Mike Smith Cosmos Posted March 7, 2015 Author Posted March 7, 2015 (edited) Mike 1. Please try and reply without needless pictures - if you have a freebody diagram please post - but endless hand-drawn sketches make the thread almost unreadable. 2. "No net force" means the following: because F=ma we get 0=ma, m cannot equal zero so therefore a must equal zero. If a equals zero then the object continues in a straight line ie v is constant. This is pretty much the first lesson in a mechanics course 3. A change in direction is a change in v - ie a non-zero a , and thus a non-zero net F. The above is pretty much axiomatic to most mechanics problems. Please confirm you are not arguing with it. Only then can we continue to explain where you are wrong about gravity 1. I am not totally certain what you are saying I can or cannot do . I personally find it almost impossible to even talk to myself ,let alone any other engineer , without a diagram . Say like my sketch of the proposed gravitational centripetal and centrifugal reactive forces. I was brought up on graphs . If you mean my drawings of engineers setting up and flying up in the air . That was intended as a ' lighten up the atmosphere of the subject, serious as it may be. Well I think it is . So which diagrams or pictures is it you are saying endless about. 2. I would assume that the v used in mvsquared /r refers to the instantaneous speed . This v surely contributes to the momentum as in angular momentum . That is why surely the motion in a circle without friction is surely infinite. The angular momentum is conserved. Like energy it cannot be destroyed only changed. Electrons buzzing about ,however they do it , conserve their angular momentum or spin , I believe. 3. I am not sure ,I can get my head round your point here . Again going back to electrons , that's why the masters of old were worried electrons would radiate their energy away and decay . There is some perfect elasticity afoot with electrons ,that appear to allow them to be,buzz,and move about ,where only their fields interact in perfect elasticity , energy is conserved perfectly , angular momentum is conserved perfectly . Radiation is not bleeding away energy ,inefficiently . I appreciate electrons are not the machine we are discussing here , but I have to use natural phenomenon as models as observations and proof that such apparently orbital systems are possible as I am discussing. It is part of the proof , part of the model. I am happy to learn , very much so, but you sound like you have a forgone conclusion that I am wrong in my proposal? Maybe I am . But when I first proposed it to the Science Research Council ( 1970's 1980's) they referred me on to various professors. In the course of developing this project I contacted Surrey University , when they were doing orbital trials with satellites , asking what accelerometers they used to check and measure in orbit . They said they ' don't' because the net force is so small practically zero , no accelerometer is sensitive enough. So their measurement of the balancing of forces in space was practically Zero. So why can my device ,all-be-it very near earth be designed to achieve practically zero by whatever method is used by satellites and electrons? Mike Edited March 7, 2015 by Mike Smith Cosmos
swansont Posted March 7, 2015 Posted March 7, 2015 .But hang about , you have introduced another equation , ( I am not saying you should not introduce another equation) , namely the force between two masses , namely a single mass m , and the earth mass M , As well as G the universal gravitational constant , as opposed to g the acceleration due to gravity. This is just stating there is a force of attraction between two masses. Where do you think the g in mg comes from? Saying mg=mv2/r is the same thing — the gravitational force is the centripetal force, which is BY DEFINITION, the net force on the object moving in a circle. There can be no other force. But by dint of the fact we are thundering off down range at 17,700 mph , all be it , for 1/4 of 1/40,000 of a second we invoke the other equation of motion that mg= mvsquared/r , then the centripetal force will invoke the reactive centrifugal force , thus neutralising the attraction between the earth and our little mass. m ( or at least one of them ) . The other one linked to it loosely is hurtling down range in the opposite direction similarly at 17,000 mph and invokes the exact same response ( but in the opposite direction of travel ) . If this were not the case, we or NASA would not have to send rockets down range at 17,700 mph to attain orbital velocity! "Invoke"? Is this magic? Action and reaction force pairs of Newton's third law do not act on the same object. You can't have the centripetal force and the reactive centrifugal force acting on the same object. That's elementary physics, and a common mistake (but not a mistake someone who taught the subject should be making) I appreciate electrons are not the machine we are discussing here , but I have to use natural phenomenon as models as observations and proof that such apparently orbital systems are possible as I am discussing. It is part of the proof , part of the model. Electrons do not have planetary orbits, so you can't compare the situations. They follow QM, so their behavior can't be part of a model that obeys classical physics. It doesn't scale up. I am happy to learn , very much so, but you sound like you have a forgone conclusion that I am wrong in my proposal? Maybe I am . You are. You have made some basic errors of physics that make this a complete fantasy. Ignoring those errors does not make them go away. (It does, however, constitute soapboxing, so stop ignoring criticism of your mistakes) 1
Mike Smith Cosmos Posted March 7, 2015 Author Posted March 7, 2015 (edited) Where do you think the g in mg comes from? Saying mg=mv2/r is the same thing the gravitational force is the centripetal force, which is BY DEFINITION, the net force on the object moving in a circle. There can be no other force. "Invoke"? Is this magic? Action and reaction force pairs of Newton's third law do not act on the same object. You can't have the centripetal force and the reactive centrifugal force acting on the same object. That's elementary physics, and a common mistake (but not a mistake someone who taught the subject should be making) Electrons do not have planetary orbits, so you can't compare the situations. They follow QM, so their behavior can't be part of a model that obeys classical physics. It doesn't scale up.You are. You have made some basic errors of physics that make this a complete fantasy. Ignoring those errors does not make them go away. (It does, however, constitute soapboxing, so stop ignoring criticism of your mistakes) Ok. I do not wish to be construed as "soap boxing " as that certainly is not my intension. And I have spent too much time on this project for it to be dismissed out of hand. What would be nice is for someone to say ! Jolly good Mike , you are a nice chap. I think you might be on to something there . Keep going . You can make it ! But be prepared to be wrong. That last phrase ,sends shudders through my very being . However I do have doubts ,sometimes in the middle of the night . Then I get fortified as I do observe and learn of such wonders both in the atom , the world , my experiences , the earth moon and stars , now galaxies , super large structures and beyond. As a Famous American once said " why shall we go to the Moon" " because it's there ! " So " why should we explore ways to overcome the confinement of Gravity ". " because we can " maybe not me , but somebody will ! " because the solution is there , somewhere " Mike Ps. However , not to sink too far down , beyond the water level . I will reflect on what ,you all are saying , and see what I can come up with. I do understand most of what you chaps are saying , BUT I still think there is something there . I wish someone else would say " you are on to something there Mike , keep going ! Mike Edited March 7, 2015 by Mike Smith Cosmos
swansont Posted March 7, 2015 Posted March 7, 2015 What would be nice is for someone to say ! Jolly good Mike , you are a nice chap. I think you might be on to something there . Keep going . You can make it ! But be prepared to be wrong. That last phrase ,sends shudders through my very being . However I do have doubts ,sometimes in the middle of the night . Then I get fortified as I do observe and learn of such wonders both in the atom , the world , my experiences , the earth moon and stars , now galaxies , super large structures and beyond. Jolly good, Mike. Nice effort. But you're wrong. This isn't a self-esteem workshop, it's science, and being nice or trying hard don't count for anything when judging the merit of the idea. I can't tell you you may be on to something. You've gotten some basic physics concepts wrong. The world doesn't work the way you are proposing. What really turns heads is experimental evidence. Come back with an experiment showing levitation based on oscillation. No more repeating how it will work — we've gone over that ground too many times already. 1
Mike Smith Cosmos Posted March 7, 2015 Author Posted March 7, 2015 Jolly good, Mike. Nice effort. But you're wrong. This isn't a self-esteem workshop, it's science, and being nice or trying hard don't count for anything when judging the merit of the idea. I can't tell you you may be on to something. You've gotten some basic physics concepts wrong. The world doesn't work the way you are proposing. What really turns heads is experimental evidence. Come back with an experiment showing levitation based on oscillation. No more repeating how it will work — we've gone over that ground too many times already. Ok. I will return with :- Experiments . Results Observations Evidence Watch this space . Mike
imatfaal Posted March 8, 2015 Posted March 8, 2015 1. I am not totally certain what you are saying I can or cannot do . I personally find it almost impossible to even talk to myself ,let alone any other engineer , without a diagram . Say like my sketch of the proposed gravitational centripetal and centrifugal reactive forces. I was brought up on graphs . If you mean my drawings of engineers setting up and flying up in the air . That was intended as a ' lighten up the atmosphere of the subject, serious as it may be. Well I think it is . So which diagrams or pictures is it you are saying endless about. But your pictures are NOT diagrams - they are sketches with little people drawn in. Freebody diagrams are great - an aid to understanding; sketches, photographs, your artwork is a distraction. 2. I would assume that the v used in mvsquared /r refers to the instantaneous speed . This v surely contributes to the momentum as in angular momentum . That is why surely the motion in a circle without friction is surely infinite. The angular momentum is conserved. Like energy it cannot be destroyed only changed. Electrons buzzing about ,however they do it , conserve their angular momentum or spin , I believe. What are you reading? I made a simple point about the balancing - or not - of forces. I asked you why you were ignoring Newton's First law that in the absence of external forces an object will continue at a constant velocity; you fail to even address the question and manage to end up on conservation of spin number!! Please answer this - if there is "no net force" as you have asserted - why is there a change in velocity ie an acceleration? 3. I am not sure ,I can get my head round your point here . Again going back to electrons , that's why the masters of old were worried electrons would radiate their energy away and decay . There is some perfect elasticity afoot with electrons ,that appear to allow them to be,buzz,and move about ,where only their fields interact in perfect elasticity , energy is conserved perfectly , angular momentum is conserved perfectly . Radiation is not bleeding away energy ,inefficiently . I appreciate electrons are not the machine we are discussing here , but I have to use natural phenomenon as models as observations and proof that such apparently orbital systems are possible as I am discussing. It is part of the proof , part of the model. WTF? You can't get your head around newtons second law and yet you presume to to talk about interaction of elections. "A change in direction is a change in v - ie a non-zero a , and thus a non-zero net F." Your test object is moving in an orbit - that means v (which I have bolded - it should be obvious why) is constantly changing. The change in velocity v is known as the acceleration a, according to N2L any a requires a Force F. You have consistently stated that there is "no net force" - so why is the object not going in a straight line? ...I am happy to learn , very much so, but you sound like you have a forgone conclusion that I am wrong in my proposal? Maybe I am . But when I first proposed it to the Science Research Council ( 1970's 1980's) they referred me on to various professors. In the course of developing this project I contacted Surrey University , when they were doing orbital trials with satellites , asking what accelerometers they used to check and measure in orbit . They said they ' don't' because the net force is so small practically zero , no accelerometer is sensitive enough. So their measurement of the balancing of forces in space was practically Zero. So why can my device ,all-be-it very near earth be designed to achieve practically zero by whatever method is used by satellites and electrons? Mike Accelerometers could not and would not be expected to measure any force during freefall - they measure proper acceleration (which is acceleration relative to freefall). Thus in freefall they measure zero by design and necessity. If the guys at Surrey were trying to measure the force of gravity using an accelerometer in freefall either they were working on something completely out of my ken, or they were actually trying to measure tidal or some other forces, or you misunderstood them. Yes - I am pretty convinced it won't work. a. It contradicts basic newtonian mechanics - which if you drew some decent freebody diagrams you would realise b. more importantly - It contradicts observation. c. most important - it contradicts Homer Simpson and others in that it would be a renewable source of useful energy.
studiot Posted March 8, 2015 Posted March 8, 2015 Mike here is my concern regarding your vibrating doodads. Newtonian mechanics requires a force acting towards the centre of curvature on any body following a curved path. (If you like it pulls the trajectory in towards the centre off the straight line path) Now you maintain that, as a result of your vibrational dance there is a force which counteracts gravity. So what force is acting to create the partial arcs you describe?
Mike Smith Cosmos Posted March 8, 2015 Author Posted March 8, 2015 (edited) .i. But your pictures are NOT diagrams - they are sketches with little people drawn in. Freebody diagrams are great - an aid to understanding; sketches, photographs, your artwork is a distraction. .M. As required by the science speculations forum rules, I am required to answer questioning , etc so I will do best to succinctly reply. These diagrams I am very familiar with. This quote from Wikipedia Quote " A free body diagram, sometimes called a force diagram,[1] is a pictorial device, often a rough working sketch, used by engineers and physicists to analyze the forces and moments acting on a body. The body itself may consist of multiple components, an automobile for example, or just a part of a component, a short section of a beam for example, anything in fact that may be considered to act as a single body, if only briefly. A whole series of such diagrams may be necessary to analyze forces in a complex problem. The free body in a free body diagram is not free of constraints, it is just that the constraints have been replaced by arrows representing the forces and moments they generate. " unquote I have been using these since 'A' level pure maths 'A' level applied maths , 'A' level Physics , plus 7 years at university doing electrical & electronic Engineering to Hons Degree Standard ( times 2 , twice separated by 30 years , moved from valves, through integrated circuits to software ) Worked within industry as an engineer , consultant , manufacturer, to name a few years of using engineering style diagrams. I probably have got a bit rusty and tired and forgetful in my retirement , but the overall principles remain. I think possibly , this subject of Centrifugal Force , has been controversial , so maybe this is colouring this discussion . Or some other reason ? I am very happy with these ' Freebody diagrams ' .I. What are you reading? I made a simple point about the balancing - or not - of forces. I asked you why you were ignoring Newton's First law that in the absence of external forces an object will continue at a constant velocity; you fail to even address the question and manage to end up on conservation of spin number!! Please answer this - if there is "no net force" as you have asserted - why is there a change in velocity ie an acceleration? .M. I have been using Newtons first law as an argument for the inertia that these masses have , by attempting to move in a strait line , if not acted on by an external force . Then when such a strait moving mass ,is , acted on by a force at right angles , the whole centripetal force ( with associated acceleration , from F= ma ) results in a responsive centrifugal force ( with associated acceleration from F=ma ) . Neither of these results in any movement along the radius ( in a restrained situation ) . If I understand it correctly ( which I am quite willing to accept that I am not understanding it correctly ) . That these accelerations are experienced as forces rather than increases in velocity that we are more familiar with , say with strait line motion. I am sorry that you are troubled with my dropping into electron orbits , but I must say some of the older text books explained electron orbitals as derived from electrostatic attraction electron to proton and centrifugal force pulling outwards. I appreciate as I too was educated in energy Bands , but the history of science does include the intermediate understanding. .I. WTF? You can't get your head around newtons second law and yet you presume to to talk about interaction of elections. (As above ) "A change in direction is a change in v - ie a non-zero a , and thus a non-zero net F." Your test object is moving in an orbit - that means v (which I have bolded - it should be obvious why) is constantly changing. The change in velocity v is known as the acceleration a, according to N2L any a requires a Force F. You have consistently stated that there is "no net force" - so why is the object not going in a straight line? . .M . " no net force RADIALLY , towards the centre. .M. Remember I was an electronic engineer by Degree and occupation . This bit , you clearly have a model in your head mechanically,. I must admit I find it hard to get my head around quite what is going on around this bit. Electrically and electronically , I can , but mechanically not quite so easily . As you say those diagrams , come in to their own there. I will try and dig out my original notes to Prof. Laithwait . And the patent office . Incidentally he said I had no right to make a mechanical - electrical. Analogy , but later went on to say he was the only one I could talk to at that time . Which we did. Then he died. ( nothing to do with me ) I think as Swansont says I better get some observation, experiment , evidence , to bring this to a head . .I. Accelerometers could not and would not be expected to measure any force during freefall - they measure proper acceleration (which is acceleration relative to freefall). Thus in freefall they measure zero by design and necessity. If the guys at Surrey were trying to measure the force of gravity using an accelerometer in freefall either they were working on something completely out of my ken, or they were actually trying to measure tidal or some other forces, or you misunderstood them. .M. Remember I was making this project here the subject of my Degree Final year Thesis. I did my book search , my history of research search ,including Surrey University satellite work . I did my own experiments , with results that I intend to show here as part of the evidence. The computer Simulation was done with Mathcad. All of which gave a positive result. The maths was overseen by the senior mathematician in the Dept. Some of which included on a previous post. .I. Yes - I am pretty convinced it won't work. a. It contradicts basic newtonian mechanics - which if you drew some decent freebody diagrams you would realise b. more importantly - It contradicts observation. c. most important - it contradicts and others in that it would be a renewable source of useful energy..M. Well I hope I can prove that wrong , in other words that it is possible. however if I am proved wrong it is a shame , but I would then concede defeat. Mike Mike here is my concern regarding your vibrating doodads. Newtonian mechanics requires a force acting towards the centre of curvature on any body following a curved path. (If you like it pulls the trajectory in towards the centre off the straight line path) Now you maintain that, as a result of your vibrational dance there is a force which counteracts gravity. So what force is acting to create the partial arcs you describe? Another driver . The oscillator or transducer driver . This is the device stimulating and topping up resonance , the 40,kHz 4 inch peak oscillation . Acting to produce the two masses to move in opposite directions . Remembering they are cleverly linked along the partial arc is a way to retard and feed each other , in the way ( but NOT the prong configuration ) tuning fork works in supportive resonant anti phase. I am happy to meet up with you , seeing as we live in the same Country . Even though not the same County . Mike Edited March 8, 2015 by Mike Smith Cosmos
studiot Posted March 8, 2015 Posted March 8, 2015 Another driver . The oscillator or transducer driver . This is the device stimulating and topping up resonance , the 40,kHz 4 inch peak oscillation . Acting to produce the two masses to move in opposite directions . Remembering they are cleverly linked along the partial arc is a way to retard and feed each other , in the way ( but NOT the prong configuration ) tuning fork works in supportive resonant anti phase. Perhaps you would explain how that would produce a force at right angles to the motion?
Mike Smith Cosmos Posted March 9, 2015 Author Posted March 9, 2015 (edited) Perhaps you would explain how that would produce a force at right angles to the motion?.I do not think, that is what it does or can do. As the resolution of any force is into its two right angle components (F cos theta ) and (F sine theta) . At right angles theta is 90 degrees . And cos of 90 degrees is zero . The force is produced as a reaction to centripetal gravity . Continuously changing its direction towards the centre of the Earth. Perhaps this is similar to the action of electro-magnetism , where cutting the lines of magnetic flux induces a back electro motive force ( back EMF) and an electric current . Whereas here we are crossing the lines of gravitational field lines and inducing a back gravitational force , centrifugal force . Oops! "That will have set the cat amongst the pigeons " Things like to continue to go the way they are going, and if you try to change that ,they react in the opposite direction . Hence with magnetism and current , now with gravity and inertia . Mike Quoted here for reference purposes :- Newtons three laws of motion link :- http://csep10.phys.utk.edu/astr161/lect/history/newton3laws.html The one we are dealing with here to do with centripetal ( action ) And centrifugal ( reaction ) Centripetal being Gravity ( as the action ). Centrifugal being the effect of opposing gravity ( as the reaction . Edited March 9, 2015 by Mike Smith Cosmos
swansont Posted March 9, 2015 Posted March 9, 2015 The one we are dealing with here to do with centripetal ( action ) And centrifugal ( reaction ) Centripetal being Gravity ( as the action ). Centrifugal being the effect of opposing gravity ( as the reaction . Action-reaction force pairs act on different objects. If the action force is exerted by the earth on the satellite, then the reaction force is exerted by the satellite on the earth. Action-reaction forces are always the same kind of force; in this case gravitational.
Mike Smith Cosmos Posted March 9, 2015 Author Posted March 9, 2015 (edited) Action-reaction force pairs act on different objects. If the action force is exerted by the earth on the satellite, then the reaction force is exerted by the satellite on the earth. Action-reaction forces are always the same kind of force; in this case gravitational. So following your above example , but with the bucket and the water . { using your different objects principle } We are exerting a force ( centripetal toward the Centre or operator ) by pulling on the bucket handle with a swing . Which exerts its force on the water. The water in turn exerts its centrifugal force ( away from the centre ) on the bottom of the bucket . Equivalent to what it would do ,if it was standing normally on the ground , where normal gravity pushes the water into the normal bucket. Surely what we have just achieved with the bucket ,swinging above our head , is the equivalent to what gravity normally does . But now ,above our head , we have managed to reverse the process . We have achieved producing an opposite to gravity by : Swinging a mass (water) in a partial arc ( bit above our head ) . I appreciate a lot more happens during the rest of the circle. Mike Edited March 9, 2015 by Mike Smith Cosmos
swansont Posted March 9, 2015 Posted March 9, 2015 At the top of its arc, there is a gravitational force on the water, and no other force. There is nothing pushing it up.
Mike Smith Cosmos Posted March 9, 2015 Author Posted March 9, 2015 At the top of its arc, there is a gravitational force on the water, and no other force. There is nothing pushing it up. So how is the water staying up there ? Mike
andrewcellini Posted March 9, 2015 Posted March 9, 2015 (edited) So how is the water staying up there ? Mike it's moving at the same tangential speed as the bucket that is carrying it and is falling with the bucket under gravity. Edited March 9, 2015 by andrewcellini
studiot Posted March 9, 2015 Posted March 9, 2015 At the top of its arc, there is a gravitational force on the water, and no other force. There is nothing pushing it up. I beg to differ, although I am not suggesting that these are enough to hold the water aloft, just that the applied forces are complicated in the case of the water. There are a variety of forces acting on the water. There will be air pressure, the contact forces with the bucket, surface tension, to name but a few. A good physics question might be "What are the static and dynamic pressures inside the water?"
swansont Posted March 9, 2015 Posted March 9, 2015 I beg to differ, although I am not suggesting that these are enough to hold the water aloft, just that the applied forces are complicated in the case of the water. There are a variety of forces acting on the water. There will be air pressure, the contact forces with the bucket, surface tension, to name but a few. A good physics question might be "What are the static and dynamic pressures inside the water?" It works with a brick, too, or a chunk of ice. Nothing special about it being liquid water. The air pressure doesn't matter, since it will be the same on all sides. Surface tension is negligible on this scale, for water in a bucket. And doesn't matter at all for a solid. You don't even need a bucket. Same principle for a loop-the-loop.
studiot Posted March 9, 2015 Posted March 9, 2015 It works with a brick, too ....etc What works? And of course there is no air pressure on 5 of the six sides of the water. I believe it was an emperor who told Van Guericke that 'air pressure doesn't matter'. Doesn't any fluid act differently from a solid in a centrifuge, shapewise? I think the shape of the parcel of water will be quite different from the shape of a block of ice in the same place. And are you also suggesting there is zero contact force between the water and the bucket? This must be complete nonsense. Sorry to be so pedantic but the string is tied to the bucket, not the water. The water is constantly being forced to travel in the arc by the bucket, not by the string. I could see you being able to use a string to circle a block of ice, but not a parcel of water by itself.
Strange Posted March 9, 2015 Posted March 9, 2015 I could see you being able to use a string to circle a block of ice, but not a parcel of water by itself. You could put a brick in the bucket.
studiot Posted March 9, 2015 Posted March 9, 2015 You could put a brick in the bucket. Or you could put a hole in the bucket and sing that immortal song There's a hole in my bucket, dear Lisa. I would also observe that the bucket and its contents are not in equilibrium so we should not expect the forces to "balance". This is often forgotten.
Mike Smith Cosmos Posted March 9, 2015 Author Posted March 9, 2015 (edited) Or you could put a hole in the bucket and sing that immortal song There's a hole in my bucket, dear Lisa. I would also observe that the bucket and its contents are not in equilibrium so we should not expect the forces to "balance". This is often forgotten. .But if there was a hole in your bucket ? Would a drop of water appear at the top /outward bottom of the bucket. And continue to fly upward having been liberated from the restraint of the bucket . Or now it's containment " force " is no longer in contact . Will it now just float there , just beyond the hole ? Assuming for this discussion there is no air currents or wind pressure to displace it . Mike Ps I think if everybody insists I do this experiment with the water in the bucket as well as bleeding through the hole , I can't win either way . Whether the centrifugal force , continues after the effects of the centripetal force are ' there ' or 'not there ' , when the hole appears . Either way I am going to get wet . Edited March 10, 2015 by Mike Smith Cosmos
swansont Posted March 10, 2015 Posted March 10, 2015 I would also observe that the bucket and its contents are not in equilibrium so we should not expect the forces to "balance". This is often forgotten. In this very thread, too. . But if there was a hole in your bucket ? Would a drop of water appear at the top /outward bottom of the bucket. And continue to fly upward having been liberated from the restraint of the bucket . Or now it's containment " force " is no longer in contact . Will it now just float there , just beyond the hole ? Assuming for this discussion there is no air currents or wind pressure to displace it . Water would fly out of the bucket all the way around the path, except at the very top, assuming you are at the correct speed. At the top of the arc, the force needed to move in a circle is provided by gravity, and needs no additional force from the bucket to satisfy the centripetal force condition. That's not true anywhere else along the path, because only some component of gravity can be acting toward the center (or will act away from it), depending on the angle. So for the rest of the path the centripetal force is a combination of the force exerted by the bucket and the component of gravity in (or away from) that direction. Circular orbits, of course, always have the condition that gravity is the centripetal force. What works? The behavior being described. Doesn't any fluid act differently from a solid in a centrifuge, shapewise? I think the shape of the parcel of water will be quite different from the shape of a block of ice in the same place. And are you also suggesting there is zero contact force between the water and the bucket? This must be complete nonsense. Sorry to be so pedantic but the string is tied to the bucket, not the water. The water is constantly being forced to travel in the arc by the bucket, not by the string. I could see you being able to use a string to circle a block of ice, but not a parcel of water by itself. Good thing I was not suggesting that. Which makes the rest of this moot. image.jpg You should not have the force exerted by the water on the bucket in the same free-body diagram as the forces on the water. A FBD only includes the forces exerted on the body.
studiot Posted March 10, 2015 Posted March 10, 2015 (edited) At the top of the arc, the force needed to move in a circle is provided by gravity, Sorry to harp on about this, but, under gravity, the water would be travelling in something approximating to a parabolic arc, not a circular path. It is the bucket which imposes a circular path and the bucket is being forced round by the forces in the string or whirling arm, not gravity. Suppose the bucket suddenly rotated or opened to release the water, the water would separate from the bucket and fall in its parabolic arc, whilst the empty bucket would carry on round on its tighter circular arc. Edited March 10, 2015 by studiot
swansont Posted March 10, 2015 Posted March 10, 2015 Sorry to harp on about this, but, under gravity, the water would be travelling in something approximating to a parabolic arc, not a circular path. It is the bucket which imposes a circular path and the bucket is being forced round by the forces in the string or whirling arm, not gravity. Suppose the bucket suddenly rotated or opened to release the water, the water would separate from the bucket and fall in its parabolic arc, whilst the empty bucket would carry on round on its tighter circular arc. Which is why it is only at the top of the arc that gravity provides the entirety of the centripetal force. As I stated. Maybe - just maybe - you could assume I know a little about what I'm talking about and read in that context, instead of assuming that omission of unneeded detail implies misunderstanding.
robinpike Posted March 10, 2015 Posted March 10, 2015 Ok. I do not wish to be construed as "soap boxing " as that certainly is not my intension. And I have spent too much time on this project for it to be dismissed out of hand. What would be nice is for someone to say ! Jolly good Mike , you are a nice chap. I think you might be on to something there . Keep going . You can make it ! But be prepared to be wrong. That last phrase ,sends shudders through my very being . However I do have doubts ,sometimes in the middle of the night . Then I get fortified as I do observe and learn of such wonders both in the atom , the world , my experiences , the earth moon and stars , now galaxies , super large structures and beyond. As a Famous American once said " why shall we go to the Moon" " because it's there ! " So " why should we explore ways to overcome the confinement of Gravity ". " because we can " maybe not me , but somebody will ! " because the solution is there , somewhere " Mike Ps. However , not to sink too far down , beyond the water level . I will reflect on what ,you all are saying , and see what I can come up with. I do understand most of what you chaps are saying , BUT I still think there is something there . I wish someone else would say " you are on to something there Mike , keep going ! Mike Mike, you should keep going! Challenging an accepted conclusion will always be open to... you are wasting your time, and here's why... And for want of an example from the past... What on earth were Whitfield Diffie and Martin Hellman thinking when they they thought they could devise a secure public key encryption method? It is obvious that no such method can exist, for whatever rules can be devised to encrypt a message, the rules can simply be reversed to reveal the message.
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