How to reconcile angular momentum measurements?

Imagine a particle tracing a counter-clockwise circular path on a flat table with a certain speed. The particle is tied with a massless string of length $R$ to a point $P$ at the center of the circular path. Will the particle rotate about $P$ forever at constant speed in the absence of any external force? Consider using different origins to measure the physical quantities.

Measurement Case 1 (origin $O$ at $P$):

If I choose the origin $O$ of a Cartesian coordinate system to be at $P$, the flat table is the xy-plane and the particle rotates about the point $O$ as described. Specifically, the angular velocity $\vec{\omega}$ of the particle points in the direction of the positive z-axis, the position of the particle at any time is specified by the position vector $\vec{r}$, the angle between $\vec{\omega}$ and $\vec{r}$ is always $\frac{\pi}{2}$, and the angular momentum $\vec{L}$ is as follows:

$\vec{L} = \vec{r}\times\vec{p}\;\ldots\text{ definition} \\ \hphantom{\vec{L}} = m\,(\vec{r}\times\vec{v})\;\ldots\text{ definition of linear momentum }\vec{p} \\ \hphantom{\vec{L}} = m\,(\vec{r}\times(\vec{\omega}\times\vec{r}))\;\ldots\text{ definition of linear velocity in terms of angular velocity} \\ \hphantom{\vec{L}} = m\,(\vec{\omega}\,(\vec{r}\cdot\vec{r}) + \vec{r}\,(\vec{r}\cdot\vec{\omega}))\;\ldots\text{ scalar triple product}\\ \hphantom{\vec{L}} = m\,(\vec{\omega}\,(r^2) + \vec{r}\,(0))\;\ldots\;\vec{r} \perp\vec{\omega}\\ \hphantom{\vec{L}} = (m\,r^2)\,\vec{\omega}$

That is, $\vec{L}$ and $\vec{\omega}$ have the same direction, and $\vec{L}$ does not change direction and magnitude while the particle is rotation about $P$. Therefore, the particle will rotate about $P$ forever at constant speed in the absence of any external force because the angular momentum $\vec{L}$ is conserved (the presence of a centripetal force does not affect the angular momentum $\vec{L}$ in any way because a centripetal force is always parallel to the moment arm, and therefore, does not give rise to a torque $\vec{\tau}$).

Measurement Case 2 (origin $O'$ vertically directly beneath $P$):

Now if I choose to describe the same problem by choosing an origin $O'$ of a Cartesian coordinate system to be vertically directly beneath $P$, the particle's position vector $\vec{r}'$ makes an angle $\phi$ with the z-axis and the particle's angular momentum $\vec{L}'$ also makes a $\phi$ angle with the z-axis. As the particle rotates, however, $\vec{r}'$ also rotates about the z-axis, and therefore, the angular momentum $\vec{L}'$ keeps changing direction as shown below:

$\vec{L}' = \vec{r}'\times\vec{p}'\;\ldots\text{ definition} \\ \hphantom{\vec{L}'} = m\,(\vec{r}'\times\vec{v}')\;\ldots\text{ definition of linear momentum }\vec{p}' \\ \hphantom{\vec{L}'} = m\,(\vec{r}'\times(\vec{\omega}\times\vec{r}'))\;\ldots\text{ definition of linear velocity in terms of angular velocity} \\ \hphantom{\vec{L}'} = m\,(\vec{\omega}\,(\vec{r}'\cdot\vec{r}') + \vec{r}'\,(\vec{r}'\cdot\vec{\omega}))\;\ldots\text{ scalar triple product}\\ \hphantom{\vec{L}'} = m\,(\vec{\omega}\,(r'\,^2) + \vec{r}'\,(r'\,\omega\,\cos(\angle(\vec{r}', \vec{\omega})))) \\ \hphantom{\vec{L}'} = (m\,r'\,^2)\,\vec{\omega} + (r'\,\omega\,\cos\phi)\,\vec{r}'$

Because the angular momentum $\vec{L}'$ is not conserved due to changing direction, and a changing $\vec{L}'$ requires the presence of an external force to give rise to a net torque $\vec{\tau}'$ about the point that is used to measure $\vec{L}'$, which is the origin $O'$, the particle will not rotate about $P$ forever in the absence of any external force.

But then, a contradiction arises: the same phenomenon has a different outcome depending on the choice of origin $O$ or $O'$! That is unacceptable in Physics!

How to resolve this contradiction?

• There's rather a difference between conservation of a vector quantity and conservation of its absolute magnitude. Clearly you have made error(s) in your math. Apr 20, 2014 at 12:06
• How do they differ mathematically? Apr 20, 2014 at 12:15

1 Answer

Your error lies in your last paragraph:

Because the angular momentum $\vec{L}'$ is not conserved due to changing direction, and a changing $\vec{L}'$ requires the presence of an external force to give rise to a net torque $\vec{\tau}'$ about the point that is used to measure $\vec{L}'$, which is the origin $O'$, the particle will not rotate about $P$ forever in the absence of any external force.

Everything in that sentence is factually correct, but the conclusion, "the particle will not rotate about $P$ forever in the absence of any external force", does not follow from it.

Let's step back and think about what the system actually looks like. In the first case, a particle rotates in a circle with the origin $O$ at its center, presumably due to some force (like gravity towards $O$) which keeps the motion circular.

In the second case, the origin is moved to $O'$ but the system itself is not changed, ie, the particle now rotates in a circle above the new origin $O'$ (again, due to gravity towards $O$).

In the first case, the centrifugal force is parallel to the moment arm, since the particle is attracted to $O$, which is the chosen coordinate origin. Thus $\vec{L}$ is conserved, which makes sense.

However, in the second case, the centrifugal (gravity) force is no longer parallel to the moment arm, since the particle is attracted to $O$, not $O'$. Thus there is no reason to expect that $\vec{L}'$ is conserved. Naturally this means that there is a nonzero torque $\vec{\tau}'$ with this choice of coordinates.

What is the physical significance of this nonzero torque? In the first case, the torque $\vec{\tau}$ is zero, because the particle orbits the origin $O$. In the second case, the particle no longer orbits the origin $O'$, but instead orbits a location $O$ above it, and thus from this coordinate viewpoint, there must be a nonzero torque $\vec{\tau}'$, as otherwise if $\vec{\tau}'=0$ the particle would orbit $O'$, rather than $O$!