Note: This is my best attempt at transpiling an article originally intended for latex into blog form. You can find the original here.

Rolling Without Slipping

Definition: Given an object¹ $R$ and a fixed curve $\gamma$ initially tangent to it, we say that $R$ rolls without slipping on $\gamma$ if $R$ moves² such that

• Some point on the boundary of $R$ is always tangent to $\gamma$, and
• When any point on $R$ is this tangency point, it is stationary.

Notably, we make no distinction to any "point" that the object rotates about.

Tracking a Marked Point

Let's see some implications of this definition to tracking a marked point on an object that is rolling without slipping.

Proposition: Mark a point $P$ on the object, and suppose that it moves on a path $\Gamma_P$. Furthermore, suppose that the object is tangent to $\gamma$ at $T$. Then, $PT$ is normal to $\Gamma_P$.

Proof: Let $P(t)$ denote the position of $P$ at time $t$. Let $T(t)$ denote the position of $T$ at time $t$. We consider both $P$ and $T$ to move with the object. Suppose the object is tangent to $\gamma$ at $T$ at the time $t=t_0$. By definition, we have $T'(t_0) = 0$. Now note that, $P'(t_0)$ is a tangent vector to $\Gamma_P$ at $P$ (at time $t=t_0$). In addition,

is a constant, so $P'(t_0) \cdot (T(t_0) - P(t_0)) = 0$, as desired. $\square$

Furthermore, since $\| T(t) - P(t) \|$ is does not depend on $t$, and $T(t)$ is momentarily stationary, we should be able to model the movement of $P$ when the time is near $t$ as following the circle centered at $T(t)$ passing through $P(t)$. Unfortunately, this is not exactly true (e.g. though they are tangent, the curvatures of said circle and $\Gamma_P$ are not always equal at $P$).

However, if we let $s(t)$ denote the arc length traversed by $P$ on $\Gamma_P$ from time $0$ to time $t$, and let $\theta(t)$ denote the angle rotated by the object from time $0$ to time $t$ (e.g. by marking a fixed line on the object, then tracking the angle the $x$-axis makes with it), we have

Theorem (Arc Length of Marked Points): Suppose $r(t)$ is the distance from $P(t)$ to the tangency point of the object with $\gamma$. Then

considering everything as a function of $t$.

In other words, this states that if the object rotates by an infinitesimal angle of $\mathrm{d}\theta$, then a point $P$ that is $r$ away from the tangency point of the object with $\gamma$ will move by $r \, \mathrm{d}\theta$.

Proof of the Theorem: This proof is a bit computational. We will show that, using notation from the Proposition, at time $t=t_0$

Let $r$ be the constant equal to $\| T(t_0) - P(t_0) \|$, and let $\hat{\imath}$ be the unit vector in the positive $x$ direction. Finally, define It's easy to see that $\cos^{-1} (\hat{n} (t) \cdot \hat{\imath})$ is the angle that the line $PT$ makes with the $x$-axis at time $t$, so it is off from $\theta(t)$ by a constant³. Thus, We can check that But note that $\frac{\mathrm{d}P}{\mathrm{d}s} \big|_{t=t_0}$ is the unit tangent vector to $\gamma_P$ at $P(t_0)$, so by the Proposition, it and $\hat{n}$ form an orthonormal basis of $\mathbb R^2$ ⁴. It follows that, Therefore, as desired. $\square$

Example Problems

Problem 1 (Physics): Suppose a circle with radius $r$ rolls without slipping on a line such that its center moves with velocity $v$ and the entire circle rotates about its center with an angular velocity of $\omega$. Then $v = r\omega$.

Solution: Track the position of the center. Note that it moves on a line at constant velocity and is always a distance $r$ from the tangency point. Therefore, by the Theorem,

Problem 2 (Folklore): Roll a coin around the circumference of another (fixed) coin of equal radius. How many times does the moving coin rotate?

Solution: Suppose the shared radius is $r$. Again, track the position of the center of the rolling object. Note that it moves on a circle with radius $2r$ concentric with the fixed coin. Therefore, by the Theorem,

so the coin has rotated twice. $\square$

Problem 3 (Delaunay): An ellipse with semimajor axis $a$ rolls without slipping along the $x$-axis for one complete turn. Find the length of curve traced out by one focus $F$.

Surprisingly (or unsurprisingly), this is independent of the eccentricity!

Solution: By the Theorem,

But by symmetry, for any $\phi \in [0,\frac12 \tau)$, the marked point on the ellipse tangent to the $x$-axis when $\theta = \phi$ (i.e. $T(t_1)$ for $\theta(t_1) = \phi$) is the reflection over the center of the ellipse of the marked point on the ellipse tangent to the $x$-axis when $\theta = \phi + \frac12 \tau$. In other words, the moments in which the ellipse is tangent to the $x$-axis at diametrically opposite points are a half turn away. But the sum of the distances from two diametrically opposite points to a fixed focus $F$ is simply $2a$ since the two foci and the two diametrically opposite points form a parallelogram. Therefore, It follows that curve has length $a \tau$. $\square$

Footnotes

¹ We represent objects as the interiors of smooth, simple, closed, planar curves

² i.e. moves rigidly preserving orientation. You can think about this as a path through the space of rototranslations starting at the identity.

³ Or off from $-\theta(t)$ by a constant. It's iffy because we ideally want $\theta$ to be directed, but our dot product can only detect undirected angles. We can simply pretend all rotations are $\ll \frac12 \tau$, and that $\theta(t)$ is always defined so that $\frac{\mathrm{d}s}{\mathrm{d}\theta}$ is positive.

⁴ In fact, these are the tangent and normal unit vectors of the Frenet—Serret frame!