Weird wheels in Tonyland!

There are four rotations, one around each corner of the square. In the first one, A describes a quarter-arc of the circle of radius $60$ centered at the right bottom corner. In the second one, A describes a quarter-arc of the circle of radius $60 \sqrt{2}$ centered at the right bottom corner. In the third one, A describes a quarter-arc of the circle of radius $60$ centered at the right bottom corner. In the fourth one $A$ is the right bottom corner and hence does not move.

The sum is therefore $30 \pi + 30 \pi \sqrt{2} + 30 \pi = 60 \pi + 30 \pi \sqrt{2}$ . So $a = 60, b = 30, a+b = \fbox{90}$ .

$1^{st} rotation = \frac{2\times 60\pi}{4} = 30\pi$

$2^{nd} rotation = \frac{2\times 60\sqrt{2}\pi}{4} = 30\sqrt{2}\pi$

$3^{rd} rotation = \frac{2\times 60\pi}{4} = 30\pi$

$4^{th} rotation = 0$

Total distance traveled

$D = [30\pi + 30\sqrt{2}\pi + 30\pi + 0] = 60\pi + 30\sqrt{2}\pi$

$a + b = \boxed{90}$