# Flight and the Conveyor Belt **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [February 3, 2006, 12:25am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759 "2006-02-03T00:25:10Z")\ **Posts on this page:** 20\ **Page:** 20
**Author:** ![HMS\_Irruncible](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hms_irruncible/32/7394_2.png) [@HMS\_Irruncible](https://boards.straightdope.com/u/HMS_Irruncible)\ **Post date:** [February 9, 2006, 10:33pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/381 "2006-02-09T22:33:30Z")
> [@Tenspace](#): > > Damn, and I thought this one was easy. But I am beginning to feel unsure. This problem basically has 5 phases - \*\*Denial: \*\* It’s obvious that the plane won’t move because it needs forward motion to produce lift. \*\*Anger: \*\* What are you, an idiot? I’ll explain it again slowly so you can understand it. Only lift makes airplanes move. Plane can’t move without lift. Clearly our educational system is going to hell if it’s producing dolts like you. **Bargaining:** OK, I’m going to reduce it down to the simplest possible elements. Let’s replace the jet with an ostrich and the engine with a lava lamp. Then it will be clear why the plane won’t fly. **Depression:** Somehow I’m wrong but I’m not sure why. Everybody else seems a lot smarter than me. **Acceptance:** OK, now I understand. There’s forward motion. There’s lift. The ground interaction is largely irrelevant. The question was deceptively worded. I can’t wait to unleash this on my brother-in-law the airline pilot, thinks he’s so damn smart just because he drives a Jaguar… ---
**Author:** ![Tenspace](https://avatars.discourse-cdn.com/v4/letter/t/50afbb/32.png) [@Tenspace](https://boards.straightdope.com/u/Tenspace)\ **Post date:** [February 9, 2006, 10:45pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/382 "2006-02-09T22:45:13Z")
> [@Brain Wreck](#): > > This problem basically has 5 phases - > \*\*Denial: \*\* It’s obvious that the plane won’t move because it needs forward motion to produce lift. > \*\*Anger: \*\* What are you, an idiot? I’ll explain it again slowly so you can understand it. Only lift makes airplanes move. Plane can’t move without lift. Clearly our educational system is going to hell if it’s producing dolts like you. > **Bargaining:** OK, I’m going to reduce it down to the simplest possible elements. Let’s replace the jet with an ostrich and the engine with a lava lamp. Then it will be clear why the plane won’t fly. > **Depression:** Somehow I’m wrong but I’m not sure why. Everybody else seems a lot smarter than me. > **Acceptance:** OK, now I understand. There’s forward motion. There’s lift. The ground interaction is largely irrelevant. The question was deceptively worded. I can’t wait to unleash this on my brother-in-law the airline pilot, thinks he’s so damn smart just because he drives a Jaguar… 🙂😃🙂 My god, you’ve reduced me to smilies. ---
**Author:** ![Tenspace](https://avatars.discourse-cdn.com/v4/letter/t/50afbb/32.png) [@Tenspace](https://boards.straightdope.com/u/Tenspace)\ **Post date:** [February 9, 2006, 10:46pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/383 "2006-02-09T22:46:56Z")
> [@Brain Wreck](#): > > There is a Bernoulli principle and it does contribute to airplane lift, but its contribution is very very minor compared to the Coanda effect. Somebody better tell the FAA. Seriously, I won’t comment further until I can digest the many posts on Bernoulli vs Coanda. It’s another bruised and battered horse, apparently. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 9, 2006, 11:20pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/384 "2006-02-09T23:20:03Z")
> [@zut](#): > > 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. If I assume a wheel on a belt and the belt goes from zero to some velocity I find that the total input work is all stored in the rotation of the wheel with none left over to translate the wheel. [Here’s the problem worked out.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=46cdre2.jpg&.src=ph) ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 9, 2006, 11:48pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/385 "2006-02-09T23:48:42Z")
> [@David Simmons](#): > > If I assume a wheel on a belt and the belt goes from zero to some velocity I find that the total input work is all stored in the rotation of the wheel with none left over to translate the wheel. > > [Here’s the problem worked out.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=46cdre2.jpg&.src=ph) But you’re begging the question though: you’re claiming that the final angular velocity is 2[symbol]p[/symbol]. The no-slip condition requires the contact point on the bottom of the wheel to be going at the same speed as the belt. In your example, that could (and, in fact, is) accomplished by ending with a wheel that is _both_ spinning and translating. You’re not allowing this case. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 10, 2006, 12:41am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/386 "2006-02-10T00:41:33Z")
> [@zut](#): > > But you’re begging the question though: you’re claiming that the final angular velocity is 2[symbol]p[/symbol]. The no-slip condition requires the contact point on the bottom of the wheel to be going at the same speed as the belt. In your example, that could (and, in fact, is) accomplished by ending with a wheel that is _both_ spinning and translating. You’re not allowing this case. Well, work out a case that shows this. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 10, 2006, 2:27am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/387 "2006-02-10T02:27:59Z")
> [@zut](#): > > But you’re begging the question though: you’re claiming that the final angular velocity is 2[symbol]p[/symbol]. I’m not _claiming_ any such thing. What I assume is that the wheel does not slip on the belt. That is, when the belt advances 1" the rim of the wheel also moves 1". As [this drawing](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=4aa7scd.jpg&.src=ph) shows, when the belt has advanced a distance of 2πr the wheel has made one revolution. Since it did that in one second the velocity is 2π at that instant. That isn’t a mere claim, it’s a result of the wheel not slipping and doesn’t beg the question at all. > [@](#): > > The no-slip condition requires the contact point on the bottom of the wheel to be going at the same speed as the belt. In your example, that could (and, in fact, is) accomplished by ending with a wheel that is _both_ spinning and translating. You’re not allowing this case. I proposed a problem and a mechanism as to why the problem represents the actual case and then worked the problem showing that the input energy is all in the wheel rotation. Go thou and do likewise with a case of the wheel slipping on the belt so as to both rotate and be translated on the belt, along with a mechanism as to why the wheel should slip. ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [February 10, 2006, 3:54am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/388 "2006-02-10T03:54:12Z")
> [@David Simmons](#): > > Well, work out a case that shows this. How do you know the wheel will be rotating at 2π rad/sec after one second? It seems like your entire proof depends on this condition being met but you never provide any reason why this should be the case and not (for example) that the wheel would spin more slowly _and_ translate left. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 10, 2006, 3:54am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/389 "2006-02-10T03:54:30Z")
> [@David Simmons](#): > > I’m not _claiming_ any such thing. What I assume is that the wheel does not slip on the belt. That is, when the belt advances 1" the rim of the wheel also moves 1". As [this drawing](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=4aa7scd.jpg&.src=ph) shows, when the belt has advanced a distance of 2πr the wheel has made one revolution. Since it did that in one second the velocity is 2π at that instant. That isn’t a mere claim, it’s a result of the wheel not slipping and doesn’t beg the question at all. Your drawing shows a detached wheel, sitting on the belt, and the wheel rotating, but not translating, right? Explain, with a free body diagram, how a single unopposed applied force can result in angular acceleration, but not linear acceleration. > [@David Simmons](#): > > I proposed a problem and a mechanism as to why the problem represents the actual case and then worked the problem showing that the input energy is all in the wheel rotation. You show all the energy being in rotation because you assumed _only_ rotational motion. That’s begging the question. If you think it’s not, explain why the case of combined rotation and forward motion is disallowed. > [@David Simmons](#): > > Go thou and do likewise with a case of the wheel slipping on the belt so as to both rotate and be translated on the belt, along with a mechanism as to why the wheel should slip. Uh…why do you want the wheel to slip? I thought we were talking about a no-slip condition. I’ll assume that’s what you meant. So. Let’s assume no slip between the belt and the wheel, and a belt speed that goes from zero to 2[symbol]p[/symbol]r ft/sec. The relationship between angular and linear velocity depends on the relationship between mass and moment of inertia. The wheel center could, for example, be translating forward at [symbol]p[/symbol]r ft/sec, and the wheel could be rotating backward at [symbol]p[/symbol] rad/sec. In this case, the bottom of the wheel moves at 2[symbol]p[/symbol]r ft/sec and no slip occurs between the belt and the wheel. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 10, 2006, 4:31am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/390 "2006-02-10T04:31:02Z")
> [@zut](#): > > Your drawing shows a detached wheel, sitting on the belt, and the wheel rotating, but not translating, right? Explain, with a free body diagram, how a single unopposed applied force can result in angular acceleration, but not linear acceleration.You show all the energy being in rotation because you assumed _only_ rotational motion. You’re the one who claims that the wheel will translate. Why don’t you explain why it does translate and and work out the math? ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [February 10, 2006, 4:48am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/391 "2006-02-10T04:48:17Z")
What is the precise mathematical relationship between linear acceleration and angular acceleration… I.E for a = kA what is the value of the proportionality constant k? a = linear acceleration A = angular acceleration F = force m= mass I = moment of inertia r = distance from center of mass, i.e radius. T = torque F = ma T = Fr T = IA I = mr²/2 _(for a solid disc)_ Start substituting… T = Amr²/2 Fr = Amr²/2 mar = Amr²/2 Start simplifying… ar = Ar²/2 a=Ar/2 So k = r/2. Which makes sense intuitively, I guess. If r = 0, the force is applied to the center of mass and there is no rotation. The further from the center of mass the force is applied the more torque is generated and the more rotation you get. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 10, 2006, 5:18am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/392 "2006-02-10T05:18:30Z")
> [@1010011010](#): > > What is the precise mathematical relationship between linear acceleration and angular acceleration… I.E for a = kA what is the value of the proportionality constant k. There is no relationship between angular and linear acceleration. Those two are seperate and completely independent of one another. All that matters for linear acceleration is the net force (and mass) on the object. It doesn’t matter in the least bit where that net force is located. All that matters for angular acceleration is the net torque (and I) on the object. In some cases, this one for instince, they are related by the physical relationships of the system. For the rolling case this relationship is that there is no relative velocity between the tire and the surface at the point of contact of the two. That means the linear edge velocity of the tire (r_omega) must equal the velocity of the center of mass relative to the surface (v). Since this is true for all time the change in velocity (acceleration) of those two have to be equal. Thus, r_alpha=a. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 10, 2006, 5:38am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/393 "2006-02-10T05:38:25Z")
> [@David Simmons](#): > > If I assume a wheel on a belt and the belt goes from zero to some velocity I find that the total input work is all stored in the rotation of the wheel with none left over to translate the wheel. > > [Here’s the problem worked out.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=46cdre2.jpg&.src=ph) You are missing angular work. The total work is force_distance+torque_angle. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 10, 2006, 12:27pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/394 "2006-02-10T12:27:42Z")
> [@David Simmons](#): > > You’re the one who claims that the wheel will translate. Why don’t you explain why it does translate and and work out the math? I already did explain it. It’s in post 161. You even quoted post 161 above. In fact, you even said, “If the force is 100% tangential the wheel **both spins and translates**. [my bold]” You said that. So either you don’t believe what you said you believe, or you don’t think it’s applicable in this case. So, since I’ve _already explained it,_ I’d like you to explain, with a free body diagram, how a single unopposed applied force can result in angular acceleration, but not linear acceleration. If you’d like some clarification of my explanation, or you don’t understand some parts of it, please ask. Seriously, I’m willing to work out the belt problem here, but it’s gonna be based on what I’ve already posted, and I don’t feel like spending a couple hours coding it up if we can’t even agree on basic principles like F=ma. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [February 10, 2006, 2:30pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/395 "2006-02-10T14:30:08Z")
> [@zut](#): > > I already did explain it. It’s in post 161. You even quoted post 161 above. In fact, you even said, “If the force is 100% tangential the wheel **both spins and translates**. [my bold]” You said that. So either you don’t believe what you said you believe, or you don’t think it’s applicable in this case. I was merely quoting you and then asking you what part of the force resulted in translation and what part resulted in rotation. A question you never answered. ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [February 10, 2006, 2:56pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/396 "2006-02-10T14:56:17Z")
> [@Tenspace](#): > > Somebody better tell the FAA. Seriously, I won’t comment further until I can digest the many posts on Bernoulli vs Coanda. It’s another bruised and battered horse, apparently. Suffice it to say (since it’s not the subject of this thread) that the standard “Bernoulli effect/curved wing surface” explanation you find everywhere from “Mr. Wizard’s Science Secrets” to “Airplanes for Dummies” fails completely in view of the demonstrable fact that airplanes can fly upside-down. It’s not that the Bernoulli effect isn’t real (it is), but that it isn’t a complete explanation for what happens with real airplanes. ---
**Author:** ![Tenspace](https://avatars.discourse-cdn.com/v4/letter/t/50afbb/32.png) [@Tenspace](https://boards.straightdope.com/u/Tenspace)\ **Post date:** [February 10, 2006, 4:51pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/397 "2006-02-10T16:51:09Z")
> [@John W. Kennedy](#): > > Suffice it to say (since it’s not the subject of this thread) that the standard “Bernoulli effect/curved wing surface” explanation you find everywhere from “Mr. Wizard’s Science Secrets” to “Airplanes for Dummies” fails completely in view of the demonstrable fact that airplanes can fly upside-down. It’s not that the Bernoulli effect isn’t real (it is), but that it isn’t a complete explanation for what happens with real airplanes. As I said eariler, I need to digest the Coanda information before commenting. Maybe you could enlighten with a source of information on the efficacies of the Coanda effect, specifically lift ratios, optimum shapes for cohesion, and how it ties in with laminar flow. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 10, 2006, 5:46pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/398 "2006-02-10T17:46:04Z")
> [@David Simmons](#): > > I was merely quoting you and then asking you what part of the force resulted in translation and what part resulted in rotation. A question you never answered. Actually, I did answer it. I said this: > [@](#): > > Well, I think that’s a bit misleading to think of it in that way. The force F produces a torque Fr around the center of mass, right? But the wheel still accelerates translationally at same acceleration it would if there was only a force F and no torque. And the wheel accelerates rotationally the same as if there were a total torque of Fr, with no total translational force. So _all_ the force is used to accelerate the wheel linearly, and _all_ the force (torque) is used to accelerate it rotationally. If you find that unclear, I’d be happy to expand on it, but please don’t claim that I never answered your question. It hurts my feelings. I’ve been answering all your questions, perhaps you could answer one of mine? For the third time: I’d like you to explain, with a free body diagram, how a single unopposed applied force can result in angular acceleration, but not linear acceleration. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 10, 2006, 7:21pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/399 "2006-02-10T19:21:50Z")
> [@David Simmons](#): > > I was merely quoting you and then asking you what part of the force resulted in translation and what part resulted in rotation. A question you never answered. There is no part of any force that results in translation and there is no part that results in rotation. 100% of the force is responsible for translation and 100% of the force is responsible for rotation. Linear and angular acceleration are completely independent of each other. ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [February 10, 2006, 7:49pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/400 "2006-02-10T19:49:33Z")
> [@treis](#): > > There is no part of any force that results in translation and there is no part that results in rotation. 100% of the force is responsible for translation and 100% of the force is responsible for rotation. Linear and angular acceleration are completely independent of each other. Sorry, but in terms of physics, that’s nonsense (though it’s a pretty good starting point for a perpetual-motion machine). [Previous page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=19) [Next page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=21)