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Originally Posted By: paul

conservation of momentum is the philosophic or


OK so we don't use momentum at all anymore in this discussion.

Quote:

I repeat a 4000N force can accelerate a 1000 kg mass
to 4 m/s/s in 1 second , that mass will move a distance of
2 meters.


I agree.

Quote:

also , there is no additional force applied to the pipe
in the direction that would reverse the pipes direction.


During that acceleration? Yes I agree to that too.

Quote:

you do agree that the 100kg mass traveling at 40 m/s
passing through the 1st turn will apply a 4000N force
to the 1st turn , correct?


It should be a negative force applied to the turn. And its magnitude is free to be whatever you choose, so yes I agree it can be -4000N.

Quote:

and that this 4000N force will apply to the pipe, correct?

Yes but in the negative direction.


Quote:

so if you agree on the above then the pipe will slow
from 4 m/s velocity to 0 m/s velocity in 1 second , correct?

Yes I agree on that too.


Do agree that the same 1st turn will also apply a 4000N force to the mass, in the opposite direction to its motion?

If you agree on that, then you should also agree that the mass will slow from -40m/s velocity to 0m/s velocity in 1 second, correct? Here's how:

F=ma
4000N = 100kg * 40m/s^2
So it slows from -40m/s to 0 in 1 second.

Now the pipe's stopped, and the mass has stopped too. So the turnaround isn't complete. The mass needs to get back up to 40m/s in the opposite direction. Doing that will start moving the stationary pipe too.

.
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Quote:
Do agree that the same 1st turn will also apply a 4000N force to the mass, in the opposite direction to its motion?

If you agree on that, then you should also agree that the mass will slow from -40m/s velocity to 0m/s velocity in 1 second, correct? Here's how:

F=ma
4000N = 100kg * 40m/s^2
So it slows from -40m/s to 0 in 1 second.

Now the pipe's stopped, and the mass has stopped too. So the turnaround isn't complete. The mass needs to get back up to 40m/s in the opposite direction. Doing that will start moving the stationary pipe too.


I cant agree that the 100 kg mass will stop.

theres no force that is applied to stop it.

in order for the pipe to apply a force to the mass
the pipe must move in that direction first.

as the 100 kg mass travels through the 1st turn
it is pressing against the pipe in the (-) direction.

the 100 kg mass never presses against the pipe
in the (+) direction as it passes through the 1st turn.

so the pipe never has a (+) force placed on it , that
would cause the pipe to move in the (+) direction other
than the friction between the mass and the 1st turn.

and these cancel out also.

ie..
the (+)friction through the 2nd half of the 1st turn cancels
out the (-) friction through 1st half of the 1st turn.

--------------

now as for the mass stopping , there is no force applied
to the mass that would stop the mass.

the mass is only being turned around by the 1st turn.

think of a pendulum , it swings on a string but it doesnt
stop as it reaches the bottom at the end of the
1st downward swing and it will only stop at the top of
the upward swing.

yet if you put the pendulum on a (non elastic) scale
that measures downward force and the mass of the ball
of the pendulum is 100 kg and the string is 4.9035 m long
it will swing to the bottom in 1 second.

the force applied is 980.7N = 100 kg * 9.807 m/s/s gc.
its acceleration is 9.807 m/s/s
its initial velocity is 0 m/s.
its average velocity is 4.9035 m/s.
its final velocity is 9.807 m/s.
at the bottom of the swing.

so the ball does not stop at the bottom of the swing , it
keeps going , just like the mass in the 1st turn keeps going.

the only reason the ball stops at the top of the up swing
is the 9.807 gc force pulling on it.

the force applied is 980.7N = 100 kg * 9.807 m/s/s gc.
its acceleration is 9.807 m/s/s
its initial velocity is 9.807 m/s.
its average velocity is 4.9035 m/s.
its final velocity is 0 m/s.
at the top of the swing.

but the (non elastic) scale that measures downward force will
read a force of 980.7 N as the ball swings past the bottom
on its way to the top of the up swing.

so after the mass passes through the 1st turn it still has
its velocity of 40 m/s assumming no friction.

--------------












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I noticed a mistake when I included a distance
ro the turnarounds.
Quote:

5) we can use 1 second for the time it takes for the mass to pass through each turn.

the mass has a final velocity of 40 m/s after it leaves the accelerator in the first cycle.

so we can use a 20 meter travel distance through each turnaround.

the mass will pass through the turnaround in 1 second.


make that 0.5 seconds , sorry.
or we can make the turnarounds 40 meters in lenght and keep the 1 second.




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We can safely assume friction is zero. In reality we can get arbitrarily close to zero and add a bit of energy every now and then to compensate. The mass could even be an electron travelling through a superconductor, where there's exactly zero friction.

Originally Posted By: paul

the 100 kg mass never presses against the pipe
in the (+) direction as it passes through the 1st turn.

so the pipe never has a (+) force placed on it , that
would cause the pipe to move in the (+) direction other
than the friction between the mass and the 1st turn.


Yes I agree. The _pipe_ never has a + force on it in the 1st turn. But the mass does. That's from Newton's 3rd law: action and reaction are equal and opposite. You used this law to find that both the pipe and mass experience a force in the accelerator.

When I say 'stop' I mean stop in the direction we're considering. Sure it keeps going at a speed of 40m/s, but as its velocity changes from backwards to forwards, the longitudinal component of it must obviously pass through 0.

It's the same with a pendulum. The _vertical_ component of velocity reaches 0 at the bottom of the swing. Sure it keeps moving, but it's stopped in the vertical direction. If you remove all forces at that moment, then it'll remain stopped in the vertical direction - and fly off in a straight line sideways.

If you're saying the mass doesn't experience a force in the turnaround, then you're saying Newton's 1st, 2nd and 3rd laws are all wrong.

The 1st law says it should keep going in a straight line, since there's no force acting on it. But it's turning.

The 2nd law says F=ma: zero force = 100kg * zero acceleration. But it's clearly accelerating because its direction is changing.

The 3rd law says if it applies -4000N to the pipe, then the pipe applies +4000N to the mass.

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Quote:
When I say 'stop' I mean stop in the direction we're considering. Sure it keeps going at a speed of 40m/s, but as its velocity changes from backwards to forwards, the longitudinal component of it must obviously pass through 0.


ok , are you still saying that the pipe will stop and then
move in the (+) direction?

because we need to clear that up first.

durring the time that the mass passes through the 1st turn.

I am saying that the pipe slows and stops.
and
I am saying that the mass still has a velocity of 40 m/s.
what are you saying?

please finalize your answers by simply stating them as I did above.

Quote:
If you're saying the mass doesn't experience a force in the turnaround, then you're saying Newton's 1st, 2nd and 3rd laws are all wrong.



if these laws are correct then the math will back them up.

if these laws are not correct then the math will not back them up , and they should be disposed or corrected.

we cannot anlalyze this concept using laws.


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Originally Posted By: paul

ok , are you still saying that the pipe will stop and then
move in the (+) direction?

No, we're both saying the accelerator causes the pipe to start moving in the + direction. I'm saying when the mass turns around, it applies enough force to first stop the pipe, then start it moving in the - direction.

Quote:

I am saying that the pipe slows and stops.
and
I am saying that the mass still has a velocity of 40 m/s.


That's a contradiction. The mass changes direction, therefore its velocity must change. Because its velocity changes it must have a force applied to it. It's impossible to change the direction of a moving object without applying a force.

I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.

I'm also saying the mass changes direction during the 1st turn, maintaining a _speed_ of 40m/s. Of course its velocity changes from -40m/s to +40m/s, so it's accelerating as it passes through the turn.

It doesn't matter that the mass moves sideways in the middle of the turn. It goes into the turn at -40m/s and comes out at +40m/s. Whatever happened during the turn caused an acceleration (here change in direction) and applied a force to the mass (by the 1st law), as well as a force to the pipe (by the 3rd law).

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Quote:

That's a contradiction. The mass changes direction, therefore its velocity must change. Because its velocity changes it must have a force applied to it. It's impossible to change the direction of a moving object without applying a force.

I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.


kallog , the mass is pressing against the pipe and that causes the mass to change direction.

through the 1st half of the 1st turn the mass is moving in the (-) direction , and pressing against the pipe
in the (-) direction.

then

through the 2nd half of the 1st turn the mass is moving in the (+) direction , and pressing against the pipe
in the (-) direction.

the mass never presses against the pipe in the (+) direction in the 1st turn.

so its not a contradiction.
the mass does change direction but the direction that its force is applied is always in the (-) direction as it passes through the 1st turn.

there is never any force applied to the pipe in the
(+) direction by the mass as the mass passes through the 1st turn.

so there is no available force to reverse the direction of the pipe.

the change in direction of the mass is not a force.
and a force is required to change the direction of the pipe.

think of a 360 degree range of motion.

the (+) direction is 360 degrees.
the (-) direction is 180 degrees.

the mass enters the 1st turn at 270 degrees.
it travels through the turn to 180 degrees.
then it travels through the turn to 90 degrees.


all the time it travels through the 1st turn it places
a force outward that has a direction that can be plotted from the center of the turn through the mass to the outermost part of the turn.

these forces are less at 270 degrees and greatest at 180 degrees and less at 90 degrees , and if you add up every force applied in the (-) direction they will all add up to -4000N.

never does the force applied to the turn transit into
the (+) side , which would be from
90 degrees to 360 degrees to 270 degrees.

momentum changes are not forces.
but momentum changes require forces.








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Originally Posted By: paul

the mass never presses against the pipe in the (+) direction in the 1st turn.


Paul, are you making an effort to misunderstand me? I never said that, I even clarified that fact when you mentioned it before. Please read my messages more carefully. I never said there is a force in the + direction applied to the pipe during the 1st turn.

We both agree there is a force in the - direction applied to the pipe. Correct?

Do you also agree that Newton's 3rd law says there must also be a force in the + direction applied to the _mass_? <<<---- Note! that says mass, not pipe.

Either there really is such a force, or Newton's 3rd law is wrong. Which way do you want to go?

Or try to visualize it. Imagine you're riding on that mass, will you feel anything as it goes through the turn? Could you black out from the g's? Could you get a sore ass from scraping against the turn? Anything at all? Or would you be unaware of the turnaround if you shut your eyes?

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Originally Posted By: kallog
Paul, are you making an effort to misunderstand me? I never said that, I even clarified that fact when you mentioned it before. Please read my messages more carefully. I never said there is a force in the + direction applied to the pipe during the 1st turn.


LOL , dont try to maneuver your way out of that one, you have always said through the entire discussion that the pipe would somehow move in the (+) direction because the mass goes through the 1st turn.

it shows clearly in your graphs that you put up
showing the (+) direction indicating that the pipe doesnt just stop.



as is shown in the above graph you posted , the pink line represents the pipe displacement , you always return the pipe to ZERO because of the 1st turn.

that reversal would require +8000N and you used the total change in the momentums direction +8000N to accomplish that reverse in the pipes direction.

kallog concerning the above graph, in order for the pipe to reverse direction you must first
apply a +4000N force to stop the pipe and then to get the pipe back to the ZERO line you must apply another force.

THERE IS NO OTHER FORCE!!!!

YOU CANNOT USE THE TOTAL CHANGE IN DIRECTION OF MOMENTUM AS A FORCE ie the 8000N you were using because the mass changes from +4000 to -4000 , you were adding up the
2 4000 to get 8000

Originally Posted By: kallog
I'm saying the pipe slows, then stops, then starts moving backwards (-ve direction) during the 1st turn.



you are using (-4000N) to slow and stop the pipe
then using (-4000N) to reverse direction.

but the imaginary force you use does not exist.

my entire post was to try and get you to realize that the
pipe will only stop , it will not reverse direction because
any available force that could reverse the pipe direction is used up stopping the pipe.

you have been claiming that the force used to change the momentum of the mass was adding the extra force to reverse the pipes direction.

thats extremely evident in this thread as is in others where this
has been discussed.

momentum is not a force and the change in momentum is not a force and that is what I have been trying to tell you all along.

so stop trying to put the blame on others , accept your mistakes and own up to them.

have you ever watched a nascar race , where cars travel
around a track , they use gas engines because the turns do not propel the cars around the track the way you think the mass is being propelled because it changes direction.

the nascar cars also change direction but they still have to use up gasoline to go around and around , the turns of the race are not like a magic carpet.

and the turns in this concept also are not like a magic carpet.









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Originally Posted By: paul

have always said through the entire discussion that the pipe would somehow move in the (+) direction because the mass goes through the 1st turn.


Yes, until just after I posted that graph, when we agreed on the opposite sign convention. Let's just stick to this way instead of wasting time:

Originally Posted By: paul

lets just use (+) for pipe movement in the direction that the accelerator moves the pipe the first time the mass is accelerated.





Quote:

you are using (-4000N) to slow and stop the pipe
then using (-4000N) to reverse direction.

but the imaginary force you use does not exist.

my entire post was to try and get you to realize that the
pipe will only stop , it will not reverse direction because
any available force that could reverse the pipe direction is used up stopping the pipe.


You assumed -4000N is only applied for 1s. Show me your calculations that say it must be applied for 1 second. Here's my calculation that shows it must be applied for 2 seconds:

4000N = 100kg * a
a = 40m/s^2
That says the velocity of the mass increases by 40m/s each second. Since it started at -40m/s, after 1s it's reached 0m/s. After 2s it's reached +40m/s. That's the velocity it must have after leaving the turn, so it took 2s.

Either agree or identify the exact point where you think I made a mistake.


Quote:

momentum is not a force and the change in momentum is not a


We're no longer using momentum at all. You can ignore my older posts where I used momentum. Because you don't trust the law of conservation of momentum, the entire concept of momentum has no use, and I can accept that doubt. We don't need momentum, it's only a convenience to make maths easier.

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I want to super-clarify our positions in case we're still talking at cross-purposes. Is this right?

We both agree:
- We're only talking about the 1st time through the 1st turn.
- The mass begins with a velocity of -40m/s
- The pipe begins with a velocity of +4m/s
- The mass finishes with a velocity of +40m/s

I say:
- The pipe finishes with a velocity of -4m/s

You say:
- The pipe finishes with a velocity of 0m/s

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Quote:
Here's my calculation that shows it must be applied for 2 seconds:

4000N = 100kg * a
a = 40m/s^2
That says the velocity of the mass increases by 40m/s each second. Since it started at -40m/s, after 1s it's reached 0m/s. After 2s it's reached +40m/s. That's the velocity it must have after leaving the turn, so it took 2s.

Either agree or identify the exact point where you think I made a mistake.


the lenght of the turnaround is what determines the time it takes for the mass to pass through it.

the mass is no longer being forcefully accelerated , it is only passing through the turnaround.

I realize there are forces inside the turnaround however.

if the turnaround is 40 meters in lenght the mass will pass
through the turnaround in 1 second because the mass has a velocity of 40 m/s.


if the turnaround is 20 meters in lenght the mass will pass
through the turnaround in 0.5 seconds because the mass has a velocity of 40 m/s.

I think we should use a turnaround lenght of 40 meters.

if that will assist our discussion.

and if we need to the lenght of the accelerator can be increased also.



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kallog
Quote:
I want to super-clarify our positions in case we're still talking at cross-purposes. Is this right?

Quote:
We both agree:
- We're only talking about the 1st time through the 1st turn.

yes
Quote:
- The mass begins with a velocity of -40m/s

yes
Quote:
- The pipe begins with a velocity of +4m/s

yes
Quote:
- The mass finishes with a velocity of +40m/s

yes
Quote:
I say:
- The pipe finishes with a velocity of -4m/s

yes, thats what you say , but I dont agree nor does physics math.
Originally Posted By: kallog
You say:
- The pipe finishes with a velocity of 0m/s

yes, thats what I say , and physics math backs me up.

the pipe stops !

the math

the action
100kg * 40 m/s = 4000N ---->mass
the reaction
1000kg * 4 m/s = 4000N ---->pipe

since the pipe has a velocity of 4 m/s and a mass of 1000 kg it will stop in 1 second if a force of 4000N is applied
for 1 second in the opposite direction that the pipe is moving in.

a=f/m
4m/s=4000N/1000kg

the pipe will stop in 1 second as the mass applies its
-4000N force to the pipe , and the pipe will continue to move in the (+) direction until it stops.

at this point the initial -4000N force has transfered to
the pipe as a -4000N force.

and the 100kg mass can now be described as having a +4000N force due to its direction and its mass * its velocity of 40 m/s.

call it what you will , say that it violates the laws of physics , conservation of momentum , conservation of energy
or conservation of laws if you choose , it does not matter , the above is what physics math shows will happen , it cannot happen any other way.

a 500N force applied for 8 seconds to a 100kg mass accelerated both the pipe and the mass , that same
force stopped the pipe and the mass still has a velocity
of 40 m/s , when the mass reaches the 2nd turn it will accelerate the pipe again to 4 m/s/s , when it exits the 2nd turn it will still have a velocity of 40 m/s.

1)free energy = true
2)conservation of momentum = false
3)conservation of energy = false

laws 1,2,3 gone...

but only the way they are used today , or should I say the
definitions that the laws have been rearanged to represent.














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Quote:

the mass is no longer being forcefully accelerated , it is only passing through the turnaround.

You really really need to understand the terminology. _All_ accelerations are forceful, and anything changing its direction of motion is being accelerated - with force. The speed can remain a constant 40m/s, but it is still accelerating. I can't believe you still haven't looked this up. Again I have a sudden realization that I'm talking in a foreign language to you.


Quote:

if the turnaround is 40 meters in lenght the mass will pass
through the turnaround in 1 second because the mass has a velocity of 40 m/s.


speed=length/time
40m/s = 40m / 1s
Indeed. So a 40m, 1s turnaround is possible. And it can only have one constant force being applied to the mass.

Quote:

if the turnaround is 20 meters in lenght the mass will pass
through the turnaround in 0.5 seconds because the mass has a velocity of 40 m/s.

40m/s = 20m / 0.5s
Yep, so a 20m, 0.5s turnaround is also possible.

Quote:

I think we should use a turnaround lenght of 40 meters.

OK. Let's stick to the 40m, 1s turnaround.

The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N

Again, please either agree or point out the exact part(s) of my calculation that you don't agree with. You didn't last time I asked, but we're not going to get anywhere unless we can isolate our disagreement to its exact root.

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Originally Posted By: paul

the action
100kg * 40 m/s = 4000N ---->mass

This isn't a valid equation. The dimensions on the left (mass * velocity) are different to those on the right (force). How about find a reference for this equation, or explain it in terms of something we know we agree on.


However, since you concluded the mass experiences a 4000N force (I assume in the +ve dirction), I'll continue..

force = mass * acceleration
4000N = 100kg * acceleartion
acceleration = 40m/s^2

That means after 1s in the turnaround, it's reached a velocity of 0m/s (in the longitudinal direction, of course it's still doing 40m/s speed, but now sideways).

0m/s in the longitudinal direction isn't turned around, so that 4000N force is incompatible with a 1s turnaround time.

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Quote:
The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N


momentum is not a force !

the masses velocity does not change.
-----
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
-----
its momentum changes , momentum is not a force
momentum is just mass multiplied by speed.

so you cannot say that a 8000N force is placed on the
pipe by the change in momentum of the mass.

if you are so certain that the pipe will experience a
-8000N force then where does the force come from , it must
come from somewhere.

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N

4 m/s/s * 1000 kg = 4000N

note: I added time as you require

where does the extra 4000N come from?

have you ever seen this happen in the real world , could
you give an example of this extra force in a turn?

does a ball dropped from 10 feet bounce to a height of 20 ft because of this principle?

because if that extra force is available in a turn then the ball bouncing off of the ground should also get a extra ummph because the earth is hard to move with a ball.


or does the earth also feel a force that is twice the
mass of the ball * its acceleration?

the reason I say twice the mass is because we know the velocity does not increase , so the mass of the ball
must increase.

am I right in assumming that that is what happens in the turn?

if we go in the middle of the turn and catch the mass we could have twice the mass? we could get rich by using gold masses.

at which point in the turn should we capture the gold before it has a chance to shrink its mass?

Quote:
That means after 1s in the turnaround, it's reached a velocity of 0m/s (in the longitudinal direction, of course it's still doing 40m/s speed, but now sideways).

0m/s in the longitudinal direction isn't turned around, so that 4000N force is incompatible with a 1s turnaround time.



the mass travels all the way through the 40 meter
turn in 1 second because it has a velocity of 40 m/s
40 meters /40 m/s =1s , so in 0.5 seconds the mass is half way through the turn.

its traveling longitudinal and pressing against the
turn.

the velocity/speed of the mass does not decrease.

again you are using the change in momentum but describing it as a change in velocity.

I thought we were going to leave momentum out of this
discussion , so why are you impersonating momentum with velocity?

momentum is not a force so lets not pretend that it is.
velocity is also not a force , btw.
and a change in velocity is also not a force.

just use the actual forces that would actually apply
a force to the pipe or the 100kg mass.







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Originally Posted By: paul

the masses velocity does not change.
-----
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s


Please please please look up 'velocity' on google or wikipedia or a text book, or anything. I thought I'd explained this before, but it seems you didn't look it up then either.

The _speed_ doesn't change, that remains at 40m/s throughout the turn. But the _velocity_ changes from "40m/s backwards" to "40m/s forwards". That's a huge change and can't just happen on a whim, it requires enough force.

Quote:

if you are so certain that the pipe will experience a
-8000N force then where does the force come from , it must
come from somewhere.

It comes from the mass pressing against the pipe as it turns around. If the mass didn't press against the pipe, it would be unable to turn around.


Quote:

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N

4 m/s/s * 1000 kg = 4000N

note: I added time as you require

4m/s/s is an acceleration not a velocity. If the pipe did accelerate at 4m/s/s then you'd be right, it'd need a 4000N force to do so.

Quote:

have you ever seen this happen in the real world , could
you give an example of this extra force in a turn?


Drive along, then brake hard. You feel the force of the seatbelt/floor/etc pushing your body backwards (of course you move forwards, but the backwards force stops you going through the windscreen).

As soon as you've stopped, go into reverse and floor it. Now you still feel a force pushing your body backwards (the same direction!).

That second application of force is an example of the extra 4000N I'm talking about. You wouldn't get it if you just stopped and stayed stopped, it was the direction-change that caused it.


Quote:

because if that extra force is available in a turn then the ball bouncing off of the ground should also get a extra ummph because the earth is hard to move with a ball.

It does get an extra oomph, enough to send it flying upwards again. Without that extra force it'd just stop when it hit the ground.

The extra oomph is powered by the stored elastic energy in the ball. If the ball is deflated it won't bounce because it can't store that elastic energy.


Quote:

velocity does not increase , so the mass of the ball

The velocity does increase. I hope by now you've looked up 'velocity', then you'll clearly see that it's increased even tho the speed has remained at 40m/s.

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Quote:

the pipe only has a velocity of 4 m/s its mass is only
1000kg so its total force available is only 4000N


This statement worries me. Are you sure you were thinking about what you said? Let me give you a clear example of why the whole concept is wrong.

A bungee jumper falls 50m before taking up the slack in the bungee cord and beginning to slow down. He feels some force, but it's comfortable.

A suicide jumper falls 50m and is caught by the ground. He feels a much higher force which is fatal.

Both jumpers were in free-fall for 50m. Therefore they both had the same speed at 50m down. Both were stopped by a force. But one experienced a much higher force than the other.

This really is common sense. It's why soft-soled sneakers are more comfortable for running in than Doc Martins.

My point here is that the force required to stop (or reverse) something isn't only determined by the thing's mass and velocity. It also depends how quickly it's stopped (or reversed).

You've said many times that force is mass * velocity. But clearly it depends on more than just mass and velocity.

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you didnt comment on my question , does the earth feel a force
that is twice the mass of the ball * its acceleration?

that is a very important question that you need to answer.

lets use the 100kg mass to figure this out.

the mass has fallen for a distance and now
has a acceleration of 40 m/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?

I recall earlier you said that you were treating
the turn as a spring.

in that case the force is absorbed by the spring but the pipe
also would be pushed by the force , then as the spring expands
the pipe would again be pushed by the force exerted by the spring.

is this how you are describing the turns still?

I would still like your answer to the force that the mass
would apply to the earth.












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Quote:
A bungee jumper falls 50m before taking up the slack in the bungee cord and beginning to slow down. He feels some force, but it's comfortable.

A suicide jumper falls 50m and is caught by the ground. He feels a much higher force which is fatal.



the bungee jumper feels a smaller force for a longer time.
the suicide jumper feels the total force all at one time.

both forces are the same exact force the the earth feels
and both forces are the same force felt by the two jumpers.

only the forces are spread out over time.


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