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  4. What is centrifugal force?
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What is centrifugal force?

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Offline Vasyl

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Re: What is centrifugal force?
« Reply #220 on: 23/12/2018 19:57:04 »
Quote from: David Cooper on 23/12/2018 19:36:54
Why wouldn't that be possible, and why are you relating that to centrifugal force?
If it is possible than we can create artificial "asteroids" and attach a spaceship to them to explore space. Or is it not possible?
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Offline David Cooper

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Re: What is centrifugal force?
« Reply #221 on: 23/12/2018 19:59:54 »
Quote from: rmolnav on 23/12/2018 12:13:08
Quote from: David Cooper on 22/12/2018 19:35:40
No one should be rejecting the existence of reactive centrifugal force, and no one should be asserting the existence of centrifugal force of the other kind (??) because it is only an artefact of an abstraction - it is the latter kind that has been rejected by science, and the two should not be mistaken for each other.
I´m not going to continue to discuss with you some "details" we have long discussed on the "tides" thread, to no avail ...

The problem there was that you wanted to have some kind of centrifugal force involved in cases where the centripetal force is gravity, but there is absolutely no centrifugal force of any kind involved in that other than in the abstraction of rotating-frame analysis.

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But please kindly note that you can find definitions of "centrifugal force" considering it ONLY as a "fictitious" force ...

Inadequate definitions exist all over the place, but we don't need to trip over them. The whole point of working to acquire a deep understanding of the physics in places like this forum is to become better able to tell whether what you're reading is correct or faulty. Often the faults are in omission where someone has written something with only one thing in mind and hasn't thought to clear up other issues, in this case because of an ambiguity. Different contributors to an encyclopaedia can use ambiguous words in different ways, all intending to provide good information (and doing so if you read them the right way), but the different ways in which they word things can appear to produce contradictions.

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Though, curiously, further down they themselves give links to other britannica articles, among others precisely the "centrifuge" one where you can read:
"Centrifuge, any device that applies a sustained centrifugal force—that is, a force due to rotation.

You're first Britannica quote dismisses centrifugal force entirely (because the writer is only thinking about the fictitious kind of centrifugal force of the kind that provides the illusion of existing in rotating frames), but your second one which refers to a sustained centrifugal force is actually talking about reactive centrifugal force which is a real force - it is just referred to in a more economical way (by missing out the word "reactive") that depends on the reader applying correct understanding to it. That way of wording things is the standard way of doing so when talking about a centrifuge. It is your job to catalogue the information correctly in your head, filing it under "reactive centrifugal force" rather than "centrifugal force".

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and, logically, that "centrifugal force" has to be REAL ... They just say "due to rotation", without any reference either to centripetal force (the one exerted by the spinning vessel on the liquid to make it follow a circular path), or to the inertial outwards reactive forces exerted by the liquid, both internally between cylindrical layers of liquid, and on the inner side of the vessel ...

Your understanding of physics should enable you to link the thing they're calling "centrifugal force" to the correct real phenomenon and not to be misled into thinking there is some other centrifugal force involved in the situation.

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And those forces are "centrifugal" and real: otherwise pressure would not increase outwards, as explained on my last posts !!

The centripetal force is higher further out, and the reactive centrifugal force goes up to match that - the thing being called centrifugal force in a centrifuge is the reactive force responding to the centripetal force.

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By the way, what happens in the centrifuge case in some aspect is similar to what in the case of a ball being made rotate with a sling, "pushing" it inwards (on its outer side), instead of "pulling" from a hook on its inner side (as when hammer throwing) ... The sling rotation (together with inertia) produces compressive forces inside the ball, the outer the more, as in the centrifuge the vessel spinning and inertia cause the increase of pressure, the outer the more too !!

And I suppose you think I didn't say that in the post you're attacking. Note where I talk about tension and pressure in relation to how the centripetal force is applied. Ball on string --> tension; mouse in hamster wheel --> pressure.
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Offline David Cooper

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Re: What is centrifugal force?
« Reply #222 on: 23/12/2018 20:04:27 »
Quote from: Vasyl on 23/12/2018 19:57:04
Quote from: David Cooper on 23/12/2018 19:36:54
Why wouldn't that be possible, and why are you relating that to centrifugal force?
If it is possible than we can create artificial "asteroids" and attach a spaceship to them to explore space. Or is it not possible?

Are you sure you're posting in the right thread? Why would you want to handicap a space probe by tying an artificial or real asteroid to it? The less mass (other than fuel) that you have to accelerate, the easier it is for the probe to go exploring, and the higher the speed you'll get it to.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #223 on: 28/12/2018 08:40:53 »
Quote from: David Cooper on 23/12/2018 19:59:54
your second one which refers to a sustained centrifugal force is actually talking about reactive centrifugal force which is a real force - it is just referred to in a more economical way (by missing out the word "reactive") that depends on the reader applying correct understanding to it.
As I´ve already said, it´d be useless to continue here our so long discussion
on the thread about tides ...
But I want to repeat I don´t agree with some of your "basic" dynamics ideas, e.g., what quoted ...
Centrifugal forces, the ones really acting in nature, are always manifestations of inertia when physical objects are accelerated "centripetally" (therefore, moving along a curved trajectory), whatever the essence of the required centripetal force and rest of circumstances.
The "fictitious" ones only "exist" inside our minds, when we use rotating frames of reference for our calculations ... In that "virtual" scenario, as actually rotating objects kind of "stop" their rotation (--> no centripetal acceleration and no inertial effects), we "need" to introduce a "fictitious" external centrifugal force, that causes effects IDENTICAL to the ones actually caused by the sheer fact of the kind of "hidden" rotation ...   
But those effects do exist in the real, natural scenario.
By the way, this very morning I saw a clip on Discovery Max tv, where a CD was made rotate at the r.p.m. of a hoover ... The CD suddenly breaks into hundred of small pieces, obviously due to huge tensile internal stresses due to both centripetal and centrifugal forces (the former causing the rotation, and the later the way inertia manifests itself, because all CD particles are being hugely accelerated).
And that, happening in nature, has nothing to do with frames of reference !!
 
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #224 on: 31/12/2018 08:45:23 »
Quote from: rmolnav on 28/12/2018 08:40:53
By the way, this very morning I saw a clip on Discovery Max tv, where a CD was made rotate at the r.p.m. of a hoover ... The CD suddenly breaks into hundred of small pieces, obviously due to huge tensile internal stresses due to both centripetal and centrifugal forces (the former causing the rotation, and the later the way inertia manifests itself, because all CD particles are being hugely accelerated).
And that, happening in nature, has nothing to do with frames of reference !!
 
Before the year ends ... Happy New Year! ...
...and an additional comment to what above.
In the case of the CD, as in the cases of the centrifuge, the hammer throwing, the sling ..., "solid" stuff is what exerts the centripetal force on the considered rotating "object" (either solid or liquid), and centrifugal forces (inertial as always) are exerted by the rotating "objects" on what exerts the centripetal force (internal stresses included).
Some people think that when gravity is involved (as cause of the curving of the trajectory), without any solid device (such as a string, wire, centrifuge vessel, and own CD material), no centrifugal force exists ...
But quite similar to the CD case is the daily earth spinning ... At its angular speed of 2π radians/24 h. it is causing a relatively huge equatorial bulge (tens of km), affecting both solid and liquid parts of our planet.
Earth solid parts are stretched as in the CD case, and also would explode at much higher velocity.
But what makes water rotate (instead of "following the tangent) is only gravity, exactly a component of water weight vector at each location: the one perpendicular to earth´s axis of rotation ...
At the equator water "lightens" much more than closer to poles, because centrifugal forces are proportional to the radius (mω²r) ...
If spinning velocity increased, the bulge would get bigger, water would evaporate more ... and eventually something similar to the case of the CD at vacuum cleaner spinning rate would happen !!
Centripetal and real centrifugal forces are necessary for all that to happen, and, again, whatever the movement of any considered observer, and the reference system he or she chose...
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Offline David Cooper

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Re: What is centrifugal force?
« Reply #225 on: 31/12/2018 23:04:14 »
Quote from: rmolnav on 31/12/2018 08:45:23
Before the year ends ... Happy New Year! ...

Happy Calendar-Replacement Day!

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In the case of the CD, as in the cases of the centrifuge, the hammer throwing, the sling ..., "solid" stuff is what exerts the centripetal force on the considered rotating "object" (either solid or liquid), and centrifugal forces (inertial as always) are exerted by the rotating "objects" on what exerts the centripetal force (internal stresses included).

... which is known as reactive centrifugal force (the only kind of centrifugal force that actually exists).

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Some people think that when gravity is involved (as cause of the curving of the trajectory), without any solid device (such as a string, wire, centrifuge vessel, and own CD material), no centrifugal force exists ...

... and those people are certainly right in cases where there is no contact between them.

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Centripetal and real centrifugal forces are necessary for all that to happen, and, again, whatever the movement of any considered observer, and the reference system he or she chose...

An object sitting on a non-rotating planet will press down on the surface of the planet and a reactive force will be generated in the opposite direction to oppose it. Neither of those forces can be called centripetal or centrifugal because there is no rotation. If you begin to spin the planet and keep increasing the speed of the rotation, those two forces will both lessen and reach zero when the object becomes weightless. Any analysis which asserts that a centrifugal force is going up as the rotation gets faster is working with an abstraction which does not represent what the real forces are doing. The gravitational force is constant until the object lifts off the ground, and the other real forces are both going down while the imaginary centrifugal force in the abstraction goes up.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #226 on: 01/01/2019 11:36:55 »
Quote from: David Cooper on 31/12/2018 23:04:14
An object sitting on a non-rotating planet will press down on the surface of the planet and a reactive force will be generated in the opposite direction to oppose it. Neither of those forces can be called centripetal or centrifugal because there is no rotation. If you begin to spin the planet and keep increasing the speed of the rotation, those two forces will both lessen and reach zero when the object becomes weightless. Any analysis which asserts that a centrifugal force is going up as the rotation gets faster is working with an abstraction which does not represent what the real forces are doing. The gravitational force is constant until the object lifts off the ground, and the other real forces are both going down while the imaginary centrifugal force in the abstraction goes up.
1) "Neither of those forces can be called centripetal or centrifugal because there is no rotation. If you begin to spin the planet and keep increasing the speed of the rotation ..." then the "reason" that there is no rotation would not apply, some force would have to function as centripetal force, and "inertial effects" would appear because massive stuff is being accelerated ...
2) "Any analysis which asserts that a centrifugal force is going up as the rotation gets faster is working with an abstraction which does not represent what the real forces are doing":
Wrong, as shown below.
3) "The gravitational force is constant until the object lifts off the ground, and the other real forces are both going down while the imaginary centrifugal force in the abstraction goes up".
Wrong ... Long before any of us could "levitate" on the equator, the equatorial bulge would get much, much bigger, and even, as I said:
 
Quote from: rmolnav on 31/12/2018 08:45:23
Earth solid parts are stretched as in the CD case, and also would explode at much higher velocity.
And earth´s crust couldn´t deform so much, let alone break into pieces, without exerting on it strong opposite REAL forces, in our case centripetal forces (the ones which make earth stuff rotate at so high speed), and centrifugal forces (always due to inertia, or "reactive" as you say ...). 
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Offline David Cooper

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Re: What is centrifugal force?
« Reply #227 on: 01/01/2019 20:26:34 »
Quote from: rmolnav on 01/01/2019 11:36:55
then the "reason" that there is no rotation would not apply, some force would have to function as centripetal force, and "inertial effects" would appear because massive stuff is being accelerated ...

When you start the rotation, you can play games by renaming forces that were already acting before there was any rotation, but they go down in strength while your imagined centrifugal force goes up. Even if you're prepared to accept that these real forces go down instead and you decide to call one of them centrifugal, you would have to call it reactive centrifugal force, which is the only real kind of centrifugal force, acting in response to centripetal force and being equal to it. Then, at the point where the reactive centrifugal force reaches zero and the object becomes weightless, the centripetal force must also have reached zero, revealing that the gravitational force which you also like to call centripetal force in situations with orbiting objects is not the same as the centripetal force that was acting in the opposite direction to the reactive centripetal force that was present up until the object became weightless.

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2) "Any analysis which asserts that a centrifugal force is going up as the rotation gets faster is working with an abstraction which does not represent what the real forces are doing":
Wrong, as shown below.

Repeatedly claiming that correct things are wrong does not make them wrong.

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3) "The gravitational force is constant until the object lifts off the ground, and the other real forces are both going down while the imaginary centrifugal force in the abstraction goes up".
Wrong ... Long before any of us could "levitate" on the equator, the equatorial bulge would get much, much bigger...

If the bulge gets bigger, the real forces go down even more, but your imaginary centrifugal force in the maths of the abstraction rises more to match.

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, and even, as I said:
Quote from: rmolnav on 31/12/2018 08:45:23
Earth solid parts are stretched as in the CD case, and also would explode at much higher velocity.
And earth´s crust couldn´t deform so much, let alone break into pieces, without exerting on it strong opposite REAL forces, in our case centripetal forces (the ones which make earth stuff rotate at so high speed), and centrifugal forces (always due to inertia, or "reactive" as you say ...). 

Trying to muddying the waters with other forces which bind material together isn't going to wash - you are trying to sell a force that doesn't exist, and I'm not buying it.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #228 on: 07/01/2019 12:13:50 »
Quote from: David Cooper on 01/01/2019 20:26:34
When you start the rotation, you can play games by renaming forces that were already acting before there was any rotation, but they go down in strength while your imagined centrifugal force goes up. Even if you're prepared to accept that these real forces go down instead and you decide to call one of them centrifugal, you would have to call it reactive centrifugal force, which is the only real kind of centrifugal force, acting in response to centripetal force and being equal to it. Then, at the point where the reactive centrifugal force reaches zero and the object becomes weightless, the centripetal force must also have reached zero, revealing that the gravitational force which you also like to call centripetal force in situations with orbiting objects is not the same as the centripetal force that was acting in the opposite direction to the reactive centripetal force that was present up until the object became weightless
With your last post one can tell you haven´t grasped what centripetal force actually is yet!!
As I´ve said many times, any rotation, or any movement of massive particles along a curved path, REQUIRES a net amount of centripetal force exerted on it, that can be provided by different sources … Otherwise, due to inertia the movement would be along a straight line at constant speed, whatever the way it initially got that speed ... That "simple" !!
The only detail not that simple is that FUNCTION as centripetal force can be exerted by either the total amount of a single force, or part of it (in size, or a certain fraction of one of its orthogonal components), or a combination of different forces …
Imaging a sling rotating in a vertical plane. When the string is horizontal, only its tension functions as centripetal force: gravity either increases or decreases linear velocity. But when at upper location gravity, added to string tension, functions as centripetal force, and our hand need not to pull down so much … if we wish to keep a constant angular speed (almost impossible though …). The opposite happens at lower position, because gravity is kind of centrifugal at that point, and string tension (necessarily much higher) minus gravity is what functions as centripetal force …
And at intermediate locations inward radial components of gravity vector, in each case with its sense, added to string tension, are which function as centripetal force.
What is much more complex is how the object continues to show its inertial "tendency" to follow straight, somehow opposing to the centripetal force which is "forcing" it to bend its path (certainly not always as a “real” centrifugal force). That depends on other details of the scenario, mainly on dynamic details such as grade and type of "freedom" the object has to "reply" acting centripetal force. 
You go from the "no rotation" scenario (gravity and internal forces do exist, but no centripetal force required), to an imaginary “perfect” orbiting situation (by the way, quite different scenario from the one I brought up), without considering any intermediate situation (the ONLY ones which could actually happen, as the current spinning at 2π radians/24 h angular speed).
By the way, you say:
Quote from: David Cooper on 01/01/2019 20:26:34
... forces that were already acting before there was any rotation, but they go down in strength while your imagined centrifugal force goes up
Previously "acting forces" were gravity and internal compressive forces. Gravity, as a result of the stretching that causes the equatorial bulge, does decrease a little ... But required centripetal force (mω²r) increases enormously, and bigger and bigger fraction of gravity pull has to function as centripetal force. And any inertial effect, such as centrifugal forces if any, increases more and more ... 
And you also say:
"... you can play games by renaming forces that were already acting before there was any rotation, ..."
It is not a question of "renaming": if earth starts spinning, inertia makes its particles try to follow a straight path (the tangent) ... Required centripetal force at each location only can come from gravity, and from internal interactions that are partially due to gravity ...
Curiously at earth solid core what happens is similar to "your" case of an object hanging from a string and a pole, made rotate with a bat: the material of the core, enormeosuly compressed due to gravity, if somehow made rotate, doesn´t let the particles to follow the tangent, and it gets radially stretched, what can provide the necessary increase of centripetal force: bigger and bigger fractions of gravitational forces (directly or indirectly) exert the FUNCTION of centripetal force.
From there to earth crust, the stuff is liquid, a quite different scenario ... Now it is like a centrifuge. Due to inertia, each particle tries to follow the tangent ... But earth crust doesn´t allow them to do so, and that induces an increase in pressure under earth crust. Initially earth´s crust and gravity provide the required centripetal force (exerted on the magma), and the magma exerts an equal but centrifugal force on the crust (as currently happens).
Own earth crust inertia also causes its deformation (inertial effects on lower latitude parts are much higher, and internal centrifugal forces appear), also tending to an earth equator diameter increase ... 
But at much higher angular speed earth´s crust can´t withstand all internal stresses caused that way, and would break into pieces (long before any "levitation" of outer objects).
By the way, "your" orbiting "weightless" object scenario would be also quite different from the case of a pair of comparable celestial objects rotating around their common center of mass, because that object wouldn´t affect earth dynamics (though theoretically would also exert an equal but opposite pull on earth, as inertial reaction according Newton´s 3rd Motion Law).
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Offline David Cooper

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Re: What is centrifugal force?
« Reply #229 on: 07/01/2019 20:17:12 »
Quote from: rmolnav on 07/01/2019 12:13:50
With your last post one can tell you haven´t grasped what centripetal force actually is yet!!

With your last post one can tell that your thinking is all over the place, leading to you failing to pin anything down. We're looking at a simple scenario here. We start with a planet with an object sitting on it and the planet not rotating. The planet pulls at the object, let's make it a 1kg rock, and the rock applies a compressive force which is resisted by a reactive force of equal strength in the opposite direction.

When we start to make the planet rotate (with the rock at the equator), that compressive force goes down, and so does the reactive force resisting it. The gravitational force remains almost the same, but it will go down a little as the equator bulges out a bit. That compressive force could now potentially be labelled as centripetal force and the reactive force opposing it would then be called reactive centrifugal force, though I don't know if science actually applies those labels to these forces in such a case.

The important point here though is that the imaginary centrifugal force which you're trying to sell is a (fake) force that goes up in strength as the planet rotates faster, and while it does this, the real forces go down - the rock presses down less hard and the ground has less to resist against. The gravitational force goes down a little as the equator bulges a fraction. No real force is going up, but your imaginary centrifugal force is growing stronger as the rotation speed goes up.

You keep attempting to keep this imaginary force as a real force by confusing people with other forces that do exist, applying incorrect names to them to pretend that some of them qualify as the kind of centrifugal force that doesn't exist. I don't know why you're playing such games. Maybe you're just so confused that you can't untangle it all and see the simple reality that lies beneath the unnecessary complexity that's blinding you.

When we get to the point where the rock becomes weightless, it is in orbit, even if it's still in contact with the ground. At this time, the rock is not pressing down on the planet and the ground is not pushing back, so even if you want to call those forces centripetal and reactive centrifugal, they are both down to zero. And now that the rock is in orbit, you want to call the entire gravitational force acting on it centripetal force, so it becomes clear that any attempt to call the compressive force centripetal is incompatible with this other kind of centripetal force that's now applying on the rock to hold it in orbit. Also, once the rock is in orbit, there is no centrifugal force in action - the only force there is the gravitational pull which you call centripetal. The imaginary centrifugal force that you want to assert is real here is just an artefact of the abstraction of analysing events using a rotating frame of reference - it isn't a real force. The existence of other forces that can be called centrifugal (and which should actually be called "reactive centrifugal force") are not evidence for the existence of the imaginary centrifugal force that you keep trying to sell.

This is why forces ought to be more carefully named to keep different kinds of force in separate compartments so that people are less likely to conflate them and come to incorrect conclusions.
« Last Edit: 07/01/2019 20:21:05 by David Cooper »
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #230 on: 12/01/2019 11:09:28 »
Quote from: David Cooper on 07/01/2019 20:17:12
And now that the rock is in orbit, you want to call the entire gravitational force acting on it centripetal force, so it becomes clear that any attempt to call the compressive force centripetal is incompatible with this other kind of centripetal force that's now applying on the rock to hold it in orbit. Also, once the rock is in orbit, there is no centrifugal force in action - the only force there is the gravitational pull which you call centripetal. The imaginary centrifugal force that you want to assert is real here ...
I was referring to what happens at different parts of our planet: the solid core, the liquid magma, and the crust, where real centrifugal forces, "reactive" if you like, do exist.
But you only refer to a loose rock on earth surface, reaching a "weightless" condition thanks to an enormous angular speed increase, something that would´t ever happen because earth crust would break up before, precisely due to above mentioned centrifugal forces, that previously would also have increased more and more the equatorial bulge size.
Nevertheless, in that particular  scenario, what you say "once the rock is in orbit, there is no centrifugal force in action" would be correct, but I´m afraid you don´t fully understand all the dynamic details of what happens, let alone what I say:
Quote from: rmolnav on 07/01/2019 12:13:50
What is much more complex is how the object continues to show its inertial "tendency" to follow straight, somehow opposing to the centripetal force which is "forcing" it to bend its path (certainly not always as a “real” centrifugal force). That depends on other details of the scenario, mainly on dynamic details such as grade and type of "freedom" the object has to "reply" acting centripetal force.
.
And when the rock gets "weightless", it is orbiting around earth C.G. (at a fixed point), quite "free" to maintain its linear speed, and experiencing (the rock as a whole) only earth´s gravitational pull, perpendicular to its speed ...
That´s the "essence" of a circular, uniform movement. And the whole earth´s pull is what "forces" the rock to follow a circular path, instead of its natural, inertial tendency to keep constant its velocity vector.
That force is a centripetal force, by definition, not just the way I "want" to call it  ...
The "reply" of the stone to being "forced" that way, in this case is what I also said:
Quote from: rmolnav on 07/01/2019 12:13:50
... that object (the rock) wouldn´t affect earth dynamics (though theoretically would also exert an equal but opposite pull on earth, as inertial reaction according Newton´s 3rd Motion Law)
and, on the one hand, if what the rock actually does is to "exert" that force, it couldn´t be considered a centrifugal force (though rock particles keep their "tendency" to "resist" given centripetal acceleration: if the surface were wet, water would move outwards, outdoing water surface tension). And on the other hand, as that force doesn´t affect earth´s dynamic whatsoever (as e.g. happens to the "couple" moon-earth), and the size of the rock is negligible, the case can be considered only as a "free" fall of the rock in a fixed gravitational field ...
Another thing would be if the hight of the rock were not negligible (compared to earth radius). In that case gravitational pull on each rock´s particle would clearly differ from required centripetal force at its location, and they wouldn´t actually be in a "proper" free fall ... Internal stresses (in pairs, centripetal and centrifugal) would appear to compensate those differences, stretching the rock radially.
(I do know you don´t agree with that, long discussed on the thread about tides, but I repeat it here for the sake of other possible readers ...)
As far as I can understand, what you say:
Quote from: David Cooper on 07/01/2019 20:17:12
The important point here though is that the imaginary centrifugal force which you're trying to sell is a (fake) force that goes up in strength as the planet rotates faster, and while it does this, the real forces go down - the rock presses down less hard and the ground has less to resist against. The gravitational force goes down a little as the equator bulges a fraction. No real force is going up, but your imaginary centrifugal force is growing stronger as the rotation speed goes up.
is rather confusing.
At current angular speed (or other "intermediate" scenario), only a fraction of gravitational pull FUNCTIONS as centripetal force: just mω²r ... And that fraction does increase with angular speed ...
As you say, " ... the rock presses down less hard and the ground has less to resist against". Please note that the difference is precisely that value (mω²r), to satisfy Newton´s 2nd Motion Law. And the rest of gravitational pull keeps earth and rock pressing onto each other (not to be called centripetal or centrifugal forces, because, as said above, they are equal to the fraction of gravitational pull which DOESN´T cause any acceleration ...).
Those action and reaction "compressive" forces, though certainly due to gravity, shouldn´t be confused either with it or with part of it: gravitational pull is exerted between all earth´s and rock´s particles, and mentioned compressions are pushes interchanged by only the particles in contact ...

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Offline David Cooper

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Re: What is centrifugal force?
« Reply #231 on: 12/01/2019 22:24:18 »
Quote from: rmolnav on 12/01/2019 11:09:28
But you only refer to a loose rock on earth surface, reaching a "weightless" condition thanks to an enormous angular speed increase, something that would´t ever happen because earth crust would break up before, precisely due to above mentioned centrifugal forces, that previously would also have increased more and more the equatorial bulge size.

I referred to a planet rather than to the Earth, and I did so for a good reason. When dealing with thought experiments, the normal approach is to simplify things to remove all extraneous factors, and there's no call for them to be brought into it by someone who just wants to sow confusion. The planet in my thought experiment could be solid rock all the way through with no hot core, and rotation at the speed that will render the rock weightless would not break the planet up at all.

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Nevertheless, in that particular  scenario, what you say "once the rock is in orbit, there is no centrifugal force in action" would be correct, but I´m afraid you don´t fully understand all the dynamic details of what happens, let alone what I say:

I suspect I understand the dynamic details in considerably more depth than you do. What I'm doing is simplifying things down with thought experiments to remove the stuff that keeps confusing you and sending you off in the wrong direction. The equatorial bulge is not pushed out or held up by centrifugal force - that is the key point here.

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Another thing would be if the hight of the rock were not negligible (compared to earth radius). In that case gravitational pull on each rock´s particle would clearly differ from required centripetal force at its location, and they wouldn´t actually be in a "proper" free fall ... Internal stresses (in pairs, centripetal and centrifugal) would appear to compensate those differences, stretching the rock radially.
(I do know you don´t agree with that, long discussed on the thread about tides, but I repeat it here for the sake of other possible readers ...)

It's a diversion away from the issue in point. If you want to discuss stresses within the rock, discuss that in its own right and then work out what the forces are that are acting there. You won't find the imaginary kind of centrifugal force acting there either, but you might be able to justify calling one of them reactive centrifugal force.

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As far as I can understand, what you say:
Quote from: David Cooper on 07/01/2019 20:17:12
The important point here though is that the imaginary centrifugal force which you're trying to sell is a (fake) force that goes up in strength as the planet rotates faster, and while it does this, the real forces go down - the rock presses down less hard and the ground has less to resist against. The gravitational force goes down a little as the equator bulges a fraction. No real force is going up, but your imaginary centrifugal force is growing stronger as the rotation speed goes up.
is rather confusing.
At current angular speed (or other "intermediate" scenario), only a fraction of gravitational pull FUNCTIONS as centripetal force: just mω²r ... And that fraction does increase with angular speed ...
As you say, " ... the rock presses down less hard and the ground has less to resist against". Please note that the difference is precisely that value (mω²r), to satisfy Newton´s 2nd Motion Law. And the rest of gravitational pull keeps earth and rock pressing onto each other (not to be called centripetal or centrifugal forces, because, as said above, they are equal to the fraction of gravitational pull which DOESN´T cause any acceleration ...).
Those action and reaction "compressive" forces, though certainly due to gravity, shouldn´t be confused either with it or with part of it: gravitational pull is exerted between all earth´s and rock´s particles, and mentioned compressions are pushes interchanged by only the particles in contact ...

Why are you confused by it? I was spelling out what the available forces are and showing you that centrifugal force (the kind that isn't reactive centrifugal force) doesn't come into play anywhere. Your centripetal force is an abstraction - you're attributing to centripetal force an increasing fraction of the gravitational force as the rotation speeds up, but the real gravitational force acting on the rock is either constant or is going down as the planet changes shape a little. That centripetal force is not opposed by any kind of centrifugal force. It therefore makes no sense to attribute the existence of the equatorial bulge to centrifugal force.
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Re: What is centrifugal force?
« Reply #232 on: 22/01/2019 14:35:51 »
Quote from: mona2200 on 19/01/2019 12:25:21
I have seen it expressed now and again that gravity is an invented power. That mysterious changes in geometry cause gravity. In this way I trust it is vital to nail down a definition.
I was going to  send a reply to quoted "mona2200" post of three days ago, but it has disappeared !! ??
Though gravity as an invented "power" sounds awful, I don´t think the post deserved removal ... After all, Einstein´s theory of curving of time-space, due to the presence of massive objects, as essence of gravity may be considered kind of "mysterious (?) changes in geometry" ...
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Re: What is centrifugal force?
« Reply #233 on: 06/03/2019 11:41:52 »
Since my last post I´ve frequently found myself ruminating on our issue, trying to find better ways to convey my stand.
On this thread and on “Why do we have two high tides a day”, many times I´ve referred to the different ways inertia manifests itself, depending on the type and degree of “freedom” to move considered material stuff actually has …
The more I ruminate on the issue, the clearer I find that the root of the confusion is that the term “centrifugal force” is used too broadly, even in cases where certainly a “centrifugal effect” does exist, but not as a real newtonian force … But in other cases that centrifugal effect appears, totally or partially, as a real force.
A general term to cover all scenarios could be “centrifugal inertial effect” (CIE?), that always is present if the trajectory of any massive stuff is curved, whatever the cause of that curved path … I´ll try and elaborate as follows.
We all know Newton´s Motion Laws. But I´m afraid not all keep in mind those laws are just the consequence of the basic Physics phenomenon of INERTIA: massive objects (and any part of them) always have a TENDENCY to maintain constant its current velocity vector, and they show a RESISTANCE to any agent trying to accelerate considered massive stuff. Those laws put it in terms of forces: f=ma (2nd law), being 1st law when f=0, and 3rd one the necessary consequence when considering two directly interacting objects.
We can analyze any possible case starting directly from INERTIA phenomenon, instead of using the “tool” of Newton´s Laws, not breeching them though: it´s a kind of other side of the coin …
In some cases, that RESISTANCE shows up as a real FORCE, but certainly not always.
The term CENTRIFUGAL always refers to an outward “tendency” to move, implying the existence of a “center”: the center of a circular path followed by an object, or at least the center of curvature of its CURVED PATH.
It´s convenient to separate cases with direct physical connection between interacting objects (A), from cases when gravity is involved (B).
A) Hammer throwing: the “hammer”, as a whole, is being centripetally accelerated. INERTIA tends to make it go on the tangent, the cable (and the athlete) don´t let it move straight, and the inertial RESISTANCE to being accelerated inwards appears as a real CENTRIFUGAL FORCE exerted by the hammer on the wire´s end … (I´m not considering now internal forces, that would be different if we had a sling instead, and form a “field” of real centrifugal forces, exerted between contiguous particles …).
It´s what David Cooper calls “reactive centrifugal force”. Similar things can be said about other cases such as wheeled wagons on a railway, vehicle rubber tires on road surface (with or without banking), etc., 
B) When gravity is involved, as it changes with distance, it´s paramount to distinguish cases when those changes are practically null (due to the rotating object negligible size, compared with distance to the object causing the gravitational field - e.g.: artificial satellites), from the rest.
B.1) In the first case the objects are in a pure “free fall”. All their particles are accelerated the same. Inertial RESISTANCE to being accelerated (proportional to mass and given acceleration) is precisely what makes necessary the existence of the gravitational pull f=ma: otherwise the object would continue straight.
But now we don´t have even a “reactive” centrifugal force: acting centripetal force, the gravitational pull at that location, is independent from the object´s inertia … If in some moment F were not equal to ma, the object would be free to change orbit (certainly a case quite different from hammer-throwing).
B.2) The simplest case is our moon rotation around earth-moon barycenter.
INERTIA makes every moon´s particle tend to keep moving straight, but all those particles are forced to follow circular paths.
The further the particle, the bigger the radius, and the bigger the acting centripetal force mω²r.
Inertial RESISTANCE to being accelerated is proportional to ma. If the particles were in a real free fall, they would be free to adjust their orbits to the acting gravitational pull at their location, that varies inversely to the square of the distance.
But that is not possible. If, e.g., we transversely “cut” the moon into too halves, the further one is being centripetally accelerated more than what earth´s pull would cause on that “hemimoon” if it were really free to move. Therefore, ALWAYS existing inertial RESISTANCE to being accelerated is only PARTIALLY compensated by earth´s gravitational pull, the unique force “external” to the moon. That fraction of that inertial RESISTANCE, as on case B.1, doesn´t cause any additional centrifugal effect, let alone force.
But the inertial RESISTANCE not “compensated” that way is still present, and, similarly to what in the hammer throwing case (A), it causes an outward pull on adjacent inner half moon. That is a real CENTRIFUGAL FORCE, quite similar to the one exerted by the hammer on the wire´s end. It “forces” closer half of the moon to keep the common orbit, instead of a smaller one that would match with the stronger earth´s pull on closer “hemimoon”.
On any other transverse section similar things happen, and the moon is stretched in the direction of the straight line earth-moon, what is also called “tidal effect”. By the way, directly connected with the fact that the moon is “tidal locked” to earth (closer and further mentioned halves don´t change, apart from some very tiny “oscillation”).
So, I consider quite opposite stands:
1) In all cases with curved paths real centrifugal forces are present,
2) The fact of having a curved path doesn´t imply the existence of real centrifugal "effects", and real centrifugal forces only appear in cases similar to hammer throwing (never when gravity is involved),
are both erroneous.
And the "invention" of a fictitious centrifugal force for cases when a rotating frame of reference is used doesn´t help diminish confusion ... That just adds something to cause the REAL inertial effects that, logically, disappear when rotation "ceases", as actually happens relatively to mentioned type of frame (precisely called "non inertial" !!).   

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Offline David Cooper

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Re: What is centrifugal force?
« Reply #234 on: 06/03/2019 23:08:20 »
Quote from: rmolnav on 06/03/2019 11:41:52
But the inertial RESISTANCE not “compensated” that way is still present, and, similarly to what in the hammer throwing case (A), it causes an outward pull on adjacent inner half moon. That is a real CENTRIFUGAL FORCE, quite similar to the one exerted by the hammer on the wire´s end. It “forces” closer half of the moon to keep the common orbit, instead of a smaller one that would match with the stronger earth´s pull on closer “hemimoon”.

If we think about an object sitting near a black hole, the gravity pulls on the object, and it does so more strongly on the near side of that object than the far side, so a tension force appears within it (which might pull it apart in the process known as spaghettification). That tension force is not any kind of centrifugal force. That force will be opposed by an opposite force which is likewise not any kind of centrifugal force.

If the planet is going round the black hole rather than falling towards it, these opposed tension forces are still acting within it, but their cause is exactly the same as in the non-rotating case. If you want to call the gravitational force centripetal force, what are you going to call the tension force that acts in the same direction? Is it centripetal force too? Maybe it is - it's just the force being converted to a different from and transferred on, so yes, and that logically requires you to call the opposing tension force reactive centrifugal force.

However, all these forces are driven by direct gravitational pull and the rotation aspect is a complete irrelevance to them, so if you want to provide people with a real understanding of what's going on, you have a duty to avoid using words like centripetal and centrifugal in the explanation.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #235 on: 07/03/2019 12:29:08 »
Quote from: David Cooper on 06/03/2019 23:08:20
the gravity pulls on the object, and it does so more strongly on the near side of that object than the far side, so a tension force appears within it (which might pull it apart in the process known as spaghettification). That tension force is not any kind of centrifugal force. That force will be opposed by an opposite force which is likewise not any kind of centrifugal force.
After many months of "dynamical" discussion, you keep not considering the real phenomenon of INERTIA, as if it didn´t exist, and therefore ignoring its "effects" ...
If it didn´t exist, the different pulls (because of the different distances) you talk about wouldn´t be able to generate any "opposite force" and cause spagettification. In that kind of impossible case, the object would get an "average" acceleration (though that is also directly connected to the phenomenon of inertia) ...
What actually causes the "opposite force" is the fact that the smaller pull on far side of the object is insufficient to "balance" inertial "resistance" to accelerate there that much. If a particle there were really "free", it would accelerate less. But the inertial resistance of the particle to accelerate is proportional to the acceleration actually given to it.
That difference between external pull on a particle, and inertial "resistance" to accelerate more than f/ma, is actually the appearing "opposite force". 
As I´ve said many times, far side particles have no way to "know" how much near particles are pulled, let alone how to "calculate" any difference between pulls on distant particles, to react accordingly !!
But all particles "feel" external gravitational pull at their location, and kind of internal inertial "resistance" to beeing accelerated, what, if not in balance, causes a real "opposite force" (exerted by the particle on adjacent particles, as the hammer on the wire) ...
And, as I´ve also said many times, that inertial effect is due to exactly the same reason as in cases were the movement is not along a straight line, but along a curved path ...
The only difference is that on curved ones, gravitational pull and relevant accelerations have not the same direction as the object velocity ... Due to that, the gravitational pull (or at least its component perpendicular to actual velocity vector), ACTING as centripetal force, causes the velocity vector change direction ...
And, similarly to the straight line case, the object offers an inertial "resistance" to being centripetally accelerated, what if not in balance with actual gravitational pull on the particle (e.g., ma>f), equally causes an "opposite force". And a force opposite to a centripetal force is, logically, a centrifugal force !!
But in the straight line case forces and accelerations have not any kind of relation with curved movements, necessary for the existence of the concept of centripetal forces, centripetal accelerations, and centrifugal forces, and where a "center" of curvature does exist ...   
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Re: What is centrifugal force?
« Reply #236 on: 07/03/2019 21:05:06 »
Quote from: rmolnav on 07/03/2019 12:29:08
After many months of "dynamical" discussion, you keep not considering the real phenomenon of INERTIA, as if it didn´t exist, and therefore ignoring its "effects" ...

Inertia is tied up in everything I've said. Things keep moving (or not moving) in the same way so long as there's no force being applied to them. When you apply a gravitational force to them and that force is stronger at one end than the other, the nearest end is pulled more strongly than the far end and you get a tension force appearing in the object. I have no problem with you using the word inertia as a result and have never objected to it.

Quote
As I´ve said many times, far side particles have no way to "know" how much near particles are pulled, let alone how to "calculate" any difference between pulls on distant particles, to react accordingly !!

Who needs the particles to know anything? They simply continue to move as they are already moving unless a force is applied to them, whether that's through gravitational pull or by a more local pull or push from adjacent material.

Quote
The only difference is that on curved ones, gravitational pull and relevant accelerations have not the same direction as the object velocity ... Due to that, the gravitational pull (or at least its component perpendicular to actual velocity vector), ACTING as centripetal force, causes the velocity vector change direction ...

The only change to the movement is along the direction in which the force is applied.

Quote
And, similarly to the straight line case, the object offers an inertial "resistance" to being centripetally accelerated, what if not in balance with actual gravitational pull on the particle (e.g., ma>f), equally causes an "opposite force". And a force opposite to a centripetal force is, logically, a centrifugal force !!

In any case where the force is still acting when the orbital movement is removed, the use of the word "centripetal" was misleading, no matter how firmly established that usage of the word is in science. It is when we stop the object and see that the forces continue to operate in full that we realise that they are not genuinely centripetal because they are still fully in play while the label is no longer valid in any way. That is how you get to a better scientific understanding - you look for the places where your labels break and then you realise that you are dealing with something more fundamental. With something going round on a string, the centripetal and reactive centrifugal labels don't break when you remove the orbital movement because the forces disappear - they genuinely are generated by the rotational movement.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #237 on: 08/03/2019 12:15:43 »
It would be utterly absurd (and useless) we to continue here our so long last summer and fall discussion on thread "Why do we have two high tides a day?" ...
But, for the sake of the possible interest of other readers, I´m going to put here something directly related to what you are saying now, root of your erroneous stand, as far as I can understand:
Quote from: David Cooper on 07/03/2019 21:05:06
In any case where the force is still acting when the orbital movement is removed, the use of the word "centripetal" was misleading, no matter how firmly established that usage of the word is in science. It is when we stop the object and see that the forces continue to operate in full that we realise that they are not genuinely centripetal because they are still fully in play while the label is no longer valid in any way. That is how you get to a better scientific understanding - you look for the places where your labels break and then you realise that you are dealing with something more fundamental.
I´ve said here time and time again that "centripetal force" is neither an "essentially" new force, nor just a  label we can give any force "at will" !! It is just a FUNCTION, which many essentially different types of forces can EXERT, if they are causing the bending of the trajectory of a moving object ...
A pilot, after landing the airplane and going home, is not EXERTING as pilot whatsoever ... Do you consider we should not "label" him or her as a "pilot" because we "are dealing with something more fundamental" (a human being) ??
Quote from: David Cooper on 07/03/2019 21:05:06
With something going round on a string, the centripetal and reactive centrifugal labels don't break when you remove the orbital movement because the forces disappear - they genuinely are generated by the rotational movement.
Many times I´ve told you that is erroneous ... Movement doesn´t actually "generate" forces, the opposite occurs !!
You say "With something going round on a string" ... ?? What follows comes from our last year discussion of that case, if the initial movement were caused hitting the ball with a bat.
Hitting a ball produces a transference of momentum, through forces as always. At the very initial instant some deformations (of the ball and the bat) occur, what produces opposite pushes on each other, which then change both speed vectors … The ball gets a speed, and its inertia tries to make it go straight … If a string attached to a pole prevent that to happen, an initial string tension has to exist for that to occur (otherwise the ball would continue straight). That initial string tension is already an initial centripetal force. What the ball´s inertia does (though only over a very short time) is to further tighten the string, and that means the centripetal force increases (exerted by the string on the ball), and subsequently centrifugal “reactive" force (exerted by the ball on the string) also increases …
The initially rectilinear movement is logically necessary to get it converted into a circular one, but it isn´t actually the cause of of that “conversion” and occurring forces … An independent initial tension of the string is required, that could be called “initial centripetal force” as soon as the rectilinear path changes into curved one (not exactly circular at the beginning, because the elasticity of the string, and the fact that initially could´t be fully tight …). And its further increase, and also increasing centrifugal forces, are not directly caused by the initial rectilinear movement, but a consequence of ball´s inertia, manifesting itself that way after the ball being “forced” to bend its path by initial string tension, initial centripetal force ...
 
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Re: What is centrifugal force?
« Reply #238 on: 08/03/2019 19:50:17 »
Quote from: rmolnav on 08/03/2019 12:15:43
I´ve said here time and time again that "centripetal force" is neither an "essentially" new force, nor just a  label we can give any force "at will" !! It is just a FUNCTION, which many essentially different types of forces can EXERT, if they are causing the bending of the trajectory of a moving object ...

All I'm asking you to do is distinguish between the different cases and to make clear that when you're dealing with the gravitational cases, the orbital aspect is not generating the forces that are being being applied, while in the case of something going round on the end of a wire, the orbital aspect is generating the forces. That is something you're still refusing to do. Stop the ball on the string by absorbing all the energy that's making it move perpendicular to the string and the tension forces in the wire disappear. Do the same with a ball orbiting a planet and the forces continue to act in full. I don't know why you're incapable of agreeing with that, but something is blocking your thinking on that point, and you would gain a lot by fixing it.

Quote
Many times I´ve told you that is erroneous ... Movement doesn´t actually "generate" forces, the opposite occurs !!

And every time you've been wrong. All you need to do is trace the cause and effect order of events. The tension in the string never causes the object on the end of it to start moving round in circles, and cutting the string to remove that tension does not make the ball stop. It is starting the ball moving that generates the tension in the string and stopping the ball that removes the tension in it. No amount of elaborating on the components of these actions can change that - the big picture remains as I have stated it.
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Offline rmolnav

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Re: What is centrifugal force?
« Reply #239 on: 09/03/2019 12:29:07 »
I repeat: it would be useless to repeat here our past year discussion on the other thread ... But:
 
Quote from: David Cooper on 08/03/2019 19:50:17
The tension in the string never causes the object on the end of it to start moving round in circles, and cutting the string to remove that tension does not make the ball stop.
I never said the tension makes the object "start moving round in circles" ... But if an object initially in rectilinear movement start being pulled perpendicularly to its velocity vector, the movement CHANGES into a curved one, whatever the way it is pulled (string, gravity, road/tyre friction, etc).
That force is always what is "accelerating" the object inwards, and CAUSES the movement to be circular (or just curved), instead of rectilinear.
And it is called CENTRIPETAL FORCE by any physicist you may find, whatever your "problems" with the term, certainly a "great area" for you as you said long ago ...
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