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

[Mike Smith Cosmos]

The moment of inertia at any one point , or all points ,on or around the circle or the curve .. Is symmetrical , thus the energy and the momentum is conserved. This momentum before existed in a straight line. It was converted to a circle , the previous linear momentum was transferred both ( momentum and energy ) were conserved so symmetry was present .

?..........Now look at moments of inertia and it's applications to rotation.

http://en.m.wikipedia.org/wiki/List_of_moments_of_inertia

.

.

Ouch !

 

Mike

 

Ps there must be easier ways than wading through , knee deep in " 3 d rotational Tensors "

 

Surly I can throw a bucket of water on the end of a rope either in a circle over my head or around horizontally . I can see and feel the force ?

 

Mike

Edited May 2, 2015 by Mike Smith Cosmos

[Mordred]

There is via Newtons three laws. You don't require tensors. Look at each moment in time in terms of net force per that specific moment in time.

 

Key words Net sum of force per moment (aka moment of inertia)

Moment A string is attached there is mechanical force causing a change in direction

 

Moment B string is let go, no force towards the center. Rock flies at a tangent (straight line from curve ) http://en.m.wikipedia.org/wiki/Tangent from that moment according to Newtons laws of inertia.

Think of a catapult, after the rock is released "which way does it fly"

Ignoring gravity it would fly in the last angle, direction upon release. In a straight line from that moment of direction.

Edited May 2, 2015 by Mordred

[swansont]

That reduction in the normal force is called the centrifugal force.

 

The reduction in the normal force is erroneously called the centrifugal force. Care to try again?

[Robittybob1]

 

The reduction in the normal force is erroneously called the centrifugal force. Care to try again?

Well take a weight and make it orbit on a string. There is a force in the string called the centripetal force but the mass does not move in the direction of the force. This is because the centripedal force equals the centrifugal force.

[swansont]

Well take a weight and make it orbit on a string. There is a force in the string called the centripetal force but the mass does not move in the direction of the force. This is because the centripedal force equals the centrifugal force.

 

And we're back to square one. If the forces cancel, then there would be no net force. By Newton's first law of motion, the weight must move in a straight line. But since it's not, the assumption that the centripetal force and centrifugal force cancel must be false.

 

Further, as I had suggested to TAR, do the math. To move in a circle, and object must feel a force toward the center of the circle. It's a simple matter of subtracting one vector from another.

 

This is a great example of why one cannot rely on naive intuition to solve problems.

[physica]

Well take a weight and make it orbit on a string. There is a force in the string called the centripetal force but the mass does not move in the direction of the force. This is because the centripedal force equals the centrifugal force.

Your statement comes from a misunderstanding of high school physics. I'm not saying this to put you down but it's very helpful if posters get an understanding of where their knowledge base is. Think about force.... what is force?? It's the rate of change of the rate of change multiplied by the mass. Basically if you want to change the direction that the ball is moving in you have to exert a force. Now lets look at a snap shot. If I swing a ball on a string above my head we can look at when the ball is directly in front of my face, the centripetal force is acting on the ball towards my head and the velocity of the ball is orthogonal to this. We when take another snap shot when the ball is beside my head... there has been a change in the direction of the ball therefore the centripetal force has acted and changed the direction of the ball. However, the direction of the force is now orthogonal to the previous force vector. There is no change in radius because the direction of the forces and the velocity of the ball is constantly changing.

 

Now for contrast look at what you've said, if there was a centrifugal force that cancelled out the centripetal force would result no net force (like what swansont said), this means that there will be no change in the direction that the ball is travelling, as a result the the call wouldn't travel in a circle but would travel in a straight line.

[Mike Smith Cosmos]

Having read many of the links provided by the experts and moderators on this science forum , concerning the issues of straight line motion and accelerated motion , as well as dredging up what relevant instruction I received in University , I have tried to pull together what I have read. And as such , if we were to acquiesce , and say o.k. We have created an accelerating frame which is circulating , by whatever means about a centre. Speed and vector velocity are very relevant , as is the issue of acceleration . I think?

 

Say gravity in the case of the Earth, or .... a hand ,in the case of a rotating bucket , on the end of a rope , with water in the bucket ,.... or a motor bike , going round a corner that ends up as a circle , say 1/4 mile in radius . I hope ?

 

If we look at the forces in operation about a radius rotating about a centre. Then we have a centripetal force acting on the mass so as to produce a vector trajectory down the radius, aimed at the centre. Left un impeded the mass would accelerate toward the centre, down the radius , and ultimately arrive at the centre. I would say ?

 

However, we find that the developed ' centrifugal force ' causing an opposite force thus acceleration, outward from the centre.( depending on the value of the centrifugal force , which itself depends on angular velocity , will dictate the resultant acceleration.) { namely excess in , moving motion toward centre! excess out , moving toward outer . Balanced forces settle at a radial distance } this whole process like a tug of war . The net result of these two forces in the accelerating , rotating frame can be zero . So the mass neither accelerates away from the centre getting further away from the centre or accelerating towards the centre , and thus getting nearer the centre. I think ?

 

We created this rotating frame in the first place , by taking straight line momentum/ inertia and by an applied inward force ( centripetal force ) down a potential radius , with sufficient energy , converted the straight line momentum/ inertia ( inertial frame ) into a rotating accelerating frame , creating , angular momentum , with its accelerating frame . The whole transformation holds true for both symmetry and conservation of energy , and conservation of momentum . I think symmetrical as the same holds true through all 360 degrees of movement ?

 

 

Mike

 

Appendum

 

It has to be said , that if the radius link is cut or broken or lost then the rotating frame stops , is lost , and the mass will immediately revert to an inertial frame , whereby the mass or satellite ( gravity cut somehow ) the mass will fly off along a straight line ( or tangent at the point of cutting gravity. Or the bucket, complete with water will fly off down the tangent at cutting the rope. Or the motor bike will vier off the road , over the hedge , into the field along a straight line ,running from the point of separation . I would guess?

 

I guess there must sometimes/ mostly be a mismatch between energies in orbit and energies in straight line trajectories. Hence the need for quantised energies with atomic interactions . I think ?

 

In all three cases momentum, energy, and inertia will be conserved . Certainly ! I am not sure about symmetry . Probably symmetry will be broken?

 

Mike

.

Edited May 2, 2015 by Mike Smith Cosmos

[Mordred]

No symmetry isn't broken, GR accounts for energy momentum. This has symmetry via SO(3.1) Lorentz group representation. Newtons laws would also follow symmetry not sure which group though later on I may try to calc the group.

 

Just checked linear and angular momentum both fall under the Lorentz group. ( lol should have been obvious to me, brain fart)

Edited May 2, 2015 by Mordred

[Robittybob1]

 

And we're back to square one. If the forces cancel, then there would be no net force. By Newton's first law of motion, the weight must move in a straight line. But since it's not, the assumption that the centripetal force and centrifugal force cancel must be false.

 

Further, as I had suggested to TAR, do the math. To move in a circle, and object must feel a force toward the center of the circle. It's a simple matter of subtracting one vector from another.

 

This is a great example of why one cannot rely on naive intuition to solve problems.

If the centripetal force was greater than the centrifugal force the string would get shortened or if the centrifugal force got greater than the centripetal force the string would stretch or even break and the weight would fly off on a tangent depending on the velocity at the time.

Your statement comes from a misunderstanding of high school physics. I'm not saying this to put you down but it's very helpful if posters get an understanding of where their knowledge base is. Think about force.... what is force?? It's the rate of change of the rate of change multiplied by the mass. Basically if you want to change the direction that the ball is moving in you have to exert a force. Now lets look at a snap shot. If I swing a ball on a string above my head we can look at when the ball is directly in front of my face, the centripetal force is acting on the ball towards my head and the velocity of the ball is orthogonal to this. We when take another snap shot when the ball is beside my head... there has been a change in the direction of the ball therefore the centripetal force has acted and changed the direction of the ball. However, the direction of the force is now orthogonal to the previous force vector. There is no change in radius because the direction of the forces and the velocity of the ball is constantly changing.

 

Now for contrast look at what you've said, if there was a centrifugal force that cancelled out the centripetal force would result no net force (like what swansont said), this means that there will be no change in the direction that the ball is travelling, as a result the the call wouldn't travel in a circle but would travel in a straight line.

it has to do with frames of reference. In your case the center point is not rotating with the mass, but if it was the ball would always be in front of your face and at the same distance.

Edited May 2, 2015 by Robittybob1

[Mike Smith Cosmos]

Having read many of the links provided by the experts and moderators on this science forum , concerning the issues of straight line motion and accelerated motion , .................

 

We created this rotating frame in the first place , by taking straight line momentum/ inertia and by an applied inward force ( centripetal force ) down a potential radius , with sufficient energy , converted the straight line momentum/ inertia ( inertial frame ) into a rotating accelerating frame , creating , angular momentum , with its accelerating frame . The whole transformation holds true for both symmetry and conservation of energy , and conservation of momentum . I think symmetrical as the same holds true through all 360 degrees of movement ? Is this whole analysis Right or Wrong or 1/2 right 1/2 wrong?

Mike

.

There appears to be a tug of war of forces going on in the accelerating frame . Centripetal down ( in towards centre ) in this snap shot . And Centrifugal force up ( away from centre ) . It seems a net force , which can be zero ( when in balance then NO radial movement in or out ) , alternatively a residual force thus acceleration toward the centre centripetal or a residual away from the centre centrifugal.

 

Again , Is this whole analysis Right or Wrong or 1/2 right 1/2 wrong?

 

 

Mike

Edited May 2, 2015 by Mike Smith Cosmos

[physica]

If the centripetal force was greater than the centrifugal force the string would get shortened or if the centrifugal force got greater than the centripetal force the string would stretch or even break and the weight would fly off on a tangent depending on the velocity at the time.

If there was a centrifugal force the acting radially outwards then the object wouldn't fly off at a tangent if the string snapped.

 

 

it has to do with frames of reference. In your case the center point is not rotating with the mass, but if it was the ball would always be in front of your face and at the same distance.

Frames of reference do not change the laws of physics. You can pick a bad frame of reference and not encapsulate the fully system. If there is a net force that equates to zero there will be no change in the direction of the ball. No frame of reference will change this. No frame of reference will change the fact that the ball is changing its direction. I'm going to be a bit more direct now considering that you failed to comprehend what I said and you're brush off simply wasn't up to the mark. Just read some basic physics textbooks, right now you'd fail a high school physics exam. Once you've understood at least the basics then you'll see how wrong you are.

[Robittybob1]

If there was a centrifugal force the acting radially outwards then the object wouldn't fly off at a tangent if the string snapped.

Do you think there is a speed of orbit that will snap the string? What is the centripetal force acting on as the rate of orbit goes up?

The mass stays the same.

centripetal force F = m * v^2 / r.

[tar]

Thread,

 

I think my problem stems from not understanding why the inertia of an object in motion is not considered a force, in and of itself.

 

If an object was not moving, a force would be required to overcome the stationary inertia. Once moving, a force would be required to deaccelerate the object and bring it to a stop.

 

In the case of a orbiting object, the thing keeps falling toward the Earth...and missing. If it would not miss, then I could understand the force toward the center, the centripical force, was the only real one being applied. But the fact that it does miss, indicates to me, that there is a force being applied on the object that is causing it to miss the center. An outward force. A tangentially outward force. Like the tangent to the lifesaver, taking you outward, away from the life saver, 90 degrees later.

 

I have been musing alot with a 1/4lb. sphere of clay in front of me. With a toothpick, drawing lines and angles, great circles intersecting at various angles. Interesting to note that a sphere can be divided into 24 60/60/90 triangles. I mention this, because two lines drawn at a 60 degree angle to each other wind up crossing each other exactly once on the diametrically opposed side of the sphere. Same thing happens if you draw two great circles intersecting each other at 90 degrees.

 

Point being, what is true on one side of the sphere if continued to the other side, is upside down and backward, but exactly the same relationship, halfway round the sphere, as long as you are dealing in great circles.

 

So...when talking about a simple circle, and something tangent to it, or normal to the tangent, and you consider the direction a force is acting in, a quarter period later, what was tangent is now radial, and what was radial is now tangent.

 

When you are swinging the bucket on the rope, the water does not come toward the rope, it goes toward the bottom of the bucket, which might be because it wants to go tangent to the circle, and the bottom of the bucket keeps getting moved into that path...or it wants to go radially outward, and the bottom of the bucket is alway positioned radially outward from the rope.

 

Regards, TAR

[Robittybob1]

Newton's third law: http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law

 

 

For every action, there is an equal and opposite reaction.
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

So what is the force paired with the centripetal force?

[swansont]

If the centripetal force was greater than the centrifugal force the string would get shortened or if the centrifugal force got greater than the centripetal force the string would stretch or even break and the weight would fly off on a tangent depending on the velocity at the time.

 

No, that's wrong. If the forces cancel, then there would be no net force and the object moves in a straight line. Newton's first law.

Newton's third law: http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law

So what is the force paired with the centripetal force?

 

The centrifugal force.

 

As I explained in one of these threads recently, and as your quote shows, Newton's 3rd law forces act on different objects. F_{a on b} = -F_{b on a}

 

But the motion of object a only depends on the forces acting on it. So the centrifugal force is not exerted on object a.

[tar]

SwansonT,

 

Sorry about wheigh and centripical. I realize those kinds of mistakes are annoying and tend to highly devalue any thought I might be trying to express.

 

I should be more correct.

 

I was looking this morning at pictures of the magnetosphere, trying to get the directions of radially outward and tangentially positioned, straight in my mind.

 

When thinking about the circles tranversed by a object sitting still on the surface of the Earth, it occurred to me that the tops of our 36,000km towers inscribed differerent size circles in the sky. The one on the equator with the largest radius and the one on the North Pole, with hardly any radius at all, and the one built at latitude 45 degrees North with a radius somewhere inbetween.

 

The 'lines of force' that eminate from the Earth are going in what direction, would you say? Radially outward, or tangential? Or some combination.

 

The circle inscribed by the 36,000km tower at 45 degrees has a center, but it does not correspond to the center of the Earth. The center is inside the Earth but halfway or so toward the North pole. Radially outward lines from this cirlce would fall all in the same plane as tangent lines from this circle, but the plane of this circle is parallel to the plane of circle inscribed by the top of the equatorial tower. The tangent lines and radially outward lines from the North pole circle get hard to envision, since we are back to the headache of envisioning lines tangent to a point and the circle is a point, which either exists or doesn't depending on the way you look at it.

 

What does it mean for a circle to have a radius of 0? And can you go radially outward, if you have no radius?

 

Regards, TAR

[swansont]

The 'lines of force' that eminate from the Earth are going in what direction, would you say? Radially outward, or tangential? Or some combination.

Is this a trick question? Gravity acts toward the center of the earth. Radially inward.

 

The circle inscribed by the 36,000km tower at 45 degrees has a center, but it does not correspond to the center of the Earth. The center is inside the Earth but halfway or so toward the North pole.

I don't believe you. I've been at the 45º latitude, and as I recall I was standing straight up (perpendicular to the surface) and did not have any lean at all to keep my balance.

 

Radially outward lines from this cirlce would fall all in the same plane as tangent lines from this circle, but the plane of this circle is parallel to the plane of circle inscribed by the top of the equatorial tower. The tangent lines and radially outward lines from the North pole circle get hard to envision, since we are back to the headache of envisioning lines tangent to a point and the circle is a point, which either exists or doesn't depending on the way you look at it.

 

What does it mean for a circle to have a radius of 0? And can you go radially outward, if you have no radius?

 

Regards, TAR

There is nothing about this problem that requires you to envision a circle of radius zero.

[tar]

SwansonT,

 

I have no problem standing straight up in New Jersey either, but the circle I travel around the axis of Earth every day is a smaller circle than the circle traveled by someone that lives on the equator.

 

If I would visit the North pole, and be on the axis, I would not travel a circle, I would just face 360 different degrees, all of them facing South.

 

We are here, in this problem, considering why there is an apparent "outward" force, that counters gravity in an opposite gravity direction, caused by the angular velocity, or moment of inertia of an object traveling in a circle, along with the surface of the Earth. This effect, if present, should be different on the equator where the circle is large, than at the North Pole, where the circle is so minimal as to not be considered a circle.

 

I have seen various figures that indicate a mass weighs less when spinning with the equator, than when hardly spinning at all on the North Pole. Some say a 1 in 200 difference, some say 1 in a 1000, but all the figures indicate the spinning thing weighs less which suggests a force "lifting" in the opposite direction as gravity. Speculative interpolation would make me guess that the 45 degree measurement would come out heavier than the equatorial measurement, and lighter than the pole measurement.

 

The "force" though, caused by the spin at 45 degrees is not opposite gravity, like it is on the equator. It is pointing South of normal.

 

Speculation would have me guess that although you have no problem standing at 45 degrees latitude, the force "countering" gravity at the equator is not countering it at 180 degrees in New Jersey, but countering it at 135 degrees. At the North pole the force of gravity is being slightly countered at 90 degrees.

 

Regards, TAR

[Robittybob1]

 

No, that's wrong. If the forces cancel, then there would be no net force and the object moves in a straight line. Newton's first law.

 

The centrifugal force.

 

As I explained in one of these threads recently, and as your quote shows, Newton's 3rd law forces act on different objects. F_{a on b} = -F_{b on a}

 

But the motion of object a only depends on the forces acting on it. So the centrifugal force is not exerted on object a.

That was really good. So can we say the centripetal force acts on the weight orbiting the center and the centrifugal force acts on the center?

On shortening the string, or speeding up the rotation, both the centripetal and centrifugal forces increase.

[Mike Smith Cosmos]

The rotational accelerating frame .

 

Surely if you set up a large rotating platform . And you had ten lusty men pulling on a rope passing through the centre. On the opposite side pulling against these ten lusty men are 20 frail ,yet enthusiastic girls / women , pulling against the men. ( in a rotational tug-of-war ) , Then there would probably be a slight yet definite mismatch , where the force pull from the men was slightly less than the 20 enthusiastic girls/women . Here the women would be accelerating gently toward the outside of the rotating platform . The men would be , being accelerated gradually , drawn by the superior force towards the centre.

 

Nobody would fall off the platform at a tangent . Quite what the end result would be , is anybodies guess. Possibly a skirmish of some sort !

 

Mike

Edited May 3, 2015 by Mike Smith Cosmos

[swansont]

That was really good. So can we say the centripetal force acts on the weight orbiting the center and the centrifugal force acts on the center?

On shortening the string, or speeding up the rotation, both the centripetal and centrifugal forces increase.

 

Yes.

SwansonT,

 

I have no problem standing straight up in New Jersey either, but the circle I travel around the axis of Earth every day is a smaller circle than the circle traveled by someone that lives on the equator.

 

You said force emanating from the earth. That's gravity. It is directed toward the center of the earth.

 

We are here, in this problem, considering why there is an apparent "outward" force, that counters gravity in an opposite gravity direction, caused by the angular velocity, or moment of inertia of an object traveling in a circle, along with the surface of the Earth. This effect, if present, should be different on the equator where the circle is large, than at the North Pole, where the circle is so minimal as to not be considered a circle.

And it's been answered. You can't trust yourself as an instrument of measurement

 

I have seen various figures that indicate a mass weighs less when spinning with the equator, than when hardly spinning at all on the North Pole. Some say a 1 in 200 difference, some say 1 in a 1000, but all the figures indicate the spinning thing weighs less which suggests a force "lifting" in the opposite direction as gravity. Speculative interpolation would make me guess that the 45 degree measurement would come out heavier than the equatorial measurement, and lighter than the pole measurement.

No, that "suggestion" stems from a misunderstanding of physics, namely that rotational motion and linear (inertial) motion are somehow identical. They aren't.

 

The "force" though, caused by the spin at 45 degrees is not opposite gravity, like it is on the equator. It is pointing South of normal.

There is no outward force, though. However, the centripetal force would indeed be toward the center of the circle, and not the center of the earth.

 

Speculation would have me guess that although you have no problem standing at 45 degrees latitude, the force "countering" gravity at the equator is not countering it at 180 degrees in New Jersey, but countering it at 135 degrees. At the North pole the force of gravity is being slightly countered at 90 degrees.

 

Except that there is no outward-directed force. The physics discussion can't proceed as long as you continue to insist that there is.

 

[pzkpfw]

... which suggests a force "lifting" in the opposite direction as gravity. ...

The little mental adjustment needed, is to realise there's no force '"lifting" in the opposite direction as gravity', instead, the scales are dropping away from the thing being weighed.

 

(Most at the equator, least at the poles.)

[Robittybob1]

 

Yes.

....

There is no outward force, though. However, the centripetal force would indeed be toward the center of the circle, and not the center of the earth.

 

 

Except that there is no outward-directed force. The physics discussion can't proceed as long as you continue to insist that there is.

 

If the centripetal force and the centrifugal are equal and opposite, how can one be real but not the other? The centrifugal force is acting on the center. It is acting on the center, not on itself. It is a force pulling the center to itself.

Edited May 4, 2015 by Robittybob1

[Mike Smith Cosmos]

Ref picture to my previous post :- ...# 120

 

 

 

In this case the forces are not equal. Say men at 50 newtons per person : ladies at 25 newtons per person .

 

The men are dragged shamefully towards the centre. The women having a larger total combined force , are working their way to the edge of the table, screaming with victorious joy. The rotation is assisting their pull by centrifugal force. Odd ones are clinging on the end of the rope , like skaters forming themselves in a rotating chain on an ice skating rink. While still holding on , they will rotate in a larger circle than the platform . They cling on in a flying mode, as if the centrifugal force is holding them in a flying mode.

 

ONLY when the rope breaks do the frail ladies ' en mass ' fly off tangentially to the point at break.

 

Is this a correct interpretation of this particular rotating , accelerating frame of reference ???

 

Mike

 

Ps the difference in force ( men to ladies ) results in a modest acceleration of the overall tug-of-war toward the centre. ( the team forces are NOT equal. ) so in this case the centripetal force does not equal the centrifugal force .the centrifugal force is greater , hence end ladies flying off the rotating platform , I think ( is that correct ) ?

Edited May 4, 2015 by Mike Smith Cosmos

[Robittybob1]

The little mental adjustment needed, is to realise there's no force '"lifting" in the opposite direction as gravity', instead, the scales are dropping away from the thing being weighed.

 

(Most at the equator, least at the poles.)

How can the scales be dropping away when they are firmly on the surface?