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Plane vs. Treadmill Solved (free body diagram)

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Uploaded by on Nov 22, 2007

This is the physics behind my experiment

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Entertainment

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  • likes, 3 dislikes

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Uploader Comments (neodocneodoc)

  • you're forgetting something: trust has a limit for the plane. The friction force is speed dependant

  • From SKF The rolling frictional moment is calculated from the equation Mrr = Grr (ν n)^0,6 where Mrr = rolling frictional moment, Nmm Grr = a variable that depends on -- the bearing type -- the bearing mean diameter dm = 0,5(d + D), mm -- the radial load Fr, N -- the axial load Fa, N n = rotational speed, r/min ν = kinematic viscosity of the lubricant at the operating temperature, mm2/s (for grease lubrication the base oil viscosity).
  • So it is not linear, squared, or cubed, but it is a function of the 1.67th root of the speed (at least for SKF bearings). This means that the frictional moment increases, but not as quickly as the speed. For example look at the 1.67th root of 10, 100, 1000

  • The third version of the question is where the belt moves at such a speed that it equals the forward speed of the wheels. This would instantly accelerate the belt and the wheels beyond the speed of light at which the wheels melt and the plane disintegrates.

  • I'm not sure if I agree with the outcome of this scenario. Its been awhile since I designed a motor control system...maybe in the future.

  • The second version is where the belt moves backwards at such a speed that it holds the plane in one position.  This would theoretically prevent the plane from flying, but the belt would quickly reach such a speed that it melts the wheel bearings. In reality, the plane would disintegrate itself.

  • Sounds like this bastardization was created by no-flyers. If the plane is held still (melted bearings, whatever) it won't fly :)

Top Comments

  • Again you don't understand what's going on. It doesn't matter how fast you spin the treadmill, once the Force of thrust is greater than the frictional force in the bearings of the wheels, the plane will roll forward regardless of what the treadmill is doing. The treadmill can be going 1000 time faster than the plane and the plane would still take off (given the wheels themselves don't fail and the plane crashes and burns on the treadmill. You can't bust the myth.

  • one problem, ur not incorporating the most important force, DRAGONFORCE. lol

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  • HIJACK

  • @subductionzone Sorry. At g*sin(theta) where theta is the angle of the ramp. A solid cylinder will accelerate a (2/3)g*sin(t) and a hollow one will accelerate at (1/2)g*sin(t). Yes, I agree with your simplified constant speed treadmill that the plane will take off, but that was not the original question. The original was an untestable thought problem that was designed to make you think of the effect of the moment of inertia. It takes a huge acceleration to keep the plane in place.

  • neodoc, the oldest known version of this problem is the one that you claimed no fly's invented. That is the one where the velocity as measured at the wheels is the same as the belt. By definition the plane stays in place. But is there a force that will keep the plane in place? It turns out there is. It is not friction, it is the resistance to rotational acceleration that keeps a plane in place. This is the same force that keeps a rolling cylinder from accelerating down a ramp at gsin.

  • Simple but great explanation

  • This is quite needed, you don't know what your talking about.

  • NO SMOKE AND MOYERS

  • as far as I can tell, the root of the original question comes from mistaking the way locomotion in a car works with the way locomotion in an airplane works. Since the plane's movement is not dependent on force generated via contact with the ground, the treadmill will only make the aircraft's time to liftoff slightly longer, as he's not starting from a stop. With frictionless landing gear, there is no theoretical limit to the treadmill's velocity - the plane will still lift off in time.

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