How do I add "rotation" vectors (pseudo-vectors?)

[snoopies622]

I'm tutoring an intro to meteorology pupil and learning about the conservation of potential vorticity, and realizing that I don't understand some basic rotational mechanics. For example, suppose I stand on the North Pole and hold a wheel such that the wheel's axis of rotation is parallel to the Earth's axis of rotation, and the wheel is spinning (relative to the stars) at twice the Earth's angular speed. Now if I were to carry my spinning wheel southward along any meridian while keeping its axis of rotation pointed directly toward the Earth's center, would its rotation speed relative to the surface of the Earth change? My impression (based on the conservation of potential vorticity equation) is that it would increase while I walk south, then decrease were I to turn around and walk north again. But I don't understand in a deeper way why this is true. Is there a law of rotational mechanics that would apply here?

 

[snoopies622]

I can see how it makes sense in this case: As I walk southward, every particle in my wheel is pushed to the right by the Coriolis force, and since the rear (northern) part of the wheel is already moving to the right while the front (southern) part of the wheel is moving to the left, and the Coriolis force is pushing the northern particles slightly harder (since they have a slightly higher latitude), then this should cause the wheel rotation to accelerate. And vice versa as I carry the wheel back toward the pole.

But there must be a simpler way to arrive at this conclusion? Some kind of vector equation with the Earth's rotation vector as one term and perhaps my walking southward as another term?

 

[hutchphd]

I personally like Feynman:
https://www.feynmanlectures.caltech.edu/I_20.html
But it is not simple.

 

[snoopies622]

Thanks, Hutch! Will watch over dinner.

(Edit: Read - thought it was a video.)

 

[robphy]

Thanks, Hutch! Will watch over dinner.

(Edit: Read - thought it was a video.)

You can now listen to audio,
via a link in the upper right corner.
Blackboard photos are linked in the upper left corner.

 

[snoopies622]

Thanks, Rob. The lecture was helpful, especially toward the end when Feynman shows that an object's rotational velocity vector and angular momentum vector need not be parallel ~ it reminded me of something I had long forgotten: that moment of inertia is properly a tensor quantity rather than a scalar. I'm now trying out some calculations with a hypothetical asymmetric object that's even simpler than Feynman's disk on a rod to see what I can learn. Hopefully I'll eventually get to the bottom of the thought experiment I posed in entry #1.

 

[ergospherical]

It's a bit late and I'm a little tired, but I tried to calculate the moment of the Coriolis force experienced by a rotating ring of mass. Let $\{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}$ be a vector basis with $\mathbf{k}$ in the direction of the local vertical at a point on the Earth's surface at polar angle $\theta$, and $\{ \mathbf{i}, \mathbf{j} \}$ tangent to the surface. The wheel has an angular velocity of approximately $\boldsymbol{\omega} = \omega \mathbf{k}$ so long as $\omega \gg \Omega$.

Let $\hat{\mathbf{z}}$ be a unit vector aligned with the rotation axis of the Earth, then $\boldsymbol{\Omega} = \Omega \hat{\mathbf{z}}$. Let $\mathbf{r}$ be the radius vector from the centre of the ring to an arbitrary point on the edge. The moment of the Coriolis force on the wheel about its centre is\begin{align*}

\mathbf{M} &= -2\int_C \mathbf{r} \times (\boldsymbol{\Omega} \times (\boldsymbol{\omega} \times \mathbf{r})) dm \\ \\

&= -2 \int_{0}^{2\pi} \mathbf{r} \times ((\boldsymbol{\Omega} \cdot \mathbf{r}) \boldsymbol{\omega} - (\boldsymbol{\Omega} \cdot \boldsymbol{\omega})\mathbf{r}) \rho r d\varphi \\ \\

&= -2 \int_{0}^{2\pi} (\boldsymbol{\Omega} \cdot \mathbf{r}) (\mathbf{r} \times \boldsymbol{\omega}) \rho r d\varphi \\ \\

&= 2 \rho \int_0^{2\pi} (\Omega r \cos{\alpha} ) (r\omega \hat{\boldsymbol{\varphi}}) r d\varphi \\ \\

&= 2 \rho \Omega \omega r^3 \int_0^{2\pi} \cos{\alpha} \hat{\boldsymbol{\varphi}} d\varphi

\end{align*}where $\cos{\alpha} = \sin{\theta} \sin{\varphi}$. Therefore\begin{align*}
\mathbf{M} &= 2 \rho \Omega \omega r^3 \sin{\theta} \int_{0}^{2\pi} \sin{\varphi} \hat{\boldsymbol{\varphi}} d\varphi \\ \\

&= 2 \rho \Omega \omega r^3 \sin{\theta} \left[ -\mathbf{i} \int_0^{2\pi} \sin^2{\varphi} d\varphi + \mathbf{j}\int_0^{2\pi} \sin{\varphi} \cos{\varphi} d\varphi\right] \\ \\

&= -2\pi \rho \Omega \omega r^3 \sin{\theta} \mathbf{i}

\end{align*}If I didn't screw up the calculation, then $\mathbf{M}$ points in the $- \mathbf{i}$ direction [tangent to the surface, anti-parallel to the direction of motion of the local surface of Earth w.r.t a space-fixed frame]. Thus it doesn't seem as if the Coriolis torque points in the right direction to change $\omega$...

... but I reckon I should actually have written $\mathbf{v}' = \boldsymbol{\omega} \times \mathbf{r} + V \hat{\boldsymbol{\theta}}$, to account for the small velocity parallel to the lines of longitude as the ring is moved slowly Southward, which might give rise to a term in $\mathbf{M}$ parallel to $\mathbf{k}$, or maybe not? I'll check tomorrow! Perhaps we actually require an external couple $\mathbf{C}$ applied to the axis.

 

[snoopies622]

---

thanks, ergospherical! i was wondering about a possible couple, too. perhaps we can imagine the wheel hanging from a string, so gravity is doing all the work of holding the wheel's axis of rotation normal to the ground.

 

[snoopies622]

So giving this some more thought, I arrive at the along-the-Earth's-axis-of-rotation component of the wheel's angular momentum being

<br /> L^z = m(R sin \theta)^2 \Omega + I(\omega + \Omega) cos \theta<br />

where m and I are its mass and moment of inertia, R the radius of the Earth, \theta the angle from the North pole, and \Omega and \omega are the Earth's rotational velocity and the wheel's rotational velocity with respect to the ground underneath it, respectively.

If I start the wheel at the North pole at rest relative to the ground and then move it down to the 60 degree latitude line (making up some simple values for I and m and keeping L constant) I end up with it spinning in the clockwise direction at around 100 million revolutions per second.

It makes sense to me that this direction is clockwise since its compensating for the angular momentum in the opposite direction it acquired by moving further away from the Earth's rotational axis. But using the conservation of potential vorticity law as explained so well here



moving southward should cause a cylinder of water/air to accelerate its rotation in the counterclockwise direction instead.

I haven't looked into how the law is derived, but on the surface it appears to be about conserving angular momentum as well.

edit: I suppose one could say that it makes no sense to assume that the wheel's \vec {L} vector is constant since its axis of rotation changes direction as it moves, and so gravity (and whatever else is holding the wheel in its "flat" position) must be applying a torque. But then, wouldn't that also be the case for a rotating cylinder of water or air?

 

[snoopies622]

Thought: In the formula in the video
<br /> \frac<br /> {\xi + f}{D}<br />
the D represents the height of a cylinder of fluid, and it's implied that the cylinder has constant volume, so when D increases, r^2 decreases proportionally and vice versa. If we replace 1/D with r^2 the result looks very similar in both structure and meaning to the second term of my equation in entry#9

I(\omega + \Omega) cos \theta.

But I'm puzzled why something analogous to the first term m(R sin \theta)^2 \Omega doesn't appear as well. Moving towards or away from the Earth's axis of rotation should certainly change an object's angular momentum about that axis. To simplify my question in entry one, If I stand on a large rotating disk holding a bicycle wheel "flat" (that is, with its axis rotation parallel to that of the disk) and I walk either towards or away from the disk's center, wouldn't that affect the wheel's rate of rotation with respect to an inertial frame?