# Does angular momentum of rotating body depend on its position?

This question might be silly. But angular momentum $$\vec{L} = \vec{r} \times \vec{p}$$. It always confuses me whether or not angular momentum depends on the origin because the position vector does depend on the origin. But in most books for rotating bodies this formula is used $$\vec{L} = I \times \vec{\omega}$$ where I is calculated using the radius of the circle. So it doesn't depend on the origin or its position. And if a body is both rotating and has linear motion then its angular momentum would be $$\vec{L} = I \times \vec{\omega} + \vec{r} \times \vec{p}$$ so angular momentum does depend on the position for translational motion but doesn't depend on position for rotational motion. But why? For any frame of reference, the angular momentum due to rotation stays the same?

If r is considered as a position vector then we can derive the angular momentum for a rotational body $$\vec{L} = \vec{r} \times \vec{p}\\ = \vec{r} \times (\vec{\omega} \times \vec{r})m=mr^2\vec{\omega}-m(\vec{\omega} \cdot \vec{r})\vec{r}=I\vec{\omega}-m(\vec{\omega} \cdot \vec{r})\vec{r}$$ So angular momentum would be $$I\vec{\omega}$$ only if the angular velocity is perpendicular to the position vector in other words if the the axis of rotation is on the origin. So it wouldn't be a valid formula for other cases. I might have some lackings or misconceptions.

This question first bothered me when I was solving a problem of a sphere that was rotating as well as going in a linear motion and It was asked to find its angular momentum.

Take a look at this diagram. How should I find the angular momentum of the rotating body about axis $$\omega$$ with respect to the main frame of reference O`(consider it as a circular disc with mass m rotating about axis $$\omega$$)? Because here I can't use $$\vec{L} = I \times \vec{\omega}$$ directly

• Yes, angular momentum depends on the position, which means it even depends on the choice of coordinate system, explicitly via the $\vec{r}$ in the definition. This is true for purely rotating systems too, except that it's conventional to choose the center of rotation as the origin of the coordinate system. May 25, 2023 at 17:32
• $\vec{v}\neq \vec{r} \times \vec{\omega}$ May 25, 2023 at 17:53
• There is a problem with notation as $\times$ is reserved for the cross product. Fix as $$\vec{L} = {\rm I} \, \vec{\omega} + \vec{r} \times \vec{p}$$ Sep 12, 2023 at 16:09

Both $$\mathbf L=\mathbf r\times\mathbf p$$ and $$\mathbf L=I\boldsymbol\omega$$ have this dependency; the latter is obtained from the former.

You should look into the more general inertia tensor and see how it does depend on the coordinate system. Typically in introductory physics specific bodies and coordinates are chosen such that $$I$$ is diagonal and $$\boldsymbol\omega$$ is along a principle axis so that we get something like $$L_z=I_{zz}\omega_z$$ as being the entire angular momentum vector $$\mathbf L=L_z\,\hat z$$

• Okay. Can you look at this diagram and tell me how should I find the angular momentum of the rotating body about the axis of omega with respect to the main origin O-prime May 25, 2023 at 17:57

In the diagram, that represents a rigid body, $$\mathbf L = \int_V \mathbf r' \times \rho(\mathbf r') \frac{\mathbf dr'}{dt} dV = \int_V \rho(\mathbf r + \mathbf R) \times \frac{\mathbf d(\mathbf r + \mathbf R)}{dt} dV$$.

Expanding: $$\mathbf L = \int_V \rho \mathbf r \times \frac{\mathbf {dr}} {dt}dV + \int_V \rho \mathbf r \times \frac{\mathbf {dR}} {dt}dV + \int_V \rho \mathbf R \times \frac{\mathbf {dr'}}{dt} dV$$

$$\mathbf R$$ and $$\frac{\mathbf {dR}} {dt}$$ can be taken out of the integral because they are constant for the integration over the volume: $$\mathbf L = \int_V \rho \mathbf r \times \frac{\mathbf {dr}} {dt}dV + \int_V (\rho \mathbf r dV)\times \frac{\mathbf {dR}} {dt} + \mathbf R \times \int_V \rho \frac{\mathbf {dr'}}{dt} dV (1)$$

The coordinates of the centre of mass are by definition: $$\mathbf R = \frac{\int_V \mathbf r' \rho dV}{\int_V \rho dV}$$

So: $$\frac{\mathbf {dR}}{dt} \int_V \rho dV = \int_V \frac{\mathbf {dr'}}{dt} \rho dV$$

The first term in $$(1)$$ is the spin angular momentum, and the expression $$\mathbf L_S = \mathbf I \times \boldsymbol \omega$$ can be derived from it. The second term is zero, because the integral is proportional to the coordinate of the COM in a frame where the origin is the COM itself. The third term is the external angular momentum with respect to the origin O'. So, the total angular momentum depends on the position of the origin with respect to the body.

• So why $\vec{L} = I \times \vec{\omega}$ is used in every frame of reference disregarding the position of the rotating body? Why total angular momentum of translating and rotating body is written like this $\vec{L} = I \times \vec{\omega} + \vec{r} \times \vec{p}$? As you have shown in your answer that angular momentum due to rotation is not just $\vec{L} = I \times \vec{\omega}$ May 26, 2023 at 6:08
• My answer shows the total angular momentum of the rigid body. Part is its movement around the COM: $I\times \omega$, and part is the movement of the COM with respect to the origin of the system of coordinates: $r_c \times p_c$ May 26, 2023 at 10:02
• If the COM is not moving does the third term cancels out because only then it can be true L=I×ω? If yes, can you explain why? Can you elaborate on this expression $\frac{dR}{dt} \int{p dV}=\int{\frac{dr'}{dt}p dV}$? How did you get it? Jun 12, 2023 at 8:11