# 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:** 23
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 3:19am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/441 "2006-02-12T03:19:22Z")
> [@David Simmons](#): > > The problem of a wheel on a conveyer belt isn’t so difficult that it should take up all of this time. I went back to my old, old, physics text to review on rotary motion. Having done that I reproduced the development of the relationship between torque and angular acceleration. The development, of course, makes use of F = ma, some kinematic relationships between linear and angular acceleration and the definition of torque as a force at a distance from a center with the force being always at right angles to the line between the center and the point of application of the force. > > [The result is in this one page paper.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=8b0ascd.jpg&.src=ph) > > Having satisfied myself, if no one else, that a torque produces only rotation, I’ll apply that to a wheel on a conveyer belt in a later post. **David Simmons** , does the text say that the point P is fixed? Did you see my post about your energy analysis and that you missed torsional work? ---
**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 12, 2006, 4:54am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/442 "2006-02-12T04:54:14Z")
> [@treis](#): > > **David Simmons** , does the text say that the point P is fixed? > > Did you see my post about your energy analysis and that you missed torsional work? No. I don’t think the point P has to be held down. The mass m moves sideways and up to start with as the force rotates. For one revolution the force assumes all possible directions around P so there is no net lateral force on P. There are constantly changing lateral and vertical forces that add up to zero for one complete rotation around P. Remember, no one is saying that the mass doesn’t move laterally in response to a lateral force. I also moves diagonally, vertcally and in all other directions in response to the F = ma input. It is just that the **center of mass doesn’t move laterally.** In the case of the wheel, elements of the wheel are simultaneously moving laterally, vertically, diagonally and so on with the center remaining stationary. Yes I saw it. I didn’t leave anything out of my energy analysis. The total **input** work is the force on the belt and the distance the belt travels. The total **output** work is the rotation of the wheel. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [February 12, 2006, 5:11am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/443 "2006-02-12T05:11:01Z")
> [@epdnyc1](#): > > This assumes that moving the belt automatically causes the wheels to spin while the plane remains stationary. Right. That’s how I understood the “belt matches the wheel speed” version of the story. If we do not assume, as I was, that when the conveyer moves, the wheels move at a complementary speed, then what happens? It looks to me like the plane remains stationary. Does that sound right to you? But if the belt movement _does_ transfer movement to the wheels, then what I was saying is right: the situation can not exist as described because there is no determinate speed at which the wheels move, correct? Now in fact, whether the belt transfers motion to the wheels or not is a function of the friction of the wheel/belt surface and the wheel/axel surface. If these frictions are such that _any_ motion at all would be transfered to the wheel by the belt, then the impossible situation arises. And I think that in fact, at least a miniscule amount of motion _is_ so transferred in any situation like the one described. -FrL- ---
**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 12, 2006, 5:45am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/444 "2006-02-12T05:45:47Z")
OK. I think it has been established that circular motion will result if a force acts on a body and remains at right angles to a line from that body to a point at a distance from the point of application of the force. The point will be the natural center of the rotation and it doesn’t have to be bolted down. Now lets take a wheel and apply a force at some point so that the wheel moves in some fashion. [Here is the analysis of that case.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=13eascd.jpg&.src=ph) So what, you say? Now we will allow the line of action of the force to move down the wheel from the center to tangential. [Here is what happens in that case.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=57b7scd.jpg&.src=ph) Now I ask you, how can a force that is tangential to the wheel, as is the case with the wheel on a conveyer belt, exert a lateral force on the wheel when the lateral force is zero? A free body diagram would be acceptable in response. But wait. As a matter of fact in the actual situation there is a lateral force on the wheel. How can that be? Tune in tomorrow for the next chapter in this enchanting story. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 5:58am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/445 "2006-02-12T05:58:00Z")
> [@David Simmons](#): > > No. I don’t think the point P has to be held down. The mass m moves sideways and up to start with as the force rotates. For one revolution the force assumes all possible directions around P so there is no net lateral force on P. There are constantly changing lateral and vertical forces that add up to zero for one complete rotation around P. There is a centripetal force on the mass and by action and reaction there must be a force with the same magnitude on P. If that is the only force then there is an unbalanced force on point P and therefore there must be acceleration. > [@David Simmons](#): > > Remember, no one is saying that the mass doesn’t move laterally in response to a lateral force. I also moves diagonally, vertcally and in all other directions in response to the F = ma input. It is just that the **center of mass doesn’t move laterally.** In the case of the wheel, elements of the wheel are simultaneously moving laterally, vertically, diagonally and so on with the center remaining stationary. Right but the “a” in F=ma refers to the acceleration of the center of mass. I am having trouble coming up with a way to explain this but I am coming up blank. Lets try an example: I have a wheel of mass m and moment of inertia I on ice (frictionless surface) that I impart momentum to. That wheel is now sliding across the ice at some velocity v. Now, it moves onto a concrete surface with friction and begins rolling. There is some force from friction “F” from the surface on the wheel that acts at the base. You claim that this force does nothing to slow down the linear velocity of the wheel correct? Well, lets look at energy. I imparted some energy into the wheel in the form of kinetic energy and that has the value of 1/2_m_v[sup]2[/sup]. Now, when it moves onto the concrete surface it begins rotating. Since, as you claim, the linear velocity is constant I still have the kinetic energy from linear velocity. But now that its rotating I also have rotational energy with the value of 1/2_I_w[sup]2[/sup] where w=rotational velocity. How can it be possible to simply create energy out of thin air like this? > [@David Simmons](#): > > Yes I saw it. I didn’t leave anything out of my energy analysis. The total **input** work is the force on the belt and the distance the belt travels. The total **output** work is the rotation of the wheel. Like I said the total work input is not the force on the belt times the distance the belt travels. The total work input is the force on the belt times the distance the belt travels and the torque on the wheel times the angle of travel. Try you analysis again with other shapes like a sphere (I=2/5_m_r[sup]2[/sup]) or a hoop (I=m\*r[sup]2[/sup]). ---
**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 12, 2006, 6:11am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/446 "2006-02-12T06:11:53Z")
> [@treis](#): > > Like I said the total work input is not the force on the belt times the distance the belt travels. The total work input is the force on the belt times the distance the belt travels and the torque on the wheel times the angle of travel. The torque on the wheel arises from the force that is pulling the belt. Calling the torque an input is counting the force on the belt twice. I think the centrifual force would require that there be a rigid connection between the mass and the point but it doesn’t mean that the point moves. Here again, over one complete revolution the centrifugal force points in all directions and adds up to zero. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 6:25am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/447 "2006-02-12T06:25:15Z")
> [@David Simmons](#): > > The torque on the wheel arises from the force that is pulling the belt. Calling the torque an input is counting the force on the belt twice. Right and thats what you have to do. It takes more energy to accelerate a body linearlly and rotationally than it does just linearlly. Try your analysis again with the different shapes I provided. You will find that the rotational energy and the work done by the force are not equal. > [@David Simmons](#): > > I think the centrifual force would require that there be a rigid connection between the mass and the point but it doesn’t mean that the point moves. No rigid connection is necessary for objects in tension. To be clear here, we are talking about centripetal force not centrifugal force becuase the latter doesn’t exist. If I have a pole with a mass on a chain rotating around it do you believe that there is no force on a pole? > [@David Simmons](#): > > Here again, over one complete revolution the centrifugal force points in all directions and adds up to zero. Not exactly. Its true that through one complete revolution the forces add up but thats not a particularly relevent value. The relevent value for motion is force\*time. In other words, while F acts at a constant value around a revolution it acts for a longer time through the first quadrent of the motion than the last. This results in a corkscrew movement of the center of mass. Sort of, ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 6:38am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/448 "2006-02-12T06:38:19Z")
> [@treis](#): > > Try your analysis again with the different shapes I provided. You will find that the rotational energy and the work done by the force are not equal. In fact, for the hoop case you will find that the rotational energy is higher than the work if you do your analysis. ---
**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 12, 2006, 6:44am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/449 "2006-02-12T06:44:55Z")
> [@treis](#): > > Right and thats what you have to do. It takes more energy to accelerate a body linearlly and rotationally than it does just linearlly. But my whole point is that the object isn’t accelerated linearly. > [@](#): > > No rigid connection is necessary for objects in tension. You’re right. I misspoke when I used the term “rigid.” A string will do. > [@](#): > > Not exactly. Its true that through one complete revolution the forces add up but thats not a particularly relevent value. The relevent value for motion is force\*time. In other words, while F acts at a constant value around a revolution it acts for a longer time through the first quadrent of the motion than the last. This results in a corkscrew movement of the center of mass. > Sort of, This has a striking resemblence to arm waving. I’ve provided what I consider a valid analysis of the case we are talking about, a wheel on a conveyor belt. If you disagree, then you provide your own analysis of your model. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 6:58am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/450 "2006-02-12T06:58:32Z")
> [@David Simmons](#): > > But my whole point is that the object isn’t accelerated linearly. And the whole point is that you are wrong in saying this. > [@David Simmons](#): > > This has a striking resemblence to arm waving. Its not. If you aren’t convinced find Fx and Fy as a function of time and integrate from 0 to infinitiy to see what the final velocity is. > [@David Simmons](#): > > I’ve provided what I consider a valid analysis of the case we are talking about, a wheel on a conveyor belt. If you disagree, then you provide your own analysis of your model. Why? So you can just ignore it like every other argument? Right now I have about a dozen questions I have asked of people outstanding in this thread. In the last hour I have asked you two thought experiments: the wheel on ice to concrete and a mass rotating around a pole. On top of that I have pointed out that if you do your analysis with different shapes you get energies that don’t equal the work input, including one that results in a higher energy than the work inputted. None of these posts recieved any reply and were apparently just roundly ignored. Don’t sit here and tell me to provide my own analysis when you consistently ignore everything I say. ---
**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 12, 2006, 7:00am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/451 "2006-02-12T07:00:59Z")
On further thought I’m not at all sure that the string is necessary. I believe that if you introduce a constant speed hydrogen nucleus into a transverse magnetic field that it will travel in a circle with no strings attached. I know that in a cyclotron the particles travel in a spiral but that’s because they are accelerating. Now that’s interesting. If the belt is accelerating does that make a difference? It’s hard to see how it would. The input force is multiplied by the cosine of 90[sup]o[/sup] to get the force causing lateral motion. And that cosine is equal to zero when the input force is tangential. So unless the force is infinite, and maybe even then (who knows?), the lateral motion force on the wheel is zero. ---
**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 12, 2006, 7:13am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/452 "2006-02-12T07:13:05Z")
> [@treis](#): > > Why? So you can just ignore it like every other argument? Right now I have about a dozen questions I have asked of people outstanding in this thread. Why should I be chastized because other people didn’t answer? > [@](#): > > In the last hour I have asked you two thought experiments: the wheel on ice to concrete and a mass rotating around a pole. On top of that I have pointed out that if you do your analysis with different shapes you get energies that don’t equal the work input, including one that results in a higher energy than the work inputted. None of these posts recieved any reply and were apparently just roundly ignored. I can’t find any questions that you asked me about ice and wheels or concrete on a pole. And I simply don’t accept that with different shapes there are different energies based solely on your _ispe dixit._ My time is worth just as much as yours. > [@](#): > > Don’t sit here and tell me to provide my own analysis when you consistently ignore everything I say. Feeling a little cranky? ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [February 12, 2006, 7:24am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/453 "2006-02-12T07:24:32Z")
> [@David Simmons](#): > > Why should I be chastized because other people didn’t answer? Becuase you are one of those people? > [@David Simmons](#): > > I can’t find any questions that you asked me about ice and wheels or concrete on a pole. > [@treis](#): > > I have a wheel of mass m and moment of inertia I on ice (frictionless surface) that I impart momentum to. That wheel is now sliding across the ice at some velocity v. Now, it moves onto a concrete surface with friction and begins rolling. There is some force from friction “F” from the surface on the wheel that acts at the base. You claim that this force does nothing to slow down the linear velocity of the wheel correct? > > Well, lets look at energy. I imparted some energy into the wheel in the form of kinetic energy and that has the value of 1/2_m_v2. Now, when it moves onto the concrete surface it begins rotating. Since, as you claim, the linear velocity is constant I still have the kinetic energy from linear velocity. But now that its rotating I also have rotational energy with the value of 1/2_I_w2 where w=rotational velocity. How can it be possible to simply create energy out of thin air like this? > [@treis](#): > > If I have a pole with a mass on a chain rotating around it do you believe that there is no force on a pole? > [@David Simmons](#): > > And I simply don’t accept that with different shapes there are different energies based solely on your _ispe dixit._ My time is worth just as much as yours. Jeez Louise, I didn’t realize plugging in different values of I was such an ardous task. Wi is still M(pi\*r)[sup]2[/sup] Ww=1/2Iw[sup]2[/sup] For w=2pi: Sphere: I=2/5m\*r[sup]2[/sup] Wi=4/5M(pi\*r)[sup]2[/sup] Hoop: I=m\*r[sup]2[/sup] Wi=2M(pi\*r)[sup]2[/sup] If you note that in neither case Ww=Wi and in the case of the Hoop Wi\>Ww. In that case you must be getting energy from somewhere. > [@David Simmons](#): > > Feeling a little cranky? Yes, I have two people (although John seems to have bowed out) that are completely wrong telling me that I am wrong and ignoring everything I say. Its very frustrating. ---
**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 12, 2006, 8:16am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/454 "2006-02-12T08:16:40Z")
> [@treis](#): > > There is a centripetal force on the mass and by action and reaction there must be a force with the same magnitude on P. If that is the only force then there is an unbalanced force on point P and therefore there must be acceleration. I don’t agree with this. As I said in a post just up from this, a constant speed hydrogen ion introduced into a transverse magnetic field will travel in a circle about a fixed center. > [@](#): > > try an example: > > I have a wheel of mass m and moment of inertia I on ice (frictionless surface) that I impart momentum to. That wheel is now sliding across the ice at some velocity v. Now, it moves onto a concrete surface with friction and begins rolling. There is some force from friction “F” from the surface on the wheel that acts at the base. You claim that this force does nothing to slow down the linear velocity of the wheel correct? Aha, found the questions. I inadvertently skipped a page when I was looking for them. And no I don’t claim that. And I certainly make any claim about that case off the top of my head and without an analysis. > [@](#): > > Well, lets look at energy. I imparted some energy into the wheel in the form of kinetic energy and that has the value of 1/2_m_v[sup]2[/sup]. Now, when it moves onto the concrete surface it begins rotating. Since, as you claim, the linear velocity is constant I still have the kinetic energy from linear velocity. But now that its rotating I also have rotational energy with the value of 1/2_I_w[sup]2[/sup] where w=rotational velocity. How can it be possible to simply create energy out of thin air like this? Well, I don’t know that I claim that the linear velocity is constant and again, I wouldn’t make any claim on the spur of the moment. > [@](#): > > Try you analysis again with other shapes like a sphere (I=2/5_m_r[sup]2[/sup]) or a hoop (I=m\*r[sup]2[/sup]). Look, we are dealing with an airplane on a belt and I would like to stick to that. And I’ve got to say that your continued throwing up hypotheticals and telling me to solve them gives me the idea that you think that I am some kind of student who is to be given assignments to do. I made an analysis and I think that analysis should be directly addressed on its own merit if it is addressed at all. If it is wrong it is wrong, but I know the analysis done in developing the torque, angular acceleration formula is correct. And I know that there is no other force involved in the development besides the one on the mass and that one is all devoted to circular motion. Either try to refute my argument about that and the following post about the wheel or don’t bother to do it. In any case, don’t hand me any more homework assignments and direct that I do them. ---
**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 12, 2006, 10:04am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/455 "2006-02-12T10:04:58Z")
> [@treis](#): > > I have a wheel of mass m and moment of inertia I on ice (frictionless surface) that I impart momentum to. That wheel is now sliding across the ice at some velocity v. Now, it moves onto a concrete surface with friction and begins rolling. There is some force from friction “F” from the surface on the wheel that acts at the base. You claim that this force does nothing to slow down the linear velocity of the wheel correct? That’s right, the force doesn’t cause the wheel to slow down, but the wheel does slow down. Here’s my take on it. The two cases are not the same. The wheel on the belt has a continued input of energy from whatever is driving the belt. In your case you give the wheel and initial impulse which imparts a certain, fixed momentum to it. When the wheel hits the paving and starts to rotate because of the tangential frictional force and it builds up angular momentum. Since there is no other source this has to come from the original store of linear momentum. So the linear momentum decreases and that results in a decrease in linear velocity. ---
**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 12, 2006, 10:07am UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/456 "2006-02-12T10:07:20Z")
> [@David Simmons](#): > > That’s right, the force doesn’t cause the wheel to slow down, but the wheel does slow down. I guess it would be more accurate to say that the tangential force upon hitting the pavement doesn’t directly cause the wheel to lose linear velocity because of an opposing linear force. ---
**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 12, 2006, 2:00pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/457 "2006-02-12T14:00:35Z")
> [@John W. Kennedy](#): > > Note to moderators. I obviously cannot continue to debate when my position is misrepresented thus. I therefore withdraw myself from this thread, with this note only, that “treis’s” original remark, “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,” continues to be nonsense. This thread isn’t about scoring debate points. This thread is about discussing physical principles. Your total contribution to this thread so far has been to make fun of other people because _you_ don’t understand the physics. You’ve produced no analysis, answered no questions, and apparently haven’t even read half of the things that would explain what’s going on, despite repeated requests. You come out with little gems like: “Fine. 1 = 1 + 1. How could I have been so blind? And just think how this solves the energy crisis…” And yet, you have the gall to say that someone else is misrepresenting _your_ position. I suggest your statement, even if it were true, is more than a little hypocritical. To be honest, it’s a bit annoying to keep attempting to explain something and be ignored and made fun of. If you simply didn’t get it, and asked questions and produced apparent counter-arguments, then that would be perfectly fine. However, you’ve adopted an attitude of aggressive, in-your-face ignorance, and frankly that rubs me the wrong way. My suggestion is that, in the future, you put some effort into actually understanding what’s going on instead of making unsupported snarky comments and walking off in a huff. ---
**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 12, 2006, 4:46pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/458 "2006-02-12T16:46:18Z")
> [@David Simmons](#): > > The problem of a wheel on a conveyer belt isn’t so difficult that it should take up all of this time. I went back to my old, old, physics text to review on rotary motion. Having done that I reproduced the development of the relationship between torque and angular acceleration. The development, of course, makes use of F = ma, some kinematic relationships between linear and angular acceleration and the definition of torque as a force at a distance from a center with the force being always at right angles to the line between the center and the point of application of the force. > > [The result is in this one page paper.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=8b0ascd.jpg&.src=ph) > > Having satisfied myself, if no one else, that a torque produces only rotation, I’ll apply that to a wheel on a conveyer belt in a later post. The analysis you are presenting is of a mass connected to a string with a force always applied tangential to the motion of the mass, right? (Just restating your picture in words.) This analysis is used to develop an expression for the angular acceleration of the radius line (the string, in other words). The analysis is used to _introduce_ the concept of angular acceleration. It’s not _equivalent_ to the rotating wheel scenario for two reasons: 1. The force is always through _the center of mass_ in the analysis you presented. 2. The point P (the center of rotation) is _fixed_, and thus it’s prevented from moving anywhere. As a consequence, there is a tension force transmitted through the string to the mass, so there are _two_ forces applied to the mass: the force F and the string tension T. The string tension is, in fact, equal to m[symbol]w[/symbol][sup]2[/sup]r, where [symbol]w[/symbol] is the angular velocity of the string at that point. So this analysis boils down to the following free body diagram (after, say, 90 degrees of rotation, just to pick a number): ```auto ******* **** **** * * * * * * T<----* O * * * * * * * **** **** ******* ^ | | | | F ``` Both forces act through the center of mass of the mass m, acceleratiing the mass angularly (from force F) and radially (from force T). What’s more, _the tension T is transmitted back to the ground through the point P_. We could also look at a diagram of the entire system, instead of just the mass m. In that case, we have this free body diagram (also shown after mass m has travelled 90 degrees: ```auto P m T<-- X==========O ^ | | F ``` where the tension in the string is transmitted to the ground at point P. This is an important point, because if you’re using this analysis as an analogy to the wheel situation, the wheel center is _not_ fixed. Now, just to keep this in perspective, let me post a free body diagram of a wheel on a treadmill. The wheel has some weight W, opposed by a normal force N. (We could just ignore these forces, if you like, since they always cancel out) and a tangential force F: ```auto ******* **** **** * * * * * * * O * * * * W * * | * **|** **V** ******* <-------F ^ | | N ``` Now, the whole point is this: _The force F is unbalanced. It **must** produce an acceleration because **F=ma** _. That is essentially my argument: F=ma. ---
**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 12, 2006, 5:42pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/459 "2006-02-12T17:42:19Z")
> [@David Simmons](#): > > No. I don’t think the point P has to be held down. It most certainly does. Think backwards for a second: if the force P does not need to be held down, that would mean there is no force applied to it: no tension in the string, in other words, right? If there’s no tension in the string, you could cut the string, because it’s not doing anything, right? And now, if you cut the string, how does the mass “know” that it’s supposed to follow a circle of radius r? Why doesn’t it follow a circle of radius 2r? What it _would_ do, in fact, would be to travel in a straight line, with acceleration F=ma. Therefore, since having the string _does_ matter, there must be a tension in the string, and the tension must be transmitted to point P. If there’s a force on point P, it must move…unless you hold it down. > [@David Simmons](#): > > Now lets take a wheel and apply a force at some point so that the wheel moves in some fashion. [Here is the analysis of that case.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=13eascd.jpg&.src=ph) You can’t cancel forces e and d because, although they are equal and opposite, they are applied in different locations and form a couple. In the end, 90 degree case, that point doesn’t matter; however, in the extreme 90 degree case, you simply get a single tangential force which is nothing different from what we’ve talked about before, and F = ma. > [@treis](#): > > Like I said the total work input is not the force on the belt times the distance the belt travels. The total work input is the force on the belt times the distance the belt travels and the torque on the wheel times the angle of travel. Careful here. The total _energy_ input _is_ the force on the belt times the distance the _belt_ travels with respect to the ground–which is not in this case the same as the distance the _wheel_ travels, or the distance the belt travels with respect to the wheel. (This, of course, is then equal to the combined output kinetic energy in both rotation and translation.) > [@David Simmons](#): > > On further thought I’m not at all sure that the string is necessary. I believe that if you introduce a constant speed hydrogen nucleus into a transverse magnetic field that it will travel in a circle with no strings attached. Not a good analogy: a charged particle moving in a magnetic field produces a transverse force that substitutes for the tension in the string. Determine what would happen with an uncharged particle. This whole discussion _still_ boils down to F=ma. Whether it’s a particle or a pendulum or a wheel, if the object is changing velocity (_vector_ velocity) there _must_ be an unbalanced force responsible. If there’s an unbalanced force, there _must_ be a resultant acceleration. You are arguing that 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 12, 2006, 5:46pm UTC](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759/460 "2006-02-12T17:46:21Z")
> [@zut](#): > > The analysis you are presenting is of a mass connected to a string with a force always applied tangential to the motion of the mass, right? (Just restating your picture in words.) This analysis is used to develop an expression for the angular acceleration of the radius line (the string, in other words). > > The analysis is used to _introduce_ the concept of angular acceleration. It’s not _equivalent_ to the rotating wheel scenario for two reasons: 1. The force is always through _the center of mass_ in the analysis you presented. 2. The point P (the center of rotation) is _fixed_, and thus it’s prevented from moving anywhere. As a consequence, there is a tension force transmitted through the string to the mass, so there are _two_ forces applied to the mass: the force F and the string tension T. The string tension is, in fact, equal to m[symbol]w[/symbol][sup]2[/sup]r, where [symbol]w[/symbol] is the angular velocity of the string at that point. No. In [this post](http://boards.straightdope.com/sdmb/showpost.php?p=7094959&postcount=451) I back off my agreement that a string was necessary. The operation of the cyclotron proves this. In the cited post I said that the particles travel in a spiral. This isn’t correct. They travel in circles with each successive circle being slightly larger than the last because they are accelerated in passing from one half the the circulation chamber (D ring) to the other. They move in circles as the development in my post that you cited says because the force is always at right angles to their motion. The center of that circular motion is the center of gravity of the particle stream and\*\* that center does not move.\*\* The particles emerge from the machine and strike the target under investigation with great precision. > [@](#): > > So this analysis boils down to the following free body diagram (after, say, 90 degrees of rotation, just to pick a number): > > ```auto > > ******* > **** > **** > * * > * * > * * > T<----* O * > * * > * * > * * > **** > **** > ******* > ^ > | > | > | > | > F > > ``` > > Both forces act through the center of mass of the mass m, acceleratiing the mass angularly (from force F) and radially (from force T). What’s more, _the tension T is transmitted back to the ground through the point P_. > > We could also look at a diagram of the entire system, instead of just the mass m. In that case, we have this free body diagram (also shown after mass m has travelled 90 degrees: > > ```auto > > P m > T<-- X==========O > ^ > | > | > F > > ``` > > where the tension in the string is transmitted to the ground at point P. This is an important point, because if you’re using this analysis as an analogy to the wheel situation, the wheel center is _not_ fixed. > > Now, just to keep this in perspective, let me post a free body diagram of a wheel on a treadmill. The wheel has some weight W, opposed by a normal force N. (We could just ignore these forces, if you like, since they always cancel out) and a tangential force F: > > ```auto > > ******* > **** > **** > * * > * * > * * > * O * > * * > * W * > * | * > **|** > **V** > ******* <-------F > ^ > | > | > N > > ``` > > Now, the whole point is this: _The force F is unbalanced. It **must** produce an acceleration because **F=ma** _. That is essentially my argument: F=ma. Your free-body diagrams are of a static system. We are dealing with a dynamic system. I maintain that a force that is always tangent to a rotating wheel is equivalent to a rotating force that is always at right angles to the radius line. Your diagrams don’t convince me that this equivalence is untrue. If I can be convinced that the claim of equivalence is faulty then I might even join your team, who knows? But until then I’ll stand by the claim. And while I’m here I might as well go on with my statement in one post that the belt actually does exert a force on the wheel in the direction of motion of the belt. [This picture says all that I need to say.](http://pg.photos.yahoo.com/ph/dwsimm2000/detail?.dir=8cf7&.dnm=7426scd.jpg&.src=ph) I’ll be back later to sum up. I appreciate that you actually responded to my analysis. What I expect of a critique is not a demand that I answer a bunch of question about other postulated systems. I expect to hear about errors in my assumptions, my analysis or in my interpretation of results. That’s what you did and that permits a methodical answer. [Previous page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=22) [Next page](https://boards.straightdope.com/t/flight-and-the-conveyor-belt/342759.md?page=24)