jmc

In its initial position, suppose the semi-circle touches the bottom line at $X$, with point $P$ directly above $X$ on the top line. Consider when the semi-circle rocks to the right. Suppose now the semi-circle touches the bottom line at $Y$ (with $O$ the point on the top of the semi-circle directly above $Y$, and $Z$ the point on the top line directly above $Y$) and touches the top line at $Q$. Note that $XY=PZ$.

$Q$ is one of the desired points where the semi-circle touches the line above. Because the diagram is symmetrical, the other point will be the mirror image of $Q$ in line $XP$. Thus, the required distance is 2 times the length of $PQ$.

Now $PQ=QZ-PZ=QZ-XY$. Since the semi-circle is tangent to the bottom line, and $YO$ is perpendicular to the bottom line and $O$ lies on a diameter, we know that $O$ is the centre of the circle. So $OY=OQ=8$ cm, since both are radii (or since the centre always lies on a line parallel to the bottom line and a distance of the radius away).

Also, $OZ=4$ cm, since the distance between the two lines is 12 cm. By the Pythagorean Theorem (since $\angle QZO=90^{\circ }$), then $$Q{Z}^2=Q{O}^2-Z{O}^2=8^2-4^2=64-16=48$$so $QZ=4\sqrt{3}$ cm.

Also, since $QZ:ZO=\sqrt{3}:1$, then $\angle QOZ=60^{\circ }$.

Thus, the angle from $QO$ to the horizontal is $30^{\circ }$, so the semi-circle has rocked through an angle of $30^{\circ }$, ie. has rocked through $\frac{1}{12}$ of a full revolution (if it was a full circle). Thus, the distance of $Y$ from $X$ is $\frac{1}{12}$ of the circumference of what would be the full circle of radius 8, or $XY=\frac{1}{12}(2\pi (8))=\frac{4}{3}\pi$ cm. (We can think of a wheel turning through $30^{\circ }$ and the related horizontal distance through which it travels.)

Thus, $PQ=QZ-XY=4\sqrt{3}-\frac{4}{3}\pi$ cm.

Therefore, the required distance is double this, or $8\sqrt{3}-\frac{8}{3}\pi$ cm or about 5.4788 cm, which is closest to $\boxed{55}$ mm.