jmc

Suppose the cube rolls first over edge $AB$.

Consider the cube as being made up of two half-cubes (each of dimensions $1\times 1\times \frac{1}{2}$) glued together at square $PQMN$. (Note that $PQMN$ lies on a vertical plane.)

Since dot $D$ is in the centre of the top face, then $D$ lies on square $PQMN$. Since the cube always rolls in a direction perpendicular to $AB$, then the dot will always roll in the plane of square $PQMN$. So we can convert the original three-dimensional problem to a two-dimensional problem of this square slice rolling.

Square $MNPQ$ has side length 1 and $DQ=\frac{1}{2}$, since $D$ was in the centre of the top face.

By the Pythagorean Theorem, $M{D}^2=D{Q}^2+Q{M}^2=\frac{1}{4}+1=\frac{5}{4}$, so $MD=\frac{\sqrt{5}}{2}$ since $MD>0$. In the first segment of the roll, we start with $NM$ on the table and roll, keeping $M$ stationary, until $Q$ lands on the table. This is a rotation of $90^{\circ }$ around $M$. Since $D$ is at a constant distance of $\frac{\sqrt{5}}{2}$ from $M$, then $D$ rotates along one-quarter (since $90^{\circ }$ is $\frac{1}{4}$ of $360^{\circ }$) of a circle of radius $\frac{\sqrt{5}}{2}$, for a distance of $\frac{1}{4}\left(2\pi \frac{\sqrt{5}}{2}\right)=\frac{\sqrt{5}}{4}\pi$.

In the next segment of the roll, $Q$ stays stationary and the square rolls until $P$ touches the table. Again, the roll is one of $90^{\circ }$. Note that $QD=\frac{1}{2}$. Thus, again $D$ moves through one-quarter of a circle this time of radius $\frac{1}{2}$, for a distance of $\frac{1}{4}\left(2\pi \frac{1}{2}\right)=\frac{1}{4}\pi$.

Through the next segment of the roll, $P$ stays stationary and the square rolls until $N$ touches the table. This is similar to the second segment, so $D$ rolls through a distance of $\frac{1}{4}\pi$.

Through the next segment of the roll, $N$ stays stationary and the square rolls until $M$ touches the table. This will be the end of the process as the square will end up in its initial position. This segment is similar to the first segment so $D$ rolls through a distance of $\frac{\sqrt{5}}{4}\pi$.

Therefore, the total distance through which the dot travels is $$\frac{\sqrt{5}}{4}\pi +\frac{1}{4}\pi +\frac{1}{4}\pi +\frac{\sqrt{5}}{4}\pi$$or $$\left(\frac{1+\sqrt{5}}{2}\right)\pi ,$$so our final answer is $\boxed{\frac{1+\sqrt{5}}{2}}$.