Airplane physics question

You obviously failed 10th grade physics, because that is not in any physics book I have ever seen. The first thing to move is NOT the wheel. It's the system! The thrust accelerates the system, the wheel rotation is a byproduct.

 

ahem

"pseudo voodoo"???? Lets keep this a technical discussion, friend!

the problem clearly states "conveyor belt matches the speed of the aircraft relative to it in opposite direction"

so: (if v_plane(t) is velocity of plane relative to conveyor)

v(t) = v_plane(t) - v_conveyor(t) = 0

according to the problem statement! As I pointed out, nothing will break since the engine energy is converted into rotational energy of tire spin.

 

I thought pseudo voodoo had a nice ring to it.

I must ask, how you jump from the planes velocity relative to the conveyer is 0 to the planes velocity relative to the air, or ground is 0?

 

Pseudo voodoo seems appropo here, as there is no way to get the energy from the engine into the spinnin' wheels. Can't work. Won't work.

Even Kandy Kane with the blonde landing strip understands this.

 

The system has inertia.Things within the system can move independently, before the inertia is over come. And when you apply energy, they do move first.

When you open a door, the house doesn't swing aside, does it? It's because the house has a lot of inertia. The door, resting on a greased hinge, doesn't. It moves. When you lock it to the house and try to swing it open, what happens? Nothing moves, right? Now it is locked to the house and will only move with the house. That takes a lot of energy/thrust/force. You are not providing enough. That's why it doesn't move. That's why you don't try to swing open a locked door. You unlock it from the house first. It still stays attached, but it can swing open with very little force. The wheels stay attached too. They rotate with very little force. That's why they rotate at such high speeds...to dissipate the high force/thrust levels coming from the engine.

 

Kandy Kane's busy. Lots of circular motion, no forward movement. Impossible for you. Not me.

 

Simply because the conveyor is not moving forward or backward relative to the ground. It is sitting there on its legs, raised up from the ground floor. If you are not moving forwards or backwards relative to the conveyor belt, you are not moving relative to the ground below or the air above. Gah!

 

Actually I did pretty well in school. College was a different story. Law school was a disaster!

 

I guess you are calling every physics text in the world magically delicious jibber jabber?

That is not the way to debate this, my friend. What equations are you referring to? I hope they are not different from Newton's...these aren't for sure. Just go ask any physics professor. You would accept their word, wouldn't you?

 

Art, that could possibly be the worst example I have ever seen. Can things in the system move independently, sure. Of course. Can the wheels move without motion of the aircraft. NO. Not within the confines of this problem.

Back to your house example. The only reason your door is opening is because your apply a force directly to it. Now close the door, I want you to bring the door to me, without applying a force directly to it. You'll have to push the entire house to me, it will move with the system. Otherwise, your SOL. Now a house is pretty hard to push. It seems by your logic you feel if you push hard enough eventually the door will just open for me because the force will somehow magically make its way to the door, rather than applying itself to the whole system.

 

I'll try one last time:

as stated in the problem, the conveyor belt speed always and magically
precisely matches the relative forward speed of the plane.

let omega(t) be the rotational velocity of the tires, r the radius of the wheel.

then

v_plane_relative_to_conveyor(t) = omega(t) * r

however

v_conveyor(t) = - omega(t) * r

according to the stated problem!!!!!!

therefore

v_plane(t) = v_plane_relative_to_conveyor(t) + v_conveyor(t) = 0.

However, omega(t) can be calculated based on the equation I gave above,
neglecting friction. this will explain how the engine energy gets (indirectly!) converted into rotational energy of the wheels. Once again, note there is no energy transfer from conveyor belt to the plane.

 

I forgot to mention that I have a PhD in Theoretical Particle Physics...
not that it would matter here...

 

I can bring the house to you if it's on rollers or wheels. If I can push it hard enough. That's a force I will be applying. The force will result in acceleration. How much? This much:
F=ma.
.: a= F/m. Agreed?

Acceleration will result in speed. Agreed?
How much speed? Assuming constant force, the speed varies with time.
You can find out how much with this:
velocity = initial velocity plus acceleration times time
v=u+at.
We know initial velocity is zero. So v=at. Simple as that. We also know from above that a=F/m. so v=(F*t)/m.

BUT the house won't go anywhere if it's on a treadmill going in the opposite direction at the same speed at all times as the house's speed going forward. It won't budge from its position. Same with the plane. No speed means no airflow. No airflow means...

 

I think they would believe you if you said you could do magic. They like magic on this board.

 

This doesn't make any sense, particle physics or not.

My first question would be, why do you insist on calibrating the conveyer to the rotation of the wheel? we're not doing that, it's calibrated to the translational speed of the airplane/wheel, not the rotational velocity of the wheel.

Second question, why do you want to calibrate it to the motion relative to the conveyer belt, instead of ground or airspeed?

 

Lol!!!! This is the easiest thing in the world to prove with a skateboard, long extension cord, fan and treadmill. First, put the long extension cord on the fan, attach the fan to the skateboard, let it get going for about 5-10 yards till it stops accelerating and then time it over 30 yards. Keep the fan speed down to make sure things don't happen too fast. You have the speed of your fanboard now. Convert it to miles per hour, since this is what your treadmill will probably be in.

Now fire up the treadmill. Get it to the same speed. See if you can calibrate it by putting a pedometer on yourself and running on it for a minute. Multiply the distance you have run by sixty to get the speed. Speed of treadmill calibrated accurately? Is it the same as what the control on the treadmill says? Can you get it to go at the same speed as your fanboard? Good.

Now, put your skateboard on, fire up the fan, keep the entension cord out of the way, and hold the skateboard on the treadmill for a while (to allow the fan to accelerate). After a while, if I am right, you should be able to leave it and it should sit there happily. If you are right, it will start to move forward. Step away from your steady state model, but don't trip over the extension cord.

Do you agree that this will prove that you can have the wheels rotating without the object moving forward?

 

I agree the skateboard will move forward while the wheels spin twice as fast as they did on the run on the ground.

 

i give up. out of time. sorry I thought I could help you guys out!
good luck to 2000yellow360 with this one. Fortunately in science
the majority does not determine who is correct.

 

You are getting mixed up. Angular or rotational velocity is measured in degrees per second, or radians per second. You can never "calibrate" a linearly moving object's speed (measured in linear distance per unit time) to an angular speed (measured in portions of a circle per unit time). Except by using a constant radius factor. Because wheels of different radius will travel at different speeds for the same linear speed, you need to know the wheel radius in your situation. As soon as you do this, you end up converting it to linear speed. You can only compare like to like. When people say wheel speed, they mean the outermoste surface of the tire, as it travels linearly (which can be measured in linear distance per unit time). Think of it as measuring how much ground is whizzing by per unit time.

So, it is always the translational speed you calibrate to. You can't calibrate it to the rotational speed even if you wnated to. For a given radius, there is a constant relationship, so it can be backed into by observation, or once the specific radius is known. But it cannot be done any other way on paper except how reinerkaiser says. Simply because the units of measurement are different, until translated by a constant radius factor. Any way you phrase it, the question CAN only mean the linear speeds. If it meant rotational, it would have to specifiy the radius of both the plane's wheels (assuming they are all the same size) and the radius of the conveyor belt's rollers (assuming they are all the same size and there's no gears and pullies involved).

All of this detail can be ignored, however. You just need to know that ultimately, the linear speed of the tires and the linear speed of the conveyor belt matches in opposite directions. That's all the question could have meant. And that's all it meant. That's not why I am saying it was poorly worded. Rather, I dont' think we have tires that can go at the high wheel speeds that would result. Simple as that.

 

By saying that, you are not agreeing with me. You are disagreeing with me. Try it and see who is right. If it stays put after the treadmill's calibrated speed is the same as the skateboard's (in the opposite direction), then I am right. If it moves forward over the treadmill, you are right (and on your way to a Nobel prize). You are right about the wheel speed...it will be twice as fast. But the skateboard won't go anywhere. Try it.

 

In your experiment, try to eliminate or minimize friction by rolling the skateboard on a very smooth floor with no seams. And try to eliminate or minimize friction on the treadmill by tilting it downward just enough so that if you put the skateboard and fan on it (no power to the fan) then it sits still, but the slightest nudge gets it to roll quite a long way. Lube the skateboard wheel bearings with plenty of WD40 and lithium grease as well.

 

You're right, I missread his equations. He is calibrating to translational velocity. However, he is still calibrating it relative to the conveyer belt. Which you cannot do. You have to calibrate it to a point independent of both bodies. A spot on the ground say. The airplanes gets X groundspeed, the conveyer gives a speed of -X.

 

yes, I know, I don't agree with you.

 

Are you going to do the experiment?

 

No. I'm right. I don't need to do it. I might try to model the system over the holidays using some sort of mechanical systems simulation software.