# Find Angular Momentum about any point

How do I find the angular momentum of a body about any point?
We know that $L=I\omega$ for a body rotating in space, where $L$ denotes the angular momentum, $I$ denotes the moment of inertia and $\omega$ denotes the angular velocity. However, this is only applicable for Fixed Axis of Rotation, Instantaneous Axis of Rotation and Center of Mass.
Can somebody state and prove the value of the angular momentum of a body about any point? (If a formula for that exists).
Thanks!

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Couldn't you use the Parallel Axis Theorem? –  Kyle Kanos Mar 7 '14 at 15:07
@KyleKanos The Parallel Axis Theorem is used only for moment of inertia. –  Samurai Mar 7 '14 at 15:18
You've got an equation for the angular momentum that uses $I$, no? Why do you think that the PAT wouldn't apply here? –  Kyle Kanos Mar 7 '14 at 15:20

Let's suppose I have some system and I know $M$, the system's total mass, $\vec{r}_{cm}$, the system's center of mass position and $\vec{L}_{cm}$, the systems angular velocity in the frame where the center of mass is the origin. How do I find $\vec{L}'$, the angular momentum with respect to some other origin, say $\vec{r}_{cm} + \Delta \vec{r}$, which is moving at a velocity $\Delta \vec{v}$ with respect to the origin? That is the question I will answer.

First I will consider the case $\Delta \vec{v}=0$. We know that the object's angular momentum can be decompose into an angular momentum due to the object's rotation about its center of mass (given by $I_{cm} \vec{\omega}$, and an angular momentum due to the revolution of the center of mass about the origin (given by $\vec{r}_{cm} \times \vec{p}$, where $\vec{p}$ is the object's momentum). If you move the origin by an amount $\Delta \vec{r}$, then the center of mass appears to move by a translation $-\Delta \vec{r}$, and so the change in angular momentum is $-\Delta \vec{r} \times \vec{p}$. Of course in the center of mass frame, the momentum will be zero, so the angular momentum does not change and we have $\vec{L}' = \vec{L}_{cm}$ in this case.

Now let's consider the case where $\Delta \vec{v} \ne 0$. We do an active galilean boost of velocity $-\Delta \vec{v}$ to our system. This does not change the object's angular momentum about its center of mass, but it does change its momentum by $-M \Delta \vec{v}$. Since the object's center of mass is displaced by $-\Delta \vec{r}$ from the origin the angular velocity from its revolution about the origin is $-\vec{r} \times -M \Delta \vec{v}$ Thus we have that $\vec{L}' = \vec{L}_{cm}+\Delta \vec{r} \times M \Delta \vec{v}.$

This can also be computed in a straightforward way. We know that $\vec{L}_{cm} = \int (\vec{r}-\vec{r}_{cm})\times \vec{v} \,dm$. Now \begin{aligned} \vec{L}' &= \int (\vec{r}-(\vec{r}_{cm} + \Delta \vec{r})\times (\vec{v}-\Delta \vec{v}) \,dm \\ &=\int (\vec{r}-\vec{r}_{cm})\times (\vec{v}-\Delta \vec{v}) \,dm - \int\Delta \vec{r} \times (\vec{v}-\Delta \vec{v})\,dm \\ &=\vec{L}_{cm} - \int (\vec{r}-\vec{r}_{cm})\times \Delta \vec{v} \,dm - \Delta \vec{r} \times \int \vec{v}\,dm +\Delta \vec{r} \times \int \Delta \vec{v}\,dm \\ &= \vec{L}_{cm} - \int (\vec{r}-\vec{r}_{cm}) \,dm \times \Delta \vec{v} - \Delta \vec{r} \times 0 + \Delta \vec{r} \times M \Delta \vec{v}\\ &=\vec{L}_{cm} - (\int \vec{r} \,dm-\vec{r}_{cm})\times \Delta \vec{v} + \Delta \vec{r} \times M \Delta \vec{v}\\ &=\vec{L}_{cm} - (\vec{r}_{cm}-\vec{r}_{cm}) \times \Delta \vec{v} + \Delta \vec{r} \times M \Delta \vec{v}\\ &= \vec{L}_{cm}+ \Delta \vec{r} \times M \Delta \vec{v}. \end{aligned}

# Relation to parallel axis theorem

You might think that you use the parallel axis theorem here. The parallel axis theorem is actually a special case of this where the displacement of the origin is perpendicular to the axis of rotation, and your new origin is some point embedded in the object (assumed to be rigid). By embedded in the object, I mean that the new origin is moving at the same velocity of the object at that point so that $\Delta \vec{v} = \vec{\omega} \times \Delta \vec{r}$.

The equation we derived in this answer then predicts \begin{aligned} \vec{L}' &= \vec{L}_{cm} + M \Delta \vec{r} \times \vec{\omega} \times \Delta \vec{r}\\ &= \vec{L}_{cm} + M((\Delta r)^2\vec{\omega} - \Delta \vec{r} \Delta \vec{r}\cdot \vec{\omega})\\ &=\vec{L}_{cm} + M(\Delta r)^2\vec{\omega}. \end{aligned}.

On the other hand, the parallel axis theorem would tell us to make the subsitution $I_{cm} \to I_{cm} + M (\Delta r)^2$. Thus we would have $$I_{cm} \vec{\omega} \to (I_{cm} + M (\Delta r)^2)\vec{\omega} = I_{cm} \vec{\omega} +M (\Delta r)^2\vec{\omega}.$$ So that $\vec{L}_{cm} \to \vec{L}_{cm} + M (\Delta r)^2\vec{\omega}$. I.e., $\vec{L}' = \vec{L}_{cm} + M(\Delta r)^2\vec{\omega}$. Thus we see how the answer we get are the same in this special case, and the parallel axis theorem can be used. However your question concerns more general transformations.

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Thanks for a very nice proof! –  Samurai Mar 7 '14 at 17:35