Apparent angular velocity (inclined orbit)?

[Andy Resnick]

TL;DR Summary

What (changing) angular velocity will I observe?

Here's the problem setup, my student and I are stuck.

A disk is rotating at constant angular velocity ω, and we are watching a point on the rim, parameterized by the angular position θ, move. Because we are observing the motion from an inclination angle Ψ, we do not always observe the velocity ω but instead a transformed angular velocity that varies with angle θ(t) [and ψ]. This is 'simple geometry' and we figured that we could look up the formula in an astronomy context ('inclined orbits'), but for whatever reason we can't seem to figure it out. Help, please, and thanks in advance!

 

[Andy Resnick]

Summary:: What (changing) angular velocity will I observe?

Here's the problem setup, my student and I are stuck.
View attachment 284903

A disk is rotating at constant angular velocity ω, and we are watching a point on the rim, parameterized by the angular position θ, move. Because we are observing the motion from an inclination angle Ψ, we do not always observe the velocity ω but instead a transformed angular velocity that varies with angle θ(t) [and ψ]. This is 'simple geometry' and we figured that we could look up the formula in an astronomy context ('inclined orbits'), but for whatever reason we can't seem to figure it out. Help, please, and thanks in advance!

Oops- never mind, we figured it out (the angular velocities are the same). Can't delete the thread, sorry!

 

[ergospherical]

Oops- never mind, we figured it out (the angular velocities are the same). Can't delete the thread, sorry!

I don't think they are the same. If the trajectory projected onto a plane parallel to the disc is $(x(t),y(t)) = (\cos{\theta(t)}, \sin{\theta(t)})$, where $\theta(t) = \omega t$, then the trajectory projected onto a tilted plane [hinged along the $x$ axis, say] will be of the form$$(\tilde{x}(t), \tilde{y}(t))= (\cos{\theta(t)}, \lambda \sin{\theta(t)})$$for parameter $\lambda \in [0,1]$. The observed angle $\tilde{\theta}(t)$ satisfies\begin{align*}
\tan{\tilde{\theta}(t)} = \lambda \tan{\theta(t)} \implies \dfrac{d\tilde{\theta}(t)}{dt} &= \lambda \omega \dfrac{\cos^2{\tilde{\theta}(t)}}{\cos^2{\theta}(t)} \\

\tilde{\omega}(t) &=\lambda \omega \cos^2{\tilde{\theta}(t)} \left(1 + \dfrac{\tan^2{\tilde{\theta}}(t)}{\lambda^2} \right)

\end{align*}It is worth to note that $\tilde{\omega}(t)$ is a function of time.

 

[FactChecker]

Oops- never mind, we figured it out (the angular velocities are the same). Can't delete the thread, sorry!

There is a difference between the angular velocity and the apparent angular velocity to an observer not on the axis of rotation. That is where @ergospherical 's result applies.

CORRECTION: Even to an observer on the axis of rotation, there is a difference between the actual angular velocity and the observed angular velocity (angular rate versus the observer).

 

[ergospherical]

CORRECTION: Even to an observer on the axis of rotation, there is a difference between the actual angular velocity and the observed angular velocity (angular rate versus the observer).

A camera on the axis, corresponding to orthogonal projection onto a plane with $\lambda = 1$, results in $\tilde{\omega}(t) = \omega$ so there will be no difference in this case.

 

[FactChecker]

A camera on the axis, corresponding to orthogonal projection onto a plane with $\lambda = 1$, results in $\tilde{\omega}(t) = \omega$ so there will be no difference in this case.

I agree. It is not clear to me exactly what the OP was asking about.
1) The true angular acceleration, not considering an external observer.
2) The orthogonal projection from an observer.
3) The azimuth/elevation from the viewpoint of an observer.

 

[ergospherical]

I am certain it is not proposition (1) because the "true" angular velocity of the disc, with respect to a set of basis vectors $\mathcal{B} = \{ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \}$ fixed in the lab, is the vector $\boldsymbol{\omega}$ such that any vector $\mathbf{a}$ rigidly embedded in the disc satisfies* $\dfrac{d\mathbf{a}}{dt} \bigg{)}_{\mathcal{B}} = \boldsymbol{\omega} \times \mathbf{a}$. As such it is independent of any "observer".

Proposition (2) is what I had in mind. The orthogonal projection onto a plane represents what an observer actually sees (after their brain has "flipped" the real image formed on the retina).

*N.B. $\dfrac{d\mathbf{a}}{dt} \bigg{)}_{\mathcal{B}} := \dfrac{da_1}{dt} \mathbf{e}_1 + \dfrac{da_2}{dt} \mathbf{e}_2 + \dfrac{da_3}{dt} \mathbf{e}_3$

 

[FactChecker]

I am certain it is not proposition (1) because the "true" angular velocity of the disc, with respect to a set of basis vectors $\mathcal{B} = \{ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \}$ fixed in the lab, is the vector $\boldsymbol{\omega}$ such that any vector $\mathbf{a}$ rigidly embedded in the disc satisfies* $\dfrac{d\mathbf{a}}{dt} \bigg{)}_{\mathcal{B}} = \boldsymbol{\omega} \times \mathbf{a}$. As such it is independent of any "observer".

Yes, but that is what he concluded was the answer in post #2. So it is not clear to me if that is what he had in mind.

 

[Andy Resnick]

I don't think they are the same.

Yes, but that is what he concluded was the answer in post #2. So it is not clear to me if that is what he had in mind.

Our tentative logic for the angular velocities being the same goes like this:

The amount of time for the point to execute 1 complete orbit (or a half-orbit, or a quarter-orbit...) is the same for observing at any angle ψ. Since the angular velocity is never observed to be faster than the 'proper' angular velocity, the angular velocity must be the same always.

Does that make sense?

Edit: another fact supporting our argument is that you can purchase watches without a circular faceplate- the numbers are not equispaced but the minute and hour hands both move at constant angular velocity.

 

[ergospherical]

The amount of time for the point to execute 1 complete orbit (or a half-orbit, or a quarter-orbit...) is the same for observing at any angle ψ. Since the angular velocity is never observed to be faster than the 'proper' angular velocity, the angular velocity must be the same always.

The error is that the observed angular velocity $\tilde{\omega}$ is in fact larger than $\omega$ for a subset of the motion (which depends on $\lambda$ and $\omega$).

With the configuration as in my post #3, $\tilde{\omega}$ exhibits maxima with $\tilde{\omega} > \omega$ at $\tilde{\theta} = \left(n+\dfrac{1}{2} \right)\pi$ and exhibits minima with $\tilde{\omega} < \omega$ at $\tilde{\theta} = m\pi$, for integral values of $m$ and $n$.

 

[FactChecker]

Our tentative logic for the angular velocities being the same goes like this:

The amount of time for the point to execute 1 complete orbit (or a half-orbit, or a quarter-orbit...) is the same for observing at any angle ψ. Since the angular velocity is never observed to be faster than the 'proper' angular velocity, the angular velocity must be the same always.

Does that make sense?

That logic makes sense if we assume that we are talking about the average and the range is always the same.

 

[ergospherical]

Edit: another fact supporting our argument is that you can purchase watches without a circular faceplate- the numbers are not equispaced in angle but the minute and hour hands both move at constant angular velocity.

Just to comment on this edit; the situation you describe is quite impossible. Either the numbers are equally spaced in angle, or the minute and hour hands do not rotate at a constant rate.