Motion
in [Physics]
|
Prev: Free fall
Next: 50% OF POPULATION BELOW AVG IQ!
From: Randy Poe on 19 Oct 2005 13:42 TomGee wrote: > Randy Poe wrote: > > The mass of the ship is not a constant. > > > > > That's what I said. Therefore F = ma is not usable. > I guess you thought I would fall for your little > setup, eh? It wasn't a "setup". You said that F = ma is true in general, and F = dp/dt is not. I propose a situation in which F = ma is incorrect, and F = dp/dt is correct. It's a response to your direct claim. Since you choose not to show why you think F = ma is correct for rockets (my "trap": I ask you to back up your claim with an example), then let me "trap" you by asking you for some other situation in which F = ma applies but F = dp/dt does not. Because, after all, F = ma is true in general but F = dp/dt is not, right? > I have passed all the tests you have put up for me, You have? You always say "no". Did blank papers get passing grades when you were in school? > but it seems they will never end. Yes, every time you make a statement like that F = dp/dt doesn't apply, I'm going to ask you for a situation where F = dp/dt doesn't apply. Since you always say no, I see no reason not to keep asking for justification of that claim. > You have been shown to be wrong from the > beginning of your objections to my model, I have? Can you state a single instance where I have "been shown to be wrong"? Just one? - Randy
From: stephen on 19 Oct 2005 13:44 In sci.math Randy Poe <poespam-trap(a)yahoo.com> wrote: > TomGee wrote: >> Randy Poe wrote: >> > 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 I think the first person who tries to use math loses according to the TomGee rules of debate. Or maybe that is according to the MSN Encarta rules of debate. :) Stephen
From: PD on 19 Oct 2005 13:54 step...(a)nomail.com wrote: > In sci.math Randy Poe <poespam-trap(a)yahoo.com> wrote: > > > TomGee wrote: > >> Randy Poe wrote: > > >> > 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 > > I think the first person who tries to use math loses > according to the TomGee rules of debate. Or maybe > that is according to the MSN Encarta rules of debate. :) > > Stephen I have yet to discern TomGee Rules of Debate. So far it's all bluster and dither, rope-a-dope waffling. It reminds me of a boxer who dances and sniffs and feints but never actually throws a punch. With blood coming out his nose and out of a deep gash over his eye, he continues to splutter through his mouth-guard, "You never touched me. I've got you down. I'm counting to ten." PD
From: TomGee on 19 Oct 2005 14:28 Just look up any one of 'em, Randy. I have shown you to be wrong in all of 'em. You keep flinging in your strawman derivative cause you have nothing valid to counter with, and that has nothing to do with the topic but still you continue to use it like a robot that is stuck in its motion, or like a broken record that's stuck at one line of a song.
From: Randy Poe on 19 Oct 2005 15:23
TomGee wrote: > Just look up any one of 'em, Randy. I have shown you to be wrong in > all of 'em. In other words, none. If there is one, you should have no problem coming up with one, should you? They should be all over this thread. Just pick any one and demonstrate that you "showed me wrong". I predict there will be no content in your reply. - Randy |