Let us say there is a rigid body M which is rotating about an axis which is not passing with an angular velocity $\omega$ through it's centre of mass and simultaneously translating with a velocity $v$. What would be the expression for the angular momentum of this body about a point P located in space outside of the body at a distance $r$ from the Centre of mass of the rigid body?

I have figure out how to write the expression for angular momentum for the case where the rigid body is translating and rotating about an axis passing through it's centre of mass with a certain angular velocity

  • $\begingroup$ If an object is moving freely under the action of no external forces, then the center of mass must move in a straight line. Any rotation must be about the center of mass. Any off center rotation would require an externally constrained axle, and angular momentum about an external point would not be conserved. $\endgroup$ – R.W. Bird Sep 24 '19 at 17:35
  • $\begingroup$ @R.W.Bird - The combined rotation about center of mass and translation of the center of mass means the instant center of rotation is going to be away from the center of mass. $\endgroup$ – John Alexiou Sep 24 '19 at 17:36
  • $\begingroup$ Please edit the post and show your work so we can contribute under the same framework. $\endgroup$ – John Alexiou Sep 24 '19 at 17:37
  • 1
    $\begingroup$ Minor comment: Angular velocity is not 'W'. It is a Greek omega. $\endgroup$ – Qmechanic Sep 24 '19 at 18:14
  • 1
    $\begingroup$ Type math inside dollar signs. So that $x+1$ shows as $x+1$. For greek letters use \omega, \alpha ... For cross product use \times and fractions use \frac{a}{b} => $\frac{a}{b}$. Read more here $\endgroup$ – John Alexiou Sep 24 '19 at 18:47

enter image description here

The angular velocity of point p is:

$$\vec{\omega}_p=\frac{\vec{u}\times \vec{v}_p}{\vec{u}^T\,\vec{u}}$$

where $\vec{v}_p$ is the velocity of point p:

$$\vec{v}_p=\vec{v}+\vec{\omega}\times \vec{u}$$

and $\vec{u}=\vec{r}-\vec{u}_1$

The angular momentum is :


with $I_p$ the inertia tensor at point p


where :

$$\tilde{r}=\begin{bmatrix} 0 & -r_z & r_y \\ r_z & 0 & -r_x \\ -r_y & r_x & 0 \\ \end{bmatrix} $$

and $I_{\text{COM}}$ the inertia Matrix (tensor) at the COM

$$I_{\text{COM}}=\begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \\ \end{bmatrix} $$

  • $\begingroup$ I am just in high school and haven't done tensors ,so I have no clue what you wrote $\endgroup$ – Schwarz Kugelblitz Sep 24 '19 at 21:00
  • $\begingroup$ @SchwarzKugelblitz I add more equation for you $\endgroup$ – Eli Sep 24 '19 at 21:06
  • 1
    $\begingroup$ It would make it more clear that your are talking about the parallel axis theorem ther if you made it $$I_p=I_{\text{COM}}+M\,\left(-\tilde{{r}}\,\tilde{{r}}\right) $$ $\endgroup$ – John Alexiou Sep 24 '19 at 21:14

Consider the general situation. A rigid body has mass $m$ and mass moment of inertia $\mathbf{I}_C$ along the world coordinate system as measured at the center of mass. The body is moving at some instant with velocity $\boldsymbol{v}_C$ as measured at the center of mass, and rotates with $\boldsymbol{\omega}$. There is also a separate point of interest $P$ away from the center of mass $C$ where quantities are measured.


  • Linear momentum (shared by the whole body) is defined sole by the motion of the center of mass and does not vary with location $$\boldsymbol{p} = m \,\boldsymbol{v}_C$$

  • Angular momentum (measured at the center of mass) is defined by the rotation of the body and the mass moment of inertia $$ \boldsymbol{L}_C = \mathbf{I}_C \boldsymbol{\omega} $$

  • Velocity at P $$ \boldsymbol{v}_P = \boldsymbol{v}_C + ( \boldsymbol{r}_C - \boldsymbol{r}_P) \times \boldsymbol{\omega} $$

  • Linear momentum in terms of the velocity at P $$ \boldsymbol{p} = m ( \boldsymbol{v}_P + \boldsymbol{\omega} \times (\boldsymbol{r}_C - \boldsymbol{r}_P) $$

  • Angular momentum about P $$ \boldsymbol{L}_P = \boldsymbol{L}_C + (\boldsymbol{r}_C-\boldsymbol{r}_P) \times \boldsymbol{p} $$ $$ \boldsymbol{L}_P = \mathbf{I}_C \boldsymbol{\omega} + (\boldsymbol{r}_C - \boldsymbol{r}_P) \times m ( \boldsymbol{v}_P + \boldsymbol{\omega} \times ( \boldsymbol{r}_C - \boldsymbol{r}_P) ) $$

  • Instant center of rotation $$ \boldsymbol{r}_{\rm COR} = \boldsymbol{r}_C + \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_C }{ \| \boldsymbol{\omega} \|^2 } $$

  • Instant axis of percussion (line where momentum goes through. Hit here with hammer to stop the body from moving and rotating) $$ \boldsymbol{r}_{\rm IAP} = \boldsymbol{r}_C + \frac{ \boldsymbol{p} \times \boldsymbol{L}_C}{ \| \boldsymbol{p} \|^2 }$$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Sep 25 '19 at 3:24

After lengthy derivation, You can use this result: $\bf{L}=\bf{L}_{cm}+\bf{r}\times M\bf{V}_{cm}$. Here, $\bf{r}$ is the position vector of centre of mass wrt the chosen frame (bonus here,type of frame doesn't matter, it can be inertial or non inertial), $\bf{L}$ is the angular momentum in the selected frame, $\bf{L}_{cm}$ is the angular momentum in CM frame.

  • $\begingroup$ So for a pure rotation angular momentum will anyways be $I\omega$ $\endgroup$ – Kashmiri Jan 10 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.