Flight and the Conveyor Belt

Cecil's Columns/Staff Reports
181

The equal mass control is an important feature of that experiment.
The comparison isn’t relevant with a full v. empty can.

The solid cylinder is first down the ramp, but how far do the solid v. hollow cyinders roll before coming to rest?
Moment of inertia and rolling resistance are not the same thing.

I’m not disagreeing that making a wheel rotate faster imparts force to the axle, even with frictionless bearings, but zut has already covered that part of the problem very well. He had diagrams and everything.

182

Bingo. I completely agree. The original intent of the question and the wording in this case don’t quite match. So some people answer the question based on their understanding of the wording, others on their understanding of the intent.

I agree with everything you say here, treis, but the experiment you propose doesn’t directly support the idea. I mean, it does support it, but it’s a bit convoluted because the inertias of the two things are different, which means the accelerations are also different, and the frictional forces are functions of both. So if someone doesn’t understand how the frictional force is transferred, I don’t think this experiment will be convincing.

183

I think you are underestimating the force that an engine would have of the free rolling wheels.

-Mike

184

If you are talking about a typical GA aircraft, or bush plane, then yes, it would take off. If a plane takes off at 60knots, and there is a 50 knot headwind then the plane will take off with a groundspeed of 10 knots. I’m not sure who would want to take off in that scenario, though.

I hear stories all the time about flying backwards in an airplane.

185

Take a piece of paper, put it on the desk in front of you. Now place a round pen (no clip, perhaps take the cover off) on it. Move the paper back and forth. What happens?

The pen rolls in the opposite direction. It does move with the paper some, but much less distance than the paper itself.

Now, blow on the pen in the opposite direction you are moving the paper. What happens? The pen rolls faster, and moves a much greater distance across the paper than before. The “wind” from blowing is similar to the force of the engine on the airplane. The plane would move across the treadmill and continue moving until the plane reached enough airspeed to take off (say 60 knots).

The fastest that treadmill is going to move is 60 knots and the plane’s wheels will be moving at about 120 knots. But the plane will still be moving forward at 60 knots and have enough airspeed to take off.

I’m not sure where the problem is here. The original question states it very clearly. The treadmill will move at the speed of the airplane in the opposite direction. It has nothing to do with frictionless bearings, or gravity, or aircraftcarriers. An engine pulls the plane. The fact that the wheels are moving in the opposite direction has absolutely no bearing on if that plane is going to move forward or not.

The plane is not dependant on the wheels to push the plane forward like a car, but the engine to pull the plane forward.

186

Sorry, I meant the treadmill, not wheels.

187

Somebody explain to me what happens when the plane takes off…give me a break, it has to be moving fast enough to fly!!! So does it remain stationary in relation to earth or does it instantly go to flying speed??? With no airspeed, no flying. I say it would bob up and down on the conveyer belt…

188

The problem here is that “the original question” is worded differently in different places. The first question in Cecil’s column states in part that the conveyer “has a control system that tracks the plane speed and tunes the speed of the conveyer to be exactly the same (but in the opposite direction).” And–let me be clear–you are absolutely correct that, in this case, the plane will roll forward and take off.

But my issue here is that this is a dumb question. There’s nothing challenging about it. The plane is allowed to go forward. No problem. In fact, if a car were on the treadmill, it would go forward. So the the statement that…

…is true, but completely irrelevant. The treadmill for this version of the question will allow either a car or a plane to move forward. There’s nothing here to argue about.

However, the more interesting (and, I think, more original) version of the question appears at the end of Cecil’s column. Here, the conveyor “is designed to exactly match the speed of the wheels at any given time, moving in the opposite direction of rotation.” The conveyor will try to keep the plane from moving forward with respect to the ground, in other words. And in this case, your assumptions about the treadmill and the wheels and friction and mass make all the difference in the world (Cecil’s crappy answer notwithstanding).

189

Post 106 should answer your question.

190

No, not correct. If the plane is facing into a 100 mph wind, then the effective current will be pushing against the back of the plane against the floats. In other words the current would be against the back of the plane not in front of the plane impeding its forward progress. Still an assist. Now if the wind were moving exactly in the same direction at the same speed as the water then it would take a little longer to get into the air.

Wait, it already is. At the equator with a calm atmosphere and calm water, the air and the water are both moving around the center of the Earth at about 1040 mph. I don’t think planes have trouble taking off there.

191

No, it supports it as much as it needs to. All I am saying is that there is a force due to friction that slows down a rolling object. This demonstration proves it nicely becuase you have two objects of the same mass that roll down the incline at different rates. That must mean there is a force opposing their motion of different magnitudes. The only force that could be is friction.

192

No. Please learn what “moment of inertia”, “torque”, and “angular momentum” mean before going any further.

193

:dubious:

And what do “moment of inertia”, “torque” and “angular momentum” have to do with linear acceleration?

194

No, John’s right. With a no-slip condition, the linear and angular motions are coupled.

Two rolling objects might have the same mass (resistance to linear acceleration), but if they have different rotational inertias (resistance to rotational acceleration), then, because these are coupled, the higher-inertia object has a greater total resistance to rolling.

A frictional force is necessary here to enforce the no-slip condition, but it’s not necessarily slowing the object down.

195

The acceleration of the cans linearally only depends on three things. Force from gravity, Force from friction and Mcan. The moment of inertia doesn’t matter, the angular acceleration doesn’t matter nor does the torque matter. That stuff deals with rotational motion which is completely independent of the linear motion of an object. In this case the boundary condition of the rolling object allows us to solve for the force of friction using rotational stuff. But make no mistake about it, the moment of inertia, angular acceleration, angular velocity or torque has no bearing on the linear acceleration of an object. All you need is the net force on an object and the mass of the object and you have completely described its linear acceleration.

196

Sorry, incorrect. The key is that the rotational motion is not independant of linear motion. If you don’t believe me, look at the page that you linked to:

An easy way to think about this problem is from an energy point of view. The rolling object starts at the top of the ramp with some potential energy (mgh). At the bottom of the ramp, it’s converted into kinetic energy. The kinetic energy is in two forms: linear (1/2mv[sup]2[/sup]) and rotational (1/2I[symbol]w[/symbol][sup]2[/sup]). These two are coupled, because v = r[symbol]w[/symbol] (no slip!), so that the total kinetic energy is (1/2mv[sup]2[/sup] + 1/2I(v/r)[sup]2[/sup]). In any case, you can see that with a larger moment of inertia, a greater proportion of the energy is used in the rotational kinetic energy term, and the velocities (both linear and rotational) drop.

Let me know if this makes sense; I’d be happy to explain further.

197

zut my friend you are not listening to me. My initial post in this thread simply states that two different cans of the same mass roll down the incline at different rates. This means that there must be a force opposing that motion and the only force that can do so is friction. To which you said:

You are bringing up moments of inertia and such stuff which is irrelevent to the point I am making. The point I am making is that becuase the cans accelerate at different rates there must be a force opposing their motion and the only force that can be is friction.

Then you responded:

While this may be true in this case it complicates the issue needlessly. The can with a higher I accelerates slower linearlly not becuase it has a higher I but becuase it has a higher force from friction against it. It is true in this case that I determines the force from friction but its important to be clear. The can accelerates linearlly slower becuase it has a higher force from friction acting on it.

198

The linear acceleration of the hollow cylinder is less because it moves linearly by rotating… and a hollow cylinder has more resistance to changes in rotation than a solid cylinder of the same mass and diameter.

This is relevant because people seem to think that the only reason rotating a wheel applies force on an axle is because of friction in the bearings resisting rotation. So if you posit “frictionless bearings” the conveyor could do whatever it wanted to the wheels and the plane wouldn’t notice… this is wrong, because the wheels themselves resist changes in rotation as an internal property… and this resistance shows up as linear force on the axle.

Frictionless bearings and massless wheels would get rid of the problem, though.

199

Ah, OK. I misunderstood your point, and was focusing on the relationship between acceleration and inertia. Carry on.

200

So, with a fair amount of guess work and some fudging to make round numbers, lets just see what it will take for us to stop a 747-400 from taking off with the belt. First our assumptions: the conveyor belt can do WHATEVER it needs to do to for the plane to take off. Friction is completely irrelevant other than that it enforces the rolling without slipping condition (neglect, for example, friction due to tire deformation, bearing friction etc)… and I think that’s mostly it, but I’m sure I’m leaving thing out. Anyway, let’s hit numbers:

747-400:
Engine thrust: ~300,000N (281,570N but lets round a little!)
Wheel mass: ~400kgx18 (This is a guess from the fact that a brake update shaved 50kg/wheel… this number is probably a little high, but I think it’s reasonable)
Wheel radius: ~.5m (again a guess, it’s probably on the small side, but a nice clean number)

The calculation:
First, we can treat all of the wheels as a single wheel with the same radius and the combined mass. Treating the wheel as a uniform disk

I=.5 (400kg*18) (.5m)^2=900 kg m^2

for the plane to remain exactly fixed, we need for the force on the wheel to exactly cancel the thrust from the engines. First,
T=r* F=Iaa
(T=torque, F=force, I= moment of intertia, aa=angular acceleration)
solving for aa
aa=rF/I=(.5m)310^5N/900 (kg m^2)
aa=166 rad/s^2

we convert the angular acceleration to the linear acceleration of a point on the surface of the tire (and hence the conveyor belt) through the relationship
a=aa*r=83m/s^2

(this is equivalent to 0-60mph in 1/3 of a second…)

The acceleration due to gravity is ~10m/s^2, and so the belt only needs to accelerate with 8g’s, which is smaller than initial intuition would have guessed, but still pretty large. At this rate, assuming that the plane has 10 hours of fuel at full throttle, the belt will be going 3,000,000m/s or 1/100 the speed of light by the end of the flight. Not too shabby!

There are, of course, a number of approximations, but the result should be correct to within an order of magnitude. If I’ve made any stupid mistakes in the calculation, please point them out! Note that the calculation was simplified by assuming an exact force balance.