Vrftsjtryk

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:

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

• 
  $S$ 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]tomathbb{R}^2 $ be some smooth curve along which a circle with radius $text{abs}(r)$ is being rolled.

Let $ text{center}:[a,b]tomathbb{R}^2 $ be the center of the circle given by:
$$text{center}(t)=c(t)+rfrac{{c_2'(t),-c_1'(t)}}{||c'(t)||_2}$$
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

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

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

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

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 $T/C=S/C+W/(2pi)$ 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:

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

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

But break it up into two integrals:
$$begin{align}
T&=int_0^Sds+rint_0^Skappa(s),ds\
T&=S+rW
end{align}$$

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 $$T=int_0^S(1-rkappa(s)),ds$$ 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.