# Does angular momentum depend on the origin?

The angular momentum of a particle rotating about a point is given by $\vec{L} = \vec{r} \times \vec{p}$.

Imagine a particle tracing a circular path on a flat table. If I put the origin of my coordinate system at the center of the circular path such that the flat table is the $xy$-plane, the angular momentum vector points in the positive direction of the $z$-axis because $\vec{r}$ lies on the xy-plane and $\vec{p}$ is always perpendicular to $\vec{r}$. Therefore, there is no change in $\vec{L}$ throughout the motion of the particle.

Now, I only move the origin vertically toward the ground. Now $\vec{r}$ makes a certain angle $\phi$ with the $z$-axis, and therefore, $\vec{L}$ also makes the same angle $\phi$ with the $z$-axis. But, now $\vec{L}$ keeps changing its direction as the particle is moving on the circular path on the flat table!

To conclude, an angular momentum $\vec{L}$ depends so much on the choice of the origin pretty much unlike the linear momentum $\vec{p}$?

Yes, angular momentum depends very much on the origin you pick. (You can see this most clearly by examining a single particle in free space with fixed velocity - the angular momentum is $0$ only if you pick the origin along the particle's line of movement.)
It's true that linear momentum is independent of your choice of origin; however $\vec p$ is still subject to some arbitrary choices. In particular, for a closed system, there always exists an inertial frame of reference in which $\vec p = 0$; the 'center of mass frame'. So the fact that we can "pick" angular momentum isn't really that strange.
Remember that your choice of frame of reference also affects other quantities, so we can't violate any laws by setting $\vec p$ and $\vec L$ independently of other quantities.