Motion
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From: TomGee on 26 Sep 2005 02:52 I guess you thought you were answering that post, Platopes, but here is the post you really answered: "What s it that keeps bodies at rest? > Is a state of rest is equivalent to a state of uniform motion? > If your worldview requires a cause for a state of uniform motion, why does > it not also require the SAME cause for a state of rest?" That's why I asked. To me, it would have been a great question if he had asked why it would not stop. His question simply reflects the standard model part of inertial motion, but that assumes something Newton never said but which has been widely attributed to him, plus it shows the low levels of reasonable thought in physics today most likely due to scientist's unwillingness to go out on a limb or to buck the trend.
From: TomGee on 26 Sep 2005 03:04 Savain, that's only because you're fencing with the automatons produced in the last two or three generations. What is so offensive from them is their apparent total inablility to have an original thought. Then they all sound like parrots over and over again. They are broken records and you yearn to hear sweet melodies. It won't happen from them, only from others who question the status quo of physical ideas being pushed off as fact today. No one can help them; they're too prideful. But such discourse affects the new generations who can and do read these ngs and who will be better armed against their teachers who have failed too many students in the past. Failed to teach them, that is.
From: Randy Poe on 26 Sep 2005 08:56 TomGee wrote: > Randy Poe wrote: > > TomGee wrote: > > It relates force and momentum, i.e., tells you how force and > > momentum are related, by telling you that force is the time > > rate of change of momentum. > > > No, that is not what it tells you. It is an equation - a math > construct It is a model telling you how the two things called force and momentum are related. The equation doesn't cause this relationship, but it is a relationship which is true. > - telling you that the force F is _equal_ to the time rate of > change of the given momentum, exactly like E=mc^2 tells you not that > energy is mass, but that the energy E is equal to the given mass > multiplied by c^2. That's a pure semantic point. Mass can be viewed as frozen energy, in the amount mc^2, all of which can be recovered in matter- antimatter interactions. So in a very real sense, matter IS energy. You are arguing some fundamental difference between "is" and "is equivalent to". I don't think it's an important distinction. As for the relation between force and momentum, what F = dp/dt tells you is that if momentum is changing, there is a force involved, and if there is a force, then momentum is changing. The two are inextricably linked. It is not telling you that sometimes there is a force but dp/dt is zero, and it is not telling you that sometimes p can change but F can be zero. It is telling you that if one is zero, the other is zero, and if one is nonzero, the other is nonzero. If you have a body with unchanging momentum but say there is a force acting on it, you have a nonzero term on the right and a zero on the left. This equation tells you that is impossible. > Here are 3 definitions of momentum, all common terms in use by > laypersons Perhaps. > and scientists Since you have never claimed to have read any scientific source, I doubt any claim that you know how scientists use any term. > alike in the various ways it can be defined. > None of them define it like you do above, but yours is as common and as > valid as the others as an equation useful for determining the amount of > force in a given momentum. You cannot avoid seeing that translates > literally into what I said, that force and momentum are equivalent. Certainly I can since that is false. > See the F on the one side of the = sign? And the other stuff on the > other side? That means, simply, that if you calculate everything on > the right side, it will tell you how much force is involved. Yes. Therefore it will tell you right away that force and momentum are not equivalent, since when you do this calculation you will find that F is not the same as p. > Therefore, if I want to know how much force is involved in a given > momentum, I have that equation handy. Did you forget the d/dt part already? It doesn't tell you how much force is "involved in a given momentum". It tells you how much force is involved in a given momentum CHANGE. It doesn't matter whether the momentum involved is a ping pong ball moving at 1 cm/sec or an elephant at escape velocity. A given force will translate into exactly the same momentum change for both, because that equation tells you that force and momentum change are equal, are equivalent, are the same thing. > "1. capacity for progressive development: the power to increase or > develop at an ever-growing pace > The project was in danger of losing momentum. This is not the meaning of p in any equation. > 2. forward movement: the speed or force of forward movement of an > object > the momentum gained on the downhill stretches of the course I agree that this dictionary definition just said that momentum is the same as "speed of forward movement" and "force of forward movement". "Force of forward movement" is not a defined quantity in physics. When somebody wants to do an analysis, the momentum, force and speed must be kept separate. > 3. physics measure of movement: a quantity that expresses the motion > of a body and its resistance to slowing down. It is equal to the > product of the body's mass and velocity. Symbol p". There you go. This is a correct classical definition of momentum, as used in Newton. I believe Newton called it "motus" or motion, but explicitly said that it is proportional to mass and velocity. The style doesn't seem to have been to define things in terms of equations. Instead, he says that the motus of an object is doubled if you double the mass, and quadrupled if you also double the velocity. The correct relativistic definition of momentum is not quite the same thing. > Microsoft® Encarta® Reference Library 2005. © 1993-2004 Microsoft > Corporation. All rights reserved. Oh gee, what a surprise. Have you considered looking at a physics book before declaring what physics books say? > Now in 1., above, momentum is used to mean a "power". It isn't used to represent a mathematical quantity at all. It's not a thing that has units or can be measured. > In 2., it is > used as a force OR a speed, Yes, two vaguely defined lay terms are equated. > and in 3., it is a quantity, just like in > your equation. It is the quantity mv, and even Encarta says that is the physics definition and does not pretend that the first two are used in physics, as you seem to be doing. > Here's another definition: A property of a moving body > that determines the length of time required to bring it to rest when > under the action of a constant force or moment. That doesn't tell me HOW it determines that, so it's useless as a definition. It doesn't tell me how to determine if object A has twice the momentum of object B or only 1.5 times as much. > Broadly impetus. That > too is an explanation of momentum as your equation uses the term. But an incomplete one. It is hinting at the existence of the F = dp/dt relation, but not telling you the relation explicitly. Momentum COULD have been defined in terms of force by your 4th definition, if it said explicitly that momentum is proportional to force and the time required to bring it to zero. The inverse of F = dp/dt, for a constant force bringing a momentum to zero, is p = F*t (in magnitude. They have opposite signs. See below for the correct expression using the correct signs). In general p(t) = integral F(t) dt. > Yet you dare to say that the only valid way to use the terms force and > momentum is other than the way they are used in my examples? Yes, I DARE to say that the only way to use force and momentum in an equation is in the single definition that said it was the physics definition, the single definition that offered an equation. > You claim > the right to demand the use of those two terms only in the way your > equation defines it? I claim the right to defend the use of those terms in physics as being the ones that Encarta told you were the physics definitions of those terms. > > By the statement F = dp/dt, I mean that force is the time rate > > of change of momentum. > > > > In what way is that not an answer to the question? > > > > It is an equation that tells you, given an expression for momentum, > > how to find out the corresponding force. It tells you, given > > a force, how to find out its effect on momentum. > > > > > High-sounding words, but they're nonsense. What is "an expression for > momentum" if not just a quantity? It's an expression that tells you momentum as a function of time, which need not be constant. > It does NOT tell you how to find out > a force's effect on momentum! Um, yes it does. It tells you that the force is the time derivative of the momentum. That doesn't leave a lot of wiggle room. If I know how force behaves in time and what my initial momentum is, then I know precisely what the momentum is at every moment in time that follows. What else could you mean by "find out a force's effect on momentum" other than "tell you exactly what the momentum will be at every time forever and ever as a result of that force?" > Tell us, your Nakedness, just how do we find out a force's effect > on momentum from F=dp/ dt? You are so funny! By integration. Would you care to ask a specific example? That is sufficient to tell you, for instance, what the effect of a rocket engine is on a rocket, even though the mass of the rocket is changing as fuel is burned. It is sufficient to tell you what gravitational force does to a ballistic object (a thrown rock, a falling mass, a satellite in orbit). It is sufficient to tell you how those things behave even when you take into account air resistance and the variability of gravity with altitude and distance. Those latter can't be solved with hand, they require computers. But the computer implementation can be very simple, and plenty of people with little physics training write computer games using F = dp/dt to model motion. Here are some simple examples of how you use F = dp/dt to tell you the effect of F on a momentum. Suppose you have a ping-pong ball of mass 0.01 kg moving at 300 m/sec (your opponent has a VERY fast serve). How long will it take a force of 10 N to bring it to a halt? p = integral of F dt = p0 + F*t I want to bring it to a stop, p = 0, so p0 + F*t = 0, t = -p0/F. The minus signs are because momentum and the force to change it in this case are in opposite directions. Momentum points forward, force points back. In terms of their magnitudes, t = |p0|/|F| = (0.01*300)/10 = 0.3 seconds. > But I knew all that already just by reading Encarta. What Encarta > failed to teach me was how to find the effect a force has on momentum > using your famous equation. Watch and learn. Now let's look at a more complicated example. I have the same ping-pong ball, but this time, the force is acting perpendicular to the original momentum. F = dp/dt is actually a vector equation. If the force is toward the East, then all the changes will be eastward whatever the actual motion is. In (2-D) vector terms, this equation is two equations: F_north = dp_north/dt = 0 (no northward component of force), so p_north = constant = 0.01*300 = 3 kg-m/sec. F_east = dp_east/dt = 10 N. p_east = p_0 + F*t. I originally had no eastward component of motion, so p_0 = 0. F = 10 N, so p_east = 10*t My total momentum as a function of time has a magnitude of sqrt(3^2 + (10t)^2) = sqrt(9 + 100t^2). The direction of this momentum vector is the direction of a vector with northward component 3, eastward component 10t. In navigation terms, which measure "bearings" clockwise from north, my bearing is arctan(10t/3) The effect of any force on any momentum can be worked out, but in general you might need a computer to do it. - Randy
From: Traveler on 26 Sep 2005 09:38 On Mon, 26 Sep 2005 00:15:26 -0600, Happy Hippy <J0HN(a)accesscomm.ca> wrote: >Pardonnez >What about mass? >State? >Or effect? Mass is a state, of course. So is energy. And state is just another name for property. Louis Savain Why Software Is Bad and What We Can Do to Fix It: http://www.rebelscience.org/Cosas/Reliability.htm
From: Herman Trivilino on 26 Sep 2005 12:13
"TomGee" <lvlus(a)hotmail.com> wrote ... > There are no bodies at rest in our universe except wrt to other > objects. What is it that keeps bodies at rest (with respect to other objects)? >> Is a state of rest is equivalent to a state of uniform motion? > No. Is a state of rest (with respect to another object) equivalent to a state of uniform motion (with respect to another object)? If your worldview requires a cause for a state of uniform motion (with respect to another object), why does it not also require the SAME cause for a state of rest (with respect to another object)? ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |