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Natural Science Forum / Physics / General Physics / July 2008circ motion question
Please help me clear up a confusion about circular motion. The net force causing the centripetal acceleration is directed toward the center of the circle. The instantaneous velocity of the object undergoing centripetal acceleration is tangential to the circle path, and therefore perpendicular to force. But if the force is perpendicular to the velocity, how can it be affecting the velocity of the object at all? For comparison, I'm thinking of projectile motion, where the horizontal velocity component of the projectile remains *constant* (assuming no air resistance) because gravity acts in a purely vertical direction (i.e., perpendicular to the horizontal velocity). On Jul 11, 1:44 pm, adgjl_...@hotmail.com wrote: > Please help me clear up a confusion about circular motion. The net > force causing the centripetal acceleration is directed toward the [quoted text clipped - 7 lines] > purely vertical direction (i.e., perpendicular to the horizontal > velocity). There is just outward weight. Move along the spokes outward and the weight builds up for a spinning wheel. A gyroscope has sideways or horizontal weight or gravity if you wish. That is why it is heavy to tip over. Mitch Raemsch | Please help me clear up a confusion about circular motion. The net | force causing the centripetal acceleration is directed toward the | center of the circle. Ok. The instantaneous velocity of the object | undergoing centripetal acceleration is tangential to the circle path, | and therefore perpendicular to force. That's where you are confused. The object has two velocities, one tangential and one radial. The tangential velocity (which is > 0) has no acceleration. The radial velocity (which is zero) has a force toward the centre. | But if the force is | perpendicular to the velocity, how can it be affecting the velocity of | the object at all? It doesn't, the instantaneous tangential velocity continues on unaffected. Only the radial velocity is affected and that was an instantaneous big fat zero to begin with. | For comparison, I'm thinking of projectile motion, | where the horizontal velocity component of the projectile remains | *constant* (assuming no air resistance) because gravity acts in a | purely vertical direction (i.e., perpendicular to the horizontal | velocity). Exactly. The horizontal COMPONENT is unaffected, but the projectile accelerates vertically. If you want a comparison, consider a satellite in orbit, or the Earth accelerating toward the Sun. > <adgjl_...@hotmail.com> wrote in message > [quoted text clipped - 31 lines] > projectile accelerates vertically. If you want a comparison, consider > a satellite in orbit, > or the Earth accelerating toward the Sun. What about when the Earth moves away from the Sun and decelerates? Gravity is not all acceleration. Mitch Raemsch > Please help me clear up a confusion about circular motion. The net > force causing the centripetal acceleration is directed toward the [quoted text clipped - 3 lines] > perpendicular to the velocity, how can it be affecting the velocity of > the object at all? Velocity is a vector. Vectors comprise two parts, magnitude and direction. The force affects the directional component. On Jul 11, 6:59 pm, "Cwatters" <colin.wattersNOS...@TurnersOakNOSPAM.plus.com> wrote: > <adgjl_...@hotmail.com> wrote in message > [quoted text clipped - 10 lines] > Velocity is a vector. Vectors comprise two parts, magnitude and direction. > The force affects the directional component. Bravo, but realize that both you and I are going to be flamed by the clueless. At least both of us seem to try, so what if we are flamed as whatever. Harry C. On Jul 11, 5:44 pm, adgjl_...@hotmail.com wrote: > Please help me clear up a confusion about circular motion. The net > force causing the centripetal acceleration is directed toward the > center of the circle. So far, so good. > The instantaneous velocity of the object > undergoing centripetal acceleration is tangential to the circle path, > and therefore perpendicular to force. Correct. > But if the force is > perpendicular to the velocity, how can it be affecting the velocity of > the object at all? Force imparts an acceleration, which is a change of velocity WITH TIME. The velocity of the object is obviously changing with time. If it weren't then the object would be moving in a straight line at a constant speed. > For comparison, I'm thinking of projectile motion, > where the horizontal velocity component of the projectile remains > *constant* (assuming no air resistance) because gravity acts in a > purely vertical direction (i.e., perpendicular to the horizontal > velocity). Wrap that analogy around an axis. The horizontal component of projectile motion (perpendicular to the applied force) becomes the tangential component of circular motion. The vertical component of projectile motion becomes the radial component of circular motion. This is exactly what happens when the "projectile" is moving at orbital velocity. Tom Davidson Richmond, VA On Jul 11, 5:44 pm, adgjl_...@hotmail.com wrote: > Please help me clear up a confusion about circular motion. The net > force causing the centripetal acceleration is directed toward the [quoted text clipped - 7 lines] > purely vertical direction (i.e., perpendicular to the horizontal > velocity). Good question, but rather trivial to explain. As you will recall from Newtons laws of motion, a body(mass) in motion, unless acted on by an external force, will contiue to travel in a straight line path. (I suspect everyone is with me to this point.) If it is acted upon by an external force at right angles to its path of flight, its path will change, because it it being acted on by a force which resulting in an acceleration resulting in its path being deflected in another direction. (You may want to look up the definition of accleration at this point, but to avoid doing a Google search, acceleration is defined as the result of a force being applied to a mass, the old F = ma thing. If the force is being applied by a rope connected to a fixed point, the mass will respond to the force and experience an acceleration towards the fixed point to which the rope is attached. The tension on the rope will at this point be mv^2/r, where r is simply the length of the rope. Picture a game of tether ball. It the ball is hit too hard, the velocity imparted with exceed the mv^2/r tensile strenght of the rope, and the ball will fly off in a straight line to land at pointsunknown. The concept is just that simple, and does not require a knowledge of vectors to grasp. The same concept applies to an orbiting mass, such as the earth around the sun or the moon around the earth. Identical principles apply. The tension on the rope is simply replaced by the froce of gravitational attraction between the two now very massive bodies. The moon orbits the earth in precisely the same way that the earth obits the sun. The only difference is that the outward radial force mv^2/r is countered by the gravitation attraction that exist between the sun and the earth, or the earth an the moon. If you look at it this way, I believe that you will agree that the concept is rather simple. Most of the confusion stems from people throwing around the terms centrifical and centrugal. Now to be honest with readers, I am a graduate physicist with some expertise in both classical theoretical mechanics and electromagnetics, and these terms confuse me, so most physicists avoid them as being terms used solely by highschool teachers who know damn little about physics and teach it as rote. Now I will tell you how most real physicists look at and orbital situation, be it in space or withing a circular particle accelerator. We draw a siimple sketch on a sheet of paper, put a dot where the mass under analysis is, and then draw little arrow to denote it tangental velocity, and both the radial and central forces acting on it. It keeps things quite simple. If it a charged particle being accelerated in an acclerator, we would label the inward directed arrow as q, (vXB). The inertial force opposing defelection of the particle from a straight path would be in the outward direction, which we would label mv^2/r. Then for a stable circular orbit we we would write the starting relationship in more formal terms: q.(VXB) = mV^2/r. Trust me, this is the starting point for any circular or semi-circular particle acclerator. Much the same thing is done in the case of orbital planetary computations, except that the q.(VXB) is replaced by the terms for gravitational attraction. Small note: When I write q.(VXB), this is vector notation, which is generally taught in second year undergraduate physics. Expanding: q.(vXB) is short for q(dot)(vV(vector cross product) B), Now here is where things start to become complicated if you do not understand vector mathematics, The vectors cross product of the vectors V and B yields a Vector forth that is a direction at right angles to the B field and the path of the particle V. The scaler (dot) product tells us the magnitude of the resulting force which is governed by the charge on the particle. Now have lost all of the 90% of the readers, except for the physicists or aspiring physicists. Ok, Nuff said. At any rate.... the terms centripical and centrifugal are rarely used by physicists, simply because they are more commonly than not a source of confusion. Harry C. p.s.. Wonder how long it is going to be before some idiot calls me 'Stupid' for sharing this basis information. > Now have lost all of the 90% of the readers, except for the physicists > or aspiring physicists. > > Ok, Nuff said. At any rate.... the terms centripical and centrifugal > are rarely used by physicists, simply because they are more commonly > than not a source of confusion. You did not lose me at all Harry, and ... Earlier in this group I explained how the "centrifical /centrifugal forces" was just a confusion of the forward force acting with a 90 degree force to the forward motion. The 10% left don't all want to be physicists. Some just like knowing "how things work" and also like to explain things in laymens terms so it can be learned by other non physicists. :)  SignatureJames M Driscoll Jr Spaceman On Jul 12, 5:24 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > hhc...@yahoo.com wrote: > > Now have lost all of the 90% of the readers, except for the physicists [quoted text clipped - 19 lines] > James M Driscoll Jr > Spaceman --------------------- lets see if all those 'knowledgable' people understand it to scratch: so here is another question: where is the conservation of energy rule kept in the above example ??? TIA Y.Porat ---------------------------- > On Jul 12, 5:24 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 37 lines] > > - Show quoted text - Dhuh! Harry C. On Jul 12, 6:45 pm, "hhc...@yahoo.com" <hhc...@yahoo.com> wrote: > > On Jul 12, 5:24 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 41 lines] > > Harry C. --------------------- hey smatry while a satellite is orbiting earth constantly there is a constant force deviating it from the straight line movement so it does constantly an alleged work of force times delta s to remain in constant r so it seems that work is invested so why is that analysis wrong ??? Y.Porat --------------------------- > On Jul 12, 6:45 pm, "hhc...@yahoo.com" <hhc...@yahoo.com> wrote: > [quoted text clipped - 56 lines] > Y.Porat > --------------------------- The vectors F and delta s are perpendicular to each other in a circular orbit. Their scalar product delta W is therefore zero. Glad to help. Tom Davidson Richmond, VA > > On Jul 12, 6:45 pm, "hhc...@yahoo.com" <hhc...@yahoo.com> wrote: > [quoted text clipped - 66 lines] > Tom Davidson > Richmond, VA --------------- thanks that is th e formal mathematical explanation but we want to understand better what happens there physically iow why is that mathematical presentation is right what is the mechanism there does it mean fo r instance that if i push a satellite outwards and perpendicular to its tangental path i dont need energy for that ?? my gut feeling is that we have to understand it a bit deeper or may be there is no tangental force at all at the sattellite case ?? but it has inertia so how can we change inertia position without paying for that in energy investment ?? (now please dont tell me that it is because curved space time (:-)) TIA Y.Porat ------------------------ say in the gravitational force case as an actor > (now please dont tell me that it is because curved space time > (:-)) OK. Goodbye. Tom Davidson Richmond, VA On Jul 11, 10:24 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > hhc...@yahoo.com wrote: > > Now have lost all of the 90% of the readers, except for the physicists [quoted text clipped - 5 lines] > > You did not lose me at all Harry, James, that does not surprise me. > and ... > Earlier in this group I explained how the "centrifical /centrifugal forces" > was just a confusion of the forward force acting with a 90 degree force > to the forward motion. You did, and that explanation is correct. Still, describing that is going on in words is often confusion. A simple sketch and some equally simply math eleminates all possible ambiguity. > The 10% left don't all want to be physicists. > Some just like knowing "how things work" and also like to > explain things in laymens terms so it can be learned by other > non physicists. I agree, and I simply try to explain things as a physicist sees them in the hope that this will help the understanding of basic principles of physics by interested people who are not physicists or student physicists. Just people that want to understand why and how things work. Physical situations expained in word can be very confusing an introduce ambiguity in the individual interpretation of the words. Thats why all physicists are taught to first draw a sketch of any mechanics problem before attempting to solve it. The path of a cannon ball fired from a cannon elevated to an angle from the horizontal at 45-degrees told to you. You are then asked to compute the distance from the cannon where the projectile will hit the ground. So, the starting point is that you would draw a simple sketch of the cannon ball flying from the muzzle of the cannon at and angle of 45- degrees to the ground. Next, you would assign two arrow to the cannon ball, One arrow would represent the velocity of the cannon ball, and the second would be a downward arrow representing the gravitational attraction of the earth. You would then label the two arrows. The arrow in the direction of the cannon balls path would be simply 'V' and the downward pointing arrow would be mg. OK, now since the cannon ball was fired at an upward angle of 45- degrees, you now need to apply a little trigonometry to determine is the horizontal and vertical components of V...etc., etc. Next, a small amount of calculus to determine where it will strike the ground. That simple sketch eliminates any possible confusion about the situation, and greatly simpifies soluton of the problem without having to do a lot of arm waving and trying to describe the situation and its soluton using hundreds if not thousands of words. I helps physicist greatly by simply drawing a sketch of problem before attempting to solve it, and I believe that it would be of immense value to non-physicists and laymen to approach a physics problem in the same way. Many basic mechanics problems can be solved by simply thinking about them, but in more sophisticated problems, such as those in Theoretical Mechanics, absolutely require a sketch to solve. For example, things get complicated very quickly when you get in problems relating to the motion of a rigid body in 3-dimensions, such as the nunation and precession of a simple gyroscope. The math becomes very complex as well. I don't believe that I need to say much more about this, because problems of this class have been known to make physics students swith to a major in business! (I know, you think I'm kidding, but this is the absolute truth!) Another method taught to physicists is, first obtain a simplified solution which focuses only on the basics. Only then do you start to throw in the secondary factors and perturbatons to the ideal, such as the effects of Special Relativity as one example. You do not start with a primary consideration on relativist effects, but usually add them in to the basic soluton to produce a more refined solution. Same thing applies frictional effects and radiation losses, or eddy current braking. Laymen attempting to learn the principles of physics from "coffee table books" appear to miss this entirely. James trust me that I can cite any number of examples where the ideal solution is only a RCH away from the real solution, and I suspect that you are are familiar with what the unit of mesurement called the RCH is (at least if you are old enough). Then too, I really don't like to deal with 2nd order differential equations in 5 variables. They created an unbelievable level of brain strain, and are best dealt with after consuming a considerable volume of Jim Beam or a sixpack. James, stay well, and realize that we both piss some people off, for very different reasons. (Have you ever viewed the movie "Good Will Hunting". See it if you haven't already.) Harry C. On 11 juil, 17:44, adgjl_...@hotmail.com wrote: > Please help me clear up a confusion about circular motion. Note that the phrase "circular motion" applies more particularly to the rotation of a body about an axis with no centripetal force involved. > The net > force causing the centripetal acceleration is directed toward the [quoted text clipped - 3 lines] > perpendicular to the velocity, how can it be affecting the velocity of > the object at all? You remarked yourself that this velocity was tangential to the orbital path. It is not affected by the centripetal force on a perfectly circular orbit. See below. > For comparison, I'm thinking of projectile motion, > where the horizontal velocity component of the projectile remains > *constant* (assuming no air resistance) because gravity acts in a > purely vertical direction (i.e., perpendicular to the horizontal > velocity). There you just reached the right conclusion. A body on a circular orbit would simply have just the right velocity for its tangential velocity vector to remain locally parallel to the surface of the sphere at all points of its orbit. André Michaud On Jul 12, 2:49 pm, srp2...@gmail.com wrote: > On 11 juil, 17:44, adgjl_...@hotmail.com wrote: > [quoted text clipped - 3 lines] > to the rotation of a body about an axis with no centripetal > force involved. You always amuse me with your odd spins (pun intended) on basic physics, Andre. Tell me, Andre, for an extended body (say, a Frisbee) that is rotating about an axis, what force is keeping the material of the rim of the Frisbee in motion in a circular path? > > The net > > force causing the centripetal acceleration is directed toward the [quoted text clipped - 22 lines] > > André Michaud > On Jul 12, 2:49 pm, srp2...@gmail.com wrote: > [quoted text clipped - 11 lines] > about an axis, what force is keeping the material of the rim of the > Frisbee in motion in a circular path? ---------------- it is called inner attraction forces now i bet you delude yourself(and others -- that you understand ANY attraction force how it works (:-) Y.Porat ------------------------------------ > > > On 11 juil, 17:44, adgjl_...@hotmail.com wrote: > [quoted text clipped - 12 lines] > ---------------- > it is called inner attraction forces Exactly. And these forces are directed toward the center. And so they are, by definition, centripetal. > now i bet you delude yourself(and others -- that you understand > ANY attraction force how it works (:-) > > Y.Porat > ------------------------------------ > On Jul 12, 2:49 pm, srp2...@gmail.com wrote: > [quoted text clipped - 11 lines] > about an axis, what force is keeping the material of the rim of the > Frisbee in motion in a circular path? Angular momentum due to initial impulse. What else. Don't tell me you can't differentiate between orbital motion due to gravity (force) and circular motion of a rigid body about its axis Shame on you PD. Not really surprised. In sync with your inability to calculate velocity with a simple relativistic equation with your own values. Tell me, do you still think that that the sum of cos^2(wt) + sin^^2(wt) is not constant ? LOL André Michaud > > > The net > > > force causing the centripetal acceleration is directed toward the [quoted text clipped - 20 lines] > > > André Michaud On Jul 14, 10:49 am, srp2...@gmail.com wrote: > > On Jul 12, 2:49 pm, srp2...@gmail.com wrote: > [quoted text clipped - 13 lines] > > Angular momentum due to initial impulse. What else. The impulse increases the tangential speed. But the material on the rim of the Frisbee isn't suddenly immune from Newton's 1st law. That is, unless there is a force acting on the material of the rim, it will travel in a tangential *straight line*. Since it is obviously not traveling in a straight line, then there must be a force that is acting on the material of the rim to bend it from the straight line. Porat pointed to the right answer. > Don't tell me you can't differentiate between orbital motion due > to gravity (force) and circular motion of a rigid body about its > axis Shame on you PD. And what do you think the difference is? > Not really surprised. > [quoted text clipped - 4 lines] > sin^^2(wt) > is not constant ? No, I do not. I think that the sum of Acos^2(wt) + Bsin^2(wt) is not constant unless there is a specific relationship between A and B. If that relationship holds (they are equal) then the answer is simply A, and there is absolutely no value in writing it as Acos^2(wt) + Bsin^2(wt). Do you now understand what I think on that matter? > LOL > [quoted text clipped - 24 lines] > > > > André Michaud > On Jul 14, 10:49 am, srp2...@gmail.com wrote: > [quoted text clipped - 17 lines] > > The impulse increases the tangential speed. So you assert that someone sending a frisbee on its way will not cause the frisbee to rotate about its axis ? > But the material on the rim of the Frisbee isn't suddenly > immune from Newton's 1st law. Definitely not. > That is, unless there is a force acting on the material of the > rim, it will travel in a tangential *straight line*. So you assert that gravitation is what holds the atoms making up the frisbee together ? Have you heard about covalent bounding and van der Waal forces and all other flavors of electromagnetic forces that bind atoms together in solids ? Apparently not. > Since it is obviously not traveling in a straight line, then there > must be a force that is acting on the material of the rim to > bend it from the straight line. Since all atoms making up the frisbee are rigidly held together and cannot move about with respect to each other by the electromagnetic links mentioned (that you apparently have not heard about yet), and that keep the frisbee together whether it is rotating or not, the only thing that can get the frisbee rotating about its axis is the part of the initial impulse in excess of that which determines its longitudinal velocity. Elementary mechanics. No centripetal force is involved in maintaining this type of rotation. > Porat pointed to the right answer. At least you have Porat to support your interestingly uninformed view. Great reference. > > Don't tell me you can't differentiate between orbital motion due > > to gravity (force) and circular motion of a rigid body about its > > axis Shame on you PD. > > And what do you think the difference is? Do I was right. You don't know the difference. I suggest you go back to intro refs on classical mechanics. Seems to me that you skipped at least one chapter on this aspect of physics too, on top of trigonometry. > > Not really surprised. > [quoted text clipped - 6 lines] > No, I do not. I think that the sum of Acos^2(wt) + Bsin^2(wt) is not > constant unless there is a specific relationship between A and B. Hey. Progress at last! The little light has started glowing. > If that relationship holds (they are equal) then the answer is > simply A, and there is absolutely no value in writing it as > Acos^2(wt) + Bsin^2(wt). Do you now understand what I > think on that matter? I see that you finally reread the intro trig chapter that you had skipped. Good show. Now if you succeed in calculating relativistic velocities with a simple equation, you will have caught up. Keep up the good work. André Michaud > > LOL > [quoted text clipped - 24 lines] > > > > > André Michaud On Jul 14, 11:59 am, srp2...@gmail.com wrote: > > On Jul 14, 10:49 am, srp2...@gmail.com wrote: > [quoted text clipped - 20 lines] > So you assert that someone sending a frisbee on its way > will not cause the frisbee to rotate about its axis ? No, I'm not asserting that at all. Please pay attention. When someone throws a frisbee and causes the frisbee to rotate about its axis, AFTER the frisbee has left the thrower's hand, there must be a force acting on the material of the rim of the frisbee that keeps it from continuing, at any given instant, in a *straight line*. This is because Newton's first law says that, independent of how the frisbee acquired its rotation, at any given instant a bit of material in motion will continue traveling in a *straight line* with that motion UNLESS there is a force that is acting on that bit of material. Since the bit of material obvious does NOT continue traveling in a straight line but has a curved path, there is obviously a force involved. This force is directed to the center of the frisbee, and it is therefore centripetal. > > But the material on the rim of the Frisbee isn't suddenly > > immune from Newton's 1st law. [quoted text clipped - 6 lines] > So you assert that gravitation is what holds the atoms > making up the frisbee together ? No, I'm not asserting that at all. Are you just making up nonsense questions, or do you really not comprehending anything I'm telling you? > Have you heard about covalent bounding and van der Waal > forces and all other flavors of electromagnetic forces that > bind atoms together in solids ? Yes, and that's exactly what Porat pointed to, which is what I in turn pointed out for you. And those electromagnetic forces act to provide a *centripetal* force on the material of the rim of the Frisbee. Please pay attention. This is not hard. > Apparently not. > [quoted text clipped - 10 lines] > the part of the initial impulse in excess of that which > determines its longitudinal velocity. We're not talking about the force that GOT it in motion. The question is whether there is a centripetal force present AFTER it got its motion. > Elementary mechanics. > > No centripetal force is involved in maintaining this type of > rotation. Bullshit. If a satellite is sent into orbit and rockets are fired to get it rotating, then *after* the rockets are turned off, there is centripetal force required to keep the material of that satellite from not flying off tangentially per Newton's first law. If there were not structural forces holding the thing together by *providing* centripetal force, then every bit of the satellite *would* fly off tangentially, per Newton's first law. When something is rotating in uniform circular motion, you know *automatically* that there is centripetal force acting. THAT is elementary mechanics. If you have an elementary physics book handy, tell me which one it is, so that I can give you a section reference for you to read to refresh your cheese- cloth grip on it. > > Porat pointed to the right answer. > [quoted text clipped - 8 lines] > > Do I was right. You don't know the difference. And what do you think the difference is? Honestly, Andre, I can't believe you would have wasted all that time worrying about relativity when you don't even have something as simple as this down. > I suggest you go back to intro refs on classical mechanics. > Seems to me that you skipped at least one chapter on [quoted text clipped - 12 lines] > > Hey. Progress at last! The little light has started glowing. This is precisely what I said in the chain of posts, which you can find by googling up the thread. Do you want some additional opportunity for embarrassment here? > > If that relationship holds (they are equal) then the answer is > > simply A, and there is absolutely no value in writing it as [quoted text clipped - 39 lines] > > > > > > André Michaud > On Jul 14, 11:59 am, srp2...@gmail.com wrote: > [quoted text clipped - 28 lines] > acting on the material of the rim of the frisbee that keeps it from > continuing, at any given instant, in a *straight line*. Yes. The electromagnetic electronic links that hold the molecules together from all direction within the solid body. No centripetal force involved. The sum of all electromagnetic interactions between each atom and all other atoms in the body that maintain each atom rigidly held in place. > This is because Newton's first law says that, independent of how > the frisbee acquired its rotation, at any given instant a bit of material > in motion will continue traveling in a *straight line* with that motion > UNLESS there is a force that is acting on that bit of material. Of course. But this applies go moving bodies. Once atoms are captive of a solid body, they become part of the moving body and then rotation of the whole rigid body becomes possible as a unit. > Since the bit of material obvious does NOT continue traveling > in a straight line but has a curved path, there is obviously a > force involved. Yes, but not a centripetal force in the case of atoms belonging to the rim of the frisbee. The sum of electromagnetic forces coming from all directions within the body that hold each atom in place. > This force is directed to the center of the frisbee, and it is > therefore centripetal. Wrong. You do not understand the difference between a rotating body whose every atom is rigidly held in place by the sum of forces coming from all other atoms in the body and an small orbiting body, which is not rigidly held to the central body and whose orbit is determined by the centripetal force from the larger body. > > > But the material on the rim of the Frisbee isn't suddenly > > > immune from Newton's 1st law. It is not. But the sum electromagnetic forces coming from all other atoms in the body (not centripetal) prevent all atoms in the frisbee to move freely in any direction. > > Definitely not. > [quoted text clipped - 7 lines] > questions, or do you really not comprehending anything I'm telling > you? I understand that you do not understand the difference between rotating motion of a rigid body and orbital motion of a small body about a large one. > > Have you heard about covalent bounding and van der Waal > > forces and all other flavors of electromagnetic forces that > > bind atoms together in solids ? > > Yes, and that's exactly what Porat pointed to, No. He was talking about a centripetal force holding all atoms of the frisbee together. > which is what I in turn pointed out for you. I know. Quite a laugh. > And those electromagnetic forces act to provide a > *centripetal* force on the material of the rim of the Frisbee. Wrong. You need to review basic physics. > Please pay attention. This is not hard. Not hard at all. > > Apparently not. > [quoted text clipped - 14 lines] > question is whether there is a centripetal force present > AFTER it got its motion. Exactly. There is no centripetal force holding the atoms of the frisbee together whether it is immobile or in motion or in rotation. > > Elementary mechanics. > [quoted text clipped - 5 lines] > centripetal force required to keep the material of that satellite from > not flying off tangentially per Newton's first law. Yes, as a unit. But if the satellite was to rotate about its axis, this would not be due to a centripetal force within the satellite. > If there were not structural forces holding the thing together > by *providing* centripetal force, then every bit of the satellite > *would* fly off tangentially, per Newton's first law. Wrong. No need for a centripetal force to hold every part of the satellite together. > When something is rotating in uniform circular motion, For this to happen, you need the body to be rigid (all atoms making it up are rigidly held in place. > you know *automatically* that there is centripetal force acting. You probably hare the only one besides Porat to believe such a no starter. > THAT is elementary mechanics. ROTFL > If you have an elementary > physics book handy, tell me which one it is, so that I can > give you a section reference for you to read to refresh your > cheese- cloth grip on it. You tell me which one you have, I will tell you which chapter to look at. > > > Porat pointed to the right answer. > [quoted text clipped - 10 lines] > > And what do you think the difference is? Explained above. > Honestly, Andre, I can't believe you would have wasted all that > time worrying about relativity when you don't even have > something as simple as this down. It so happens that I have both relativity and "this" down quite solidly. Not your case apparently. > > I suggest you go back to intro refs on classical mechanics. > > Seems to me that you skipped at least one chapter on [quoted text clipped - 16 lines] > find by googling up the thread. Do you want some additional > opportunity for embarrassment here? I remember the laugh you gave me. But you can save face if you want, as long as you stop misinforming newbees about sin^2(wt) +cos^2(wt) André Michaud > > > If that relationship holds (they are equal) then the answer is > > > simply A, and there is absolutely no value in writing it as [quoted text clipped - 39 lines] > > > > > > > André Michaud On Jul 14, 2:06 pm, srp2...@gmail.com wrote: > > On Jul 14, 11:59 am, srp2...@gmail.com wrote: > [quoted text clipped - 44 lines] > and then rotation of the whole rigid body becomes possible as > a unit. You can still treat a chunk of matter in a rigid body as a body that is subject to the forces from the rest of the body. If that chunk accelerates (as does the material of the rim as it goes around in uniform circular motion), then these forces are responsible for that acceleration. The chunk of material is NOT in mechanical equilibrium if it is going around in a circle, and as a consequence there is an *unbalanced* force present, and this is due to the sum of all those internal forces. For a case of a dime-sized chunk of material two- thirds of the way radially out on a spinning Frisbee, the sum of the internal forces holding that dime-sized chunk in the rest of the Frisbee points toward the center of the Frisbee. If this sum did NOT point toward the center of the Frisbee, the chunk of material would *have* to travel in a straight line, by Newton's first law. Being part of a larger solid body does not release the chunk of material from obeying Newton's first law. As an example of this, imagine a steel ball welded onto a steel rod, and a steel ball tied to a rope of the same length as the rod. Now whirl the steel balls around your head, hanging onto the other end of the rope or the steel rod. I think it's plain that the rope provides the centripetal force that keeps the ball traveling in a circle, and it SHOULD be plain that the steel rod provides the centripetal force that keeps the ball traveling in circle, even though the rod and the ball are one rigid body. Does the steel ball care whether what's pulling it is a rope or more steel? This applies to your further comments below. Think about it. > > Since the bit of material obvious does NOT continue traveling > > in a straight line but has a curved path, there is obviously a [quoted text clipped - 125 lines] > You tell me which one you have, I will tell you which chapter > to look at. I have two dozen and access to a dozen more. Which one do you have? > > > > Porat pointed to the right answer. > [quoted text clipped - 19 lines] > It so happens that I have both relativity and "this" down quite > solidly. Not your case apparently. Are you embarrassed yet? > > > I suggest you go back to intro refs on classical mechanics. > > > Seems to me that you skipped at least one chapter on [quoted text clipped - 21 lines] > But you can save face if you want, as long as you stop > misinforming newbees about sin^2(wt) +cos^2(wt) What I just told you is what I've always told you, Andre. > On Jul 14, 2:06 pm, srp2...@gmail.com wrote: > [quoted text clipped - 49 lines] > You can still treat a chunk of matter in a rigid body as a body that > is subject to the forces from the rest of the body. Of course. That's what I have been telling you. > If that chunk accelerates (as does the material of the rim as it goes > around in uniform circular motion), then these forces are responsible > for that acceleration. That chunk cannot accelerate separately of the whole body it is a rigid part of. It is the body as a whole that accelerates. > The chunk of material is NOT in mechanical equilibrium > if it is going around in a circle, and as a consequence there is an > *unbalanced* force present, and this is due to the sum of all those > internal forces. That chunk is in perfect mechanical rigid equilibrium within the whole body, whether the body is rotating or not. There is no change of state as it rotates unless the angular momentun becomes important enough to cause the body to disintegrate. > For a case of a dime-sized chunk of material two-thirds of the way > radially out on a spinning Frisbee, the sum of the internal forces > holding that dime-sized chunk in the rest of the Frisbee points > toward the center of the Frisbee. No. The force holing every atom within the body is the sum of all forces coming from all surrounding. Forces coming from the direction of the center have no particular preponderance. > If this sum did NOT point toward the center of the Frisbee, the chunk > of material would *have* to travel in a straight line, by Newton's first > law. No since the frisbee, by structure is perfectly balenced masswise to rotate smoothly. Mass distributed equally about the center. > Being part of a larger solid body does not release the chunk of > material from obeying Newton's first law. No, but since the local electromagnetic equilibrium prevents it from obeying that law, every atom of that chunk has to first obey the predominant omnidirectional force that rigidly holds it in place within the mass of the frisbee. It will become free to obey Newton's first law if the angular momentum becomes strong enough for the frisbee to disintegrate to the point of leaving its individual move separately. > As an example of this, imagine a steel ball welded onto a steel rod, > and a steel ball tied to a rope of the same length as the rod. Now > whirl the steel balls around your head, hanging onto the other end of > the rope or the steel rod. Ok. For the comparison to hold when compared to the rotating frisbee, you must let go of the rope to see if your ball and rod and rope will still behave as the frisbee. If it does not, then your ball-rod-rope setup is non sequitur. > I think it's plain that the rope provides the centripetal force that > keeps the ball traveling in a circle, Are you all there ! So now the rope exerts a force that keeps the ball traveling in circle? It seems to me that it is the person (you ?) holding the rope that is providing the force. No ? You really nead a refresh in elementary mechanics. > and it SHOULD be plain that the steel rod provides the centripetal > force that keeps the ball traveling in circle, So not it is the steel rod that provides a centripetal force. How can this be possible ! First time I hear that a steel rod can provide a centripetal force. You should get your data straight even for you imaginary set up. It still is you providing the force. > even though the rod and the ball are one rigid body. Sure, but it is not rotating about its center of mass. Non sequitur. > Does the steel ball care whether what's pulling it is a rope > or more steel? If you let go of the rope, you will instantly see that it does care, since it is neither the rope nor the steel rod that provides the force. > This applies to your further comments below. Think about it. The day you give me something to think about (not laugh about) I will certainly oblige. Still waiting for that day. > > > Since the bit of material obvious does NOT continue traveling > > > in a straight line but has a curved path, there is obviously a [quoted text clipped - 127 lines] > > I have two dozen and access to a dozen more. Which one do you have? Give me the list, I will tell you. > > > > > Porat pointed to the right answer. > [quoted text clipped - 21 lines] > > Are you embarrassed yet? I am rather amused. Your understanding is so sketchy that one wonders how you could get your degree. You have one, don't you ? > > > > I suggest you go back to intro refs on classical mechanics. > > > > Seems to me that you skipped at least one chapter on [quoted text clipped - 23 lines] > > What I just told you is what I've always told you, Andre. That's not what I recall. You asserted the opposite for quite a while. André Michaud On Jul 14, 4:48 pm, srp2...@gmail.com wrote: > > On Jul 14, 2:06 pm, srp2...@gmail.com wrote: > [quoted text clipped - 95 lines] > strong enough for the frisbee to disintegrate to the point of > leaving its individual move separately. :>) This is a fine one for Dirk's Memorable Screw-Ups page. The whole tract is, really. OK, you win, Andre, you've convinced me of several things: - You don't know what you're doing with basic mechanics - You don't have a textbook, where I could at least give you a couple section references to read - There isn't really any point in trying to correct your thinking on anything, though I may take the opportunity to poke fun at some of the stupid things you no doubt are going to say. Jeez, what a goofball. > > As an example of this, imagine a steel ball welded onto a steel rod, > > and a steel ball tied to a rope of the same length as the rod. Now [quoted text clipped - 238 lines] > That's not what I recall. You asserted the opposite for quite > a while. Look it up and show me where I asserted the opposite, Andre. It's all easily googlable. > André Michaud > On Jul 14, 4:48 pm, srp2...@gmail.com wrote: > [quoted text clipped - 101 lines] > This is a fine one for Dirk's Memorable Screw-Ups page. > The whole tract is, really. Particularly your referencing Porat as the source of your knowledge. > OK, you win, Andre, you've convinced me of several things: > - You don't know what you're doing with basic mechanics You don't know enough physics to be the judge of this despite the obvious high opinion you have of yourself. > - You don't have a textbook, where I could at least give you a > couple section references to read Same here. > - There isn't really any point in trying to correct your thinking on > anything, though I may take the opportunity to poke fun at some > of the stupid things you no doubt are going to say. I know that you are obcessing on me. I for one have no time to waste following you around. Feel free to keep posting your half baked comments if this is what keeps you feeling alive. > Jeez, what a goofball. Ditto. You should get a real job doing real work instead of spending your days gawking at your screen posting hundreds of messages each month trying to impress newbees with your sketchy "knowledge". > > > As an example of this, imagine a steel ball welded onto a steel rod, > > > and a steel ball tied to a rope of the same length as the rod. Now [quoted text clipped - 241 lines] > Look it up and show me where I asserted the opposite, Andre. It's all > easily googlable. Done more than once. Easy for anyone to look up and check. André Michaud On Jul 14, 7:40 pm, srp2...@gmail.com wrote: > > On Jul 14, 4:48 pm, srp2...@gmail.com wrote: > [quoted text clipped - 131 lines] > your days gawking at your screen posting hundreds of messages each > month trying to impress newbees with your sketchy "knowledge". Oh, I've got a GOOD job, thanks very much. And I certainly have an hour and a half of free time a day to post two dozen posts a day without any effort at all. If you have the impression that everyone who posts here is a shiftless, unemployed crank, then I would ask how long you've had this problem of projecting your own situation on others? Is it your presumption that no real physicist would have the time to post to sci.physics? Sorry to crash your personal Club for Amateurs Who Can Pretend to Know More Than They Do If They Want To. PD > On Jul 14, 7:40 pm, srp2...@gmail.com wrote: > [quoted text clipped - 137 lines] > hour and a half of free time a day to post two dozen posts a day > without any effort at all. You certainly have a good job for your employer to tolerate, (does he only know) your chain postings all through the day. I sample checked the time you posted a fair sample of your output, and it is not true that you post only early morning and at lunch time. > If you have the impression that everyone who posts here is a > shiftless, unemployed crank, Not everyone. You however, definitely appear to have an addiction problem to re-reading the reams of your own output on the web. The more the better apparently. > then I would ask how long you've had this problem of > projecting your own situation on others? I don't project. I almost exclusively read only questions by newbees, and have no opinion about most posters, and this only once in a while. Not hard however to size up trollers like you who regularly show up with a chip on their shoulder on my posts. Particularly since your typical comments are not even right with respect to classical physics. Instead of having spent all of your spare time gawking at your screen and speed typing half-cooked comments you should have use it to really finish your education. That's what real physicist do. That's how they eventually become knowledgeable in the field. > Is it your presumption that no real physicist would have > the time to post to sci.physics? Real physicists do post here once in a while. The difference between you and them is that they at least come out in sync with classical physics. > Sorry to crash your personal Club for Amateurs Who > Can Pretend to Know More Than They Do If They > Want To. Not hard to know more than you. Even first year undergrads seem to have gotten further than you even on classical mechanics. Not even talking about electromagnetism. You are a pitiful and ignorant keyboard addict. I suggest you consult to get over it. André Michaud On Jul 14, 7:59 pm, srp2...@gmail.com wrote: > > On Jul 14, 10:49 am, srp2...@gmail.com wrote: > [quoted text clipped - 60 lines] > At least you have Porat to support your interestingly > uninformed view. Great reference. --------------------------------- Hey Andre!! please dont underestimte me expecially after you got my book (many years ago ..) about 'the model of the Atom and the nuc' !! (:-) 2 it was you who discovered to me **that the mass of the quark is about 3 persent of the mass of the proton and neutron** and i thank you for that!! now just have a little look about WHAT HAVE** YOU** DONE WITH THAT INFORMATION... AND WHAT HAVE **I ** DONE WITH THAT INFORMATION !! and may be then you can see the little difference between me and you .... keep well Y.Porat --------------------- On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > Please help me clear up a confusion about circular motion. The net > force causing the centripetal acceleration is directed toward the [quoted text clipped - 7 lines] > purely vertical direction (i.e., perpendicular to the horizontal > velocity). Do the same thing conceptually to both. - If you break the circular motion into tangential and radial components, as you did the projectile motion, then the centripetal force does nothing to the tangential component, and simply adds a radial piece -- exactly like what happens to horizontal and vertical components in projectile motion. OR - If you treat projectile motion like you do circular motion, then gravity acting perpendicular to a horizontally fired projectile does in fact change the velocity vector of the projectile by changing its *direction* -- exactly like what happens in projectile motion. PD > On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > [quoted text clipped - 23 lines] > > PD ---------------- so can we conclude that changing direction of movenet of a mass that moves in a constant velocity does nor need energy ?? Y.Porat -------------------------- > > On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > [quoted text clipped - 28 lines] > of a mass that moves in a constant velocity > does nor need energy ?? No, it does not require energy. PD > > > On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > [quoted text clipped - 32 lines] > > PD ------------------ if we deal with a mass in the gravitational field you are right ! and it is an experimental fact ! if it is in open space without any influence of an attraction field you are wrong !! : if i have a mass moving in ** net space** and i want to deviate its direction say by hitting it with another mass that is even( perpendicular to the line of movement of the tested one) i do need energy !! (because a change of the above ** innertic** movement needs energy ) change in inertia need energy ) so imho we need to make a difference between the two cases the constant orbital movement is a special case !! btw to be honest i never before thought about that problem and i wonder how many people di d it before ie from energy point of view ?? TIA Y.Porat ------------------ > > > On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > [quoted text clipped - 32 lines] > > PD What a double whammy goofball! Still pretending you know any physics ? Just ask the shuttle pilot if he does not need energy to change direction as he proceeds to align with the ISS for docking. André Michaud On Jul 15, 10:43 am, srp2...@gmail.com wrote: > > > > On Jul 11, 4:44 pm, adgjl_...@hotmail.com wrote: > [quoted text clipped - 40 lines] > to change direction as he proceeds to align with the > ISS for docking. You are charming, Andre. Here is a simple question that is addressed in the work chapter in virtually any textbook: Tommy carries a 2 kg book horizontally 5m at constant speed. Does he do any work as he does this? What's your answer, Andre, yes or no? And what textbook do you have, so that I can point to the work chapter and the pages you should read? > André Michaud > On Jul 15, 10:43 am, srp2...@gmail.com wrote: > [quoted text clipped - 44 lines] > > You are charming, Andre. So no answer ? So you assert that the shuttle will expend no energy to change direction to align with the ISS for docking? How much propellant before, and how much after should give you a hint. Or maybe not. I am positive that Nasa will be very interested in knowing that no energy is required for the operation. Why not contact them and inform them of your groundbreaking finding. Very simple question. Or are you afraid of answering it ? Show us your real knowledge. Do you even know what I am talking about ? André Michaud On Jul 15, 2:36 pm, srp2...@gmail.com wrote: > > On Jul 15, 10:43 am, srp2...@gmail.com wrote: > [quoted text clipped - 62 lines] > > Do you even know what I am talking about ? You didn't answer my question, either. Do you think that expending fuel means doing physical work? Do you know what work is? PD > On Jul 15, 2:36 pm, srp2...@gmail.com wrote: > [quoted text clipped - 66 lines] > > You didn't answer my question, either. I asked first. Aren't you supposed to be a real physicist ? Show us that you are not yet one of more of those ignorant imbeciles addicted to their keyboard that you really appear to be with your chain postings on a variety of newsgroups, that try to pass as real physicists. Hundreds of posts per weeks !! I could hardly believe that in february 2006 you posted 1400 (By god , 50 per day, ONE EVERY 15 MINUTES NON STOP on a very conservative 12 hour shift), and topping the 1000 mark for 5 months in a row. I guess that you find more reasonable now to hover in the 800 - 900 range (a more reasonable 2 per hour on a 12 hour shift) which seems to be your size this year. What do you do in life besides posting and sleeping ? Now, if you are not the keyboard addicted idiot on the block you are making yourself to be but a real physicist, explain to us all how the shuttle pilot can change direction to align with ISS for docking without expending one drop of propellant. André Michaud On Jul 15, 4:10 pm, srp2...@gmail.com wrote: > > On Jul 15, 2:36 pm, srp2...@gmail.com wrote: > [quoted text clipped - 89 lines] > > What do you do in life besides posting and sleeping ? You vastly underestimate my speed. I can post 50 posts in 1.5 hrs, which I typically do in the early morning and several more over lunch. Just because YOU take 15 minutes to form a thought and hunt and peck your way laboriously through a post doesn't mean you should project that on others. There, you see? That only took 20 seconds. PD > Now, if you are not the keyboard addicted idiot on the > block you are making yourself to be but a real physicist, > explain to us all how the shuttle pilot can change direction > to align with ISS for docking without expending one drop > of propellant. I didn't say that. I said no WORK is involved. Can't help if it you don't know what work is and equate it with fuel expended. Tell me, Andre, if I am a helicopter burning fuel at a gallon a minute while hovering in midair, am I changing the energy of the helicopter? Another 18 seconds. Ta. PD > André Michaud > On Jul 15, 4:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 120 lines] > > > André Michaud ------------------ Hey Guys 'both of you sent out of track ' and it is a pitty fo rboth of you that are generally Gentlemen (a rare characteristic in this ng ) so you see waht can personal instincts do to even positive civilized people so here is my understanding about some of your debate 9i admit my guilt tosatrt your wrating by my introducing the energy aspect in circular motion so : we must make a difference between a mass being moving at* any* distance form the mother mass and a case in which the mass is orbiting by 'no energy more investment' that last one is jsut a very specific distance r from 'mother mass' and only there if you what to change that specific diatance you must invest energy (even because of the formal reason that while changing the 'equilibrium distance r' we changed the potential energy of the 'daughter mass' and let me tell you that even the question about 'why thereis no need for energy to orbit at the 'equilibrium distance'? is much more complicated than a trivial answer !!! it belongs tothe still unknown secrets of any attraction force !! it belongs to the mechanism of the **attraction agents* and the way they work !!! so changing constantly the direction at the equilibrium distance is only due to those attraction agents !!! **and it i snot at all **at any straight line movement at open space were there is no gravitational force !! in that case ***there *is* a need for changing direction of a straight line moving object !!*** 2 a side remark about movement and energy: if we move a mass horisonlally without friction there is no need to add energy yet if me and you carry a mass even on a perfectly horisontal path *there is again need for investment of energy* because we are not a block of mass we have legs and muscles that has to do work stretch and release etc etc and ower body even on a horisontal path walk is sweinging ' up and down' constantly etc etc even if an helicopter is hoving at a consatnt level i t needs investment of energy because it is not ORBITING but hoving and the chopper if 'fighting constantly against his falling position the chopper is not in the specific unique situation of equilibrium that is gained by help of gravity ONLY AT ONE SPECIFIC DISTATANCE FROMEARTH AND NOT AT ANY CASUAL DISTANCE my conclusion : we have still to analyse the specific distance from earth ie in the gravitatinal field AND FIND OUT THE SECRET WHY IS IT THAT AN EQUILIBRIUM POSITION IS GAINED BY THE HELP OF 'GRAVITONS ' and gravitons behaviour !! AND WHY ONLY IN THAT CASE IS AN EQUILIBRIUM OF ENEREGY IS ACHEVED ie no need in energy investmet is needed in order to 'orbit' ATB Y.Porat ---------------------- Y Porat Spinning gives constant inertia(no coasting) To duplicate spin an object has to accelerate all the time. Bert > Y Porat Spinning gives constant inertia(no coasting) To duplicate spin > an object has to accelerate all the time. Bert Inertia is independent of "spin", Herb. Babble babble! Sam Spin creates gravity,and inertia and gravity are the same thing. Think man think Bert > Sam Spin creates gravity,and inertia and gravity are the same thing. > Think man think Bert No Herb. We have a really good model for gravitation. Mass-energy warping space time. It works remarkably well! Ref: Hartle, "Gravity: An Introduction to Einstein's General Relativity", Addison Wesley (2003) "A few properties of the gravitational interaction that help explain when gravity is important can already be seen from the gravitational force law F_grav = G m_1 m_2 / r_12^2 o Gravity is a universal interaction in Newtonian theory between all mass, and, since E = mc^2, in relativistic gravity between all forms of energy. o Gravity is unscreened. There are no negative gravitational charges to cancel positive ones, and therefore it is not possible to shield (screen) the gravitational interaction. Gravity is always attractive. o Gravity is a long-range interaction. The Newtonian force law is a 1/r^2 interaction. There is no length scale that sets a range for gravitational interactions as there is for the strong and weak interactions. o Gravity is the weakest of the four fundamental interactions acting between individual elementary particles at accessible energy scales. The ratio of the gravitational attraction to the electromagnetic repulsion between two protons separated by a distance r is F_grav G m_p^2 / r^2 G m_p^2 -------- = -------------------- = ------------- ~ 10^-36 F_elec e^2 / (4 pi e_0 r^2) (e^2/4pi e_0) where m_p is the mass of the proton and e is its charge. These four facts explain a great deal about the role gravity plays in physical phenomena. They explain, for example, why, although it is the weakest force, gravity governs the organization of the universe on the largest distance scales of astrophysics and cosmology. These distance scales are far beyond the subatomic ranges of the strong and the weak interactions. Electromagnetic interactions COULD be long range were there any large-scale objects with net electric charge. But the universe is electrically neutral, and electromagnetic forces are so much stronger than gravitational forces that any large-scale net charge is quickly neutralized. Gravity is left to govern the structure of the universe on the largest scales. Want to know more? See: The General Relativity Tutorial http://math.ucr.edu/home/baez/gr/gr.html >> Sam Spin creates gravity,and inertia and gravity are the same thing. >> Think man think Bert [quoted text clipped - 47 lines] > See: The General Relativity Tutorial > http://math.ucr.edu/home/baez/gr/gr.html Sam makes a good point! "General Relativity works remarkably well " in providing cushy, secure jobs to the hand full of GTR Cult Leaders on the taxpayer dole, unlike the DNA model that is used by thousands of folks in the free market to improve medical care, fight crime, improve crops, improve livestock, reconstruct history, etc. SignatureTom Potter http://www.geocities.com/tdp1001/index.html http://notsocrazyideas.blogspot.com http://www.flickr.com/photos/tom-potter/ http://tdp1001.wiki.zoho.com http://groups.msn.com/PotterPhotos http://www.androcles01.pwp.blueyonder.co.uk/dingleberry.htm > >> Sam Spin creates gravity,and inertia and gravity are the same thing. > >> Think man think Bert [quoted text clipped - 53 lines] > in providing cushy, secure jobs > to the hand full of GTR Cult Leaders on the taxpayer dole, ---------------- well said Potter you hit exactly the main point!! peole are not interested in physics they are interested in private bussiness and Wormley is a wonderful example for that at the good case is is only GOOD a crackparroter GR didnt prdict anything it just fiddles in data to formulas the most it can do is to extrapoalte or interpolate data as any other mathematics machine !! but the bottom line is it doers not solve the enigma of gravity !!! and even eorse because of the illusion case prevents real advance in physics ATB Y.Porat ---------------------------------- > .uk/dingleberry.htm > On Jul 15, 4:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 99 lines] > > There, you see? That only took 20 seconds. No wonder the bulk of your messages are half-cocked, like the answer you gave Porat. Did you at least study physics before starting pontificating your stupidities on physics ngs ? > > Now, if you are not the keyboard addicted idiot on the > > block you are making yourself to be but a real physicist, [quoted text clipped - 3 lines] > > I didn't say that. I said no WORK is involved. NO. That's not what you said Quote: Porat: "so can we conclude that changing direction of move net of a mass that moves in a constant velocity does nor need energy ??" PD "No, it does not require energy." No mention of any work until I caught you peeing on the ivory tower wall. Now how fast are you going to deny it again ? By the way, mister lightning fast half-cocked answerer, I checked what Porat told you about the forces acting on the atoms on the rim of the frisbee. He told you exactly the same thing as I. It was you rationalizing what he said as approval of your stupid centipetal angle when only angular momentum is involved. You see, maybe you should take a minute or two reading and UNDERSTANDING what people write before you zip out your 20 seconds meaningless answers, like the pitiful keyboard addict that you are confirming yourself to be. André Michaud. On Jul 16, 9:52 am, srp2...@gmail.com wrote: > > On Jul 15, 4:10 pm, srp2...@gmail.com wrote: > [quoted text clipped - 130 lines] > No mention of any work until I caught you peeing on > the ivory tower wall. That's fine. It requires no infusion of energy. But you asked if I was saying the astronauts did not have to expend FUEL. Here's another example. You burn food calories holding up a book off the desk, stationary in the palm of your hand. But you do no mechanical work to the book and you do not change the energy of the book in doing so. So here again, you expend FUEL but do not add to or otherwise change the energy of the book. > Now how fast are you going to deny it again ? > [quoted text clipped - 12 lines] > > André Michaud. > That's fine. It requires no infusion of energy. But you asked if I was > saying the astronauts did not have to expend FUEL. [quoted text clipped - 4 lines] > book in doing so. So here again, you expend FUEL but do not add to or > otherwise change the energy of the book. PD, You really and truly need to go back and study classical physics. You have been brainwashed by SR and GR to think such thoughts at all. SignatureJames M Driscoll Jr Spaceman On Jul 16, 10:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > That's fine. It requires no infusion of energy. But you asked if I was > > saying the astronauts did not have to expend FUEL. [quoted text clipped - 7 lines] > PD, > You really and truly need to go back and study classical physics. I'm pulling this statement from classical physics. Would you like a reference to a classical physics book that points this out? How far away from you is the nearest public library? > You have been brainwashed by SR and GR to think such thoughts > at all. > -- > James M Driscoll Jr > Spaceman > On Jul 16, 10:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > wrote: [quoted text clipped - 13 lines] > reference to a classical physics book that points this out? How far > away from you is the nearest public library? You are pulling the "section" that is not going over an entire process of energy usage. You are ignoring the "rest" of the dang book. Sheesh! SignatureJames M Driscoll Jr Spaceman On Jul 16, 10:30 am, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > On Jul 16, 10:16 am, "Spaceman" <space...@yourclockmalfunctioned.duh> > > wrote: [quoted text clipped - 18 lines] > You are ignoring the "rest" of the dang book. > Sheesh! So when you drop your wallet on the ground, how much energy does it get? And have you included the energy spent by the leather sewing machine to stitch together the wallet, the chemical energy spent in tanning the leather, the energy the cow expended in creating the hide, and how much energy you had to spend to earn the money to buy the wallet in the first place? PD >> That's fine. It requires no infusion of energy. But you asked if I >> was saying the astronauts did not have to expend FUEL. [quoted text clipped - 9 lines] > You have been brainwashed by SR and GR to think such thoughts > at all. And there we have it in a nutshell; James really has no clue about even the basics of physics; he does not understand even the basic concepts of work and energy. He conflates things like velocity and force, acceleration and velocity. He doesn't understand units attached to quantities. He cannot handle the concept of negative numbers, or perform even simple mathematical manipulations. >>> That's fine. It requires no infusion of energy. But you asked if I >>> was saying the astronauts did not have to expend FUEL. [quoted text clipped - 18 lines] > quantities. He cannot handle the concept of negative > numbers, or perform even simple mathematical manipulations. I do no such thing. I am the one that puts them all together for the bigger picture unlike you that only looks at the "postage stamp" instead of the entire envolope and what is also inside. Again, you seem to love to ignore inside the box when outside, and outside the box when inside. I think your logic function is badly broken. Poor Greg. He lost logic whenever he traps himself inside the box without walls. and also locks himself out of the box without walls when he is outside it. LOL SignatureJames M Driscoll Jr Spaceman > On Jul 16, 9:52 am, srp2...@gmail.com wrote: > [quoted text clipped - 134 lines] > > That's fine. It requires no infusion of energy. So you confirm what you said to Porat ? You still assert that changing direction of motion of a mass that moves in a constant velocity does require energy ? The fact that this is an unmoderated group does not allow you to behave like an irresponsible sob. Unfortunately, many youngs here seem to trust you at your word. André Michaud > > Now how fast are you going to deny it again ? > [quoted text clipped - 12 lines] > > > André Michaud. On Jul 16, 11:23 am, srp2...@gmail.com wrote: > > On Jul 16, 9:52 am, srp2...@gmail.com wrote: > [quoted text clipped - 140 lines] > of a mass that moves in a constant velocity does > require energy ? Not what I said. It may require FUEL, but no energy is added to an object by simply changing its direction and not its speed. A force that is directed perpendicular to an object's velocity will not energy to the object, even if a rocket is fired and fuel is spent to produce that force. > The fact that this is an unmoderated group does not > allow you to behave like an irresponsible sob. > > Unfortunately, many youngs here seem to trust > you at your word. Why do you suppose that is? > André Michaud > [quoted text clipped - 14 lines] > > > > André Michaud. > Not what I said. It may require FUEL, but no energy is added to an > object by simply changing its direction and not its speed. Changing direction requires energy if you wish to keep the same speed. If you only change direction alone (causing a slowing of course) You will lose energy. You truly are lost and dillusional if you think otherwise. Totally lost in your dillusionary rubber ruler world forever. It is a total waste to try and enlighten you, since you have lost all the "energy needed to produce the light" in your brain at all anymore. Sheesh! SignatureJames M Driscoll Jr Spaceman On Jul 16, 12:17 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > Not what I said. It may require FUEL, but no energy is added to an > > object by simply changing its direction and not its speed. [quoted text clipped - 4 lines] > You will lose energy. > You truly are lost and dillusional if you think otherwise. Got a classical mechanics book handy, Spaceman? I can point to what it is you need to read from it. > Totally lost in your dillusionary rubber ruler world forever. > It is a total waste to try and enlighten you, since you have lost [quoted text clipped - 5 lines] > James M Driscoll Jr > Spaceman |
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