Currently in my last year of high school, and I have always been told that centrifugal force does not exist by my physics teachers. Today my girlfriend in the year below asked me what centrifugal force was, I told her it didn't exist, and then she told me her textbook said it did, and defined it as "The apparent force experienced towards the outside of a circle is the centrifugal force and is due to the mass of the object resisting the inward centripetal acceleration that the object is experiencing". I was pretty shocked to hear this after a few years of being told that it does not exist.

I did some reading and found out all sorts of things about pseudo forces and reference frames. I was wondering if someone could please explain to me what is going on? Is it wrong to say that centrifugal force does not exist?

This has always nagged me a bit as I often wonder that if every force has a reaction force then a centripetal force must have a reaction centrifugal force, but when I asked my teachers about this they told me that centrifugal force does not exist.

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    $\begingroup$ Note that the textbook says: "The apparent force". It's not real; what you are feeling is inertia. $\endgroup$
    – hdhondt
    Apr 22, 2014 at 11:06
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    $\begingroup$ My answer here will shed light on this matter : physics.stackexchange.com/q/108925 $\endgroup$
    – evil999man
    Apr 22, 2014 at 12:42
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    $\begingroup$ xkcd.com/123 $\endgroup$
    – evil999man
    Apr 22, 2014 at 12:43
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    $\begingroup$ @hdhondt How does one feel inertia? What you feel is the effect other objects have on you as they accelerate and you remain (momentarily) in your own inertial frame. See also this answer $\endgroup$
    – garyp
    Apr 22, 2014 at 13:39
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    $\begingroup$ Do not confuse "apparent" and "non-existant", though. The force is very real inside of a newtonian reference frame - the fact that it's just a correction does nothing to change that. After all, most of the forces you work with on a daily basis use some correction (you're not calculating the force of each of the air molecules in the wind - you're content with the mean force of the whole). The force is very real in the sense that if you don't use it in your model based on Newtonian physics (the "intuitive" mechanics), you'll get wrong results. The context matters. $\endgroup$
    – Luaan
    Apr 22, 2014 at 14:55

11 Answers 11



Centrifugal force and Coriolis force exist only within a rotating frame of reference and their purpose is to "make Newtonian mechanics work" in such a reference.

So your teacher is correct; according to Newtonian mechanics, centrifugal force truly doesn't exist. There is a reason why you can still define and use it, though. For this reason, your girlfriend's book might also be considered correct.


As you know, Newton's laws work in so-called "inertial frames of reference". However, a point on the surface of the Earth is not really an inertial frame of reference because it is spinning around the center of the Earth. (So you can think of it as a rotating coordinate system.) So Newton's mechanics don't apply if you want to describe motion and use a reference point on the Earth. This is quite inconvenient, because we mostly want to engineer things that work on the Earth.

Fortunately, there is a trick: you can use a point on the surface of the Earth as your reference and pretend that it's an inertial frame of reference, if you also pretend that some external "imaginary" (fictious) forces exist in addition to the real ones. These are the centrifugal force and the Coriolis force.

Further reading

If you are interested in more, see:



The trick is, centrifugal force is a fictitious force.

Centrifugal force exists! To everyone denying it, do this to them: xkcd.com/123. However it is a fictitious force. To quote wikipedia:

A fictitious force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

So, if you sit in a merry-go-round, you can feel a force pulling you out. You can measure it. For you this force exists but for your mother standing outside the merry-go-round, watching you, there is no centrifugal force. She can see the merry-go-round applying a centripetal force to you, so you go along with the merry-go-round and do not fall off. If it didn't, your mass makes you go in a straight line and you fall off.

The reason why the two observers observe different forces is that the merry-go-round is not an inertial frame of reference whereas the ground, upon which your mother stands, is.

In an inertial frame of reference there is no centrifugal force but there can be in a non-inertial frame of reference.

So the centrifugal force appears to be there because an observer in a merry-go-round is not in an inertial frame of reference. By changing frames of reference you can eliminate it.

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    $\begingroup$ "Where the ground, your mother stands on, is" -- well it isn't, but it's a lot less non-inertial since it's doing a lot fewer revolutions per day :-) $\endgroup$ Apr 22, 2014 at 16:51
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    $\begingroup$ You are technically totally correct ;-) So for the sake of completeness, let's assume the mother's ground is at rest. $\endgroup$ Apr 22, 2014 at 16:57
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    $\begingroup$ To be technically correct (which is the best kind of correct), even if the earth were at rest, an earth-based frame of reference is not inertial, as it contains gravity as a fictitious force. $\endgroup$
    – Tristan
    Apr 22, 2014 at 17:19
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    $\begingroup$ Well, gravity is fictitious only in general relativity, not in Newtonian mechanics. $\endgroup$
    – Ruslan
    Apr 22, 2014 at 19:43

As I disagree with all the answers I am going to try to explain some of the fundamentals of science: Science in its very essence can not explain why things happen the way they do, it simply tries to model reality based on observations in the past to predict events in the future. In other words, defining a centrifugal force is possible as for example your girlfriend's book does, but it is redundant in the greater scheme of the physics model of reality as other aspects of the physics model can be used to describe the same events without the need for such a force.

Now, to address your original question: Does centrifugal force exist?

  • No, because just like gravitational force/gravitation/"space warping" or even things like the electroweak interaction it doesn't exist in any way except as a term to describe an observed pattern in the past we expect to happen in the future as well. Science can never* claim to explain anything, it can just build more and more efficient and abstract models to predict future events.
  • No, because in the most generally accepted model(s)** of physics the force is not used/defined. See the other answers for this.
  • Yes, in the sense of it being useful at a certain level of prediction to model certain things without too much abstraction. Similarly certain things can be explained in chemistry with certain "laws" which physics can predict in a more complex and abstract form. This doesn't mean the chemical laws do not exist, they simply are less abstract.

* <small> well, just to be perfectly correct here something like theology is considered by some a science and it at least at a fundamental level has the right to make the claim, regardless of the question whether it can or can not explain anything </small>

** Models if you take for example the Newtonian model and the newer quantum physical model as separate models with different levels of abstractions rather than the quantum physical as simply a better newer version of the Newtonian model.

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    $\begingroup$ Finally, an answer that presents both points. Although I think you have the abstractions wrong - the deeper you go, the less abstractions you have, not the other way around. Abstractions are handy to protect you against unnecessary complexities - e.g. a friction force is an abstraction of something that ultimately boils down to the electromagnetic force (and beyond). But calculating friction based on EMF? Oh boy. Oh, and science explains things alot - it just doesn't guarantee that it's the 100% exact explanation. That's why it's constantly being updated for new observations :) $\endgroup$
    – Luaan
    Apr 22, 2014 at 14:59
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    $\begingroup$ @Luaan: Yeah, it is mostly a thing of definition in which direction you go with "abstraction". I - and the philosophy of science if memory serves me right - tend to see abstraction as going farther from the simple observations made in reality. This makes sense if you look at the development of science from a macroscopic perspective through time, but seems odd to someone who is taught forces like the centrifugal force form the beginning as a simpler perspective on the "real deal". Regardless, how could science explain things? It can only say: we saw this and this happen statistically $\endgroup$ Apr 22, 2014 at 15:25
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    $\begingroup$ I don't think this answer is useful within the context of the question. $\endgroup$
    – garyp
    Apr 22, 2014 at 16:09
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    $\begingroup$ @garyp: It is the only answer which explains the concept of abstraction and explains why his girlfriends book isn't simply wrong. $\endgroup$ Apr 22, 2014 at 16:26
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    $\begingroup$ David, I think this is a totally awesome answer, I enjoyed reading it :) But I also agree with @garyp in the sense that maybe it's a bit too abstract here, because the OP asked for an answer "in high school student terms". $\endgroup$
    – Venemo
    Apr 22, 2014 at 19:26

In Newtonian physics, objects continue moving in a straight line unless a force acts on them, therefore if an object is not moving in a straight line, a force must be acting on it.

Consider planets. Why don't they just fly off into space on a straight line? Because the sun pulls them.

Consider a rock at the end of a string. Why does it not fly off when spun? Because the line holds it in place.

When you get into something that is spinning (such as merry go round), you will fly off in a straight line because you weren't holding on to anything.

If you do hold on to something, you will feel as if something is pulling you outwards. Actually, your arms are pulling you inwards (which is what stops you from flying off), and you are feeling the reaction to your action (also Newton's laws). Because motion is relative, you can define some clever reference points which make it seem like there is a force pushing you outwards, but in the end it doesn't make a lot of sense.

If you hold an accelerometer while on the merry go round, you will see that you are experiencing acceleration (which implies a force, because of $F=ma$). If you spin fast enough, you will also feel yourself the various effects of having large forces applied to you, such as blood draining from parts of your body. If you let go and fly off, you will notice that the accelerometer shows no acceleration, and you are not feeling any, even if you were spinning very quickly. Where did the "centrifugal force" go? It went away as soon as you stopped applying a centripetal (towards the center) force on yourself by letting go.

This is why it's called a fictional force, it only seems to exist if you use a frame of reference that allows fictional forces to appear.

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    $\begingroup$ "Where did the "centrifugal force" go? It went away as soon as you stopped applying a centripetal (towards the center) force on yourself by letting go." No, in the rotating frame of reference this is the force that flung you away from the merry go round. The reason you don't keep rotating around the centre while moving exponentially outward is that once you start moving out there is a Coriolis force that makes you fall behind, and once you start falling behind the Coriolis force slows your outward motion. The centrifugal force is not the only fictitious force you need. $\endgroup$
    – k-l
    Feb 16, 2016 at 17:50
  • $\begingroup$ @KiranLinsuain There is no force that flings you away from the merry go round. $\endgroup$
    – Superbest
    Feb 16, 2016 at 21:43
  • $\begingroup$ It is not a real force in the same way as $i$ is not a real number, in the non-inertial reference frame it must be included in free body diagrams. When you get flung away from the merry go round, the reference frame keeps rotating with it and fictitious forces must still be included; someone watching from the merry go round will see you follow a curved path. Someone watching from the ground does not need any fictitious forces. @Superbest $\endgroup$
    – k-l
    Feb 17, 2016 at 0:42
  • $\begingroup$ excellent explanation, easily understood by undegrad or high school kid without confusing the OP with correct but overwhelming unnecesary details. $\endgroup$
    – Vortex
    Dec 25, 2022 at 2:40

The key to the conundrum is that for the purpose of explaining the apparent forces on someone to whom a rotating frame of reference appears to define stationary, for example all human beings everywhere, centrifugal force may need to be taken into consideration since it appears to be there. Although it may be small depending on the speed of rotation. Which is what your girlfriend's textbook says. For the purpose of stating Newton's laws of motion in an inertial frame of reference, which is what your teachers were doing, there is no such thing as this centrifugal force. You might reasonably think that since they contradict, one of those points of view must be so stupid that nobody would ever say it. But that's not the case.

"Centrifugal force" and "Coriolis force" exist as terms in the equation of motion of an object relative to a rotating frame of reference:

  • Consider "rotating with the frame of reference" to be stationary. That's just what "frame of reference" means.
  • Consider an object initially "at rest" (that is to say, at some instant in time it is rotating with the frame of reference), but nothing is in place to keep it rotating. Like spinning something on a string in a circle and then releasing just at the instant we start to calculate its motion.
  • Let time run from that initial point.

Then the object moves away from the centre of rotation. In an inertial (non-rotating) frame of reference it was not initially at rest. In that frame of reference we'd say that it moves in a straight line. In the rotating frame of reference it accelerates away from the centre -- the initial acceleration is directly away but it starts to follow a curved path.

The "force" that causes the initial "acceleration" is called "centrifugal force", and the "force" that acts on a moving body in a rotating frame and causes the the curve is called "coriolis force".

But the equations of motion in a rotating frame of reference are horrible, and the equations of motion in an inertial frame of reference are really simple. So, who cares about rotating frames of reference, to the point of giving a name to a "force" that doesn't exist in the inertial frames of reference that we prefer for calculation? People who live on a planet, is who. These non-existent forces have to be taken into account if you want to accurately compute the flight of a sufficiently long-range artillery shell or the movement of weather, relative to the ground instead of relative to some fixed point in space through which the earth rotates.

Do they exist? If you take a rotating frame of reference then they can be observed, just like any other force, and for that matter we subjectively experience them when we spin fast enough. If you take an inertial frame of reference then there's no such term in the laws of physics. So yes they exist, you can measure them if you're standing on a planet. No, they don't really exist, they're just a by-product of the frame of reference you chose. A bit like gravity in general relativity is a by-product of choosing an "unnatural" frame of reference, one that fails to follow the curvature of space-time ;-)

Note that in both frames of reference, inertial or rotating, an object that remains stationary in the rotating frame (and rotates in a circle in the inertial frame) necessarily experiences a "centripetal force" (a force towards the centre). Thus gravity causes things to orbit, and the tension in a piece of string causes a spinning poi to follow a circular path. So what "centrifugal force" really is, in a rotating frame of reference, is the term you need in order to satisfy the requirement that a stationary object must experience 0 net force. In an inertial frame that object describing a circle is not stationary, so it does not experience 0 net force, so there is no balancing term.

You are on to something when you speak about equal and opposite forces. Since the spinning object experiences a centripetal force, the object exerting that force necessarily must experience an equal and opposite force. The moon pulls the earth in its direction, and the string of the poi pulls your hand in the direction of the poi. This is not what is usually called "centrifugal force", but it is away from the centre, and it "really" does exist.

  • $\begingroup$ In the last paragraph you talk about equal and opposite forces. Is the force towards the outside what is called the "reactive centrifugal force"? or is this something else altogether? $\endgroup$
    – Dylan
    Apr 22, 2014 at 19:16
  • $\begingroup$ @DylanCleaver: yes, that's exactly it. I just hadn't heard that phrase (or if I have heard it didn't remember it). en.wikipedia.org/wiki/Reactive_centrifugal_force points out a subtlety that I missed, which is that I could have distinguished the string from the post/hand a bit better and considered all forces on both. $\endgroup$ Apr 22, 2014 at 19:24

Centrifugal force is force that pulls rotating object away from the center of rotation, Centrifugal is part of Newtonian mechanics and it's derived from Newton's Second law $$F=ma$$ Where $F$ is force in newtons, $m$ is mass of an object and $a$ is acceleration. In circular motion acceleration is $a=\frac{v^2}{r}$ and full equation for centrifugal force is $$ F=\frac{mv^2}{r} $$ Where $v$ is speed and $r$ is radius. Here's sample image of circular motion and centrifugal and centripetal force acting on an object:

enter image description here
(source: explainthatstuff.com)

Same happens with pilots of jet planes:

enter image description here

They experience centrifugal force and if that force is enough they can lost consciousness.


Imagine a bus and someone who is standing in it, and suddenly the bus driver turns the bus, because the car changed its direction of velocity it has accelerated, while someone who was standing was continuing moving in old direction, because of that he/she will fall, but because of bus driver is sitting very hardly he will experience centrifugal force. It's easier to imagine a spaceship, if he/she was floating relative to spaceship and spaceship has changed its direction of motion, it will accelerate, while he/she will be continuing moving in old direction so he/she won't experience centrifugal force, while spaceship will. (I have found nice animation about that situation in spaceship Click Here.)

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    $\begingroup$ The reason I have voted this down is that this answers seems to intentionally conflate inertial pseudo-forces (in this cases the effects of using a rotating frame of reference) with real forces (those that appear when you use a inertial frame of reference). It is certainly true that there are mathematical mechanisms for doing physics in non-inertial frames, but the are a complicating factor; it should be emphasized that the laws of physics have their simplest form in inertial frames. $\endgroup$ Apr 22, 2014 at 13:20
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    $\begingroup$ I agree with @dmgee here. The first sentence says "Centrifugal force is force that pulls rotating object away from the center of rotation". Yet, in an inertial frame, the object accelerates towards the centre. There is no force pulling the object away from the centre, there is a force pulling it toward the centre; this force is centripetal. $\endgroup$ Apr 22, 2014 at 23:57
  • $\begingroup$ @dmckee, do you still have that view about the answer after the EDIT part having the example of turning the bus? But I don't know how the animation is showing centrifugal force and maybe this is what you are talking about in your comment. $\endgroup$
    – user22180
    Aug 21, 2014 at 17:01

Let's see how inertial pseudoforces (like the centrifugal pseudoforce) arise in the theory of Newtonian mechanics.1

Rule: Newton's laws pre-suppose that you are working in an inertial frame.

The first rule can be regarded as a way of defining or identifying those frames (assuming you can identify forces, anyway).

Those three laws don't give any direct advice about doing physics in non-inertial frames.

Way to identify real forces

If you examine a single physical situation from several frame of reference,2 some "forces" that you see may change their direction or magnitude between frames, while others will remain stubbornnly the same.3 The ones that are always the same are are "real".

Observation: Sometimes it's nice to do physics in non-inertial frames.

If you are sitting in a vechicle that is moving and put a nice cup of coffee onto a tray. It sits there, at rest relative you. In class we'd use that kind of observation ("It's just sitting there.") to identify things that are in equilibrium, and then we would assert (by way of the second law) that the sum of the forces acting on it is zero.

And if your vehicle is in uniform motion that identification would be correct. but if your vehichle is accelerating (changing speed, going arouind a curve, both...) it's formmaly incorrect. The cup is also accelerating.

But we might want to go ahead with our usual analysis anyway. That's where intertial pseudofoces come in.

Plan: let's lash it up!

Our lash-up scheme is very simple. We start with the physics given to us by Newton's laws; move any inconvenient accelerations from the RHS to the LHS; and call the new terms on the LHS "forces".

That's the whole shebang.

Round-about example

Consider for concretness a car driving $20\,\mathrm{m/s}$ around a curve with radius $10\,\mathrm{m}$. We suppose that the thust provided by the drive train balances the drag and rolling friction and so that the only unballanced force is static friction directed toward the inside fo the circle.

  1. Set-up Newton's laws in an inertial frame \begin{align} \sum_i \vec{F}_i &= m \vec{a} \\ \vec{F}_\text{thrust} + \vec{F}_\text{drag} + \vec{f} &= m \frac{v^2}{r}\left(-\hat{r}\right) \\ \vec{f} &= -m \frac{v^2}{r} \hat{r} \;. \end{align}

  2. Decide we want things "at rest" in our non-inertial frame to be "in equilibrium", so move that inconveinent term to the other side.4 \begin{align} \vec{f} + m\frac{v^2}{r}\hat{r} &= 0 \;. \end{align} This is apurely formal, mathematical operation.

  3. Give the term we just moved a name $\vec{F}_\text{centrifugal} = m \frac{v^2}{r}\hat{r}$, so that the equation now has two "forces" it in \begin{align} \vec{f} + \vec{F}_\text{centrifugal} &= 0 \;. \end{align} Now, because of the subtraction process this newly created "fake" force has the opposite direction tha the real acceleration had.

    Tradiational, however we write this force in terms of the rotational velocity $\Omega$ of the frame $\vec{F}_\text{centrifugal} = m r \Omega^2 \hat{r}$.

More generally

There is a standard, fully generaly way to deal with complicated non-inertial motion. It develops four pseudoforces, each related to a specific kind of behavior.

  • A pseudoforce related to observers with straight-line acceleration $\vec{A}$ with respect to an inertial frame (oddly this one has no tradiational name; I sometimes call it the "seatbelt pseudoforce"): $$ \vec{F}_\text{seatbelt} = -m\vec{A} \;.$$

  • The "centrifugal" or "centrifical" pseudoforce related to observer in rotating with angular velocity $\Omega$ with respect to an inertial frame. This affects all objects including those at rest with respect to the observer. We just computed it5 $$ \vec{F}_\text{centrifugal} = -m \vec{\Omega} \times (\vec{\Omega} \times \vec{v}_b) \;. $$

  • The "Coriolis" pseudoforce which is also related to rotations but is observed only for objects that move wth velocity $\vec{v}_b$ in the non-inertial frame. $$ \vec{F}_\text{Coriolis} = -2m \vec{\Omega} \times \vec{v}_b \;, $$ where $\vec{v}_b$ is the velocity observed in the non-inertial frame.

  • The "Euler" pseudoforce which arises for observers experiencing angular accelration relative an inertial frame. $$ \vec{F}_\text{Euler} = -m \frac{\mathrm{d} \vec{\Omega}}{\mathrm{d}t} \times \vec{x}_b \;.$$

You can find detailed mathematical developments of this stuff in typical upper-division or graduate mechanics textbooks.

But ... do pseudoforces exist, already?!?

Yes? no? Depends?

This is, fundamentally, a philosophical question that turns on how you understand "existing". Their magnitude and direction depend on the frame from which you view a physical interaction, which definitely sets them apart from those "real" forces that don't have that property.

For myself, I make a point of maintaining a strong distinction between "real" and "pseudo" forces. But I am perfectly happy to work in non-inertial frames when that makes my life easier.

Bonus: What are people going on about gravity for?

Did you notice that all the pseudoforce definitions I gave have exactly one factor of the object's mass in them? That means that all objecs experince the same "centrifugal" or "Coriolis" accelration, which is supiciously like the rule that everything falls with the same acceleration.

As it turns out, when Einstien found with way through the weeds to general relativity he found tha he had created a theory in which gravity was also an inertial pseudoforce (and it such as heck is "real" enough for day to day purposes, isn't it?).

In general relativity when you are standing at rest in the lab you are observing the world from a non-inertial frame. The inertial frame is what you would see standing on (or ratehr, floating next to) the ball the instructor just dropped.

1 In this discussion I'm going to consistently identify that class of apparent forces tha arise in non-inertial frams as "pseudoforces" simply to set them apart from those "real" forces that appear in all frame inertial or otherwise.

2 It is important that I emphasize: one set of physical events as seen by observers with different states of motion. Not multiple events characterized by different motion of the participants.

3 Here the direction and magnitude are to be identified as their intrinsic values. Don't worry abrout changing components, just about changing nature.

4 The 'at reast" and "equilibrium" that appear here are purely for motivation. You shouldn't read any implication that this analysis only applies to things at rest in the non-inertial frame. Indeed the Coriolis forces is only interesting for this in motion in the non-inertial frame.

5 So far, I've written the centrifugal force in it's simplest form (figuring the directions has been easy because they are just "in" or "out"). For consistency with what follows I'm using $$ \vec{F}_\text{centrifual} = -m \vec{\Omega} \times (\vec{\Omega} \times \vec{x}_b) $$ with $\vec{\Omega}$ being the vector angular velocity and $\vec{x}_b$ begin the position of the object in a coordinate system with it's origin at the axis of rotation. Trust me. It's the same thing.


Your girlfriend's book is wrong.

....is due to the mass of the object resisting the inward centripetal acceleration that the object is experiencing"

Centrifugal force is not due to the resistance. The resistance towards acceleration is called "Inertia". Centrifugal force only occurs in non-inertial rotating frame of reference.


Here's my addition for what it's worth.

Imagine you are floating in space in a big box (like an elevator). At first you and the box are just floating around. You can move from any side of the box to the other, spin around etc. There's no sense of up or down. At some point in time the box starts to accelerate in a straight line in a direction perpendicular to one of the sides of the box at $9.8 ms^{-2}$ (the acceleration of gravity that we all know and love). The acceleration is caused by an external force on the box, perhaps a rocket attached to the outside of the box. This force is real. Lets call it the "gravipetal" force. Due to this acceleration you feel as if you have fallen to one of the sides of the box (one of the two sides that are perpendicular to the direction of acceleration). You hit the side. You notice you are stuck/attracted to that side. You try to move and realize that you can stand up on that side. You now feel that you are experiencing gravity. This is not gravity. This is a fictitious force experienced due to the acceleration of the reference frame (the box). Let's call it the "gravifugal" force. (Did I just invent a word?). It is not real. You do not know you are accelerating and therefore mistake this feeling for a force. But you can measure the force because it is the force that would give the feeling of this acceleration of your reference frame.

Now we return to a body in circular motion. It is the same idea. The body is accelerating towards the centre of motion. This acceleration is due to the tension in the string which is the centripetal force. It is real. There is tension in the string. The object doesn't know it is accelerating. Why not? The same reason humans took so long to realize the Earth is moving. From our perspective we are stationary and we can see everything else go around us. The body does all its calculations whilst in this accelerating reference frame. It feels the effect of the acceleration of the reference frame but mistakes it for a force on itself. It calls it the centrifugal force. The body has misinterpreted the physics. This force does not exist.


There is no such thing as centrifugal force. Take the example of a hanging chain in a gravitational field that's stationary on Earth's frame of reference. Its acceleration due to Earth's rotation is much tinier than Earth's gravitational acceleration so let's ignore it. The bottom link has the 2 forces acting on it, the force of gravity and an equal and opposite force by the link second from the bottom. Now the force the bottom link exerts on the link second from the bottom is equal and opposite to the force the link second from the bottom exerts on the bottom link and is the weight of a link pulling down. There is also a gravitational force on the link second from the bottom that's the weight of the link. We know that its net acceleration is 0 which means the link third from the bottom is exerting a force on the link second from the bottom which is an upward force of the weight of 2 links.

Now consider the example of a chain in a uniformly accelerating nonrotating frame of reference accelerating at 1 g such as a rocket accelerating in the direction from its bottom to its top like rockets normally do, where there's literally extremely little gravity such as the space between galaxies. I'll just state the result and then later explain why it makes sense. It will be hanging in the direction opposite the acceleration which is towards the bottom of the rocket. Force equals mass times acceleration. The link closest to the bottom will be accelerating with 1 g which means the total force acting on it is the weight of a link. The only force exerted on the link closet to the bottom of the rocket is the force the link second closest to the bottom of the rocket is exerting on it which is the force of the weight of a link in the direction from the bottom to the top of the rocket. The link second closest to the bottom of the rocket is also accelerating at 1 g so the total force acting on it must be the weight of a link in the direction from the bottom of the rocket to the top of the rocket. By Newton's third law, since the link second closest to the bottom of the rocket is exerting a force of the weight of a link on the link closest to the bottom in the direction from the bottom to the top, the link closest to the bottom must be exerting a force on the link second closest to the bottom that's equal to the weight of a link in the direction from the top to the bottom. Now we know the total force acting on the link second from the bottom is the weight of a link in the direction from the bottom to the top of the rocket so the link third closest to the bottom of the rocket is exerting a force on the link second closest to the bottom of the weight of 2 links in the direction from the bottom to the top of the rocket.

This makes you unable to distinguish between being in a gravitational field without accelerating and being in something accelerating in the absense of a gravitational field. Now consider the example of a chain hanging near the edge of something spinning in the absense of a gravitational field. Going in circles is a type of acceleration. Velocity can be defined in terms of position component wise. Similarly, acceleration is also defined in terms of velocity component wise. I really think experiment shows that it's consistent that force is extremely close to mass times acceleration using that definition of acceleration for speeds that are low enough for relativistic effects to have an extremely low unmeasurable effect. Otherwise, I would have found out otherwise. I've sometimes tossed an egg into the air and air resistance is too tiny for me to detect with the naked eye and it appears to be accelerating uniformly. So in the case of the chain hanging in the spinning frame of reference, each link is going in circles and is therefore accelerating and the only forces acting on it are those by the links on both sides of it. I know the acceleration is not the same for each link in the spinning object.

According to the textbook I had for Grade 12 physics, you can take an accelerating object as a frame of reference and decide to define the force in that frame of reference to be the acceleration of an object in that frame of reference divided by the mass of that object in that frame of reference so when they're describing stuff in terms of a spinning frame of reference, you can say there is a centrifugal force acting on the object in the frame of reference of the spinning object. Now if you calculate how will an object moving at constant velocity move in the frame of reference of a nonaccelerating uniformly counterclockwise spinning object, what do you get? It's acceleration in the frame of reference of the spinning object is its distance from the center times the angular frequency squared in the direction away from the center plus an acceleration of magnitude twice the velocity in the spinning frame of reference 90° clockwise of the velocity in the spinning frame of reference. The component of the acceleration that is dependent on the position in the spinning frame of reference is called the centrifugal force and the component that is dependent on the velocity in the spinning frame of reference is called the Coriolis force.


Centrifugal force does exist... it's clearly the force that makes a centrifuge work:


Can't spin an object in a centrifuge without a centrigual force, right? =)

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    $\begingroup$ "Clearly" has no explanatory value. $\endgroup$ Apr 22, 2014 at 17:09
  • $\begingroup$ I downvoted this answer. Downvotes are meant to be for when an answer can't be fixed up into an answer that adds something to the already existing answers without writing it in a fundamentally different way. It's better to not write an answer than to write that kind of answer. I'm not trying to criticize you but am trying to help teach you how to do a better job of contributing. People defined things in such a way that centrifugal force doesn't exist. This answer is inaccurate. $\endgroup$
    – Timothy
    Mar 2, 2020 at 3:10

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