Lets say a, b, c, and d are in meters (because I personally like meters).
Corner 1 has a speed of ω√(b²+c²) in meters per second and angle of arctan(c/b)+π/2 in radians.
Corner 2 has a speed of ω√(a²+c²) in meters per second and angle of arctan(c/a)+π/2 in radians.
Corner 3 has a speed of ω√(a²+d²) in meters per second and angle of arctan(d/a)+π/2 in radians.
Corner 4 has a speed of ω√(b²+d²) in meters per second and angle of arctan(d/b)+π/2 in radians.
Work: (Yes I'm lazy)
http://i.imgur.com/xCsYXzw.jpg?1
For the sake of simplicity, we'll call "n" the line segment that ends at any corner of the robot and the point of rotation. Create a right triangle with n as its hypotenuse. We can call the angle in the right triangle closest to the robot, the arctangent of the opposite leg divided by the adjacent leg. Then we add pi/2 radians to find the angle of the ray perpendicular to n.
Finding the rotational speed, we first create a circle with its center at the point of rotation and n as its radius. With the right triangle from the last paragraph, n can be found with the Pythagorean Theorem, where n is the square root of the sum of the squares of the two legs. Because n is the radius of the triangle, and we want the speed of the robot in ω radians per second, we can multiply ω by the radius to get the tangential speed.