The root cause of these arguments is poor definition. What is meant by "velocity of the wheel". Normally I would take that to mean velocity of the centre of gravity, which would be the velocity of the axle. But it could mean velocity of the top surface of the wheel, velocity of the bottom surface of the wheel. None of these cases pose a particular theoretical problem.
Originally Posted by Andy!
Let Vp = velocity of the plane. Lets also assume this direction is "positive" velocity.
Let Vc = velocity of the plane.
Let Vw = velocity of the wheel, and Vl = velocity of the lower surface of the wheel, Vu = the velocity of the upper surface of the wheel and Va = velocity of the centre (axle) of the wheel.
Case 1: Vw = Vl.
Then Vl = Vc (since it is in contact) and Vw = Vc.
But it is given that Vw = -Vc, Vc = -Vc and therefore Vc=0.
This is just a normal take off from a stationary surface.
Case 2: Vw = Va.
Then Vw = Vp (assuming the axle is attached to the plane)
and is is given that Vw = -Vc so Vp = -Vc. The conveyor belt speed is equal and opposite to the plane speed. At take off the wheels angular velocity will be twice normal but otherwise no problem.
Case 3: Vw = Vu.
Vw = Vu = Vp + (Vp -Vc).
and it is given that Vc = -Vw so
Vc = -2Vp + Vc
And therefore Vp = 0.
So case 3 is the only one in which the plane doesn't take off. There is no need to imagine infinities or control systems or any such thing. This interpretation of the problem has only one physical solution. This is also not what I would normally take "the speed of the wheel" to mean though.
You could imagine what a system designed to do the impossible would do when the planes engines applied thrust, but that is not productive. Hypothetical impossible systems can do anything. Given an impossible specification a control system which does nothing at all (ie a stationary conveyor belt AKA a runway) would perform as well (or arguably better) than one which blows up.