Observation about the rotation of a disc

[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!

 

[A.T.]

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 $\omega_y$?

It might add clarity to strictly distinguish between two types of angular velocity:
- spin, changing orientation of a rigid body
- tangential motion of a point

The tangential motion of $P$ about the $x$ & $y$ axes is not even constant, so it's obviously not what $\omega_y$ & $\omega_y$ are supposed to represent.

 

[etotheipi]

It might add clarity to strictly distinguish between two types of angular velocity:
- spin, changing orientation of a rigid body
- tangential motion of a point

The tangential motion of $P$ about the $x$ & $y$ axes is not even constant, so it's obviously not what $\omega_y$ & $\omega_y$ are supposed to represent.

Yeah, nicely put. Really everything becomes very clear once one has seen how the angular velocity vector arises from orthogonal rotation matrices, but it's not possible to cover such things with secondary-school students. So how to interpret this object $\boldsymbol{\omega}$ can be a little hard to explain

 

[Ibix]

A neat little demo of the oddity of angular velocity is to consider a wheel reflected in a mirror. Compare what happens to the axle's direction vector and the angular velocity "vector" under reflection in the cases where the rotation is in the plane of and perpendicular to the mirror. It isn't rigorous, but it's an easy-to-see way to see that there's some problems thinking of angular velocity as a vector (because it isn't one).

 

[etotheipi]

Yeah, it is the pseudovector $\tilde{\omega}_i := (\star \omega)_i = \frac{1}{2} \epsilon_{ijk} \omega_{jk}$ dual to the anti-symmetric rank-2 tensor $\omega$

 

[wrobel]

To avoid questions like this, one needs to carefully explain three theorems about the angular velocity of a rigid body to the student:

1. Euler's formula: $\boldsymbol v_B=\boldsymbol v_A+\boldsymbol\omega\times\boldsymbol{AB}$;
2. Let $Oxyz$ be a fixed Cartesian coordinate system. Suppose a rigid body rotates about the $Oz$ axis by an angle $\psi = \psi(t)$, measured counterclockwise from the perspective of an observer looking down the positive $Oz$ axis. Then the angular velocity of the rigid body is given by the formula $\boldsymbol \omega=\dot\psi\boldsymbol e_z$.
   
   Such a rotation means that there is a right-handed frame $\boldsymbol e_1\boldsymbol e_2\boldsymbol e_3$ connected to the rigid body such that
   
   $$\boldsymbol e_3\equiv \boldsymbol e_z,\quad \boldsymbol e_1=\cos\psi \boldsymbol e_x+\sin\psi\boldsymbol e_y,$$ and $$ \boldsymbol e_2=-\sin\psi \boldsymbol e_x+\cos\psi\boldsymbol e_y .$$
3. Assume that a coordinate system $A\xi\eta\zeta$ moves relative to $Oxyz$ with an angular velocity $\boldsymbol \omega_e$ and the rigid body moves relative to the system $A\xi\eta\zeta$ with an angular velocity $\boldsymbol \omega_r$. Then the angular velocity of the rigid body relative to $Oxyz$ is calculated as follows: $\boldsymbol\omega=\boldsymbol \omega_r+\boldsymbol \omega_e$.

In fact, these three theorems, together with Poisson's formulas $\boldsymbol{\dot e}_i=\boldsymbol\omega\times \boldsymbol e_i$, are all the student needs to know about angular velocity itself.

Euler's formula can be taken as the definition of the angular velocity.

 

[yjqwh]

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!

This is an interesting demonstration of the non-commutative property of finite rotations. It is easy to overlook that infinitesimal rotation around the x-axis will also contribute to the angular change around the y-axis. Your derivation clearly explains why the effective rotation rate about the y-axis exceeds \(\omega_y\).