don't understand $\mathbf{L}_A = I_{cm} \mathbf{\omega}-m \mathbf{r}_{cm} \times\mathbf{r}_{cm} \times \mathbf{\omega} + m \mathbf{v}_{A}$. As to $I_{cm}$, is it a matrix? The only definition I know, used by my book, Gettys-Keller-Skove's Physics, of moment of inertia $I$ is a scalar $I:=\sum_im_iR_i^2$ -and $I=\int_V\rho R^2dV$ for a continuous body- where $R$ is the distance of the point from the axis. May 13, 2015 at 7:23
• $I_{cm}$ is the mass moment of inertia tensor. It is defined as $$I_{cm} = \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix}$$ (See farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html). It contains the components of inertia for each axis in the diagonal, and cross terms on the off diagonal. As the body rotates (with a 3×3 rotation matrix $E$) the components of $I_{cm}$ change also. This is done with $$I_{cm} = E I_{\rm body} E^\top$$. May 13, 2015 at 12:17
• Very, very interesting. I can't wait to study it! Thank you very much again! May 13, 2015 at 12:49

If we look at $CP_i X m_i (\omega \times CP_i)$ we can say that the cross product in the parenthesis gives the component of the vector $CP_i$ along the direction of $\omega$. Let us call that component $R_i$. (note: $R_i$ is the perpendicular distance between the particle in the system of particles in which we are interested in and the axis of rotation of the system of particles)

Now we have:

$CP_i \times m_i R_i \omega$ = $m_i R_i^2 \omega$

(We are taking the cross product of $CP_i$ with $R_i$ which is in the direction perpendicular to both $\omega$ and $CP_i$ which will again give us $R_i$. )

Thus we have:

$L_{cm} = (\sum_i m_i R_i^2 ) \omega$

• Oh! sorry. My mistake. I have made the correction in the answer. May 12, 2015 at 14:01
• Your question made me think about the basic definition of the direction of $\omega$. It is always defined with respect to the position vectors $r$ and $r+\Delta r$ where: $\Delta r = r\|\Delta \theta\|\hat\theta$, $\theta$ being the angle traversed by the vector $r$ in time $\Delta t$ and $\hat\theta$ being the tangent direction to the path of motion of the particle (Which is circular). I am really thankful to you @Self-teachingDavide for asking this question which led to the fundamental definition of direction of $\omega$. May 12, 2015 at 18:10
• Coming back to the question, it can be seen that for taking the cross product $\omega \times r$, the direction of $\omega$ cannot be taken as $\hat k$. In order to get the direction of the angular velocity, one has to know the path of the particle, get the unit vector of the tangent ($\hat \theta$) at $r$ and find the cross product: $\hat r \times \hat \theta$. Please correct me if I am wrong. May 12, 2015 at 18:23
• The math is pretty difficult and I have not seen such a proof presented anywhere. All my college courses dealt with 2D problems where $r$ was always in the X-Y plane and hence $\omega$ was always in the $\hat k$ direction. Also, most books present the proof only for planar rigid bodies and mention that the same is applicable to all rigid bodies. May 12, 2015 at 18:26
• I am sorry to have put up so many comments. Please check this link for a complete description of the angular velocity via Euler angles: physics.stackexchange.com/questions/73961/… May 12, 2015 at 18:39