It isn’t nonsense in the least bit, you just don’t know what you are talking about.
And why would you say that? Is there a specific part of treis’s statement you object to, or is the whole thing bunk? Do you have a counter argument? Can you show in any way how the statement would be equivalent to a perpetual motion machine?
100% + 100% = 200%. Vectors don’t sum like that.
That isn’t the way it works. The relative magnitude of translational and rotational acceleration do not depend on each other. There’s no “proportion” of the force you can assign to one or the other; the question is meaningless. A force F produces an acceleration equal to F/m regardless of whether rotational acceleration exists.
Try this one: How much of your foot is in your shoe, and how much of your foot is in your sock?
Well then its a good thing I am not summing vectors.
Go here: http://i18.photobucket.com/albums/b119/treis/Circles.jpg
“F” is equal in all cases, in case 1 it always acts at the bottom of the circle, in case 2 the F acts through the center of mass and in case 3 one F acts through the center of mass and the other acts on the bottom edge. All circles are identical and are free to rotate or translate.
Does Case 1 accelerate linearlly any differently from Case 2? (Assume that the force does not rotate with the circle, i.e. it always acts on the bottom). If so, please state exactly what the net force in the x direction is for each case.
Does Case 1 accelerate rotationally any different from Case 3? If so, please state exactly what the torque around the center of mass is for each case.
This is insane. This thread ceased being an informative discussion of the airplane / conveyor belt problem long ago; it is now an interesting case study in
a.) information overload
and
b.) signal-to-noise ratio (in this case, there is an awful lot of noise)
My astonishment at the volume of logical fallacies and flaky physical theories that have popped up on this thread is exceeded only by my suprise at the tenacity with which people will stick to them.
I’m not going to drag myself into a flame war, and probably won’t even post another message after this. In the interests of helping those still genuinely confused by this whole thread, however, I’d like to point out a couple of things:
1.) The lift of an airplane is determined by the speed with which it moves relative to the air around it – **and ** vice-versa (the speed with which the air moves relative to the airplane). Doesn’t matter which principle you cite, it’s this movement of air that’s important. I think we can all agree on that. Airplanes don’t fly in vacuums, regardless of how much thrust they have.
2.) The ground and the air do not always move in tandem or at the same speed. Go stand outside on a windy day for a demonstration of this.
3.) An airplane needs to achieve a certain airspeed in order to take off (airspeed = speed the airplane is moving relative to the air and vice versa).
4.) When taking off in an airplane, you determine your airspeed with the following, time tested formula:
Va = Vg - Vw
where
Va = Airspeed
Vg = Groundspeed, meaning the speed at which the airplane is moving relative to the ground beneath its wheels
Vw = Windspeed component, or the speed of the wind expressed negatively for a headwind (opposite direction of travel) and positively for a tailwind (same direction).
So, if you’re travelling along the ground beneath you in an airplane at 20 mph and a headwind is blowing at 30 mph, your airspeed is 50 mph (20 - -30). Period, end of discussion. Your speed, relative to the air around you, is 50 mph. If that is what your airplane requires in order to take off, you will take off. If your airplane requires an airspeed of 70 mph, you will not take off. If you doubt this, you’re probably not a pilot, and if you are a pilot, I hope to God I never ride in your plane.
5.) Given that airspeed determines lift, and airspeed at takeoff can be calculated by groundspeed and wind effects (roughly), you should be able to understand this whole thing without getting in the minutiae of thrust, angular momentum and all the other clutter being thrown about on this thread. Those things are relevant only in discussing how a plane reaches a given speed relative to the ground beneath it. An airplane with an groundspeed of 70mph has that groundspeed regardless of how it achieves said speed – whether it uses thrust (the expulsion / accelleration of matter), whether it’s carried along the back of a moving car, whether it’s being dragged down the runway by a team of clydesdales. The means for achieving a groundspeed of 70 mph are 100% irrelevant as far as the wing is concerned.
6.) Now, imagine you’re standing still on a miles-long conveyor belt runway on a windless day. This runway / conveyor is so long that we could run it at a very high speed for a very long time, perhaps an hour or so, without hitting the end of it. The conveyor belt turns on and gradually increases its power. Relative to the actual belt, you are not moving. If you stare at a painted spot on the belt beneath your feet, it stays in the same place, no matter how much the belt increases its power.
However, relative to both the ground next to the belt and the air around you , you are now moving, because you are standing on a moving belt. Go to the gym and stand on a treadmill if you disagree with me on this. Let’s suppose, then, that the belt is moving its occupants at 70 mph relative to the rest of the world. If you put a blindfold on, the air around you moving relative to you and the belt would feel identical to a 70-mph wind whipping down the runway-belt from one end.
Now imagine that there’s a plane next to you, and its nose is pointed in the same direction that the “wind” is moving. That airplane is a jet, and requires a 70-mph airspeed to achieve lift & takeoff. Because its nose is pointed in the same direction that the wind is moving, that plane can be said to have a 70-mph tailwind. Here’s a little diagram for you:
Belt is moving in this direction ---->
---------------------------------------------
<<< <
X <<<<<<<<
<<< <
You Airplane (note direction)
____________________________
<------- Air is moving in this direction (relative to belt and its occupants)
You get inside the plane. You power it up. You apply enough thrust to your engine to move along the belt at a groundspeed of 70mph – in other words, if a car was driving next to you on the belt, and its speedometer said “70 mph”, the plane and the car would be moving down the belt in tandem.
Let’s go back to the formula.
Groundspeed = 70 mph
Wind component = +70 mph (70 mph tailwind, so positive number here)
Airspeed = 70 - 70 = 0 mph
Do you take off?
You’re going 70 mph. The driver in the car next to you can confirm this. You are thrusting, your wheels are grinding against the conveyor belt, and a stationary observor on the belt would see you whiz by at 70 mph. But, because of the 70 mph “tailwind” that you feel when you stand on the belt, your plane does not take off. Once again, if you disagree, go try to take off with a tailwind without compensating for it. You will have problems.
7.) Instead of being the pilot of the plane or an observor on the belt, what if you witnessed the entire scene from the area next to the belt? What would it look like?
Well, initially, when the belt was fully powered up and neither the plane nor the person on the belt were moving, they’d pass by you at 70 mph. You would also observe that it was a calm, windless day. You would be sitting there in the calm, windless air and would see this plane and this dude go moving past you really quickly on this absurd, incredibly long conveyor belt.
Suppose, however, that instead of observing the scene at that moment, you placed yourself so that you would be at the belt at the exact momemnt that the plane reached a groundspeed of 70mph. What would it look like?
Well, you’d see plane approaching on the belt, nose pointed the other direction, but the speed of its approach to your position would gradually get slower as it reached the point where you stood. Eventually, it would come to a halt right in front of you. It would sit there, right in front of you, engines screaming, wheels spinning. It would look pretty weird. If you held your hand above the conveyor belt, you would note that the air was moving at 0 mph around it. That’s an airspeed of 0 mph.
8.) Jump back to the jet, same moment in time. Your engines are blasting, your thrust is moving your tires and your plane along the conveyor belt at 70mph, but you’ve got this ferocious 70 mph tailwind whipping up from behind you. Your airspeed remains zero. When you stick your hand out the window, the air moves around it at 0 mph. There’s also a man standing next to the conveyor belt, watching you, waiting to see when you take off.
He will be disappointed. The plane, moving along a conveyor belt at the same groundspeed as the conveyor belt moves in the opposite direction, will never take off, not on a windless day.
9.) If you’re about to get into some argument about how the acceleration of the plane allows it to gradually move faster than the belt and how the belt is always playing catch up, please do me a favor and stick a sock in it. All those holds true in the case of the initial question – if the plane accelerates from zero and belt is tuned to increase its own speed at the same time, it still doesn’t take off.
We can take Va = Vg - Vw and graph it over time as the plane accelerates. Remember, on a windless day, Vw = the speed of the conveyor belt, as the belt moves objects along it at a speed relative to the air around it. Wind is a relative phenomemon. As long as the belt’s acceleration doesn’t lag behind the plane too much, the airspeed never gets much above zero. Va represents the difference between the plane’s movement down the belt and the belt’s movement relative to the rest of world at any moment in time.
In short: Cecil done got it wrong. Like many of you, he confused the way the scene looks from two very different frames of reference – the world on the conveyor belt and the world off. The only thing these two share is the air. When the belt is off, the two worlds move at the same speed relative to the air around them – 0 mph. When the belt is on, the air moves at different speeds. One experiences a world with no wind; the other experiences world with a 70-mph unidirectional wind. As a result, when you stick a plane on a conveyor belt and accelerate the plane and the belt in opposite directions …
… the plane does NOT take off.
You’re wrong. You need to look at the problem in terms of a free body diagram and all of the forces acting on the plane, not in terms of relative velocities. Looking at the problem in terms of relative velocities has led you into a fallacy.
Your explanation goes astray at this point that I have quoted above, when the pilot starts the engines. Everything that follows after that is incorrect.
I’m going to throw in some assumptions to simplify matters as much as possible.
Let’s assume that:
[ul]
[li]the wheel bearings are frictionless.[/li][li]the belt is going backwards at a constant speed of 70 mph, as you stated.[/li][li]the plane is initially sitting on the belt with the engines off, so it is also going backwards at 70 mph.[/li][/ul]
OK, the pilot starts up the engines.
At this point, the plane will begin to accelerate forward with respect to the belt and the ground. With respect to the belt, the plane’s velocity will increase in the forward direction. With respect to the ground, the plane’s velocity will gradually decrease, and eventually reach zero, as you stated.
Now here’s the rub: at this point, the belt is moving at constant velocity, and the wheel bearings are frictionless, so the belt cannot be exerting any force on the plane. If the pilot maintains thrust, he will continue to accelerate to takeoff speed. His velocity with respect to the belt will increase to 140 mph. His velocity with respect to the ground (and the air) will increase to 70 mph, at which point he takes off.
In real life, with wheel bearings that have friction, a small bit of thrust would be counteracted by friction, but not much more than that of a normal takeoff. All that would differ is that at takeoff speed of 70 mph with respect to the ground (and the air), the wheels would be turning at 140 mph (linear speed).
While I agree Cecil’s answer wasn’t very good, the “answer” to the question depends on a few different things. The first of these things is just what question you’re answering. You’ll note that there are at least two wordings in Cecil’s column: the original wording where the belt speed matches the plane fuselage speed, and the the wording in the final paragraph where the belt speed matches the wheel speed. And then there are the assumptions you’ve got to make (because, let’s face it, the question isn’t worded that well. What does wheel speed mean? Are there restrictions on the accelertation of the belt? Friction? And so forth. The final answer depends heavily on your assumptions.
One thing you may have overlooked in your analysis is that very little force would be transferred to the plane through a belt that’s going a constant speed, AND, therefore, it would take very little thrust to hold a jet steady (at zero airspeed) on a constant-speed belt. So the scenario you propose would require the plane to dial engine thrust way back so as to not accelerate forward. Not, I think, the original intent of the problem.
Then either you’re the King of Pointland or you’re not doing physics, Q.E.D., because, everywhere but Pointland, force is a vector.
No, I’m not wrong. I’m very, very right. You don’t need to look at any of things, You need to figure out if a plane that accelerates along a runway will take off if that runway moves in the opposite direction at the the same rate of acceleration. And frankly, this question is all about relative velocities. Planes fly when their air moves at a certain speed relative to their wings.
This is where you fall into a fallacy. The plane is not moving at 70 mph. The plane is moving at 70 mph relative to a observer standing next to the belt. To a person standing next to the plane on the belt, the plane is not moving.
OK, the pilot starts up the engines.
That is correct, and in no way differs from what I’m arguing. But the whole point of this question is that the belt does not stop accelerating. It continues to match the plane’s own groundspeed relative to itself. So when the plane is moving at 90 mph relative to the belt in one direction, the belt is moving 90 mph to the rest of the world in the other direction.
So on what exactly are you disagreeing with me?
You’re missing the point. The stationary jet exercise was intended to show the how the world on the belt and the world off the belt are two different reference points. The original question says the belt continues to increase its speed as the pilot adds more thrust, and as I point out later, all of what’s true for a plane moving along the belt at a constant speed is true of a plane accelerating down an accelerating belt.
Jesus, John.
He’s not saying force isn’t a vector, he’s saying you don’t sum them. If you don’t understand someone’s point, just ask. I would be happy to explain further, but I just don’t understand what you’re not getting about this. You don’t partition the force into linear and angular acceleration sections. F=ma, always.
Well, it depends on your asumptions. I’ve posted some of those in post #106 of this thread, if you want to take a look. But IF the belt keeps the plane at zero airspeed, the belt really needs to accelerate: a constant-speed belt won’t work in this scenario. I believe that’s pretty much what robby is saying also.
You might already understand the difference between velocity and acceleration, but people misunderstanding the effects of acceleration have contributed a fair amount to low signal-to-noise ratio you commented on earlier.
Right and I am not summing forces. You have a wrong interpetation of forces and torques. Go to my link and answer the questions about the cases.
Fine.
1 = 1 + 1.
How could I have been so blind?
And just think how this solves the energy crisis. All we have to do is apply some given force to a skillion different objects. Since 100% of the force will affect each object individually, we can multiply the original energy input by any quantity we want, and one well trained hamster can power the entirely national electric grid.
Energy!=force.
Why don’t you answer my question about the 3 cases?
Hold on a minute. I thought you stated in your example (and I repeated) that the belt was going backwards at a constant speed of 70 mph when the pilot started his engines. Now you are saying that the belt is actually accelerating?
I don’t disagree with this at all. In your example, before the pilot starts the engines, the plane is going 70 mph backwards relative to the ground and the air. The plane is motionless with respect to the belt.
Well again, is the belt moving backwards at a constant speed, or is it accelerating backwards? These are two different scenarios.
Also, is your belt trying to match the plane’s velocity with respect to the ground, or is it trying to match the planes’s velocity with respect to the belt?
In the first case, at takeoff, the plane is traveling at 70 mph with respect to the ground (and the air), and the belt is moving backwards at 70 mph with respect to the ground. The plane at takeoff is traveling at 140 mph with respect to the belt. (This corresponds to situation A in post #106.)
In the second case, you end up with a situation where the plane has a constant thrust and the belt accelerates backwards without limit to ridiculous speeds. (This corresponds to situations B through L in post #106.)
Why don’t you read post #106? 
In any event, we’re back to rehashing the same old stuff. I thought we were discussing a new, simplified version of the problem where the belt was simply moving backwards at a constant speed of 70 mph.
Force. Is. Not. The. Same. Thing. As. Energy.
In fact, if you had read my original post on this particular subject (number 360), you would see that I suggested comparing the energy quantity associated with translation and rotation instead of comparing the force; that would create a more sensible and meaningful measurement, since the proportion of force allocated is essentially meaningless.
I’ll be happy to explain this farther, but to be honest, your snide comments are a little off-putting. I suggest you might put a little more effort into understanding the physics and a little less into the snark.
Generically, sure. For the system under consideration, not so much.
And when te object is the same object and the force and torque in question are caused by the same source… the two have a relationship.
Knows answer when told. Gotcha.
How many people are going to bother to look up what it means when you say “r omega” or “r alpha”? Don’t use jargon or shorthand without explicitly defining it at first use. Basically all you’ve done is restated what I said in my post.
Here’s the question:
A 3kg solid disc 30cm in diameter sits at rest relative to a conveyor. The conveyor begins accelerating at 1m/s per second. How fast is the disc moving at time t, relative to the conveyor? Show your work. State all assumptions. Define all base equations.
The conveyor applies a force to the disc. You may be tempted to whip out F=ma and say “m=3kg and a=1m/ss therefore F=3kgm/ss” and you would be wrong. The force applied to the disc will also make it rotate, so that, relative to the disc, the conveyor is no longer accelerating at 1m/ss. What is the conveyor’s acceleration, relative to the disc? SYW, SAA, DABE.
This is all relevant to finding out how much force a treadmill will transmit to an airplane’s axles if we assume no-slip tires with frictionless wheel bearings.