Does angular momentum of rotating body depend on its position?

Question

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

Answer 1

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 $\mathbf L=L_z\,\hat z$

Answer 2

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.