Correct! You're starting to get it!
Easy. If after 1 second of steady acceleration the mass is moving at 10m/s, then the acceleration was 10 m/s^2.
The Swing of a Pendulum
- Thread starter Motor Daddy
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Correct! You're starting to get it!
Easy. If after 1 second of steady acceleration the mass is moving at 10m/s, then the acceleration was 10 m/s^2.
Motor Daddy
Valued Senior Member
10 m/s compared to what? The distance between the two rockets remains the same at all times. There is no measurable change in your relative velocity.
They are moving with respect to some set of coordinates. I am moving relative to a car on the road and the car is moving relative to me. If said car doesn't physically exist then I am still moving with respect to the inertial frame. The problem is your failure to grasp the concept of an inertial frame. Each and every time it comes back to your inability to move beyond your absolute motion ignorance.
Since you can only handle specific numbers, not algebra, I'll give an example in the Newtonian case. Car 1 and Car 2 are a distance 100m apart and at rest with respect to one another. They both start accelerating with rate $$a = 10m/s^{2}$$ for 10 seconds before cutting their engines.
We define frame F by the coordinates (x,t) where Car 1 is initially at $$x_{1}(0) = 0$$ and Car 2 $$x_{2}(0) = 100$$. They accelerate for 10 seconds uniformly, so by the SUVAT equations for $$0<t<10$$ we have $$x_{1}(t) = \frac{1}{2}at^{2} = 5t^{2}$$ and $$x_{2}(t) = 100 + \frac{1}{2}at^{2} = 100 + 5t^{2}$$. After T=10 seconds we have $$x_{1}(10) = 50$$[ and $$x_{2}(10) = 150$$. Now for $$t>10$$ we have no acceleration and velocity v =aT = 100 so $$x_{1}(t) = \frac{1}{2}aT^{2} + v(t-10) = 50 + 100(t-10)$$. Car 2 has $$x_{2}(t) = 150 + 100(t-10)$$. Throughout we have had $$x_{2}-x_{1} = 100$$.
Now let's work in F', with coordinates (x',t') where $$x' = x-vt = x - (aT)t = x-100t$$ and $$t'=t$$ by Galilean transforms. Initially car 1 is at x' = 0 and moving with speed -aT = -100. The acceleration then gives for 0<t<10 the position $$x'_{1}(t') = -100t' + \frac{1}{2}a(t')^{2} = -100t' + 5(t')^{2}$$. After 10 seconds we have t'=10 so $$x'_{1}(10) = -1000 + 500 = -500$$. The velocity is now $$-aT + aT = 0$$ so for t'>10 we have $$x'_{1}(t') = -500$$. Repeating this for car 2 we get $$x'_{2}(t') = 100 - 100t' + 5(t')^{2}$$ for 0<t'<10 and then for t'>10 we get $$x'_{2}(t) = -400$$. Again, we always have $$x'_{2}-x'_{1} = 100$$ throughout the motion.
You can recover all of these results for the F' frame by applying the Galilean transforms $$x' = x-100t$$ and $$t'=t$$ to the results from frame F, as it should be.
So both frames, F and F', describe precisely the situation you've given us, two vehicles always the same distance apart and who undergo acceleration. Despite the fact we haven't had to resort to an absolute frame we've got a consistent description of the system using relative inertial frames. I could have considered F'', an arbitrary inertial frame, and precisely the same would happen. You just change the initial position and velocities but the dynamics are otherwise unchanged (not surprising since Galilean transforms amount to (time dependent) translations in 1d examples). Someone in F would say the cars start at speed 0 and end at speed 100 while someone in F' would say they start at speed -100 and end at speed 0. Someone in a third frame, say one moving at speed 50m/s relative to F (and thus -50 relative to F') would say they start with speed -50 and end at speed +50. In this case we have F'' coordinates (x'',t'') where x'' = x - 50t and t'' = t. There doesn't need to be a third car for us to consider this frame, for us to say "The cars are initially moving at speed 50 with respect to the F'' frame". This is why your 'prove it' comment is flawed, the frame is a conceptual thing, a choice of description where the cars are moving at speed 50m/s. It doesn't need an absolute frame for us to work in the (x'',t'') coordinates in our calculations.
For any initial velocity you care to consider there is a frame where the cars start at that speed and, for Newtonian systems, after acceleration they will end up going 100m/s faster than they started, there is no need to assert the existence of some absolute speed to do any of this. It comes about by the choice of coordinates (x*,t*) where x* = x+Vt and t*=t. In this frame the cars are initially moving at speed +V. Since we have defined this conceptual frame relative to the objects in question it has physically meaning. If you were driving a third car then you can put yourself in this frame by driving at the speed where you see the cars moving at speed V relative to you. Then you'd be in that frame. Everything is done by relative comparisons, no need to do anything 'absolutely'.
This is, as I said, literally the stuff of high school mechanics problems, just putting in specific numbers to general formulae and seeing how it all fits together. I provided you with the general case and you couldn't even put in example numbers yourself, you couldn't even grasp the general concepts involved. Instead you demand I put a numerical value to their velocity, as if doing so would mean there's an absolute motion. No, I can give a specific value by saying it is with respect to a particular frame. However, someone else could say the cars move with a different initial velocity and they would be equally valid. All we need to do to convert between one anothers descriptions is apply the appropriate Galilean transform (or Lorentz if working in special relativity). The fact our descriptions would be linked by Galilean (or Lorentzian) transforms removes the need for an absolute frame. Saying "The cars started at 30m/s" doesn't imply an absolute frame, it implies the use of a specific inertial frame.
This is why I asked you to go and read and actually do some high school mechanics, you'd know all of this if you'd done so and I wouldn't have to be explaining children's level work to you. You're a grown man, don't you wish to expand your horizons, to learn new things and think about new ideas? Why do you deliberately wallow in your own ignorance? How many times must posts like this be done in response to an ill thought out and ignorant post of yours before you see the pathetic nature of your actions?
Motor Daddy
Valued Senior Member
If after 1 second of acceleration you claim the mass is moving 10 m/s then if after 1 second of acceleration the mass is moving 20 m/s, what was the acceleration
Motor Daddy
Valued Senior Member
Trying to use make believe made up acceleration rates is just plain stupid. How do you know that the velocity increased at a rate of 10 m/s per second?
You know what? I think MD is actually starting to see his mistake, so he is being deliberately obtuse.
Motor Daddy
Valued Senior Member
No, I am making a point that is irrefutable. Do you know what acceleration is? Acceleration is not force! Acceleration is the rate of change of velocity. The only velocity you know is your relative velocity between you and the other rocket, and it is not changing, so what is your basis to claim your velocity increased? It most certainly did not!
You (along with all the other Einstein groupies) want to try to play the game of "velocity has to be relative to something." Well, what is your change in velocity relative to?
Assuming steady acceleration, 20 m/s.
Motor Daddy
Valued Senior Member
Assuming an initial velocity of 0 m/s, correct? I suppose if I asked you what the acceleration was if after one second the mass was moving 100 m/s you would say 100 m/s^2, and so on...
In all of your answers you ASSUME that the initial velocity was 0 m/s, why? Why not assume that the object had a initial velocity of 99 m/s and after 1 second it had a velocity of 100 m/s, so the acceleration was not 100 m/s^2 was it?
I never assumed a zero velocity. It is so obvious that I believe even you can see it. If you were in a closed container moving at a constant velocity you would have no idea how fast you were moving. You could even be stationary and you would have no idea. If you put a marble on a flat surface the marble would just sit there wouldn't it? If the container were to accelerate the marble would roll along the surface and you would feel the acceleration. You would not know if your initial velocity before the acceleration was 0 or 1000 kph. You could do a calculation and determine how much the velocity changed but you would not know what your new velocity was would you.
Can you bring yourself to admit that this scenario is what would happen or explain why it would not?
Motor Daddy
Valued Senior Member
In order to make a statement about your own motion in space you must first measure your motion according to the light sphere. You have not measured your motion according to the light sphere so all you have is relative velocity. You can measure the distance between you and any other object over a duration of time and know the relative velocity but that says nothing as to your initial and final velocity so you have no way of using any type of unit for your acceleration, since you have no idea what your initial velocity or final velocity are.
Motor Daddy
Valued Senior Member
Origin, here is a very simple question for you. If two rockets are inside a sphere and the distance between them is increasing at the rate of 20 m/s, will they each exit the sphere at the same point on the sphere, one later than the other?
Yes, we are assuming the mass has an initial velocity relative to the rocket of 0 m/s.
Because a zero initial relative velocity is easier to measure. No need for it; you could easily start the mass at 10 m/s, accelerate at 20 m/s^2 for 1 second and observe a final relative speed of 30 m/s. Just as valid, just takes more room inside your rocket.
Sure, you could start at 99 m/s (and see a final velocity of 119 m/s) if you have a big enough rocket.
Motor Daddy
Valued Senior Member
Do you not understand that you can't just change numbers like that without causing all sorts of changes in the universe? Each set of numbers means something. Each set of numbers means a specific reality. You are changing realities when you switch the numbers like that.
That's not what I said. I said if you assumed the initial velocity was 99 m/s and the velocity was 100 m/s after 1 second of acceleration then that is not an acceleration of 100 m/s^2. Do you not understand the difference?
How do you know if your acceleration is increasing velocity or decreasing velocity?
I explicitly gave the general equations. I also explicitly said I would do a numerical value example because undoubtedly you are incapable of understanding the general algebraic version.
Well done on failing to even understand the concept of an example! My god the education system was a waste of time for you, wasn't it!
Motor Daddy
Valued Senior Member
Two rockets in space with a distance between them. How do you determine the distance between the rockets?
However one prefers. Use a ruler at rest with respect to the vehicles. Bounce light off one another and time it. Actually have vehicle 1 accelerate at a constant rate and time how long it takes to catch up with vehicle 2 and then reverse the kinematics to return to the original configuration. Take your pick, all of them are valid methods which do not require recourse to an absolute frame. Besides, the way in which it is done is a separate matter to your complaining that there are issues defining velocity due to them always being the same distance apart.
Or are you trying to change the direction of the discussion to avoid accepting the posts I've made? Please answer the following questions directly and clearly :
- Do you understand and accept that the cars can measure acceleration they experience without having to consider the distance between them?
- Do you understand and accept the numerical example, with a=10 and L=100 and T=10, I did of the 2 car example?
- Do you understand and accept the general algebraic description I did of the 2 car example?
- Do you understand that in those examples I didn't need to make reference to an absolute frame?
If you're unwilling to respond to these questions, even if to only say "I don't accept any of them" will result in me not answering your questions, given I've responded to many them and into quite some detail, often having to repeat myself. It is time you stopped bouncing from claim to claim, each time moving on to a new one when your old one is debunked.
Motor Daddy
Valued Senior Member
I choose the ruler method. We extend a ruler between the rockets and it reads 100 meters at t=0. At t=1 it reads 100 meters. At t=2 is reads 100 meters, so the distance remains unchanged and for the purposes of this entire exercise the distance between the rockets remains 100 meters.
No, I don't accept that. One car alone in space has no relative velocity to another object and you have no way of knowing the absolute velocity like I do. You have no measure of velocity or change of velocity with no relative velocity, do you??? If you do please share with the entire group exactly which velocity you are talking about if it isn't a relative velocity or a absolute velocity?
AN, in order to know an acceleration you first must have measured a change in velocity. In order to measure a change in velocity you first must have measured a velocity. In order to have measured a velocity you first must have measured a change in distance over a duration of time. You have no basis for claiming you changed distance, velocity, or even changed velocity since all your measurements are based on the other rocket. You can't just claim an object accelerates at say 9.81 m/s^2 like a beach ball at rest on a beach. The ball is not getting any closer to the center of the earth so the ball's velocity is 0 m/s towards the center of the earth. Ten minutes later when you measure it again the distance to the center of the earth from the beach ball is exactly the same, so the ball's velocity is the same, no change over a duration of time. No change in velocity over a duration of time means a zero acceleration, ie the ball is not accelerating towards the center of the earth because the velocity of the ball towards the center of the earth is not changing, it remains at 0 m/s while at rest on the beach!
No, because you did not measure a 10 m/s^2 acceleration you just claimed it. You have no measurements so how do you know what the velocity or increase/decrease in velocity was?
Do you understand that when you said the acceleration rate was 10 m/s^2 that that was an acceleration in the absolute frame, one which you have no means of measuring, so how would you know that??
Motor Daddy
Valued Senior Member
Still workin' on that torque/HP problem? Damn, it's taking you quite some time. You must really be working hard on that one. If you need assistance let me know, I'm here to help!