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 [itex]R[/itex] to a point [itex]P[/itex] at the center of the circular path. Will the particle rotate about [itex]P[/itex] 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 [itex]O[/itex] of a Cartesian coordinate system to be at [itex]P[/itex], the flat table is the xy-plane and the particle rotates about the point [itex]O[/itex] as described here. Specifically, the angular velocity [itex]\vec{\omega}[/itex] 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 [itex]\vec{r}[/itex], the angle between [itex]\vec{\omega}[/itex] and [itex]\vec{r}[/itex] is always [itex]\frac{\pi}{2}[/itex], and the angular momentum [itex]\vec{L}[/itex] is as follows:

[tex]\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}[/tex]

That is, [itex]\vec{L}[/itex] and [itex]\vec{\omega}[/itex] have the same direction, and [itex]\vec{L}[/itex] does not change direction and magnitude while the particle is rotating about [itex]P[/itex]. Therefore, the particle will rotate about [itex]P[/itex] forever at constant speed in the absence of any external force yes because the angular momentum [itex]\vec{L}[/itex] is conserved.

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

Because the angular momentum [itex]\vec{L}[/itex] is not conserved due to changing direction, and a changing [itex]\vec{L}[/itex] requires the presence of a net torque [itex]\vec{\tau}[/itex] about the point that is used to measure [itex]\vec{L}[/itex], which is the origin [itex]O[/itex], the particle will not rotate about [itex]P[/itex] forever because an external force is needed to keep [itex]\vec{L}[/itex] changing direction.

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

3. The attempt at a solution

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

http://ift.tt/1loyCJs

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