# Angular Momentum Contradiction Due To Choice Of Origin

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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 here. 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 rotating about $P$. Therefore, the particle will rotate about $P$ forever at constant speed in the absence of any external force yes because the angular momentum $\vec{L}$ is conserved.

But, if I choose to describe the same problem by choosing an origin $O$ of a Cartesian coordinate system to be $\sqrt{3}R$ vertically beneath $P$, both $\vec{r}$ and $\vec{L}$ will make a $\frac{\pi}{6}$-angle with the z-axis. But, as the particle rotates, $\vec{r}$ also rotates about the z-axis, and therefore, the angular momentum $\vec{L}$ keeps changing direction.

Because the angular momentum $\vec{L}$ is not conserved due to changing direction, and a changing $\vec{L}$ requires the presence of 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 because an external force is needed to keep $\vec{L}$ changing direction.

But then, a contradiction arises: the same problem has a different outcome depending on the choice of origin $O$ !

3. The attempt at a solution

How to solve the contradiction? By treating the rotation of the particle about $P$ in the second coordinate system as a rotation about the z-axis by projecting the $\frac{\pi}{6}$-angled $\vec{L}$ to the z-axis as suggested here?

http://ift.tt/1loyCJs