« on: 29/03/2011 16:35:53 »
I'm still hoping that someone can prove to me, once and for all, if that plane on the treadmill can lift or not?
I say it will not..
Awh, come on now, myth-busters has defined it as lifting, and the majority of that physics community I was on then seemed to think it should too?
==the exact quote;
A plane is standing on runway that can move (some sort of band conveyer). The plane moves in one direction, while the conveyer moves in the opposite direction. This conveyer has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in opposite direction).
The question is:
Will the plane take off or not? Will it be able to run up and take off?
Jump right in, but prove your point sientiflicly please, I do have some standards?
I hope people will forgive me for moving this to another thread. I want to answer this one thoroughly, and I don't want to threadjack the other discussion since it may still continue.
Right. *cracks knuckles* Get ready for a wall o' text.
One problem that bedevils most of the discussions I have seen about this topic is that the problem as stated is ambiguous. Depending on how you interpret the problem gives you a very different answer. This leads to a lot of discussion at cross-purposes. I'm going to look at the two simplest interpretations of the problem and work them out in detail so that you see what is involved. There will be a number of simplifying assumptions, but I will point them all out as I make them.
The ambiguity is in the phrase "tracks the plane's speed." We are measuring the speed with respect to what? Are we measuring it with respect to the ground or with respect to the conveyor belt? We'll take them both in turn.
(Two comments: 1. Obviously the same ambiguity applies to the speed of the conveyor belt, but I've never seen anyone interpret the conveyor's speed as being anything other than with respect to the ground, and I'll be sticking with that. 2. The two interpretations of the plane's speed I gave are not the only ones, and indeed neither is the one the Mythbusters tested. The Mythbusters tested using the interpretation the plane's speed with respect to the ground if it were on a stationary runway. This is more difficult to analyze, but I think the ones I show you will be sufficiently illustrative.)
Before we get into the conveyor belt cases let's be sure we understand the simple case of a plane taking off from a stationary runway. Now the important thing for an airplane to take off is that its wings need to be moving quickly with respect to the air, and for most practical purposes that means moving quickly with respect to the ground. We'll assume that we're dealing with dead calm so that the plane's velocities relative to the two are identical.
The force acting on the plane in the direction that it wants to go is the thrust from the engines; specifically the propellers or the jets push air backwards and the air pushes the plane forwards by Newton's third law. All other forces on the jet are either working against the thrust or are perpendicular to the thrust.
I'm making the following simplifying assumptions:
1) There is no atmospheric drag
2) The plane's wheels roll without slipping
3) There is no friction in the wheel bearings
4) There is no "rolling friction" between the wheels and the runway (rolling friction is a dissipative force due to deformation of the rolling object and the surface it rolls over)
These assumptions add up to two results
1) The only force working opposite the direction of the thrust is the force of static friction between the wheel and the runway.
2) The work done by the friction all goes into making the wheels spin.
In equation form we have
F - f = Maand
τ = Iα
Where F is the thrust, f is the force of friction, M is the mass of the entire plane, a is the plane's acceleration, τ is the torque on the wheels from the friction, I is the net moment of inertia of the wheels, and α is the angular acceleration of the wheels.
As another simplifying assumption, we will assume all the wheels are identical. This means each has the same formula for its moment of inertia, and we can treat the problem as if the plane had only one wheel. (We would actually reach this point of simplification anyway if we treated all the wheels separately, but the final formulas are ugly if the different wheels have mass distributed differently). Finally, I'm going to treat the wheels as uniform disks, so the moment of inertia is I = ½mR², where m is the mass of the wheels and R is their radius. (This is not the worst case scenario for the plane being able to take off, by the way. Worst case scenario is treating the wheels as hoops (I = mR²). I suspect the disk approximation comes closer to the truth, but you should be able to work out the worst case scenario for yourself if I succeed in explaining everything clearly. *touch wood*)
Having a single radius also lets us rewrite the torque as τ = fR and we can rewrite
τ = Iαas
fR = ½mR²αor
f = ½mRα
Right now we have three unknowns: f, a, and α. However, we assumed above that the wheels are rolling without slipping. Which gives the relationship
a = Rαso
f = ½ma
We can now solve for either f or a in terms of the knowns:
f = F / [1 + 2M/m]or
a = F / [M + ½m]
Both equations imply that good airplane design is going to involve having the wheels account for as small a fraction of the total mass as possible. This would be true even if we put back in all the dissipative forces we decided to neglect above. I don't have any figures on realistic airplane wheel masses. Below I'm going to crunch some numbers using an assumption of m = .1 M, which sounds like an overestimate to me, but as I said I have no data.
We'll get into the conveyor cases in the next post.
[MOD EDIT - TITLE RE-FORMATTED AS A QUESTION - PLEASE TRY TO DO THIS IN FUTURE, IN LINE WITH THE FORUM POLICY]