classical mechanics - What will happen if a plane trys to take off whilst on a treadmill?

So this has puzzled me for many a year... I still am no closer to coming to a conclusion, after many arguments that is. I don't think it can, others 100% think it will.

If you have a plane trying to take off whilst on a tread mill which will run at the same speed as whatever the planes tires rotation speed is will it take off?

[edited to be more clear]

The question is simple. Will a plane take off if you put this plane onto a treadmill that will match whatever speed the plane wheels are moving at. So the plane should not be able to move.

This is a hypothetical situation of course. But I am very interested.

Answer

Idealizing the plane's wheels as frictionless, the thrust from the propeller accelerates the plane through the air regardless of the treadmill. The thrust comes from the prop, and the wheels, being frictionless, do not hold the plane back in any way.

If the treadmill is too short, the plane just runs of the end of it and then continues rolling towards take off.

If the treadmill is long enough for a normal takeoff roll, the plane accelerates through the air and rotates off of the treadmill.

UPDATE: Don't take Alfred's word for it. Mythbusters has actually done the experiment.

UPDATE 2: I've been thinking about how the problem is posed (for now as I'm typing this) and it occurred to me that the constraint "run at the same speed as whatever the planes tyres rotation speed" actually means run such that the plane doesn't move with respect to the ground.

Consider a wheel of radius $R$ on a treadmill. The treadmill surface has a linear speed $v_T$ to the right. The center of the wheel has a linear speed $v_P$ to the left. The CCW angular speed of the wheel is:

$\omega = \dfrac{v_T + v_P}{R}$

If "run at the same speed as whatever the planes tyres rotation speed" means:

$\omega = \dfrac{v_T}{R}$

then the constraint requires $v_P = 0$. That is, the question, as posed, is:

If the treadmill is run such that the plane doesn't move, will the plane take off?

Obviously, the answer is no. The plane must move to take off. Looking at mwengler's long answer, we see what is happening. The rotational speed of the tires and treadmill are not the key, it is the acceleration of the treadmill that imparts a force on the wheel axles (ignoring friction for simplicity here).

So, it is in fact the case that it is possible, in principle, ( don't think it is possible in practice though) to control the treadmill in such a way that it imparts a holding force on the plane, preventing it from moving. But, once again, this force is not proportional to the wheels rotational speed, but to the wheel's angular acceleration (note that in the idealized case of massless wheels, it isn't even possible in principle as the lower the moment of inertia of the wheels, the greater the required angular acceleration).