Angular Momentum Contradiction Due To Choice Of Origin

[TadeusPrastowo]

Homework Statement

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.

Homework 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 http://fsinet.fsid.cvut.cz/en/u2052/node39.html. 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 $O$ !

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 http://fsinet.fsid.cvut.cz/en/u2052/node41.html?

 

[amind]

Your mistake here is that $\vec{r}$ is the (perpendicular) radius vector from the axis of rotation , not the origin.

 

[TadeusPrastowo]

In the first choice of origin where the origin $O$ is at $P$, $\vec{r}$ is the (perpendicular) radius vector from the axis of rotation and the origin.

In the second choice of origin where the origin $O$ is located directly vertically beneath $P$, $\vec{r}$ is not the (perpendicular) radius vector from the axis of rotation.

In both choices of origin, however, $\vec{p}$ are always tangential to the circular path.

I don't see my mistake in each separate case that considers a different origin with regard to $\vec{r}$ as you said.

 

[haruspex]

the string is taut, and this exerts a torque about O.
btw, L and r will not both make an angle $\pi/6$ to the z axis since they are at right angles to each other.

 

[TadeusPrastowo]

When the string is taut and the speed of the particle is constant, only the centripetal force $\vec{F}_c$ exists. A centripetal petal force, however, can never give rise to a torque because the angle it makes with the moment arm is always $\pi$ and $\sin\pi$ is zero. So, the contradiction is not about the taut string.

Also, when I wrote: "both $\vec{r}$ and $\vec{L}$ will make a $\frac{\pi}{6}$-angle with the z-axis", I mean that the angle between $\vec{r}$ and the z-axis is $\frac{\pi}{6}$ and the angle between $\vec{L}$ and the z-axis is also $\frac{\pi}{6}$.

 

[voko]

The force of tension in the string has zero moment about $P$. When $O$ is offset higher or lower than $P$, the moment of the tensile force about $O$ is not zero. This is the torque that rotates angular momentum in the offset frame.

 

[TadeusPrastowo]

If there is a torque due to the centripetal force, then there is an external force being applied because there is no second particle to form a pair of action-reation forces such that the centripetal force can be treated as an internal force. The presence of an external force, however, contradicts the phenomenon itself that shows no external force being present at all.

EDIT: Sorry, I overlook the presence of the point $P$ that acts as the second particle to make a pair of action-reaction forces so that the centripetal force can be regarded as an internal force. Now let me think for one other case.

 

[voko]

In fact, there is an external force in this system. This is the force that keeps the fixed point of the string fixed.

But that is irrelevant for the "paradox". If the origin of the frame is offset vertically from the fixed point, the tensile force has a non-zero rotating moment, pretty much for the same reason angular momentum rotates in the offset frame.

 

[amind]

In the first choice of origin where the origin $O$ is at $P$, $\vec{r}$ is the (perpendicular) radius vector from the axis of rotation and the origin.

In the second choice of origin where the origin $O$ is located directly vertically beneath $P$, $\vec{r}$ is not the (perpendicular) radius vector from the axis of rotation.

In both choices of origin, however, $\vec{p}$ are always tangential to the circular path.

I don't see my mistake in each separate case that considers a different origin with regard to $\vec{r}$ as you said.

Okay , I get it now.

 

[TadeusPrastowo]

Thank you very much, voko! I wrote down the math on paper and all cases that I can conceive match your description. Case closed.

 

[haruspex]

Also, when I wrote: "both $\vec{r}$ and $\vec{L}$ will make a $\frac{\pi}{6}$-angle with the z-axis", I mean that the angle between $\vec{r}$ and the z-axis is $\frac{\pi}{6}$ and the angle between $\vec{L}$ and the z-axis is also $\frac{\pi}{6}$.

No, $\vec{L}$ makes an angle $\frac{\pi}{3}$ to the z axis.

 

[TadeusPrastowo]

Ah, yes, you are right about it, haruspex! Thank you very much!