PD wrote:

Mike wrote:

Martin Hogbin wrote:

"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message
news:59J0g.376852$vE3.10805301@phobos.telenet-ops.be...

"Martin Hogbin" <goatREMOVETHIS123@hogbin.org> wrote in message news:8OKdnUm3iJ4jyt7ZRVnyjA@bt.com...

There is no need to mess with anything. You must agree that there is
some difference between the two types of 'centrifugal' force. (Please
say if you do not) Why, then, would you want to give them exactly
the same name?

No there is not. I am nit talking about the rotating frame. I indicated
to you before that you jump around dancing between reference frames. In
the inertial frame, by Newton's third law, the reaction to the
centripetal force is the centrifugal force.

This would be very bad in an inertial frame. Allow me to explain why.
Suppose we have a rock (tied to a string) that is going around in a
circle in an inertial frame. According to you, there are two forces
acting on the rock, the centripetal force (inward, supplied by the
string) and the centrifugal force (outward, supplied by ... well, you
say by absolute space).

I have never, never said that. This will violated basic physics. Look
at my posts. the centrifugal is acting on the hand or post or whatever
the rope is attached to. Come on, do you want to have a serious
discussion?

The net force on the rock, then, would be zero.
However, this would violate both Newton's 2nd law and Newton's 1st law,
since an object with a net force of zero acting on it would also have
to exhibit an acceleration of zero. But the acceleration of the rock is
plainly not zero, because it is going around in a circle which
intrinsically involves an acceleration (a change in the direction of
the velocity vector, you see). By virtue of the circular motion, we
know that the rock *cannot* be in dynamic equilibrium in an inertial
frame. And in fact, we find that the centripetal force (supplied by the
string) is *precisely* what is required to account for the acceleration
of the rock. Thus the centrifugal "force" is not required at all to
account for the observed motion in the inertial frame. And if it's not
required to account for the motion, and if we can't attribute its agent
to anything other than ... well, "absolute space", whatever that
means... then what purpose does it serve?

You wrtings come from a misunderstanding of my postings. Either you
cannot read or you intentionaly misrepresent my postings. Read again
carefully what i have written and come back. Please do not come back
with misrepresentations.



I summarize:


In inertial frame: centrifugal is reaction to centripetal as of
Newton's third law. In this frame these forces are real with measurable
effects.

In non-inertial frame. centrifugal force must impress on the rock to
explain the state of rest. This is again explained by simple force
analysis. When talking the origin to a frame co-moving with the rock,
the centrifugal force, which is a real force, must be also carried
along and subracted from the centripetal to have the state of rest. It
is just standard force analysis.


In both frames, the same real force is used.

Remeber Newton's thrid law
applies only in inertial frames that is why in the second case we are
not talking about a reaction any longer but a balancing force. Newton's
argument was that in that frame. the reactions are wrt an absolute
space, for both centripetal and centrifugal forces. This is because,
for Newton inertial frames are special and non-inertial frames are the
wrong choice of a FoR to apply his laws.


This is simple physics. Of course, some twist around the meanings for
the purpose of pushing through their own agenda. Reality and experiment
shows they are failures. Centrifugal forces are not ficticious, not
illusions, they are real forces in both inertial and non-inertial FoR.


Mike

This is very basic. If you
do not understand that we have no common basis for any discussion.

Exactly my point. It's very basic. If you do not understand it, we have
no common basis for any discussion.

You
can continue the discussion with van der Vomit which you seem to
patronize well.

Martin, whatever you manage to drag out of him at this point,
I don't think I will add another entry to the series :-)
Trop if too much, right?
Sorry - couldn't resist - do carry on :-)

I know it is hard work but one day we may find someone
who is not a complete crackpot.

You must learn basic physics before, understand Newton's laws and then
after sufficient time passes and you have done both your theoretical
and experimental duties to question yourself. Those who llok foR
crackpots as their main activity, like van der Vomit, are
psychoneurotic individuals with paranoia flashes.

Mike

Martin Hogbin

Martin Hogbin wrote:

"Herman Trivilino" <physhead@kingwoodREMOVECAPScable.com> wrote in message news:1145214635_27343@sp6iad.superfeed.net...

"PD" <TheDraperFamily@gmail.com> wrote ...


Take a look at the connection between the string and the rock. The string
exerts a centripetal force on the rock, and the rock exerts a centrifugal
force on the string. They form a Third Law pair.

The use of 'centrifugal force' for the force the rock
exerts on the string is bad terminology. No one can
argue with the fact that there is a reaction force on
the string and that this force is centrifugally directed
but to call this 'centrifugal force' is bad practice and
can cause endless confusion, particularly to the OP
of this thread.

You are one of the two or three confused here. You say it acts like
centrifugal, it smells like centrifugal but do not call it centrifugal
because someone like you might get confused. So be it. Get confused. We
are not going to mess physics so that you avoid confusion.

At least PD can come up with innovative arguments. Appealing on
confusion is a ridiculus thing to do.

Mike

Martin Hogbin

"PD" <TheDraperFamily@gmail.com> wrote ...

Mike wrote:

Tom Roberts wrote:

joe_avery_2005@yahoo.com wrote:

Centrifugal forces can be considered reactions to Centripetal forces.

No, they cannot.

When swinging a stone by a string, the force on my hand is radially
outward at radius zero, so the "centrifugal force" is zero. The force
on
my hand is quite clearly the tension in the string, not "centrifugal
force". The force on the rock is also the tension on the string.

You are wrong.

No, he's quite right. And if you look in a basic physics text, this is
precisely what you'll see.

Take a look at the connection between the string and the rock. The string
exerts a centripetal force on the rock, and the rock exerts a centrifugal
force on the string. They form a Third Law pair.

The confusion arises when a Freshman claims there is a centrifugal force on
the rock. You have to wait until you're a Junior before you can do that!

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PD wrote:

Mike wrote:

Daryl McCullough wrote:

mmeron@cars3.uchicago.edu says...

stevendaryl3016@yahoo.com (Daryl McCullough) writes:

Suppose I observe a ball's motion and plot its position x versus
t. I get a straight line, so I conclude (via Newton's laws) that
there is no force acting on the ball (at least not in the x-direction).

Now, I change coordinates to X = arctan(x) and plot X versus t.
In the new plot, I *don't* get a straight line, but I get some
curved line. Is that a "real effect"? Does it indicate the presence
of a force? No, it's obviously an artifact of my particular
choice of coordinates.

We're not talking about changing coordinates,

It doesn't matter whether I *changed* coordinates or not. If
I had some strange kind of ruler that happened to measure X
instead of x, then plotting the trajectory of the ball would
yield a curved line, rather than a straight line. That doesn't
mean that there is any force involved, necessarily, it just
means that my coordinate system is non-Cartesian.

It's not likely that I'll ever have rulers that happen
to measure arctan(x), but it isn't hard to come up with
rulers that measure X = x cos(wt) + y sin(wt) and
Y = - x sin(wt) + y cos(wt).

we're talking about measurements of postions,
velocities and accelerations. That's plain kinematics.

What does it mean to measure positions, velocities,
and accelerations? I know how to measure those things
in a particular *coordinate* system, but how do you
measure them in a coordinate-independent way?

From this point on, the interpretation dpends on the
definition of a force. If you define force as, say, "transfer of
momentum resulting from an interaction between physical entities, an
interaction which can be expressed, in the ultimate account, by an
exchange of bosons between said physical entities", then no, the above
is not a real force. But, I do not recall force being strictly defined
this way.

What is the definition of force, then? Or if we are assuming
Newton's law, F=ma, then we can ask: what is the definition
of "acceleration"?

What I would say is that acceleration is a vector (in classical
physics, anyway) defined by

a = (d/dt) v

where v is the velocity vector. In terms of components, we
can write v = v^i e_i where e_i is the ith basis vector. Then
we have

a = (d/dt v^i) e_i + v^i (d/dt e_i)

The fact that d/dt v^i = 0 does *not* imply that a=0 unless
we know that our basis vectors e_i are time-independent.

If, on the other hand, you define force as anything causing change of
meomentum than yes, the above is force.

That's what I'm disputing. How are you proposing to measure
momentum? How are you operationally deciding whether momentum
is changing are not?

You may say that "the change of momentum is only apparent
since there is no change relative to an inertial frame,

Regardless of whether you use inertial frames or not, what
does it *mean* to say that momentum is changing? How do
you operationally measure momentum?

Delta(p) = p(final) - p(initial)

p = mv

v = ds/dt

All you need is to measure position using a ruler and a clock and then
take the slope. We are talking here physics 101. Measuring change in
momentum wrt a FoR is the easiest thing to do.

but this implies another defintion, namely that "a
valid mesaurement is a measurement performed relative
to an inertial frame". Again, says who?

Well, I certainly didn't say that. You can measure things
in whatever coordinate system you like, but not all coordinate
systems are Cartesian.

So what? I think you are confused about his, maybe because you dropped
in late in the discussion. We are not talking here about motion as
observed from another frame. We are talking about measurements made in
a particular frame, in this case non-inertial, and inferences that can
be made from those measurements about forces. Of course, as I pointed
out before and Meron put it in the correct context, it all depends on
the definition of force. But assuming no particular definition of force
a priori and then talking about ficticious forces is non-sense physics
to me. And that's the main point of this IMO.

Thus, I sense that classical mechanics and Relativity have a different
conception of force and this is where I wanted to bring the discussion
to. If this is the case, any convergence of GR to Newtonian laws of
motion and gravitation at weak field limits is suspicious, to say the
least. The Newtonian definition is clear: force causes change in state
of motion and equals the rate of change of momentum. Without force
acting the state of rest or uniform linear motion is maintained.

I disagree. Newtonian laws of motion include the 3rd law in the
definition of forces. Please describe how a coriolis or a centripetal
force satisfies the 3rd law.

PD

Good point as I noted in another post. One answer is that we do not
know. Just because the answer is not known, one cannot draw
automatically the conclusion that Centrifugal --- I think you meant to
say --- and Coriolis forces are pseudoforces.

Centrifugal forces can be considered reactions to Centripetal forces.

I
do not know of any experiment to generate Centrifugal forces in the
absence of Centripetal forces.

The Coriolis case is much more complicated.


Joe Avery