Geometry or Rotation?

Spatial coordinates with respect to center (circle radius is $R$ and radius variable is $r$ ):

$x = r \, \cos\theta \\ y = r \, \sin\theta$

Normal unit-vectors associated with velocity due to rotation:

$u_x = \sin\theta \\ u_y = - \cos\theta$

Angular speed:

$\omega = \frac{v_0}{R}$

Total velocity (superposition of translation and rotation):

$v_x = v_0 + r \omega \, u_x = v_0 + r \frac{v_0}{R} \, \sin\theta \\ v_y = r \omega \, u_y = -r \frac{v_0}{R} \, \cos\theta$

Find the locus of points with velocity magnitude equal to $v_0$ :

$v^2 = v_x^2 + v_y^2 = v_0^2 + \frac{2 v_0^2}{R} \, r \, \sin\theta + \frac{v_0^2}{R^2} r^2 = v_0^2 \\ 1 + \frac{2}{R} \, r \, \sin\theta + \frac{1}{R^2} r^2 = 1 \\ \boxed{r = -2R \, \sin\theta}$

Angular range over which formula applies ( $r$ cannot be larger than $R$ ):

$-\frac{\pi}{6} \leq \theta \leq 0 \\ -\pi \leq \theta \leq -\frac{5 \pi}{6}$

Region B (area of lesser speed) therefore consists of these two regions, as well as a region $(-\frac{5 \pi}{6} \leq \theta \leq -\frac{\pi}{6})$ in which the radius is $R$ . See the attached image above.

Constant-radius sub-region area:

$B_C = \pi R^2 \Big(\frac{4 \pi / 6}{2 \pi}\Big) = \frac{\pi R^2}{3}$

Integrate in polar coordinates to find the combined area of the variable-radius region within Region B:

$B_V = \int_{-\pi/6}^{0} r^2 \, d\theta = 4 R^2 \int_{-\pi/6}^{0} \sin^2 \theta \, d\theta = R^2 (\frac{\pi}{3} - \frac{\sqrt{3}}{2})$

Total Area of Region B:

$B = B_C + B_V = \frac{\pi R^2}{3} + R^2 (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = R^2 (\frac{2 \pi}{3} - \frac{\sqrt{3}}{2})$

Total Area of Region A: $A = \pi R^2 - B = R^2 (\frac{\pi}{3} + \frac{\sqrt{3}}{2})$

Ratio of $B$ area to $A$ area:

$\frac{B}{A} = \frac{R^2 (\frac{2 \pi}{3} - \frac{\sqrt{3}}{2})}{R^2 (\frac{\pi}{3} + \frac{\sqrt{3}}{2})} \\ = \frac{ 4 \pi - 3\sqrt{3}}{ 2 \pi + 3\sqrt{3}}$

The rolling motion can always be viewed as a rotation around the point where the object touches the ground. Therefore the points of low velocity will be on a circle drawn around the touching point. When the circle contains the center of the disc all points on that circle move with velocity $v_0$ .

We want to determine the overlapping area of two circles: one is the original disk (blue and green) and the other one is the line separating the slow and fast velocities (red circle). This area $A$ is twice the area between the horizontal black line and the bottom part of the blue circle $A=2A'$ . $A'$ is the difference of the area of the blue segment of the circle, $r ^2\phi/2$ , and the area of the black triangle, $r^2 (\cos \phi/2) (\sin \phi/2)$ . Here the angle at the center of the circle is $\phi = 120^{\circ}=2\pi/3$ and we get $A'=r^2(\pi/3-\sqrt{3}/4)$ and $A=r^2(2\pi/3-\sqrt{3}/2)$ . The remaining area is $r^2\pi-A'$ . The ratio is $\frac{4\pi-3\sqrt{3}}{2\pi +3\sqrt{3}}$ .