What exactly is centrifugal force

[hms.tech]

I think that there is no such thing as centrifugal force .

Am I right ? is this force fictitious ?

 

[Studiot]

For some systems that are not in equilibrium ie moving, D'Alembert introduced a method of applying a fictitious force to bring the system to equilibrium (rest). The equations of equilibrium could then be employed to solve the system. These equations then included the fictitious force.

Centrifugal force is just such a force applied to objects executing circular motion.

 

[hms.tech]

What is the source of this force ?

For example, in gravitational circular motion, where does this force originate from ?

 

[Studiot]

It doesn't exist.
That is why it is fictitious.

However we can do the maths as though it did exist.
It is a mathematical trick.

The modern alternative is to use Newton's Laws of Motion, which I think came after D'Alembert, but I will check that.

 

[Nemanja989]

If something as a source of centrifugal force exists, then I think it is the inertia.

 

[rcgldr]

I think that there is no such thing as centrifugal force. ... is this force fictitious ?

Depends on how you're using the term centrifucal force. There's the fictitious centrifugal force in a rotating frame. A less common (for physicists) usage is "reactive centrifugal force", which is part of a pair of Newton third law forces: say a string exerts a centripetal force on an object, then the object exerts a reactive outwards force on the string. Note that this "reactive centrifugal force" is not exerted on the object itself, but on the string. In a rotating frame, the fictitious forces, centripetal and coriolis (and euler if there is angular acceleration of the frame) all appear to act on the object. Wiki articles:

http://en.wikipedia.org/wiki/Reactive_centrifugal_force

http://en.wikipedia.org/wiki/Centrifugal_force_(fictitious)

http://en.wikipedia.org/wiki/Rotating_frame

 

[A.T.]

It doesn't exist.

All forces are just mathematical models invented by humans.

That is why it is fictitious.

"Fictitious force" is just an unfortunate name choice. It gives people the wrong idea that some forces are more "real" than others. But that is just philosophy, irrelevant to physics. That is why I prefer the terms "inertial forces" and "interaction forces".

 

[hms.tech]

All forces are just mathematical models invented by humans.

"Fictitious force" is just an unfortunate name choice. It gives people the wrong idea that some forces are more "real" than others. But that is just philosophy, irrelevant to physics. That is why I prefer the terms "inertial forces" and "interaction forces".

Yes, I read this word "inertial" a lot on the Wikipedia link provided by rcgldr.

Can you explain this notion ?

 

[Studiot]

hms.tech

Before you get confusing explanations as to the different ways of regards motion under a central force or circular motion perhaps you would tell us what your course says ie which approach they take. Your exams will be based on this.

 

[hms.tech]

hms.tech

Before you get confusing explanations as to the different ways of regards motion under a central force or circular motion perhaps you would tell us what your course says ie which approach they take. Your exams will be based on this.

Centripetal and centrifugal forces

I couldn't find the term centrifugal force in any of my physics books, hence I checked out PF. Can you provide some links (non Wikipedia) about centrifugal force .

Here is what I understand about this force :
I think this term was self created by humans for their own better understanding of the "motion" of objects in circular orbit.
We, the modern physicists would call it nothing but a consequence of the fact that any object undergoing circular motion tends to move in a straight line (tangential to its path) . So it can be thought of as "inertia" , but I would stick to my claim that this force is non existent.

 

[rcgldr]

Yes, I read this word "inertial" a lot on the Wikipedia link provided by rcgldr. Can you explain this notion ?

An "inertial" frame of reference means the frame of reference is not accelerating (no linear acceleration and no rotation). Newton's laws of physics apply in an inertial frame without any modification.

If the frame was accelerating, such as a rocket in space, then to an observer inside the rocket, there would be an apparent force on any object within the spacecraft , similar to gravity on earth.

The wiki article on this goes into more detail:

http://en.wikipedia.org/wiki/Inertial_frame_of_reference

I couldn't find the term centrifugal force in any of my physics books, hence I checked out PF. Can you provide some links (non Wikipedia) about centrifugal force.

http://www.answers.com/topic/centrifugal-force

So it can be thought of as "inertia" , but I would stick to my claim that this force is non existent.

In your diagram of the car in a turn, there is a Newton third law pair of forces, an inwards centripetal force exerted by the road onto the car, and an equal in magnitude but opposing outwards reaction force that the car applies to the road. That outwards force is real, the issue for some people is using the term "reactive centrifugal force" to describe that outwards force that the car exerts to the road.

 

[nrivera1]

A more simple example might serve to make this clear:

consider a train that is accelerating with acceleration a. You are on this train and you drop a ball. The only force due to a physical interaction acting on the ball is that of gravity which is due to the ball's interaction with the Earth.

However, you see the ball accelerating away from you with the same magnitude of acceleration as that of the acceleration of the train. Thus if we consider the acceleration of your reference frame (keep in mind that since you're measuring the motion of the ball with your position as the origin, you are at rest in your reference frame) as the source of a force on the ball, it applies a force of magnitude ma, where m is the mass of the ball. This force isn't due to any physical interaction that you can observe, that's why we call it fictitious. I agree that calling it fictitious is confusing and that inertial is less confusing and also an apt name in that the magnitude of the force is proportional to the mass of the object

A similar thing happens when you're in a rotating reference frame (for instance say someone does this while on a merry go round) except that now this fictitious force is directed outwards from the center of the merry go round. This type of fictious force due to being on an accelerated (or noninertial reference frame) is called centrifugal.

A centripetal force is the net force in the radial direction due to real forces and is generally talked about in an inertial or non-accelerated reference frame. Hope that clears it up a bit

 

[Studiot]

Look at post#9 of this thread.

https://www.physicsforums.com/showthread.php?t=645048&highlight=centrifugal+force

It puts what I said in post#2 here into equations.

 

[A.T.]

Yes, I read this word "inertial" a lot on the Wikipedia link provided by rcgldr.

Can you explain this notion ?

http://en.wikipedia.org/wiki/Fictitious_force

In short: Newton based his laws of motion for inertial reference frames on interaction forces. To generalize some of the laws so they apply to non-inertial frames inertial forces were introduced.

 

[Studiot]

Gosh I'm glad I didn't meet that wiki as my first introduction to centrifugal force.

It's one saving grace is that at the very bottom it refers to the D'alembert's principle of inertia wiki.

That is much more digestible.

http://en.wikipedia.org/wiki/D'Alembert's_principle

Look towards the end of the much shorter page.

 

[stevendaryl]

Here's a mathematical way to think about fictitious forces.

Suppose you have a velocity vector \stackrel{\rightarrow}{U} which is constant (no forces acting on it). That means

\dfrac{d}{dt}\stackrel{\rightarrow}{U} = 0

Now, this is a vector equation. It's common for people to denote vectors by ordered pairs:
\stackrel{\rightarrow}{U} = (U^x, U^y), where U^x = \dfrac{dx}{dt} and U^y = \dfrac{dy}{dt}.

But that's actually a BAD notation, because what it really means to write a vector as components is that we've chosen certain vectors called basis vectors, and we've expressed \stackrel{\rightarrow}{U} as a combination of those basis vectors. So a better way to write it is

\stackrel{\rightarrow}{U} = U^x\ \hat{e_x} + U^y\ \hat{e_y}

When you write it this way, you can see that
\dfrac{d}{dt}\stackrel{\rightarrow}{U} = \dfrac{dU^x}{dt} \hat{e_x} + U^x \dfrac{d \hat{e_x}}{dt} + \dfrac{dU^y}{dt} \hat{e_y} + U^y \dfrac{d \hat{e_y}}{dt}

Of course, the most important fact about Cartesian coordinates is that the basis vectors are constant. So the terms \dfrac{d\hat{e_x}}{dt} and \dfrac{d\hat{e_y}}{dt} are both zero.

But if you switch to polar coordinates r,\theta, it's no longer the case that the basis vectors are constant. So you get additional terms having to do with the rate of change of the basis vectors. Those additional terms are the "fictitious forces".

I'm going to spare the details, but it turns out to be, in polar coordinates:

\dfrac{d}{dt}\stackrel{\rightarrow}{U} = (\dfrac{dU^r}{dt} - r (U^\theta)^2) \hat{e_r} + (\dfrac{dU^\theta}{dt} + \dfrac{2}{r} U^r U^\theta) \hat{e_\theta}

where U^r = \dfrac{dr}{dt} and U^\theta = \dfrac{d \theta}{dt}.

So, if there are no forces acting on an object, then \stackrel{\rightarrow}{U} will be constant, but that doesn't mean the components will be constant. Instead, the components will obey:

\dfrac{dU^r}{dt} = r (U^\theta)^2
\dfrac{dU^\theta}{dt} = - \dfrac{2}{r} U^r U^\theta

So it seems that there are "fictitious" forces acting on the object, causing the components of the velocity to change.

Nonconstant basis vectors give rise to nonconstant components of velocity, even in the absence of any real forces.

 

[hms.tech]

In your diagram of the car in a turn, there is a Newton third law pair of forces, an inwards centripetal force exerted by the road onto the car, and an equal in magnitude but opposing outwards reaction force that the car applies to the road. That outwards force is real, the issue for some people is using the term "reactive centrifugal force" to describe that outwards force that the car exerts to the road.

If you think about it, the reactive force is just the force exerted by the car on to the road, but this does not relate (even remotely) to the diagram .
The diagram shows that there "seems" to be a force acting on the objects which tend to move outward (in the opposite direction of centripetal force). So the force you talked about is the force exerted by the car on the road and not the objects.

Think about this :

You are sitting on the backseat of a car, and this car is turning around a corner at high speed. You would have noticed yourself moving outwards (right ?) until you reach the door of the car which then would provide the appropriate centripetal force to your body.
Notice that this situation is based on the fact that friction could not provide the centripetal force. Notice also that at all times you would be "FEELING" some outward push , (again , which in my opinion is just inertia but not centrifugal force)

Am I right in the above mentioned aspect ?

 

[A.T.]

Am I right in the above mentioned aspect ?

You switch frames back and forth and confuse the issue. Stick to an inertial frame, and there are no inertial forces. Stick to a non-inertial frame, and you always have inertial forces which act just like interaction forces (except Newtons 3rd).

Both is correct and leads to the same quantitative predictions, which is all that matters in physics. Musings about "realness of forces" is philosophy.

 

[Andrew Mason]

An "inertial" frame of reference means the frame of reference is not accelerating (no linear acceleration and no rotation). Newton's laws of physics apply in an inertial frame without any modification.

If the frame was accelerating, such as a rocket in space, then to an observer inside the rocket, there would be an apparent force on any object within the spacecraft , similar to gravity on earth.

The wiki article on this goes into more detail:

http://en.wikipedia.org/wiki/Inertial_frame_of_reference



http://www.answers.com/topic/centrifugal-force

In your diagram of the car in a turn, there is a Newton third law pair of forces, an inwards centripetal force exerted by the road onto the car, and an equal in magnitude but opposing outwards reaction force that the car applies to the road. That outwards force is real, the issue for some people is using the term "reactive centrifugal force" to describe that outwards force that the car exerts to the road.

The centripetal force produces centripetal acceleration. The force that causes centripetal acceleration produces mechanical tensions (ie chains of electromagnetic forces at the atomic level). These mechanical tensions operate radially. The net sum of these tensions is the centripetal force that produces the centripetal acceleration.

Somebody had the idea that we should call the reaction to a centripetal force that is the result of the application mechanical forces a "centrifugal reaction force". As far as I can tell, this is a concept that serves no purpose and simply causes confusion. It can produce no acceleration so it is hardly a centre-fleeing force as the term "centrifugal" would imply. It is a mechanical tension so it does not exist where the centripetal force is applied by a force operating at a distance, such as gravity or an electric field.

Calling it a centripetal reaction force also confuses the physics. The true reaction to a centripetal force is another centripetal force. One does not appreciate this in the situation of a tethered ball rotating about a pole fixed to the Earth but that is only because we do not take into account the movement of the earth. In the case of two balls tethered to each other by a rope and rotating in space about their centre of mass, the rope has a tension. Each ball pulls on the rope and, through the rope, pull on each other ie toward the centre. The ball pulls on the rope but produces no acceleration of the rope radially outward. Rather this pull is applied through the rope, via tensions, to accelerate the balls and rope toward the centre.

AM

 

[A.T.]

"centrifugal reaction force". As far as I can tell, this is a concept that serves no purpose

Because you are considering only the most trivial cases, like:

two balls tethered to each other by a rope and rotating in space about their centre of mass,

In the other thread we gave you plenty examples where a reactive centrifugal force exists.

It can produce no acceleration so it is hardly a centre-fleeing force as the term "centrifugal" would imply.

The term "centrifugal" has nothing to do with producing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force.

The true reaction to a centripetal force is another centripetal force.

Sometimes it is, and sometimes it is not.

 

[stevendaryl]

Because you are considering only the most trivial cases, like:

In the other thread we gave you plenty examples where a reactive centrifugal force exists.

If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.

 

[A.T.]

If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.

Yes, for local interactions the reaction to a centripetal force is always a centrifugal force. It's only when we invoke action at a distance (like Newton's gravity or a mass-less string that is not considered an object itself, just a means of force transmission) that we have a Newtons 3rd force pair of two centripetal forces.

 

[stevendaryl]

To me, "fictitious forces" are very ugly, because they spoil Newton's laws of motion. In a rotating coordinate system, there are "centrifugal forces" acting on absolutely every object in the universe. Where's the equal and opposite forces? There are none.

So some people would take this to mean "Newton's laws only apply in an inertial frame". I don't like that conclusion. If you view them as vector equations, then they apply in all circumstances, not just inertial frames.

If \stackrel{\rightarrow}{U} is the velocity vector, you write
m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}

this equation is always valid, regardless of what coordinate system you are using, and the force on an object always has an equal and opposite force. What changes when you go to a noninertial or curvilinear coordinate system is not Newton's laws, but the idea that you can compute the time derivative of a vector by only taking the derivative of its components. That is, it is not the case that

(\frac{\stackrel{\rightarrow}{dU}}{dt})^j = \frac{<br /> dU^j}{dt}

 

[A.T.]

To me, "fictitious forces" are very ugly, because they spoil Newton's laws of motion.

They extend the applicability of Newton's 1nd & 2nd to non-inertial frames. Which is a very useful thing, when you are actually using physics to compute something, not just muse about the beauty of laws.

Where's the equal and opposite forces? There are none.

Newton's 3rd doesn't apply to inertial forces, because momentum is not conserved in non-inertial frames. So we cannot extend Newton's 3rd to them.

 

[stevendaryl]

They extend the applicability of Newton's 1nd & 2nd to non-inertial frames.

No, they don't. As vector equations, Newton's laws hold in any coordinate system or frame. It's always the case that in the absence of external forces, the velocity vector \stackrel{\rightarrow}{U} satisfies

\frac{\stackrel{\rightarrow}{dU}}{dt} = 0

It's always the case that in presence of a (real, non-fictitious) force \stackrel{\rightarrow}{F}, the velocity satisfies

m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}

It's always the case that if one object exerts a (real, non-fictitious) force \stackrel{\rightarrow}{F} on another, then the second exerts a force -\stackrel{\rightarrow}{F} on the first.

Newton's laws, as vector equations are not changed at all by using noninertial or curvilinear coordinates. Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's needed is a more sophisticated notion of what

\frac{\stackrel{\rightarrow}{dU}}{dt}

means when the basis vectors themselves are nonconstant.

 

[stevendaryl]

Newton's 3rd doesn't apply to inertial forces, because momentum is not conserved in non-inertial frames. So we cannot extend Newton's 3rd to them.

That's not true. Momentum as a vector quantity is conserved in non-inertial frames. What isn't conserved are the components of momentum.

 

[stevendaryl]

That's not true. Momentum as a vector quantity is conserved in non-inertial frames. What isn't conserved are the components of momentum.

Actually, I should clarify this. There are two different departures from good old inertial Cartesian coordinates: (1) Using curvilinear coordinates and (2) using noninertial coordinates.

With curvilinear coordinates, the basis vectors become functions of position. With noninertial coordinates, the basis vectors become functions of time. Curvilinear coordinates can be handled with the usual Newton's laws of motion, viewed as vector equations. Noninertial coordinates can also be handled with the usual Newton's laws of motion, but only if you adopt a 4-D spacetime view.

 

[A.T.]

Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's needed is a more sophisticated notion ... basis vectors themselves are nonconstant... adopt a 4-D spacetime view...

Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.

 

[Studiot]

hms.tech

My advice is to stick with one the two methods I outlined for you, either will always get you the right answer, D'Alembert's often leads to rather less computational effort.

I'm sure you can see that a by product of introducing reference frames is to cause experts to squabble amongst themselves as a result of a great increase in complexity.

 

[Andrew Mason]

If a string pulls on an object, the object pulls back on the string. That's always the case, by Newton's third law. In the special case where the force of the string on the object is radially inward, the force of the object on the string is radially outward.

That is quite true, but you are only looking at part of the picture. If there was only the object and the string you could not have a centripetal force acting on the object. There are necessarily other forces on the string.

AM

 

[sophiecentaur]

Words like "fictitious force" and "particle" should have a warning sign on them when they're used for public consumption. If you can feel the effect then it's not really fictitious so a different word should really be used to describe it. Likewise, calling a photon a particle is just asking for the sort of misinterpretations we read every day. A term like quasi-force or quasi-particle would at least ring warning bells for people who might want to rush off and think in terms of conventional meanings.

 

[A.T.]

That is quite true, but you are only looking at part of the picture.

"Part of the picture" is fully sufficient for Newtons 3rd Law and local interactions. You don't have to consider all forces acting on a object or the net acceleration of the object, to tell that an individual force acting on it is centrifugal (points away from the rotation center) and forms a 3rd Law pair with a centripetal force acting on a different object.

 

[Andrew Mason]

Because you are considering only the most trivial cases, like:

So where is the concept of a centrifugal reaction force used? It cannot ever produce a centrifugal acceleration. So all it does is explain the tension.

In the other thread we gave you plenty examples where a reactive centrifugal force exists.

My quibble is not with the force per se but with the name "centrifugal" in conjunction with the term "force". A force is something that is capable of producing an acceleration of an object. The acceleration that this reaction force produces is always centripetal ie. opposite to the direction of the force.

The term "centrifugal" has nothing to do with producing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force.

From Wikipedia:
Centrifugal force (from Latin centrum, meaning "center", and fugere, meaning "to flee") is the apparent outward force that draws a rotating body away from the center of rotation.

There is nothing about the "centrifugal reaction force" that causes anything to flee from the centre. Nothing. The only thing that causes a rotating object and rotating rope to "flee" the centre is inertia, not a force.

AM

 

[sophiecentaur]

"Part of the picture" is fully sufficient for Newtons 3rd Law and local interactions. You don't have to consider all forces acting on a object or the net acceleration of the object, to tell that an individual force acting on it is centrifugal (points away from the rotation center) and forms a 3rd Law pair with a centripetal force acting on a different object.

Using the term 'Centrifugal Force' when I was at school was verboten. It was regarded as worse than talking about sex in front of your parents.

 

[A.T.]

So where is the concept of a centrifugal reaction force used?

It's not a concept. The concept here is Newtons 3rd Law: A pair of equal but opposite force acting at the interface of two objects: one inwards (centripetal) the other outwards (centrifugal)

From Wikipedia:
Centrifugal force (from Latin centrum, meaning "center", and fugere, meaning "to flee") is the apparent outward force that draws a rotating body away from the center of rotation.

This describes the potential effects of the inertial centrifugal force as seen in the rotating frame.

There is nothing about the "centrifugal reaction force" that causes anything to flee from the centre.

Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.

 

[sophiecentaur]

Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.

But will 'flee' along a tangent and not a radius, when you cut the string. That was why the word was not 'permitted'. Of course, they failed to mention that a tangent takes you further from the centre, too.

 

[sophiecentaur]

It is really not that confusing if you make clear which reference frame you consider.

We didn't 'do' astronauts when I was at school (they hadn't been invented - except for Dan Dare) and the term 'reference frame' was University stuff at the time.

Having tried those ideas on teenagers, not long ago, I did find it worked with some. It didn't work with others but, so what? They'll have ended up in banking or as 'TV personalities'.

 

[A.T.]

In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.

But will 'flee' along a tangent and not a radius, when you cut the string.

I was talking about about the rotating frame, where it flees along the radius.

 

[sophiecentaur]

I was talking about about the rotating frame, where it flees along the radius.

See my post - just above.

 

[stevendaryl]

Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.

That's what introducing the idea of "centrifugal force" does. It makes things unnecessarily complicated and confusing.

 

[stevendaryl]

Yeah, you can always replace a simple approach with a mathematically equivalent, but more complicated one.

Really, you have things completely backwards here. The SIMPLE approach is to use vector equations:

1. In the absence of any forces, \frac{\stackrel{\rightarrow}{dU}}{dt} = 0
2. In the presence of an external force m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}
3. If one object exterts a force \stackrel{\rightarrow}{F} on another, then the second object exerts a force -\stackrel{\rightarrow}{F} on the first.

There is nothing confusing about these rules, and there are no exceptions for "fictitious forces" or "noninertial frames". The only complicated thing is that you need to realize that a vector can change for two reasons: (1) the components change, or (2) the basis vectors change. It might be simpler to pretend that (2) never happens, but it's a falsehood, and you're doing physics in a crippled way when you do it.

 

[stevendaryl]

That's what introducing the idea of "centrifugal force" does. It makes things unnecessarily complicated and confusing.

Actually, different people disagree about what is "complicated". I consider it complicated when you have lots of ad hoc rules that apply in specific circumstances: If you are using inertial Cartesian coordinates, do this, if you're using curvilinear coordinates, do that, if you're using noninertial coordinates, do that. I'd rather have a fundamental set of principles that apply in all those circumstances, even if working out the details might be complicated.

 

[ZealScience]

It appears because of choice of coordinates and reference frame, and yes it does not exist in inertial reference frame.

 

[Andrew Mason]

It's not a concept. The concept here is Newtons 3rd Law: A pair of equal but opposite force acting at the interface of two objects: one inwards (centripetal) the other outwards (centrifugal)

Ok. But it gets confusing to students.

Example: there is no centrifugal reaction force to the weight of a car sitting on a road on the equator, even though the car is whipping around at 1000 mph. as the Earth turns and even though it's weight bears on the Earth surface. Ques. What about the normal force? Do we call that a centrifugal force? Answer: no. Ques. But it is directed away from the centre of rotation (ie. up). Answer: Yes, but we do not call it centrifugal. Ques. Why not? Answer: My answer would be: We don't call it centrifugal because it does not cause anything to flee the centre of rotation. I am not sure what your answer would be. Should we call it centrifugal?

This describes the potential effects of the inertial centrifugal force as seen in the rotating frame.

Yes, but the point is that it means "fleeing the centre".

Wrong. In the rotating frame the centrifugal reaction force can push an object outwards, just like the inertial centrifugal force can.

But that is a fictitious centrifugal force, not the "centrifugal reaction force". Give us an example - just one example where the reaction force to a centripetal force causes matter to flee the centre of rotation.

AM

 

[rcgldr]

There is no centrifugal reaction force to the weight of a car sitting on a road on the equator, even though the car is whipping around at 1000 mph.

A more generalized version of this would be 2 body system in outer space, free of external forces, each object orbiting about a common center of mass due to gravity or opposite charge. The Newton third law pair of forces are the two attractive forces between the objects and directed towards a common center of mass. If the orbits are circular, then the two attractive forces are also centripetal forces. In this situation, there are no reactive centrifugal forces.

Change this 2 body system to one where there are no attractive forces, and the two objects are connected by a string and rotate in a circular path about a common center of mass. Both objects exert a reactive centrifugal force on the ends of the string (assuming that the common center of mass is not located within one of the objects, in which case only one end of the string experiences a reactive centrifugal force).

 

[Andrew Mason]

Actually, different people disagree about what is "complicated". I consider it complicated when you have lots of ad hoc rules that apply in specific circumstances: If you are using inertial Cartesian coordinates, do this, if you're using curvilinear coordinates, do that, if you're using noninertial coordinates, do that. I'd rather have a fundamental set of principles that apply in all those circumstances, even if working out the details might be complicated.

The principles and the laws of physics are the same. The problem is that nature makes a distinction between non-inertial and inertial frames of reference.

AM

 

[Andrew Mason]

Change this 2 body system to one where there are no attractive forces, and the two objects are connected by a string and rotate in a circular path about a common center of mass. Both objects exert a reactive centrifugal force on the ends of the string (assuming that the common center of mass is not located within one of the objects, in which case only one end of the string experiences a reactive centrifugal force).

And how would the objects produce a centrifugal acceleration of the ends of the string?

AM

 

[A.T.]

It might be simpler to pretend that (2) never happens,

It is simpler, that’s why inertial forces are widely used.

but it's a falsehood, and you're doing physics in a crippled way when you do it.

It leads to the same quantitative predictions, which is all that matters in physics. The rest is philosophy.

 

[dextercioby]

In classical mechanics we treat the Coriolis force, the translation inertial force and the centrifugal force on equal footing.

I wonder why only the centrifugal force is poorly understood by people, since it's the only inertial force generating multi-page debates o PF...

 

[stevendaryl]

It is simpler, that’s why inertial forces are widely used.

No, inertial forces are NOT simpler. They may seem simpler to people who prefer to memorize formulas instead of understanding them.