Acme Posted May 12, 2015 Share Posted May 12, 2015 . . Yes I figure out its possible, but not observed as everyday happening, because either the tube or the earths surface would need to be travelling at a circumferential speed of approx .17,700 mph . Which is not an easy , everyday occurrence . It is not possible to achieve that rotation rate for Earth in the first place and even were it possible, at that speed the entire planet would start flying apart and your tube & balls would be of little consequence. But in theory it is possible and with a bit of 'gigery pokery ' it could be arranged practically! Currently ,this is the sort of trajectory NASA uses to launch satellites. Except they use a, loose rocket ,rather than a vast tube ! Mike Gigery pokery? I guess that's one phrase for it, but not the one I'd choose. Fortunately science does not need to rely on your...erhm..., well, science does not have to rely on you. Link to comment Share on other sites More sharing options...
Janus Posted May 12, 2015 Share Posted May 12, 2015 . . Currently ,this is the sort of trajectory NASA uses to launch satellites. Except they use a, loose rocket ,rather than a vast tube ! Mike No, it isn't. NASA uses a gravity turn trajectory, which is nothing like that of an object traveling in a straight tube away from the Earth. By the time a rocket reaches LEO altitude it is moving at an orbital velocity of 7.74 km/sec. The end of a straight tube extending away from the Earth at that same altitude would only be moving at 0.486 km/sec Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 12, 2015 Author Share Posted May 12, 2015 (edited) Do we observe things ascending like that? Can you figure out why?.Maybe one day . In the distant future ! . It is not possible to achieve that rotation rate for Earth in the first place and even were it possible, at that speed the entire planet would start flying apart and your tube & balls would be of little consequence.Gigery pokery? I guess that's one phrase for it, but not the one I'd choose. Fortunately science does not need to rely on your...erhm..., well, science does not have to rely on you.. Ahh. Yes, but one of my ' blue sky ' ideas will 'come up trumps ' , one of these days ! There is a place for me ! No, it isn't. NASA uses a gravity turn trajectory, which is nothing like that of an object traveling in a straight tube away from the Earth. By the time a rocket reaches LEO altitude it is moving at an orbital velocity of 7.74 km/sec. The end of a straight tube extending away from the Earth at that same altitude would only be moving at 0.486 km/sec.. Thanks Janus , perhaps maths has a place , after all! Humm! . Could they not just pop out of the tube at low earth orbit height, say 500 miles up , turn left 90 degrees to a horizontal direction and accelerate up from 0.5 kilometers per second to 8 kilometers per second , and sail off into the sunset ? I suppose that could be a bit tricky ! ---------- ------------ Humm! Yes! Well I guess , " ,it's back to the drawing board " for me , as they say ! . " I knew there was a snag somewhere ! " . . Has anybody got a ' slide rule , I can borrow ' ? Mike Edited May 12, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
Strange Posted May 12, 2015 Share Posted May 12, 2015 (edited) Humm! . Could they not just pop out of the tube at low earth orbit height, say 500 miles up , turn left 90 degrees to a horizontal direction and accelerate up from 0.5 kilometers per second to 8 kilometers per second , and sail off into the sunset ? I suppose that could be a bit tricky ! That horizontal acceleration is where pretty much all the fuel goes. So using a little bit of fuel to get up to altitude is easy than building a massive glass tube. Even if that would work - which it wouldn't: get a cardboard tube, put a ping-pong ball at the bottom and place the tube vertically on the ground. Does the ping-pong ball rise? Edited May 12, 2015 by Strange Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 12, 2015 Author Share Posted May 12, 2015 (edited) That horizontal acceleration is where pretty much all the fuel goes. So using a little bit of fuel to get up to altitude is easy than building a massive glass tube. Even if that would work - which it wouldn't: get a cardboard tube, put a ping-pong ball at the bottom and place the tube vertically on the ground. Does the ping-pong ball rise?I thought they used tons upon tons of fuel, just to get the further tons and tons of fuel, off the ground. However I , understand they need to use a lot of fuel to accelerate ' down range ' to reach orbital velocity. As regards the tube , I was never intending the earths rotation to be a source of the circumferential velocity of 17,700 mph , required. The tube is really a thought experiment. However I think I will attempt the cardboard tube , with ping pong ball, but at a much less radius, and much less velocity so that the ( m x V squared / r = mg ) equation , produces the lift in the tube I am anticipating. If r is reduced to 0.5 m , The m's cancel so are irrelevant so V should be quite manageable for me to waggle it. ( by hand ) Mike Edited May 12, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 12, 2015 Share Posted May 12, 2015 (edited) Fine. Irrelevant. Or compressed. Yes. The scale pushes on the ball. Got that? The scale pushes on the ball. Inward. Yes, indeed. Feel free. I explored the maths of the situation today. When the scale at the end of the tube registers a force on it, the spring is compressed or stretched, either way, energy is added to the spring until the reactive centrifugal forces equals the centripetal force (the normal force provided by the scale. But the "push" does not push the ball back inward but allows the ball to travel in a circular path. The circular path it travels at is a larger radius that when the ball first contacted the scale. If the speed of rotation slowed the energy in the scale springs would push the ball back toward the center. I thought they used tons upon tons of fuel, just to get the further tons and tons of fuel, off the ground. However I , understand they need to use a lot of fuel to accelerate ' down range ' to reach orbital velocity. As regards the tube , I was never intending the earths rotation to be a source of the circumferential velocity of 17,700 mph , required. The tube is really a thought experiment. However I think I will attempt the cardboard tube , with ping pong ball, but at a much less radius, and much less velocity so that the ( m x V squared / r = mg ) equation , produces the lift in the tube I am anticipating. If r is reduced to 0.5 m , The m's cancel so are irrelevant so V should be quite manageable for me to waggle it. ( by hand ) image.jpg Mike You would need to keep the tube perfectly radial as you swing the tube. Edited May 12, 2015 by Robittybob1 -1 Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 12, 2015 Author Share Posted May 12, 2015 (edited) I explored the maths of the situation today. When the scale at the end of the tube registers a force on it, the spring is compressed or stretched, either way, energy is added to the spring until the reactive centrifugal forces equals the centripetal force (the normal force provided by the scale. But the "push" does not push the ball back inward but allows the ball to travel in a circular path. The circular path it travels at is a larger radius that when the ball first contacted the scale. If the speed of rotation slowed the energy in the scale springs would push the ball back toward the center. You would need to keep the tube perfectly radial as you swing the tube. I am off to the local carpet shop , to see if I can scrounge a Capet roll centre, though that could be a bit heavy and unwieldy . I just this moment tried it with a bit of rolled up card , and put a cardboard tube into the rolled up card . Then swung it back and forward in an arc . The slug of small roll , shot up the tube , then out the top but changed direction to the anticipated tangential trajectory at that point . I then realised this is partly what I see every day . Dog walkers have a ball thrower pole. It has a loose cup at the far end. They place the ball in it , give the device a radial swing , out pops the ball tangentially , but travels a vast distance. More than they could throw without the device. I know this is not the same , as there is no going up a tube . This is more a tangential device . But back to the tube , this is an orbital increasing device , I believe? Hey ! This is table top physics ' in the raw ' , I love it . Mike Edited May 12, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 12, 2015 Share Posted May 12, 2015 I am off to the local carpet shop , to see if I can scrounge a Capet roll centre, though that could be a bit heavy and unwieldy . I just this moment tried it with a bit of rolled up card , and put a cardboard tube into the rolled up card . Then swung it back and forward in an arc . The slug of small roll , shot up the tube , then out the top but changed direction to the anticipated tangential trajectory at that point . I then realised this is partly what I see every day . Dog walkers have a ball thrower pole. It has a loose cup at the far end. They place the ball in it , give the device a radial swing , out pops the ball tangentially , but travels a vast distance. More than they could throw without the device. I know this is not the same , as there is no going up a tube . This is more a tangential device . But back to the tube , this is an orbital increasing device , I believe? Hey ! This is table top physics ' in the raw ' , I love it . Mike A paper towel roll would be easier to handle. 1 Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2015 Share Posted May 12, 2015 I then realised this is partly what I see every day . Dog walkers have a ball thrower pole. It has a loose cup at the far end. They place the ball in it , give the device a radial swing , out pops the ball tangentially , but travels a vast distance. More than they could throw without the device. I know this is not the same , as there is no going up a tube . This is more a tangential device . But back to the tube , this is an orbital increasing device , I believe? It's all the same physics. . . Yes I figure out its possible, but not observed as everyday happening, because either the tube or the earths surface would need to be travelling at a circumferential speed of approx .17,700 mph . Which is not an easy , everyday occurrence . But in theory it is possible and with a bit of 'gigery pokery ' it could be arranged practically! Currently ,this is the sort of trajectory NASA uses to launch satellites. Except they use a, loose rocket ,rather than a vast tube If you moved it at the speed where gravity is equal to the centripetal fore at the surface, nothing would happen. The ball would be weightless and in orbit, but there would be no motion to move it higher in the tube. You could rotate it faster and then release it, and it would begin with the tangential velocity as we've discussed at length. Or crawl up the tube, in that scenario. But gravity would still be pulling it back to earth. You would need to rotate it at speed of the desired orbit. Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 12, 2015 Author Share Posted May 12, 2015 (edited) It's all the same physics. If you moved it at the speed where gravity is equal to the centripetal fore at the surface, nothing would happen. The ball would be weightless and in orbit, but there would be no motion to move it higher in the tube. You could rotate it faster and then release it, and it would begin with the tangential velocity as we've discussed at length. Or crawl up the tube, in that scenario. But gravity would still be pulling it back to earth. You would need to rotate it at speed of the desired orbit. . Yes! Well of course I did the sums on that ,over 10 years ago . It Is displacement 4 inches at 40 kHz . Bingo ! We have lift off ! Mike Ps . If it works ! Edited May 12, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 12, 2015 Share Posted May 12, 2015 (edited) . Yes! Well of course I did the sums on that ,over 10 years ago . It Is displacement 4 inches at 40 kHz . Bingo ! We have lift off ! Mike Ps . If it works !image.jpg How long was the tube? How did you get the tube to shake at 40 hertz? That is 80 stops and starts per second. I doubt if you could do that manually. If you moved it at the speed where gravity is equal to the centripetal fore at the surface, nothing would happen. The ball would be weightless and in orbit, but there would be no motion to move it higher in the tube. You could rotate it faster and then release it, and it would begin with the tangential velocity as we've discussed at length..... When the G force equals the centripetal force (Fcp). Fg = Fcp or Fg - Fcp = 0 A minus centripetal force has the same value as the reactive centrifugal force (Fcf) Fg + Fcf = 0 Edited May 12, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2015 Share Posted May 12, 2015 How long was the tube? How did you get the tube to shake at 40 hertz? That is 80 stops and starts per second. I doubt if you could do that manually. When the G force equals the centripetal force (Fcp). Fg = Fcp or Fg - Fcp = 0 A minus centripetal force has the same value as the reactive centrifugal force (Fcf) Fg + Fcf = 0 You never, ever combine forces acting on different objects in a force equation. It is meaningless. Action/reaction force pairs would ALWAYS add to zero, from Newton's third law, so there is no information added by writing that. You are literally writing down that 0 = 0 Link to comment Share on other sites More sharing options...
imatfaal Posted May 12, 2015 Share Posted May 12, 2015 ! Moderator Note .Yes! Well of course I did the sums on that ,over 10 years ago . It Is displacement 4 inches at 40 kHz .Bingo ! We have lift off !MikePs . If it works !image.jpg Mike The thread on this subject was closed - do not bring it up in this new thread. As always do not respond to this moderation within the thread Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 12, 2015 Author Share Posted May 12, 2015 (edited) How long was the tube? How did you get the tube to shake at 40 hertz? That is 80 stops and starts per second. I doubt if you could do that manually. When the G force equals the centripetal force (Fcp). Fg = Fcp or Fg - Fcp = 0 A minus centripetal force has the same value as the reactive centrifugal force (Fcf) Fg + Fcf = 0 I have been out today with a carpet tube and a lead pellet filled doll with gentle fit in the tube . I am not sure how to upload the video of it. It's quite good . The lead filled rat , can be seen coming out of the top of the tube . How do I upload this from my I pad ? video player. Here are the ingredient parts of the experiment . Cardboard carpet tube held vertically and swept in arc . Also lead pellet filled toy rat . It will just rest in the tube without sliding down . But is free to move up or down under changing conditions . ( see video ) these two are still pictures . Experiment performed outside , in the middle of a park . A passer by pressed the button . On the video on I player . Although unseen inside the tube , because it was already loaded half way up the tube. It had to have moved up during the rotation. The rat then exited the tube . And carried on forward , what looked like a tangent to the circle being prescribed by the tube. Mike Edited May 12, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2015 Share Posted May 12, 2015 Although unseen inside the tube , because it was already loaded half way up the tube. It had to have moved up during the rotation. The rat then exited the tube . And carried on forward , what looked like a tangent to the circle being prescribed by the tube. And what's the point here? Is there a question? We have covered all of the physics involved; there is no outward force. At any given instant the object is trying to move along the tangent, restrained by the tube, since there is no inward force to make it move in a circle. In doing so, r increases, so it moves down the tube. Link to comment Share on other sites More sharing options...
tar Posted May 13, 2015 Share Posted May 13, 2015 Mike, You, with a tube, have very little centripedal force pushing the lizard toward the center. There is nothing radially outward from the lizard, but the air in the tube. The only thing contacting the lizard is the inside of the tube, which at all times, as you swing the arc is pushing on the lizard in a direction normal to the radial direction. I don't think it should move out the tube. Perhaps the tube is flexing, allowing the stationary inertia of the lizard to cause it to exit the top of the tube, as would happen if the tube were made of cloth. The lizard would stay still, and it would travel toward the open end of the tube as you attempted to pull the other end around in an arc. Or. Perhaps you are not holding the tube exactly on a horizontal plane as you pivot as the center point, but are pointing the end down from horizontal a little, and the lizard is merely sliding down the tube. To check this, lower the center end or raise the outside end to where there is a 10 degree slope toward the center end. The lead lizard should slide toward the middle. However, the tube idea is interesting to think about. But if you had a perfectly horizontal tube, and you spun it on its center, a lizard placed in the middle of the tube would have no reason to come out one end more than the other. Once you place it half way out the radius, and swing the arc, you are causing the tube to push against the lizard in some manner where the forces on its tail are not the same as the forces on its head. The head is traveling faster than the tail, for one, being at an increased radius. I don't know how this translates to an outward motion, but I saw it with my clay balls and you saw it with the lizard, so there might be an explanation that does not require any outward force. In fact there has to be an explanation that does not conflict with the laws of physics. Regards, TAR If you are using the inertia of the ball and just manipulating the tube to cause the inertia of the ball to take the ball toward the open end of the tube, there still is no force pushing the ball radially outward. I am convinced that physics is correct, and you and I an RobittyBob1 are not talking about this right. We are thinking about the inertia of the lizard. We are not finding any outward force acting on the lizard. We can not be finding any such force, because the way forces are defined needs the force to cause a mass to accelerate in the direction the force is being applied in. Sorry SwansonT. We cross posted. Also sorry for ignoring the well defined, understood definition of force. I get it now. What motion happens because of the inertia that a mass has, has nothing to to with a force acting on the mass to accelerate (decellerate) that mass (including accelleration in the form of a change in direction of the mass.) Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 13, 2015 Author Share Posted May 13, 2015 (edited) And what's the point here? Is there a question? We have covered all of the physics involved; there is no outward force. At any given instant the object is trying to move along the tangent, restrained by the tube, since there is no inward force to make it move in a circle. In doing so, r increases, so it moves down the tube.Yes. To much of the physics has been covered ( not all ) . There is an "" apparent "" outward with respect to the 'frame of reference' of the tube . As the lizard went ..' up ' .. The tube and out of the ' top' , not ..' down ' the tube and out of the bottom . Yes it was observed with the video that it ' the lizard ' took a trajectory of the instantaneously set tangent at exit moment . Your comments above end with " it moves "down the tube " , . it did not , it moved " up the tube " , and out the top. So to the frame of ref of the tube , the lizard moved up , and out , increasing its distance vertically away from the end of the tube. True from the external observer( the person operating the video) . The lizard was assumed to move from the centre of the tube to the top( as it came out the top ) and can clearly be observed to follow a path away from the top of the tube. Were the tube to have been following an earth radius swing this would have shown an increases in radial distance from the centre of the earth,up as far as the exit from the tube . Instantaneous tangent only . But that particular tangent would put an increasing separating distance between the end of the tube and the flying lizard , again an apparent increasing radius . Question . Even though there may be all sorts of ' apparent this' and ' apparent that ' achieving orbits and increasing orbits , why can this not be termed a movement of upward motion along an increase in radial distance ( even though not a conventional straight line force as such , if we accept that it is a fictitious force away from the centre centrifugal ) ? Mike Mike, You, with a tube, However, the tube idea is interesting to think about. Regards, TAR If you are using the inertia of the ball and just manipulating the tube to cause the inertia of the ball to take the ball toward the open end of the tube, there still is no force pushing the ball radially outward. I am convinced that physics is correct, and you and I an RobittyBob1 are not talking about this right. We are thinking about the inertia of the lizard. We are not finding any outward force acting on the lizard. We can not be finding any such force, because the way forces are defined needs the force to cause a mass to accelerate in the direction the force is being applied Tar, I think you have hit the nail on the head . All this is about is that it is the inertia that is doing all the pushing and pulling the effective radially sideways push or pull of inertia is causing the movement up the tube , which is away from the centre, so therefor linguistically centre- fugal , but in physics terms as you say is not a force radially ,in the way we normally think. It is the inertia of the mass , or lizard that slides it up the constraining tube. This does however put the ' spotlight ' on the interface between linear motion and circular motion , which I must say ,I suspected and raised a question on . Perhaps I did not phrase it correctly ? However the consequence of this inertia does provide a mechanism for a devise gaining radial height, as indeed my lizard did. And would I believe be useful ! I would guess this interface ( linear to circular ) kicks in because of the different angles of preferred motion. One circular , the other tangential . I can see a component * of the tangential going into conflict with the radial ,once the circular path 'parts company ' with the immediate tangential path . Or at least tries to 'part company ' that's where the tension lies , and that's where the inertia can win and drag the mass skyward. ( effectively ) ( upwards radially , if they have any shared interest still ) . While totally constrained in the circular orbit . The physics of circular motion dominate , balance , and are explained . When slippage occurs with inertia 'winning' and the radius is being increased by the dominant pull of the preferred tangential straight line , moment by moment ( as MORDRED . Says ) inertia . Then the laws of straight line operation prevail . Or possibly a compromise of the two ? * component . When a line or force is considered as an amalgam of two right angular components , rather than a single direction ( abSine x, and abCos x as opposed to just line ab, where x is the angle between the line and the horizontal ) Mike Edited May 13, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
swansont Posted May 13, 2015 Share Posted May 13, 2015 Yes. To much of the physics has been covered ( not all ) . There is an "" apparent "" outward with respect to the 'frame of reference' of the tube . As the lizard went ..' up ' .. The tube and out of the ' top' , not ..' down ' the tube and out of the bottom . Yes it was observed with the video that it ' the lizard ' took a trajectory of the instantaneously set tangent at exit moment . Your comments above end with " it moves "down the tube " , . it did not , it moved " up the tube " , and out the top. Up and down are relative to a chosen coordinate system. I said r increases. Was that not clear enough? (In fact, that should be more clear than up and down, since the tube can be pointed in any direction at an arbitrary moment in time. You can launch the lizard up, down, or across, depending on how you do it.) So to the frame of ref of the tube , the lizard moved up , and out , increasing its distance vertically away from the end of the tube. True from the external observer( the person operating the video) . The lizard was assumed to move from the centre of the tube to the top( as it came out the top ) and can clearly be observed to follow a path away from the top of the tube. Were the tube to have been following an earth radius swing this would have shown an increases in radial distance from the centre of the earth,up as far as the exit from the tube . Instantaneous tangent only . But that particular tangent would put an increasing separating distance between the end of the tube and the flying lizard , again an apparent increasing radius . Again, this is only covering ground we have covered several times before. Question . Even though there may be all sorts of ' apparent this' and ' apparent that ' achieving orbits and increasing orbits , why can this not be termed a movement of upward motion along an increase in radial distance ( even though not a conventional straight line force as such , if we accept that it is a fictitious force away from the centre centrifugal ) ? Because there is no force. Force has a specific definition in physics. Motion does not require a force to be present, and if there is one, the direction of motion does not have to be the same as the direction of a force. Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 13, 2015 Share Posted May 13, 2015 (edited) Mike, You, with a tube, have very little centripedal force pushing the lizard toward the center. There is nothing radially outward from the lizard, but the air in the tube. The only thing contacting the lizard is the inside of the tube, which at all times, as you swing the arc is pushing on the lizard in a direction normal to the radial direction. I don't think it should move out the tube. Perhaps the tube is flexing, allowing the stationary inertia of the lizard to cause it to exit the top of the tube, as would happen if the tube were made of cloth. The lizard would stay still, and it would travel toward the open end of the tube as you attempted to pull the other end around in an arc. Or. Perhaps you are not holding the tube exactly on a horizontal plane as you pivot as the center point, but are pointing the end down from horizontal a little, and the lizard is merely sliding down the tube. To check this, lower the center end or raise the outside end to where there is a 10 degree slope toward the center end. The lead lizard should slide toward the middle. However, the tube idea is interesting to think about. But if you had a perfectly horizontal tube, and you spun it on its center, a lizard placed in the middle of the tube would have no reason to come out one end more than the other. Once you place it half way out the radius, and swing the arc, you are causing the tube to push against the lizard in some manner where the forces on its tail are not the same as the forces on its head. The head is traveling faster than the tail, for one, being at an increased radius. I don't know how this translates to an outward motion, but I saw it with my clay balls and you saw it with the lizard, so there might be an explanation that does not require any outward force. In fact there has to be an explanation that does not conflict with the laws of physics. Regards, TAR If you are using the inertia of the ball and just manipulating the tube to cause the inertia of the ball to take the ball toward the open end of the tube, there still is no force pushing the ball radially outward. I am convinced that physics is correct, and you and I an RobittyBob1 are not talking about this right. We are thinking about the inertia of the lizard. We are not finding any outward force acting on the lizard. We can not be finding any such force, because the way forces are defined needs the force to cause a mass to accelerate in the direction the force is being applied in. .... Sounds like it is time to do an actual experiment. Hypothesis to be tested: The scale is providing a centripetal force on the ball, so the ball must be applying a centrifugal force on the scale. Setup proposed horizontal Rotating tube with a bathroom scale attached. (The reason to do it horizontally is so gravity would have any net effect on the result.) Rotate it at 10 Hz. See what the scale reads Put soccer ball down tube and re-spin the tube at the same rate, take another measurement. ... if there is one, the direction of motion does not have to be the same as the direction of a force. Example please. Edited May 13, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
imatfaal Posted May 13, 2015 Share Posted May 13, 2015 .... Example please. The trajectory of any freely moving thrown object after launch. Once the initial acceleration has finished the forces on the object are - simplistically - air resistance which is directly opposed to the direction of travel and gravity which is directed downwards. At no point in the flight will the forces align with the direction of motion. Link to comment Share on other sites More sharing options...
swansont Posted May 13, 2015 Share Posted May 13, 2015 Example please. Ever used brakes in a car? Or the very example we are discussing, centripetal acceleration. Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 13, 2015 Share Posted May 13, 2015 The trajectory of any freely moving thrown object after launch. Once the initial acceleration has finished the forces on the object are - simplistically - air resistance which is directly opposed to the direction of travel and gravity which is directed downwards. At no point in the flight will the forces align with the direction of motion. That would be due to the addition of force vectors surely? Ever used brakes in a car? Or the very example we are discussing, centripetal acceleration. I thought you might say that! Link to comment Share on other sites More sharing options...
swansont Posted May 13, 2015 Share Posted May 13, 2015 That would be due to the addition of force vectors surely? Individually, or the sum. Works in a vacuum, where there would only be one force. Link to comment Share on other sites More sharing options...
Mike Smith Cosmos Posted May 13, 2015 Author Share Posted May 13, 2015 (edited) I have just come back from a painting session ,in the countryside, traveling on my motor bike . I used the time as a further practical experiment , finding the direction of forces involved with circular motion and straight line inertia. It was a sunny day , with fine dry conditions , the open road,, all is right with the world. I am still very very conscious of the feeling of inertia , when traveling at moderate speed. That bike just wants to go straight. You can play with her, try steering the handlebars by half an inch , into a circular bend ,without leaning , NO WAY. Utterly set in a straight line momentum. O.k. The only way to turn a circular corner is to lean. Or at least lean the bike . What exactly is going on at this === straight line to curve interaction. Wow! In this instance , where is the force coming from ? Is there some pressure coming in from outside the circle ? Something is definitely there , like a wall! I am in a circle , on a bike that wants to go straight forward , in a straight line. But there is no escape, I am trapped in a circle . At every inch of the way around this circle there is an invisible line that's saying ,in effect , do not cross, as the road goes round and round. If I did nothing this machine and me would grind our way round the road barriers in a complete circle. We would scrape the barrier, my legs would get squashed by the pressure of the bike on my leg trapped against the barrier. On and on , if the road went there ,my bike and I are trapped. Trapped pushing curve of a circle after curve of a circle against the barrier. Wait a minute this straight line inertia is causing a pressure to emanate from within the circle , without let up , through the bike through my leg and into the ironwork of a circular barrier of metal . This force emanates outwards from a centre of a circle . This outward going force no doubt induces an equal and opposite reaction force , inward toward the centre of the circle . What is the name of these forces? There is no way the force is the initiative of an inward force. At this stage the steel rails have no contact with the bike or rider. ( hopefully ) The preferred, attempted , straight line inertia, of the motorbike / me combination is the origin of the initialising outgoing force ( centrifugal ) caused by us nearing and nearly touching the ring of the circle . Only if the person ,me , driving the motor bike , decides I can not allow my legs and bike to touch the steel barriers of this curve , but rather allow the pressure to go down the bike FrameMaker, down the wheel spokes , through the rubber tyre , to apply the pressure caused by this inertia ( must) go through the tyres to the road rather than my legs. The pressure is applied radially outwards ( Centrifugally ) through the tyre- road junction away or near the curve of the circle. This then invokes an equal and opposite reaction as an inward ( centripetal ) force , maybe the centripetal force is fictitious ? ( in this case) . I believe this to be demonstrated by the toy lizard going up the tube by centrifugal force , invoked by the inertia of the lizard in the tube pushing outward as a force and slipping against the cardboard of the tube . Probably met by a reactive centripetal force in the circulating cardboard tube . Mike Mike Edited May 13, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
swansont Posted May 13, 2015 Share Posted May 13, 2015 When you lean in on the bike there is an inward component to the force on the bike, exerted by the road. If you are upright and try to turn, your inertia carrying you forward would cause your center of gravity to be outside of the bike, and you tip over. Still no outward force on you, though (repeating yourself will not make you right all of the sudden). Just your tendency to go in a straight line, and the requirement that there be an inward force to move in a circle. Link to comment Share on other sites More sharing options...
Recommended Posts