Okay so far, but I still have some questions. Essentially, I would like to know what are the frictional forces and their directions relative to the wheel. Maybe my terminology isn’t correct but I’ll try to explain what I mean.
Relating to the scenario 2. I posed earlier:
All right, in this case, the decelaration comes from the tire flex. This creates a torque to the negative y-direction, so it creates an angular acceleration that opposes the rotation, thus decelarating the tyre. (This is a bit fuzzy now, maybe my terminology/understanding of the rolling resistance isn’t quite clear.) You could think of this torque to be created by the “frictional force” that slows down the tyre. How I understand this is that since the tire is rolling to the positive x-direction, from the point of view of the plane, the contact point of the tire could be considered to be trying to move to the negative x-direction. If there were no frictional forces, the tire would slip. Since we are assuming no slip, there has to a “frictional” force (a “static” frictional force, at that) that opposes this sliding, and this force has to point to the positive x-direction. And this force would result in the decelaration of angular motion, and thus also in the deceleration of the translational motion. Maybe that’s not what really happens, but I’m just trying to simplify things a bit.
And how about if we replace the tire with a perfectly rigid cylinder? Then the tire flex wouldn’t be a factor, but the cylinder should still slow down, if there’s friction between the cylinder and the plane.
I wasn’t really clear before and I’m not sure if I’m any more clear now: The acceleration of the belt exerts a force on the wheels, but shouldn’t there also be a “frictional” force between the belt and the wheels that opposes the movement of the wheels on the belt, as in the case 2? If there wasn’t any friction between the belt and the tires, the motion of the belt wouldn’t have effect on the wheels, the wheels would just slide on the belt. So there has to be a frictional force to stop the wheels from sliding. To what direction does this frictional force point? (Maybe I should write “frictional” force, since the force opposing the rotation of the tires comes from tire flex (if I understood correctly), which should create a torque opposing the rotational motion of the tires.)
Let’s go at this another way. If the wheel is a detached wheel in space sitting on the belt with the belt running under it, the peripheral speed of the wheel is equal to the linear speed of the belt if there is no slip, right?
So in order for the wheel to be forced to go in the direction of the belt wouldn’t the peripheral speed of the wheel have to be less than the linear speed of the belt?
In other words, for the wheel to be carried in the direction of the belt wouldn’t there have to be some slip between the belt and the wheel?
Good question. My sense is this: a friction coefficient between two objects doesn’t necessarily mean a frictional force between two objects. In this case, friction comes into play only to enforce the no-slip condition. If angular and linear velocity are already matched, and there are no other forces decelerating the object, the frictional force should be zero. Why would it be anything else?
However, I admit to not being an expert in rolling friction. My suggestion would be to post this question in General Questions; I suspect there’s a good half-dozen people who could give a more comprehensive answer, but they’re unlikely to see the question at the end of a seven-page thread on treadmills.
The belt and the tire are coupled through friction. If there was no friction, the belt would not be able to accelerate the wheels at all, to begin with.
Yes and no. For no slip, the peripheral speed of the wheel at the point of contact, with respect to the ground, has to equal that of the belt. This could be accomplished by having a non-rotating wheel being carried along entirely at the speed of the belt. This could be accomplished by having a wheel with a stationary hub rotating with peripheral speed at all points equal to belt speed. Or it could be accomplished by some combination.
Be a little careful here, though: we’re talking about matching velocities to enforce the no-slip condition, while the application of force produces accelerations Different physical things, different equations.
Well, what’s the end point that you guys are trying to reach? That it is possible to create enough rolling drag to keep the plane stationary relative to the ground? Give the plane 50’ diameter wheels? See my earlier post about the velcro conveyor belt. The question really doesn’t have anything to do with rolling friction, tangental forces, etc… The main intent of the question was to ask that if you put a plane on a conveyor belt, would it take off. Everyone is of a mind that the thrust of the engine(s) would be able to overcome the reverse force of the conveyor belt.
Well, yes and no. Depending on your assumption about the problem statement (very important!), the forces may or may not cancel out (see post 106). In any case, you are correct that matching fuselage and treadmill velocities will not cancel out the forces.
I’m reasonably certain that everyone agrees that airspeed is key to this question. Most of the discussion here is centered on just how a treadmill might prevent a plane from moving forward, and if that is a legitimate interpretation of the question (the question is worded in different ways, and imprecisely).
It all depends on your assumptions about the problem. Under some assumptions, the plane would take off, under other it wouldn’t. See post 106 for as many different classes of assumptions as I could think of.
Yes, for rigid bodies the problem is different. I think I finally start to understand the rotational friction a bit more. For perfectly rigid bodies that are not accelerating, the frictional forces are zero, since they only oppose the change of translational/rotational velocity. For a real tire, the tire flex provides the torque that results in the slowing down of the tire.
Yes, this is actually what I was getting at, I was just expressing myself poorly. Without friction, the belt cannot accelerate the wheels. Now my next question is, starting from zero velocity/acceleration, is it possible for the belt to accelerate so fast, that the tires start slipping? In other words, is it possible that the static friction, which normally gets the tires to rotate when a tangential force is applied, is too small compared to the force exerted on the wheels by belt? (For instance, if you try to accelerate a car too quickly and provide too much torque to the wheels, then the wheels start to slip.)
The treadmill would be complementary to the airplane’s thrust, not friction, right? In the case of a jet (which relieves us of worrying about propwash), as the thrust increased, so would the treadmill’s speed in the opposite direction, acceleration matching the torque from thrust.
A=airplane; B=treadmill; T=net thrust
T=A+B; if A=200 and B=-200 (we’re measuring velocity, not speed), you’d end up with zero net thrust. I don’t see how the treadmill’s speed could be greater than the airplane’s wheels rotational speed, since the moments of inertia are 180 deg out of phase. They’d always have to be equal for the airplane to remain stationary.
Only if the acceleration of the belt overcame the wheel’s internals ability to rotate to the equivalent speed. If the wheel had a frozen/seized bearing, for example, the tire would slip at any rotaton. Imagine a bearing full of a cold, thick oil. the resistance from viscosity would allow the wheel to slip if the belt was accelerated faster than the oil’s ability to flow.
Basically, it depends on the interpretation of the question. The question was posed in a vague way that leaves 2 possibilities open.
The first is that the linear reverse speed of the conveyor belt is controlled to always be equal to the linear forward speed of the plane. In this case, the answer is that the reverse motion of the conveyor belt cannot produce enough friction to overcome the forward thrust of the engines (assuming a realistic airplane with all parts to spec and well-maintained; i.e. no flat tires or frozen bearings). Thus the plane will move forward and reach sufficient airspeed (at a slightly longer takeoff distance).
Most reasonable people will agree that this was the teaching point, agree that it’s kind of a neat thought experiment, make a list of who they’ll spring this on later, and move on with life.
Still other people will point out that, wait a minute, in the wording of the question, the person might have meant that the *rotational * velocity of the conveyor belt is controlled to match the rotational velocity of the airplane’s wheels. This rapidly goes down a path of theoretical absurdity that is largely an exercise in how many possible ways the question could be interpreted, and what kinds of imaginary materials would be needed to withstand the resulting heat and torque that arises in these scenarios.
It’s your choice which road you want to take. And by the way, it’s not the Bernoullli effect that makes airplanes fly, it’s the Coanda effect.
That’s exactly my understanding (although I disclaim expertise here).
Maybe. Assuming we’re still talking about the case where the treadmill is trying to keep the plane stationary, it would depend on the tire coefficient of friction and the thrust to weight ratio of the airplane. (You want friction coefficient X weight to be greater than thrust, in other words.) Looking at the real world numbers from the links, it looks like you wouldn’t bark the tires of a 747, but you might be able to in an F-15.
Mmmm…I think you’re wrong. Depends what you mean. Matching velocities won’t cancel out thrust. The treadmill would have to accelerate in order to transfer enough force to the wheel (and the plane) to cancel out the engine thrust.
We could make this much simpler by specifying not a treadmill, but a dynomometer with a large drum for the wheels, where the drums rotation always matche the wheels’ rotation.
I’ll have to look that up - They taught us about Bernoulli in flight school.
"BERNOULLI’S PRINCIPLE OF
PRESSURE
A half century after Sir Newton presented his laws,
Mr. Daniel Bernoulli, a Swiss mathematician,
explained how the pressure of a moving fluid (liquid
or gas) varies with its speed of motion. Specifically,
he stated that an increase in the speed of movement
or flow would cause a decrease in the fluid’s
pressure. This is exactly what happens to air passing
over the curved top of the airplane wing.*
(Source: FAA/Pilot’s Handbook of Aeronautical Knowledge, 2003 edition)
Eh. I’d rather characterize it as an exploration of how forces and motions are related, and how they’re transferred from one part to another. I suppose you could talk about kinds of imaginary materials and so forth, but no one’s really done that yet. It’s all a function of how interested you are in the thought experiment, I guess.
Damn, and I thought this one was easy. But I am beginning to feel unsure. Thank you for not tossing me on the heaping pile of know-it-all newbies you guys are so good at shredding.
Seriously, Mr. Ten-posts-in-three-years: being willing to discuss and reinterpret problems, asking questions, and eradicating ignorance is what the Dope’s all about. Being virulently proud of one’s ignorance to the point of ignoring contrary arguments is what earns the metaphorical toss to the heaping pile. Those things are light-years apart.
In this case, you have a positive feedback loop. In this case, the rotational speed will quickly accelerate to a speed great enough to destroy the usual materials that airplanes and conveyor belts are made of. From this point it degenerates into what would happen if you used indestructable materials, what is the latency of this magic feedback coupling that makes the wheels go the same speed, what will happen when the system reaches relativistic speeds, etc. We’ve done it to death, trust me.
Please do. Again, been done to death. There is a Bernoulli principle and it does contribute to airplane lift, but its contribution is very very minor compared to the Coanda effect.
One would deduce from your posts that you are more thorough than your weak attempt at qualifying me as you have. Maybe “Mr. Eight Posts in One Day and Two Posts A Loong Time Ago That I Forgot About When I Ponied Up Fifteen Bucks This Morning” would be a more appropriate moniker.
Thank you for providing the differentiation above. I shall endeavor to meet the standards found here and nowhere else. If my ignorance is showing, please tell me so I can zip up.