Motion
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From: TomGee on 19 Oct 2005 01:53 Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > TomGee wrote: > > > > Randy Poe wrote: > > > > > In situations (b) and (c), there is a force present, either > > > > > at right angles to the motion (b) or not at right angles > > > > > to the motion (c). > > > > > > > > > Let's start from here. I don't recall your situations, but I will > > > > accept the above as being correct wrt (b) and (c). > > > > > > Goody. To refresh the conversation: > > > > > > (a) No external forces. Straight line, constant velocity motion. > > > > > > (b) Circular motion. External force at right angles to the motion > > > at all time. Constant speed, non-constant velocity (because the > > > direction is changing). > > > > > > (c) General motion. External force at arbitrary angle to the > > > motion, so it is doing work. Both speed and direction in general > > > are changing. > > > > > > > > There's not "inertial force". > > > > > > > > Yes, there is, everyone knows that. > > > > > > No, there isn't. Inertia is not a force. Can you pull a quote > > > from Tipler where he refers to the force of inertia? He's > > > a someone, isn't he? > > > > > No, > > Then you don't have any backing for your claim that "everybody > knows" inertia is a force, do you? > > > but if you're referring to the claim by some that some forces are > > psuedo-forces, > > I'm not. You stated something about inertia being a force, and > that "everybody knows it". I asked you to back that up. You > respond by changing the subject. > > > or don't really exist, they can take that up with > > Newton, as psuedo-forces exist in certain situations. > > A pseudo-force, like any other force, is one that causes > a change in velocity. They are perfectly well modeled by > Newton's laws. > > What makes them "pseudo" is that they are artifacts caused > by being observed in accelerating frames, such as the > surface of earth. An observer in a non-accelerated frame > of reference will see no changes in velocity and no force. > > A centrifuge works perfectly well, settling its contents > as if there is a centrifugal force much stronger than > gravity and pointing directly outward. Yet the situation > is equally well described by a person standing in the lab > who sees only an inward-directed, centripetal force. > > > In mechanics, > > the gravitational mass of a body is the source of a force that exists > > between it and another body and it is effectively the same as its > > inertial mass, which is what determines the motional response of the > > body to any force exerted on it. > > F = ma, or in general F = dp/dt > > Nope, that's backwards. The general case, or, the function, is F=ma and the special case, or, a derivative of the function, is F=dp/dt. > > > Yes, mass determines the acceleration caused by a force. > > That does not make mass a force. Force is the thing on > the left-hand side of those equations. You can't use > the mass as a force in those equations. That's why it's > given a different symbol. > > That little "thingy" between the two sides is a different symbol used to mean that the two sides are reversible. > > > > > > > > Everyone knows that. > > > > > > > > > > Everyone does not "know that". > > > > > Well, you're the only one who doesn't know that for a fact. > > And Tipler, apparently. I asked you for an example of Tipler > using inertia as a force. You say you can't provide one. > > So I'm not the "only one", am I? > > Yes, you are. > > > But fine, F = ma. Tell me how I use inertia for F to describe > the motion of a body. > > > > I see. And as we watch it change from nonmoving to moving, we > > > can tell that something is "resisting" this change by...? > > > > > By measuring the amount of energy required to cause it to start moving. > > The amount of energy required for it to have 1 Joule of > KE is 1 Joule. The amount of energy required for it to > have 10 J of KE is 10 J. The amount of energy required > for it to have 3.485 J of KE is 3.485 J. > > > So what? That's a tautology not worth the paper it's written on. > > > All I see is work energy being turned into KE. Where is > this additional thing, this "resistance", that requires > energy to overcome? > > >From a body's Inertial Force, according to Newton. > > > > > And the rate at which it accelerates can be determined precisely > > > from F = dp/dt. > > > > > No, that will depend on the forces and the body mass sizes involved in > > the interaction.. > > Yes. That's what it means to use F = dp/dt to determine acceleration. > > No it isn't. > > > I plug in the correct expression for F and for dp/dt, using the > correct value of mass, and I get acceleration. F = dp/dt is > the equation I use to do this. > > No. You get the rate of acceleration, not the amount of the acceleration. PD tried to warn you, but - Too Late!
From: Randy Poe on 19 Oct 2005 09:25 TomGee wrote: > Randy Poe wrote: > > F = ma, or in general F = dp/dt > > > > > Nope, that's backwards. The general case, or, the function, is F=ma > and the special case, or, a derivative of the function, is F=dp/dt. Do you understand that "general case" means true in more situations than "special case"? How do I apply F = ma to a rocket, in which reaction mass is burned and therefore m is not a constant in time? Since F = ma is the "general case", it should apply in this case, right? Here are some sample parameters: A rocket has initial mass 10 kg. It is carrying 8 kg of fuel, which it burns at 0.5 kg/sec, providing 5 N of thrust. The rocket is in deep space, far from any gravitational influence. Tell me its position, acceleration, and velocity as a function of time (or, simpler, any one of the above). - Randy
From: TomGee on 19 Oct 2005 12:29 Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > F = ma, or in general F = dp/dt > > > > > > > > Nope, that's backwards. The general case, or, the function, is F=ma > > and the special case, or, a derivative of the function, is F=dp/dt. > > Do you understand that "general case" means true in more > situations than "special case"? > > It's still backwards the way you said it. And above are you saying that "general case" is truer in more sits than "special case"? I agree. > > > How do I apply F = ma to a rocket, in which reaction mass > is burned and therefore m is not a constant in time? Since > F = ma is the "general case", it should apply in this case, > right? > > Right. You apply the force produced by the rocket fuel against the mass of the ship less the fuel being burned as the rocket accelerates. To find the rate of acceleration, if that is what you want to do, you simply derive that from the general case facts. > > > Here are some sample parameters: A rocket has initial > mass 10 kg. It is carrying 8 kg of fuel, which it burns > at 0.5 kg/sec, providing 5 N of thrust. The rocket > is in deep space, far from any gravitational influence. > > Tell me its position, acceleration, and velocity as a function > of time (or, simpler, any one of the above). > > No. You've already lost your counter-arguments - why go on like this forever? Give up, already. Throw in the towel, submit to the logic, accept the better explanations....IOWs, stop your silly nonsense!
From: Randy Poe on 19 Oct 2005 12:58 TomGee wrote: > Randy Poe wrote: > > TomGee wrote: > > > Randy Poe wrote: > > > > F = ma, or in general F = dp/dt > > > > > > > > > > > Nope, that's backwards. The general case, or, the function, is F=ma > > > and the special case, or, a derivative of the function, is F=dp/dt. > > > > Do you understand that "general case" means true in more > > situations than "special case"? > > > > > It's still backwards the way you said it. And above are you saying > that "general case" is truer in more sits than "special case"? I > agree. Good. > > How do I apply F = ma to a rocket, in which reaction mass > > is burned and therefore m is not a constant in time? Since > > F = ma is the "general case", it should apply in this case, > > right? > > > Right. You apply the force produced by the rocket fuel against the > mass of the ship less the fuel being burned as the rocket accelerates. The mass of the ship is not a constant. > To find the rate of acceleration, if that is what you want to do, you > simply derive that from the general case facts. Can you show me how that works? > > Here are some sample parameters: A rocket has initial > > mass 10 kg. It is carrying 8 kg of fuel, which it burns > > at 0.5 kg/sec, providing 5 N of thrust. The rocket > > is in deep space, far from any gravitational influence. > > > > Tell me its position, acceleration, and velocity as a function > > of time (or, simpler, any one of the above). > > > No. You can't, with F = ma. I can, with F = dp/dt. > You've already lost your counter-arguments I did? What did I lose? > - why go on like this forever? Give up, already. Just out of curiosity, what part of this discussion do you think I "lost"? - Randy
From: TomGee on 19 Oct 2005 13:21
Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > TomGee wrote: > > > > Randy Poe wrote: > > > > > F = ma, or in general F = dp/dt > > > > > > > > > > > > > > Nope, that's backwards. The general case, or, the function, is F=ma > > > > and the special case, or, a derivative of the function, is F=dp/dt. > > > > > > Do you understand that "general case" means true in more > > > situations than "special case"? > > > > > > > > It's still backwards the way you said it. And above are you saying > > that "general case" is truer in more sits than "special case"? I > > agree. > > Good. > > > > How do I apply F = ma to a rocket, in which reaction mass > > > is burned and therefore m is not a constant in time? Since > > > F = ma is the "general case", it should apply in this case, > > > right? > > > > > Right. You apply the force produced by the rocket fuel against the > > mass of the ship less the fuel being burned as the rocket accelerates. > > The mass of the ship is not a constant. > > That's what I said. I guess you thought I would fall for your little setup, eh? I have passed all the tests you have put up for me, but it seems they will never end. You have been shown to be wrong from the beginning of your objections to my model, where you said, "There is no force that keeps bodies in motion. Forces only act to change motion", to your last ploys in this post. > > > > To find the rate of acceleration, if that is what you want to do, you > > simply derive that from the general case facts. > > Can you show me how that works? > > I won't because that's what you have been trying to do since the very first time you entered this discussion - direct attention away from the topic that caused your initial kneejerk reaction. I'm sure there are others like you who just love to derive rates of acceleration all day long, why not find them and join them. That is not the topic here. > > SNIP > > > > You've already lost your counter-arguments > > I did? What did I lose? > > All your counter-arguments. > > > > - why go on like this forever? Give up, already. > > Just out of curiosity, what part of this discussion do you > think I "lost"? > > All your counter-arguments to my claims of a body's inherent force being what keeps it at CV when free of any net external forces. |