[dellagd]

.85 rad/s

The rotation speed will be achieved when the net torque on the robot becomes zero (after it was non-zero while the robot is initially accelerating). Also, there is only one force coming from each wheel, the frictional force provided by the carpet on the wheels. Therefore, the only way bring this force to zero (since it has a non-zero magnitude) is to have its line of action be in line with the center of rotation. (such that cos(theta) = cos(180) = 0 in the torque calculation).

The frictional force will also be in line with the vector sum of the wheel's tangential speed and the carpets speed.

Knowing the tangential speed of the wheel, the tangential direction of the robot's spin and the direction of the frictional force at the wanted equilibrium, the speed of the robot's spin can be solved for.

Spoiler for Calculations:

Vcarpet = Vc = opposite Vrobot
F = friction (not the actual force but friction lies along this vector, this is technically the vector sum mentioned earlier)

2 + Vcy = Fy
Vcx = Fx
Fy/Fx = -(28/12) / 2 #Known angle of vector sum/friction
Vcx/Vcy = (28/12) / 2 #Known angle of carpet velocity

Vcy = -.847
Vcx = -.988

Vc magnitude = Vrobot magnitude = 1.302 ft/s

Vrobot = r * omega

omega = 1.302 / (sqrt(2^2 + (28/12)^2) / 2) = .847 rad/s

*crosses fingers*