- #1
etotheipi
Someone that I tutor asked a simple but pretty good question today which I thought I'd share the answer to. In a tidied up form: a disc with centre at the origin and central axis parallel to a unit vector ##\mathbf{n}## in the ##xy## plane rotates with a constant angular velocity ##\boldsymbol{\omega} = (\omega_x, \omega_y, 0) = \omega \mathbf{n}## about this axis. A point ##P## on the disc revolves around this axis in a time ##T = 2\pi / \omega##. However, the time it takes for ##P## to revolve once around the ##y## axis is also clearly ##T = 2\pi / \omega##, and not ##2\pi / \omega_y## so what actually is ##\omega_y## and why does a point on the disk rotate about the ##y##-axis faster than ##\omega_y##?
By symmetry, we can restrict our attention to the first quadrant of the motion. Consider that at time ##t## the point ##P## is at polar angle ##\theta_y## about the ##y##-axis. The first infinitesimal rotation is about the ##y##-axis, and takes ##P## to ##P'##. This changes this polar angle to ##\theta_y' = \theta_y + \omega_y \mathrm{d}t##.
Now let the point ##M## be the the orthogonal projection of ##P'## onto the ##x##-axis, and consider the triangle ##OMP'##. The second infinitesimal rotation by ##\omega_x \mathrm{d}t## about the ##x##-axis takes the triangle ##OMP'## to a triangle ##OMP''##. For convenience define ##s := MP'##. Since the arc ##P'P''## is of length ##s \omega_x \mathrm{d}t##, if ##\theta_x## is the polar angle about the ##x##-axis (which satisfies ##\cos{(\theta_x)} = -y / s##) then the amount by which the ##z##-coordinate increases is ##\xi = s \omega_x \cos{(\theta_x)} \mathrm{d}t##. This can easily be related to an increment ##\delta \theta'## in ##\theta_y## by the relation ##\sec^2{(\theta_y)} \delta \theta' = \xi / x = (s \omega_x \cos{(\theta_x)} \mathrm{d}t)/x = (-y\omega_x \mathrm{d}t)/x##.
Thus the second infinitesimal rotation about the ##x##-axis actually contributes a further ##\delta \theta' = (-y \cos^2{(\theta_y)} \omega_x \mathrm{d}t)/x## of rotation about the ##y##-axis. Defining the angle of tilt of the disk relative to the ##xz## plane as ##\alpha##, we have ##y = -x\tan{\alpha}## and thus the total increment in the polar angle ##\theta_y## after both infinitesimal rotations is\begin{align*}
\frac{\mathrm{d}\theta_y}{\mathrm{d}t} = \omega_y + \omega_x \tan{(\alpha)} \cos^2{(\theta_y)}
\implies \int_0^{\pi / 2} \frac{d\theta_y}{\omega_y + \omega_x \tan{(\alpha)} \cos^2{(\theta_y)}} &= \int_0^{T/4} \mathrm{d} t \\
\frac{\pi}{2} \frac{1}{\sqrt{\omega_y^2 + \omega_x \omega_y \tan{(\alpha)}}} = T/4
\end{align*}but since ##\omega_x = \omega_y \tan{(\alpha)}## and further since ##\omega^2 = \omega_x^2 + \omega_y^2##, this gives simply ##T = 2\pi / \omega##, as expected. The main thing to notice is that the actual rate of rotation about the ##y##-axis is greater than ##\omega_y##, precisely because the infinitesimal rotation about the ##x##-axis also constitutes a little bit of rotation about the ##y##-axis.
Hope I didn't make any mistakes, I've only briefly checked it over
. Anyway hope it's useful to someone!
By symmetry, we can restrict our attention to the first quadrant of the motion. Consider that at time ##t## the point ##P## is at polar angle ##\theta_y## about the ##y##-axis. The first infinitesimal rotation is about the ##y##-axis, and takes ##P## to ##P'##. This changes this polar angle to ##\theta_y' = \theta_y + \omega_y \mathrm{d}t##.
Now let the point ##M## be the the orthogonal projection of ##P'## onto the ##x##-axis, and consider the triangle ##OMP'##. The second infinitesimal rotation by ##\omega_x \mathrm{d}t## about the ##x##-axis takes the triangle ##OMP'## to a triangle ##OMP''##. For convenience define ##s := MP'##. Since the arc ##P'P''## is of length ##s \omega_x \mathrm{d}t##, if ##\theta_x## is the polar angle about the ##x##-axis (which satisfies ##\cos{(\theta_x)} = -y / s##) then the amount by which the ##z##-coordinate increases is ##\xi = s \omega_x \cos{(\theta_x)} \mathrm{d}t##. This can easily be related to an increment ##\delta \theta'## in ##\theta_y## by the relation ##\sec^2{(\theta_y)} \delta \theta' = \xi / x = (s \omega_x \cos{(\theta_x)} \mathrm{d}t)/x = (-y\omega_x \mathrm{d}t)/x##.
Thus the second infinitesimal rotation about the ##x##-axis actually contributes a further ##\delta \theta' = (-y \cos^2{(\theta_y)} \omega_x \mathrm{d}t)/x## of rotation about the ##y##-axis. Defining the angle of tilt of the disk relative to the ##xz## plane as ##\alpha##, we have ##y = -x\tan{\alpha}## and thus the total increment in the polar angle ##\theta_y## after both infinitesimal rotations is\begin{align*}
\frac{\mathrm{d}\theta_y}{\mathrm{d}t} = \omega_y + \omega_x \tan{(\alpha)} \cos^2{(\theta_y)}
\implies \int_0^{\pi / 2} \frac{d\theta_y}{\omega_y + \omega_x \tan{(\alpha)} \cos^2{(\theta_y)}} &= \int_0^{T/4} \mathrm{d} t \\
\frac{\pi}{2} \frac{1}{\sqrt{\omega_y^2 + \omega_x \omega_y \tan{(\alpha)}}} = T/4
\end{align*}but since ##\omega_x = \omega_y \tan{(\alpha)}## and further since ##\omega^2 = \omega_x^2 + \omega_y^2##, this gives simply ##T = 2\pi / \omega##, as expected. The main thing to notice is that the actual rate of rotation about the ##y##-axis is greater than ##\omega_y##, precisely because the infinitesimal rotation about the ##x##-axis also constitutes a little bit of rotation about the ##y##-axis.
Hope I didn't make any mistakes, I've only briefly checked it over
. Anyway hope it's useful to someone!
