Posted on August 21, 2026
For her photography show, Frederica Fiddleria attaches a light to a point on the circumference of a circular wheel with a radius of 1 meter. She points a camera at the wheel and, during a single long exposure, rolls the wheel for one revolution along the ground.
Fiddler: When she develops the film, she is curious about the path the light took as the wheel rolled. What is the length of this path?
Extra Credit: Instead of placing the light on the circumference, Frederica picks a random point inside the circle (uniformly by area — any two regions of equal area are equally likely to contain the point). She again rolls the wheel for one revolution. On average, what can she expect the length of the path to be?
Put the ground along the \(x\)-axis and let \(t\) be the angle through which the wheel has rotated. Rolling without slipping means the center has traveled exactly \(t\) meters when the wheel has turned by \(t\) radians, so the center sits at \((t, 1)\). A light attached at distance \(r\) from the center (with \(0 \le r \le 1\)) traces
\[ \big(x(t),\, y(t)\big) = \big(t - r\sin t,\ 1 - r\cos t\big), \qquad t \in [0, 2\pi]. \]
This is a trochoid; the \(r = 1\) case is the familiar cycloid. Differentiating,
\[ \big(\dot x, \dot y\big) = \big(1 - r\cos t,\ r\sin t\big), \qquad v(t) = \sqrt{1 - 2r\cos t + r^2}. \]
That speed has a much better interpretation than the algebra suggests. Rolling without slipping means the contact point is instantaneously at rest — it is the instantaneous center of rotation. Every other point of the wheel is, at that moment, rotating rigidly about the contact point at angular rate \(\omega = 1\). So
\[ v = \omega \cdot (\text{distance to the contact point}) = \big|\,P - C(t)\,\big|, \]
where \(C(t)\) is the contact point. Check it: the light is at distance \(r\) from the center, the contact point is at distance \(1\) from the center, and the angle between them (measured at the center) is exactly \(t\). The law of cosines gives \(|P - C| = \sqrt{r^2 + 1 - 2r\cos t}\), matching the calculus. That reframing is the whole extra credit, so hold onto it:
Key fact: \(\displaystyle L = \int_0^{2\pi} \big|P - C(t)\big|\,dt\) — the path length is just the total distance-to-the-contact-point, integrated over one revolution.
For a light on the rim, \(r = 1\), and the distance from \(P\) to the contact point is a chord of the unit circle subtending angle \(t\), namely \(2\sin(t/2)\). (The algebra agrees: \(\sqrt{2 - 2\cos t} = 2|\sin(t/2)|\), and \(\sin(t/2) \ge 0\) on \([0, 2\pi]\).) Hence
\[ L = \int_0^{2\pi} 2\sin\!\frac{t}{2}\,dt = \Big[-4\cos\frac{t}{2}\Big]_0^{2\pi} = 4 - (-4) = 8. \]
The arch of a cycloid generated by a wheel of radius \(R\) has length \(8R\) — a genuinely surprising fact, since the wheel's own circumference is \(2\pi R \approx 6.283R\). The light travels about 27% farther than the wheel rolls, because near the top of the arch it is moving at twice the wheel's translational speed (at \(t = \pi\) the light is diametrically opposite the contact point, at distance 2, so \(v = 2\)), and it only slows to a stop at the two cusps where it briefly touches the ground.
Answer: \(\boxed{8}\) meters.
Now the light sits at a uniformly random point \(P\) inside the disk, and we want \(\mathbb{E}[L]\). The brute-force route is to note that the radial distance \(r\) has density \(2r\) on \([0,1]\) and integrate the trochoid length \(L(r) = \int_0^{2\pi}\sqrt{1 + r^2 - 2r\cos t}\,dt\) against it. That inner integral is a complete elliptic integral of the second kind, so this path leads to \(\int_0^1 8r(1+r)E(k)\,dr\) — doable numerically, but ugly.
The key fact above makes it collapse. Swap the expectation and the integral:
\[ \mathbb{E}[L] = \mathbb{E}_P\!\int_0^{2\pi}\big|P - C(t)\big|\,dt = \int_0^{2\pi} \mathbb{E}_P\big|P - C(t)\big|\,dt. \]
Here is the punchline. Work in the wheel's own frame, where \(P\) is a fixed random point and \(C(t)\) walks once around the rim. The distribution of \(P\) is rotationally symmetric, so \(\mathbb{E}_P|P - C(t)|\) is the same number for every \(t\) — it does not matter which boundary point the contact happens to be. The integral is therefore just a constant times \(2\pi\):
\[ \mathbb{E}[L] = 2\pi \cdot \mathbb{E}\big[\text{distance from a uniform point in the unit disk to a fixed point on its boundary}\big]. \]
That expected distance is a classic, and it is cleanest in polar coordinates centered at the boundary point \(C\). Seen from \(C\), the unit disk is the region \(0 \le \rho \le 2\cos\phi\) for \(\phi \in [-\pi/2, \pi/2]\). With area \(\pi\) and area element \(\rho\,d\rho\,d\phi\),
\[ \mathbb{E}|P - C| = \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\!\!\int_0^{2\cos\phi} \rho \cdot \rho\,d\rho\,d\phi = \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\frac{8\cos^3\phi}{3}\,d\phi = \frac{8}{3\pi}\cdot\frac{4}{3} = \frac{32}{9\pi}, \]
using \(\int_{-\pi/2}^{\pi/2}\cos^3\phi\,d\phi = 4/3\). Multiplying by \(2\pi\), the \(\pi\)'s cancel and everything rational survives:
\[ \mathbb{E}[L] = 2\pi \cdot \frac{32}{9\pi} = \frac{64}{9} = 7.1\overline{1}. \]
Extra credit answer: \(\displaystyle \boxed{\tfrac{64}{9}} \approx 7.111\) meters.
Two sanity checks on the value. The shortest possible path is the light at the dead center, tracing a straight line of length \(2\pi \approx 6.283\); the longest is the rim cycloid at \(8\). Our average, \(7.111\), sits between them — and closer to the cycloid than to \(2\pi\), which is right, since uniform-by-area sampling favors large radii (the median radius is \(1/\sqrt2 \approx 0.707\)).
Watch the wheel roll through one revolution and trace out the paths. The red curve is the rim cycloid (length 8); the blue curves are interior lights at \(r = 0.75, 0.5, 0.25\); the gray line is the center (\(r = 0\), length \(2\pi\)). Notice the cusps where the red curve touches the ground and momentarily stops.
Each click samples 20,000 uniformly random interior points, computes each one's exact trochoid length in closed form, and updates the running average and histogram. The running average should settle on \(64/9 \approx 7.1111\) (red line). The histogram is strongly right-skewed toward 8 — most randomly chosen points land near the rim.