I’m not sure how friction factors into this. A really wide wheel will have much more turning scrub than a thin or donut shaped wheel.

Here’s what I’ve got so far:
I broke the wheel velocity into two components, one in the direction from the wheel to the center of the bot, and the other perpendicular to this one. The inward pointing one is all lost to friction. The other one has a magnitude of 0.707 * 2 = 1.414 feet/second.

The distance from the center to the wheel is 2.828 feet. This means one rotation is 2pir=17.768 feet. This means it’s rotating at 1.414/17.768 = 0.079 rotations per second.

I’m not getting the friction part yet either, though I think you may have miscalculated the radius you used, 2.828 is the distance from wheel to wheel diagonally, making the rotation .159 rev/s. With the friction I’m guessing it will be less than this.

I would think the coefficient of kinetic friction (and robot mass, for that matter) comes into play when calculating torque required to continue to rotate at the determined steady-state speed. But this was not asked.

My first thought was 2rad/sec, but I’m pretty sure that’s with 4 omni wheels. In order to solve this you need to know how much linear distance is traveled, when 2ft of tire tread is pulled by.

Just where do you define the tangential speed? Tangent to the circle subscribed through the wheels center line? if so rotation is 2ft/s* C/8.884ft=.2251rev/sec.

If “Tangential speed” is the slip between the wheel & the ground normal to the wheel axis, then life becomes more complicated & you need weight & friction. For instance a coefficient of 0 rotational speed is uncoupled from wheel speed.

At steady-state, the net torque acting on the bot around its center of mass must be ZERO.

The only forces acting on the bot in the horizontal plane are the four friction forces (one at each wheel). The direction of these forces will start out aligned with the plane of each wheel, but will change as the robot begins to spin. Given the assumption about CoF being independent of speed and direction, the friction at each wheel will always point in the direction of the relative motion between the carpet and the wheel at the contact patch. This relative motion is the vector sum of the wheel tangential speed plus the carpet speed at the wheel due to the robot’s motion.

When the robot spin reaches the point where the net torque around the center of rotation is zero, the spin will stop changing.

The robot weight (150 lbs) and CoF (0.8) were red herrings. Given the assumptions, their values do not affect the answer (as long as neither is zero).

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.

Calculations
[spoiler]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

In this particular problem, you can get the correct answer – as you have done – by simply projecting the wheel speed unto the direction the wheel is moving.

But as the problem gets more general (and more complicated), that will no longer work.

Oh, it would seem that I made a stupid error. I swapped what I used for Wheelbase and Trackwidth. I wasnt sure, so I looked it up online and people were saying that trackwidth is the distance between wheels on a vehicle. It seems thats not what you used here.

I swapped my values in the calculations and my method gave me the correct answer. When I get home I’ll repost a correct explanation.

Yes, trackwidth is the distance between the 2 front wheels. (Or the 2 rear wheels). Here’s an easy way to remember: It’s the width of the “track” left by the vehicle when traveling forward in a straight line.

It seems thats not what you used here.
That is what I used.

Ah, well even if I did find out the dimensions correctly I must have set up the proportion between the carpets velocity and the robot’s orientation inversely. The solution below works out, so unless a mathematical equality that makes this give the right answer for the wrong reason is working here and I haven’t realized what it is, I must have accidentally flipped that value somehow. It was late, I probably used the wrong angle for being tangential to the robots rotation which lead me to the wrong proportion.

Updated solution:

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