# Centrifugal forces ' appear ' to act opposite to gravity . How is this possible?

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SwansonT,

Well if you cannot have a line tangent to a point, how about a really really tiny circle?

Would not the tangent line drawn from a tiny circle be indistinquishable from a line drawn radially outward from the tiny circle?

Regards, TAR

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SwansonT,

Well if you cannot have a line tangent to a point, how about a really really tiny circle?

Would not the tangent line drawn from a tiny circle be indistinguishable from a line drawn radially outward from the tiny circle?

Regards, TAR

That would be a matter of relativity.

Edited by Robittybob1
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SwansonT,

Well if you cannot have a line tangent to a point, how about a really really tiny circle?

Would not the tangent line drawn from a tiny circle be indistinquishable from a line drawn radially outward from the tiny circle?

Regards, TAR

A tangent line and a radial line are perpendicular. So they are basically maximally distinguishable.

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A tangent line and a radial line are perpendicular. So they are basically maximally distinguishable.

.

I think it is worth a trip down one of these tangential strait lines , when the radial line is cut ,temporarily. And also what happens as that temporary cut is re-established , and what is needed to re-establish it .

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Firstly ,in the immediate moment , the circulating mass , for that instant , could be thought of as trying to do what every well behave mass does , and that is ' travel in a strait line ' as according to Newton .

So irrelevant of what has gone before , at the moment of 'cutting ' , the strait line ( which is the tangent at that moment T1 ) , will start to continue , with that strait line . It has its tremendous inertia , so to do .

At the moment of cutting the radius is say R 1 .

If we imagine a moment later T2 , had the radius R1 had not been cut , then R2 = R1 .

However the radius has been cut and the mass has moved on a bit tangentially . It is now at T2 , a little bit down its strait line , away from the circumference by a bit delta R . Now effective R2 is greater than R1 by a factor of delta R.

If we try to restore the situation we would have to apply a considerable inward force ( centripetal force ) inwards to the centre along R2 in order to restore it to the same length of R1 .

This is no mean feat, and has the appearance of an opposing force ( centrifugal force ) . Almost like a sumo wrestler trying to push us ' away from the Centre ' namely Centrifugally ( or apparently ) . If we do not succeed to oppose this wrestler , who is on the roundabout we will have found ourself out at R2 distance from the centre. ( even though no longer connected to the roundabout ) , which is greater than R1 by a factor Delta R .

This whole experience has surely achieved , what appears to be a forced movement away from the centre . Is that not so?

If this were to be considered vertically , upward surely it would be the same . Water in bucket, satellite in orbit . Both feeling a forced movement , Apparently acting opposite to gravity. Is this also not so ?

ps Apparent centrifugal force..being like the Sumo wrestler in our story above .

Mike

Edited by Mike Smith Cosmos
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.

I think it is worth a trip down one of these tangential strait lines , when the radial line is cut ,temporarily. And also what happens as that temporary cut is re-established , and what is needed to re-establish it .

Mike - it is hard enough to follow your reasoning let alone when you start mixing non-literal construction lines with what seems like a line as in a cable. And is it too much to ask you to spell straight correctly - without the g and h it means a narrow range of water connecting two larger bodies.

Firstly ,in the immediate moment , the circulating mass , for that instant , could be thought of as trying to do what every well behave mass does , and that is ' travel in a strait line ' as according to Newton .

So irrelevant of what has gone before , at the moment of 'cutting ' , the strait line ( which is the tangent at that moment T1 ) , will start to continue , with that strait line . It has its tremendous inertia , so to do .

At the moment of cutting the radius is say R 1 .

If we imagine a moment later T2 , had the radius R1 had not been cut , then R2 = R1 .

However the radius has been cut and the mass has moved on a bit tangentially . It is now at T2 , a little bit down its strait line , away from the circumference by a bit delta R . Now effective R2 is greater than R1 by a factor of delta R.

If we try to restore the situation we would have to apply a considerable inward force ( centripetal force ) inwards to the centre along R2 in order to restore it to the same length of R1 .

This is no mean feat, and has the appearance of an opposing force ( centrifugal force ) . Almost like a sumo wrestler trying to push us ' away from the Centre ' namely Centrifugally ( or apparently ) . If we do not succeed to oppose this wrestler , who is on the roundabout we will have found ourself out at R2 distance from the centre. ( even though no longer connected to the roundabout ) , which is greater than R1 by a factor Delta R .

This whole experience has surely achieved , what appears to be a forced movement away from the centre . Is that not so?

If this were to be considered vertically , upward surely it would be the same . Water in bucket, satellite in orbit . Both feeling a forced movement , Apparently acting opposite to gravity. Is this also not so ?

ps Apparent centrifugal force..being like the Sumo wrestler in our story above .

Mike

No. Per your first sentences - the object goes off in a straight line because there is no longer a centripetal force accelerating it inwards. It is not like a stationary object in a sumo ring that gets thrust out.

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Mike - it is hard enough to follow your reasoning let alone when you start mixing non-literal construction lines with what seems like a line as in a cable. And is it too much to ask you to spell straight correctly - without the g and h it means a narrow range of water connecting two larger bodies.

No. Per your first sentences - the object goes off in a straight line because there is no longer a centripetal force accelerating it inwards. It is not like a stationary object in a sumo ring that gets thrust out.

Well I perhaps should have introduced the subjects we had recently been discussing , namely spinning around on a childhood circular 'stand on ' spinning mini roundabout , spinning buckets of water overhead, and satellites in orbit .

Sorry about the spelling . I was still taking 'O' level English for the third time in 6th form ,while doing my 'A' level physics and maths . My English teacher said to me " smith , you are an enigma !

Now as regards the diagram , I am trying to reason out, that in order for NOT ALLOWING the inertial mass to go down that supposed straight line , had it been allowed to, it would have reached R 2 a larger radius . But to almost immediately after moment of release, I have to fight this sumo wrestler of force trying to take me down the strait line ( due to inertia) , by pushing me outward radially, which would if I lost have increased my effective radius to the greater R2 . However I am overcoming this outward force with my counter centripetal force. ( inward towards maintaining R1. In the circle . ) perhaps by pulling on the rope, holding the rail bars , or using gravity as an anti-force to the 'sumo'

It would seem I on my travel down my legitimate strait line am in the right , the sumo 'takes pet ' when I am being dragged screaming into a circle by this R 1 restraining platform, rope or gravity , as the case may be . Then surely I become aware of this darn sumo wrestler pushing me to staining pitch out of the circle . As it happens the platform holds, the string holds , but anything loose in my pockets, hair, objects , water or whatever will fly off into a larger radius R2 . Is this not the operation of a centrifugal pump, centrifugal filter , and the facial mussel in an astronauts face as he experiences 2 or 3 G in his test centrifuge?

Mike

Edited by Mike Smith Cosmos
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.

I think it is worth a trip down one of these tangential strait lines , when the radial line is cut ,temporarily. And also what happens as that temporary cut is re-established , and what is needed to re-establish it .

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

Firstly ,in the immediate moment , the circulating mass , for that instant , could be thought of as trying to do what every well behave mass does , and that is ' travel in a strait line ' as according to Newton .

So irrelevant of what has gone before , at the moment of 'cutting ' , the strait line ( which is the tangent at that moment T1 ) , will start to continue , with that strait line . It has its tremendous inertia , so to do .

At the moment of cutting the radius is say R 1 .

If we imagine a moment later T2 , had the radius R1 had not been cut , then R2 = R1 .

However the radius has been cut and the mass has moved on a bit tangentially . It is now at T2 , a little bit down its strait line , away from the circumference by a bit delta R . Now effective R2 is greater than R1 by a factor of delta R.

IOW, if it continued this way it would be a spiral, just as I said.

If we try to restore the situation we would have to apply a considerable inward force ( centripetal force ) inwards to the centre along R2 in order to restore it to the same length of R1 .

This is no mean feat, and has the appearance of an opposing force ( centrifugal force )

But the force exerted BY the object doesn't matter. It's only the forces exerted ON the object that dictate its motion. And the only force on the object is, as you have stated, toward the center of the circle. Centripetal.

IOW, the forces in Newtons third law act on each other. They never both appear in a statement of Newton's second law.

But to almost immediately after moment of release, I have to fight this sumo wrestler of force trying to take me down the strait line ( due to inertia) , by pushing me outward radially, which would if I lost have increased my effective radius to the greater R2 .

If you are moving in a straight line, you are feeling no net force.

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SwansonT,

But 1/4 of the period of rotation later, a line normal to a radial line becomes parallel to the radial line drawn either 1/4 period earlier or later. And indistiguishable, direction wise, from a radial line,

In a static discription of the direction the object on a carosel travels, a vector can be considered tangent to the circle traveled and another pointing radially inward, adding up to the circle traveled. The forces acting on the body, like the friction of the seat of the pants against the carosel and the mechanical stress on the muscles of the kid and the friction of hand against steel keep the kid from flying off the thing. Whether she lets go at a particular moment and flies off tangentially or continues to fight the force that is pulling her radially outward only determines whether she stays on, or flies off. It does not make the fact that she is being pulled radially outward, false.

Regards, TAR

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QUOTE(MigL): "That's a non-answer DrP.......Why is it further from the centre of the earth.... Does the rock weigh less at the equator?"

It is further from the centre because the earth is not a sphere - it is squished. I didn't think it necessary to explain that. So, yes, the rock weighs less at the equator than it does at the poles. Seriously? have you not covered that in school yet or something?

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Enough talk.

Sorry Mike, but here is some maths.

Sorry also that my artwork is not up to your standards, you will just have to use your imagination there.

Consider a weight hanging by a string from the frustrum of a witch's hat cone.

The weight rests on the side of the cone.

Ignoring friction, the weight is supported against gravity by the tension in the string and the reaction with the surface of the cone.

Now the weight is swung so it is set in motion travelling around the surface, but not leaving it.

The weight therefore travels in a horizontal circle around the cone.

In the attached sketch I have shown two possible methods of analysis.

On the left is a dynamic analysis by Newton's laws using only the real forces that are acting.

On the right is the D'Alembert transformation to a static equilibrium analysis by the introduction of a fictious centrifugal force.

Both analyses arrive at the same results and therefore look pretty similar.

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SwansonT,

But 1/4 of the period of rotation later, a line normal to a radial line becomes parallel to the radial line drawn either 1/4 period earlier or later. And indistiguishable, direction wise, from a radial line,

And that matters how, exactly?

SwansonT,

In a static discription of the direction the object on a carosel travels, a vector can be considered tangent to the circle traveled and another pointing radially inward, adding up to the circle traveled. The forces acting on the body, like the friction of the seat of the pants against the carosel and the mechanical stress on the muscles of the kid and the friction of hand against steel keep the kid from flying off the thing. Whether she lets go at a particular moment and flies off tangentially or continues to fight the force that is pulling her radially outward only determines whether she stays on, or flies off. It does not make the fact that she is being pulled radially outward, false.

If there is a radial force outward, how is it that the person is moving in a circle? Motion in a circle requires a centripetal acceleration, not just as a component, but as the net acceleration.

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QUOTE(MigL): "That's a non-answer DrP.......Why is it further from the centre of the earth.... Does the rock weigh less at the equator?"

It is further from the centre because the earth is not a sphere - it is squished. I didn't think it necessary to explain that. So, yes, the rock weighs less at the equator than it does at the poles. Seriously? have you not covered that in school yet or something?

I think MigL ment that it is the rotation that is causing Earth to be squished in the first place...
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An object weighs less at the equator because it is farther from the center of the Earth.

I replied that...

For the surface (rocks, dirt, etc. ) of the Earth to be farther from the center, it must also weigh less ( not mass, mind you, but weight ) otherwise it would form a perfect sphere.

So why does the surface weigh less ( and everything below it ) ?

And I only wish I was back in school ( No responsibilities, spending 6 out of 7 nights in clubs, etc.; I'm getting depressed just thinking about it. Damn this middle/old age sucks. )

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An object weighs less at the equator because it is farther from the center of the Earth.

I replied that...

For the surface (rocks, dirt, etc. ) of the Earth to be farther from the center, it must also weigh less ( not mass, mind you, but weight ) otherwise it would form a perfect sphere.

The difference in gravity (and therefore weight) at the north pole and the equator is because the north pole is nearer the center of the Earth than the equator is. This is because the Earth is spinning, causing it to be flattened.

So that was not a non-answer.

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My fact about the distance to the centre of the earth at the equator compared to that of the poles was to put Tar on the right track after he was barking up the wrong tree. Things at the equator weigh more. How is that a non answer MigL? It isn't an answer at all, it was a reply to Tar getting things arse about face even though as Strange pointed out that swansonT has already covered all of this... hell, I wish I hadn't bothered. lol.

Oh, and sorry if my comment about school was a little sarcastic. I know how you feel MigL.

QUOTE, MigL: "I replied that, for the surface of rocks...." well, actually, you didn't, you replied as I quoted you.

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No offence taken DrP.

And I hope you don't take my referring to your answer as a 'non-answer' as an implication that it is wrong.

It is accurate, it just doesn't add new information.

A mass weighs less at the equatorial surface because it is further from the center of the earth. But the surface at the equator is further from the center than at the poles, because it weighs less ( weight being a force due to Earth gravity ).

A proper explanation for this would include that the centripetal force ( gravity ) at the equatorial surface has a greater inertia to overcome ( higher angular momentum due to greater distance from axis of rotation ) than at the poles. This is what leads to the flattened spheroid shape > greater radius > lower weight.

Or is this just semantics ?

In which case you could call this a non-argument.

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But the surface at the equator is further from the center than at the poles, because it weighs less ( weight being a force due to Earth gravity ).

That is what I thought you were implying but I don't understand it. The surface at the equator is further from the centre because it is being spun more, not because it weighs less.

I suppose you could argue that it weighs less because of centrifugal forces and is therefore further away and therefore weighs less. That is not a non-answer, it is a very interesting answer (in my opinion).

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That is what I thought you were implying but I don't understand it. The surface at the equator is further from the centre because it is being spun more, not because it weighs less.

I suppose you could argue that it weighs less because of centrifugal forces and is therefore further away and therefore weighs less. That is not a non-answer, it is a very interesting answer (in my opinion).

If you don't like centrifugal force you could always say it has "more inertia" at the equator therefore weighs less.

If inertial mass equals the inertial mass can I rightly say something has "more inertia"?

More momentum or even more angular momentum (wrt the center of the Earth) rather than more inertia might be more correct scientifically.

Edited by Robittybob1
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And that matters how, exactly?

If there is a radial force outward, how is it that the person is moving in a circle? Motion in a circle requires a centripetal acceleration, not just as a component, but as the net acceleration.

SwansonT,

Well the force that got the kid moving in the circle was the other kids applying a tangential push to the outside perimeter of the merry-go-round, overcoming the inertia of the device and the kid. The kid holding on is what keeps the kid from flying tangentially off the thing. If the kid started in the middle the "pull", even after everybody stops pushing, is toward the outside of the device. The kid (me in this memory) slides radially outward, in reference to the device, as his strength gives out and he pleads with the pushers to slow the thing down so he doesn't go flying off and get hurt.

Regards, TAR

Edited by tar
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SwansonT,

Well the force that got the kid moving in the circle was the other kids applying a tangential push to the outside perimeter of the merry-go-round, overcoming the inertia of the device and the kid. The kid holding on is what keeps the kid from flying tangentially off the thing. If the kid started in the middle the "pull", even after everybody stops pushing, is toward the outside of the device. The kid (me in this memory) slides radially outward, in reference to the device, as his strength gives out and he pleads with the pushers to slow the thing down so he doesn't go flying off and get hurt.

Regards, TAR

The how does the kid move in a circle? The vectors don't work. Do the math.

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I don't get it ! Everything that swings around in a circle , provided there is a bit of slop in the system , the Scotsman swinging a shot or discus around , spins out into a sort of fried egg.

Eg the moons circle the planets so it looks like a fried egg. Jupiter , earth , Saturn , etc

The galaxy spins out like an egg. , no doubt the universe is egg shaped, . The only thing that stops everybody falling off the merry go round is GRAVITY .

Why don't we all turn around and say " look! The king has got no clothes on , "

I don't get it , the spinning is everywhere , the fried eggs are everywhere, The forces pushing everything that is loose , goes out ward to make up the flattened yoke and pushed out egg white. It's the eternal battle between Gravity pulling in and something spinning causing a pushing out . We have all been there on those mini , merry go rounds . The straight bit is only the bit when we fall off. When we are loose on the roundabout we are being pushed to the barrier! Also "the wall of death" .

Mike

Edited by Mike Smith Cosmos
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Do the math. Subtract the vectors.

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Do the math. Subtract the vectors.

Surely there is a conflict here ,between what we see and feel . And what some mathematical structure somehow says " there is nothing there!" It's fictitious . When our senses tell us something that we know and feel is happening to us . If we were in a cardboard box we would feel ourself moving toward the edge of the. Roundabout, as if some other child was pushing the box . Surely by the law of equivalence , then it is so!

Mike

Edited by Mike Smith Cosmos
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Surely there is a conflict here ,between what we see and feel . And what some mathematical structure somehow says " there is nothing there!" It's fictitious . When our senses tell us something that we know and feel is happening to us . If we were in a cardboard box we would feel ourself moving toward the edge of the. Roundabout

The line we'd take as we slid off the merry go round is not a straight line but a curved line across the surface of the spinning disc.

It is a line described by the Coriolis Effect.

Edited by Robittybob1
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It's measurable and repeatable not fictitious.

It's simple f=ma vectors, really don't understand how you have difficulty with that. So it's more than just what we feel

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