Total rotation of a circle (or other closed curve) when 'rolled' along a curve in $\mathbb{R}^2$

Question

As to compute how much a circle rotates when 'rolled' along a curve in $\mathbb{R}^2$, the most intuitive way to me to find the number of rotations is:

$S/C+W/(2\pi)$

• $S$ is the arc-length of the curve
• $C$ is the circumference of the circle
• $W$ is the total curvature of the curve

However, this seems to agree with:

$T/C$

• $T$ is the arc-length of path of the center of the circle

Can anyone intuitively explain why the latter works as well?

Also I'm wondering whether $T/C$ still works if the circle is replaced by some closed curve (whereby $C$ is the arc-length of the closed curve and $T$ is the arc-length of path of the mass-center of the closed curve). Edit: After writing down the integrals, I think a sensible generalization might (rather than the mass-center) have more to do with the center of the osculating circle of the close curve at its current intersection with the curve it's being rolled on.

$\ $
Edit: In other words:
Let $\ c:[a,b]\to\mathbb{R}^2\ $ be some smooth curve along which a circle with radius $\text{abs}(r)$ is being rolled.
Let $\ \text{center}:[a,b]\to\mathbb{R}^2\ $ be the center of the circle given by: $$\text{center}(t)=c(t)+r\frac{\{c_2'(t),-c_1'(t)\}}{||c'(t)||_2}$$ That is the sign of $r$ determines on which side of the curve the circle is being rolled.

Expressed with integrals, the formulas for the total rotation are

$S/C+W/(2\pi)={\large\int_a^b}\dfrac{||c'(t)||_2}{2r\pi}dt+ {\huge\int_{\large a}^{\large b}}\dfrac{\det{\left( \begin{array}{cc} c_1'(t) & c_2'(t) \\ c_1''(t) & c_2''(t) \\ \end{array} \right)}}{||c'(t)||_2^2\cdot (2\pi)}dt$

$T/C = {\large\int_a^b}\dfrac{||\text{center}'(t)||_2}{2\,\text{abs}(r)\pi}\cdot\text{sign}\left(\dfrac{1}{r}+\dfrac{\det{\left( \begin{array}{cc} c_1'(t) & c_2'(t) \\ c_1''(t) & c_2''(t) \\ \end{array} \right)}}{||c'(t)||_2^3}\right)dt$

both of which are influenced by the signs of $r$ and the curvature determinant.

Answer 1

Can anyone intuitively explain why the latter works as well?

It's easier to understand if the curve is parametrized by arc length. Also, I find it easier to understand the relation $T=S+rW$ relating absolute lengths than the relation $T/C=S/C+W/(2\pi)$ relating counts of rotations.

So let $s$ parametrize the curve by arc length. I'll also assume that the rotating circle is along the "outside" of the curve the entire time, and that $\kappa$ is nonzero. At the point of tangency, there is the osculating circle with radius $1/\kappa(s)$. The rotating circle has radius $r$. So the center of the rotating circle is (for an infinitesimal moment) tracing a circular path with radius $r+1/\kappa(s)$.

Over a short length $ds$ within the curve, the length of the circular arc that the center travels through can be calculated using proportional reasoning:

$$dt=\frac{r+1/\kappa(s)}{1/\kappa(s)}ds=(1+r\kappa(s))ds$$

Now integrate over $s$ and you get the arc length through which the center passes. That is, $$T=\int_0^S(1+r\kappa(s))\,ds$$

But break it up into two integrals: $$\begin{align} T&=\int_0^Sds+r\int_0^S\kappa(s)\,ds\\ T&=S+rW \end{align}$$

Now divide by $C$ to get the form you have observed.

---

If the rotating circle is along the "inside", then the same reasoning changes the integral for $T$ to $$T=\int_0^S(1-r\kappa(s))\,ds$$ If the curve has zero curvature throughout, then $$T=\int_0^Sds=S$$ And lastly for more complicated curves, if they can be broken up piecewise into curves of these three types, you are set.