Mike Smith Cosmos Posted May 20, 2015 Author Share Posted May 20, 2015 (edited) So far, the answer is clearly no.Simply trying to reverse a bunch of science and going back to claiming centrifugal force is real is contributing nothing.You may as well try to argue Earth stands still and the Sun orbits us.No ! I was wondering if a simple model of circular activity , say like a sphere , could in some way come to represent what happens with forces about circular motion . This is purely ' plucked out of the air ' , but it's that sort of thing . Perhaps comparing circular motion with what happens around black holes, galaxies, suns , worlds ,planets , asteroids and meteors . After all , these things nearly all have circulating objects , can we not make observations specifically hunting for evidence of centripetal- centrifugal action . I expect you will say, " it's already being done " . But people will not be inclined to look for things that science says " does not exist " ( eg centrifugal force ) . Whereas if we posited " it does exist! " and go looking for evidence of it ? May be it's tied in with all those distortions in space ? Mike (Now 10 pm sleep . Tomorrow at 5 am I set off to fly out of here for Italy. In contact again one week. May get a peep via a dongle type device . Reception not too good in the mountains . ) Edited May 20, 2015 by Mike Smith Cosmos Link to comment Share on other sites More sharing options...
pzkpfw Posted May 20, 2015 Share Posted May 20, 2015 Unicorns. Clearly you are looking for Unicorns. I give up. Link to comment Share on other sites More sharing options...
Endy0816 Posted May 20, 2015 Share Posted May 20, 2015 If you do the math you'll find that it does not exist. Therefore we do not need to look for it because it doesn't exist. There some things you feel which are not real. Your mass resists changes to velocity and direction. That is all that is going on here. If you have math that indicates otherwise then please present it. 1 Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Yes I do understand what you are saying, I just wonder if there is not another perspective which might just handle the circular motion - centrifugal force " mental , instinctive , conflict , non intuitive " , that 'gets at ' quite a few people , not just me. No, we aren't going to reinvent physics because you don't get it. We can't, because what we have works: it describes how nature behaves. On a side note Michael Kaku appeared on panorama a few days ago saying " gravity does not pull things down " but rather " space is pushing down " No. Please dear god, no more "side notes". Stick to the discussion for once. If the centrifugal force were real, there would be a real force outwards on something being kept moving in a circle. And if so, when that force is removed (e.g. that string being cut) that thing would move outwards, not at a tangent. (The only time you appear to see outwards movement is when you constrain that movement in your tube.) I second this, and it's been asked before and NEVER addressed. If there is a force, what is doing the pushing? Every other force in these problems can be identified. What is it that pushes out? Answer that before doing anything else. 1 Link to comment Share on other sites More sharing options...
Phi for All Posted May 21, 2015 Share Posted May 21, 2015 Answer that before doing anything else. ! Moderator Note Let's make that a staff request. Discussion without learning is unproductive, and is inconsistent with our purpose. 3 Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 21, 2015 Share Posted May 21, 2015 (edited) All that's required for a transverse component is a deviation from the radial line you draw. Ok, just to be very clear, will it be ahead or behind the radial line. Even I could accept a very slight slippage back from the radial line, for the coin has to accelerate at the larger radii and that could mean a slight difference in the coin's tangential speed and the speed of the merry go round. But this will be at the most very little. So what do you think the direction will be? If it has to slide 300mm to reach the edge how much transverse direction will there be? I fail to see how it could be much simpler than it is. (I am just baffled that this, and related threads, have gone on for many hundreds of posts, with you and Rob failing to grasp even the simplest concepts.) Just wait till I run another experiment this weekend. Weather permitting of course. ... I second this, and it's been asked before and NEVER addressed. If there is a force, what is doing the pushing? Every other force in these problems can be identified. What is it that pushes out? Answer that before doing anything else. Is it true there is only the need for a centrifugal force while there is a system that is rotationally accelerating the mass? As soon as it falls off the merry go round or the string breaks there is no centrifugal force. So you only get the outward force while there is a centripetal force. What pushes it out is the difference in the rotational inertia of the material being accelerated. Matter further out will have a higher rotational inertia per unit mass than parts closer in. (My terminology might be a little wrong, but it was like inertia but in a rotating frame so I guess the factor is rotational inertia. I'll attempt the math in the weekend.) Edited May 21, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
DrP Posted May 21, 2015 Share Posted May 21, 2015 QUOTE(Rob):- "As soon as it falls off the merry go round or the string breaks there is no centrifugal force"... There never was a centrifugal force in the first place, as we have been telling you for over 400 posts. lol. ...and I don't care if you can 'feel' the force of it through the rope... you can't... what you feel is the force that YOU YOURSELF are applying to the object through the rope to keep it's motion circular - this is the centripetal force. Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 21, 2015 Share Posted May 21, 2015 (edited) QUOTE(Rob):- "As soon as it falls off the merry go round or the string breaks there is no centrifugal force"... There never was a centrifugal force in the first place, as we have been telling you for over 400 posts. lol. ...and I don't care if you can 'feel' the force of it through the rope... you can't... what you feel is the force that YOU YOURSELF are applying to the object through the rope to keep it's motion circular - this is the centripetal force. That is not what is being said time and time again, but what I find is you would accept there is a centrifugal force if the object would fly off with a radial component. "If there is a force, what is doing the pushing?" You all want to see a push, but you don't get a push. But the laws of physics tell us there are pairs of forces and the force paired with centripetal force is the centrifugal force. If you let the rope slip what pulls the rope through your hands? Edited May 21, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
studiot Posted May 21, 2015 Share Posted May 21, 2015 Hello Robittybob, I notice you have skillfully avoided my fresh look (post425). And BTW if you let the rope slip you have released the tension, and therefore the centripetal force. You have also just agreed that if there is no centripetal force there is no centrifugal force. So whatever pulls the rope through your hands is not centrifugal force. Link to comment Share on other sites More sharing options...
DrP Posted May 21, 2015 Share Posted May 21, 2015 Nothing 'PULLS' the rope through your hand. If you let go, the weight carries on moving and so does the rope....so it carries on going and leaves you hand. It is not a force that is doing this, it is already in motion and carries on going in a straight line when you let go. Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Ok, just to be very clear, will it be ahead or behind the radial line. Even I could accept a very slight slippage back from the radial line, for the coin has to accelerate at the larger radii and that could mean a slight difference in the coin's tangential speed and the speed of the merry go round. But this will be at the most very little. So what do you think the direction will be? If it has to slide 300mm to reach the edge how much transverse direction will there be? There is no way to predict the amount with the information you have provided. The fact that you apparently did not realize this speaks volumes. That is not what is being said time and time again, but what I find is you would accept there is a centrifugal force if the object would fly off with a radial component. "If there is a force, what is doing the pushing?" You all want to see a push, but you don't get a push. But the laws of physics tell us there are pairs of forces and the force paired with centripetal force is the centrifugal force. If you let the rope slip what pulls the rope through your hands? Again, reaction force pairs act on DIFFERENT OBJECTS. We are not asking about the force on your hands, we are asking about the force on the swinging mass. Where does the alleged outward force ON THE MASS come from? It's not the rope, since the rope is pulling the mass in. What's left to exert a force? Link to comment Share on other sites More sharing options...
studiot Posted May 21, 2015 Share Posted May 21, 2015 Nothing 'PULLS' the rope through your hand. If you let go, the weight carries on moving and so does the rope....so it carries on going and leaves you hand. It is not a force that is doing this, it is already in motion and carries on going in a straight line when you let go. Can't quite agree with this. The rope changes its velocity as it trails after the weight. It doesn't move bodily sideways keeping as straight as it did when it was aligned taught along the radius. And it is unfair to claim that circular motion requires an acceleration force because there is a change of velocity direction, but not say the same for the rope after release. Link to comment Share on other sites More sharing options...
Strange Posted May 21, 2015 Share Posted May 21, 2015 That is not what is being said time and time again, but what I find is you would accept there is a centrifugal force if the object would fly off with a radial component. "If there is a force, what is doing the pushing?" You all want to see a push, but you don't get a push. But the laws of physics tell us there are pairs of forces and the force paired with centripetal force is the centrifugal force. If you let the rope slip what pulls the rope through your hands? With this level of ... [i don't know what it is: denialism?] ... is it about time to close this thread. Clearly some people are just not equipped to understand the world around them, no matter how clearly and simply it is explained. Another 400 posts are not going to change that. 4 Link to comment Share on other sites More sharing options...
DrP Posted May 21, 2015 Share Posted May 21, 2015 I don't understand your problem with that Studiot...the weight flies off at a tangent and the rope goes trailing off after it... the way it would do according to the laws of physics.. seems pretty obvious to me. The circular motion DOES require a force inward... the rope just goes off following the weight as it is dragged along. Either I am misunderstanding what you are saying or you are just wrong. Link to comment Share on other sites More sharing options...
studiot Posted May 21, 2015 Share Posted May 21, 2015 (edited) I don't understand your problem with that Studiot...the weight flies off at a tangent and the rope goes trailing off after it... the way it would do according to the laws of physics.. seems pretty obvious to me. It's a nit, I know. But for the rope to trail it's velocity must change, therefore it is accelerated, therefore it must be subject to force (=pull). Nothing 'PULLS' the rope through your hand. Edited May 21, 2015 by studiot Link to comment Share on other sites More sharing options...
DrP Posted May 21, 2015 Share Posted May 21, 2015 Only from the weight as it flies off - and this is drag from the weight, NOTHING to do with centrifugal forces. This is just semantics and kinda confuses the scenario imo. You could just hold the ball in your hand and spin then let go - the effect would be identical - the ball would 'appear' to be PULLED out of your hand... but it is not being pulled, it is just continuing on in a straight line from your stationary position. Link to comment Share on other sites More sharing options...
imatfaal Posted May 21, 2015 Share Posted May 21, 2015 Guys This thread is littered with failures to understands Newton's laws and I do not think we need anymore complications. Let's stick to ideal scenarios and keep interesting diversions and nitpicking out of the thread until the protagonists have got a grip on the basics. Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Only from the weight as it flies off - and this is drag from the weight, NOTHING to do with centrifugal forces. This is just semantics and kinda confuses the scenario imo. You could just hold the ball in your hand and spin then let go - the effect would be identical - the ball would 'appear' to be PULLED out of your hand... but it is not being pulled, it is just continuing on in a straight line from your stationary position. The ball is already pulling on your hand, because your hand is pulling on the ball, which is necessary to keep it moving in a circle. We literally need to keep our eye on the ball here, because that is what is moving in a circle. Forces acting on anything else are irrelevant to why the ball is moving in a circle. The forces exerted on Uncle Jim do not affect the ball. So, if you are talking about a force on anything else, you are off-topic. What people who think there is a real centrifugal force need to answer is what is providing that force to the ball. Nothing else, especially at this point. 2 Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 21, 2015 Share Posted May 21, 2015 (edited) The ball is already pulling on your hand, because your hand is pulling on the ball, which is necessary to keep it moving in a circle. We literally need to keep our eye on the ball here, because that is what is moving in a circle. Forces acting on anything else are irrelevant to why the ball is moving in a circle. The forces exerted on Uncle Jim do not affect the ball. So, if you are talking about a force on anything else, you are off-topic. What people who think there is a real centrifugal force need to answer is what is providing that force to the ball. Nothing else, especially at this point. Can an object's circumstances generate a force on itself? This is through the influence of the velocity of a mass and then with the application of a centripetal force (which makes the mass move in a circle). Once you have those three things , a mass, with velocity, held by a force orthogonal to that motion this produces fictitious "centrifugal force". I used the word "internal force" earlier but centrifugal forces aren't normally listed as an internal force. Edited May 21, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Can an object's circumstances generate a force on itself? No. If this were possible, momentum wouldn't be conserved. Newton's third law would also fail. Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 21, 2015 Share Posted May 21, 2015 Guys This thread is littered with failures to understands Newton's laws and I do not think we need anymore complications. Let's stick to ideal scenarios and keep interesting diversions and nitpicking out of the thread until the protagonists have got a grip on the basics. The movement along the radial line experiment - a lazy Susan - might have one at work. If we can sort this one out, then that will produce one less area of contention. No. If this were possible, momentum wouldn't be conserved. Newton's third law would also fail. Can you expand those ideas a bit for it sounds wrong. Because the two forces are present at the same time momentum is conserved (and besides momentum is always conserved). Describe your objections with a bit more flesh please. Google definition: Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. I say there is a pair of forces, so Newton's Third Law is not broken, so why do you say it was? Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Can you expand those ideas a bit for it sounds wrong. Because the two forces are present at the same time momentum is conserved (and besides momentum is always conserved). Describe your objections with a bit more flesh please. Google definition: I say there is a pair of forces, so Newton's Third Law is not broken, so why do you say it was? You didn't say there was a pair of forces. You said the particle exerted a force on itself. What would the reaction force be? Another force on itself? Momentum is NOT "always conserved". (A common conceptual error). Momentum is conserved when the net force on a system is zero. If an object could start spontaneously moving because it generated a force on itself, momentum would not be conserved. The object would be moving with one velocity, and then spontaneously have a different one, with no external reason. p1 ≠ p2 Momentum not conserved. (Momentum of an object traveling in a circle is not conserved; whatever is exerting the centripetal force is changing it) Link to comment Share on other sites More sharing options...
swansont Posted May 21, 2015 Share Posted May 21, 2015 Angular vs linear momentum discussion split http://www.scienceforums.net/topic/89079-angular-vs-linear-momentum/ We are focusing on one thing here: the actual source of the purported centrifugal force. Nothing else, at this point. Link to comment Share on other sites More sharing options...
Klaynos Posted May 22, 2015 Share Posted May 22, 2015 The movement along the radial line experiment - a lazy Susan - might have one at work. If we can sort this one out, then that will produce one less area of contention. The paper I gave earlier provides the maths to answer this. It uses rolling balls on a tossing table as that's a little easier than things like coins as the free ruling assumption makes a nice simplification. Link to comment Share on other sites More sharing options...
Robittybob1 Posted May 22, 2015 Share Posted May 22, 2015 The paper I gave earlier provides the maths to answer this. It uses rolling balls on a tossing table as that's a little easier than things like coins as the free ruling assumption makes a nice simplification. I'll look back for it, but every situation will be slightly different. Rolling balls are rather frictionless so won't have a significant friction force compared to sliding object on a rotating table. It is Friday night here so with a bit of luck we'll be able to test it within 24 hours. Link to comment Share on other sites More sharing options...
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