# Is it true?

## Recommended Posts

If you construct Newton's laws in a rotating system, does a centrifugal term fall out?

##### Share on other sites

I don't know what you mean by "fall out," but yes there will be a centrifugal force term.

##### Share on other sites

I don't know what you mean by "fall out," but yes there will be a centrifugal force term.

Wanna show me?

##### Share on other sites

Take a sharp turn in your car and see for yourself.

I can't do it today, but I will come up with something in the next day or two if nobody else does it first.

##### Share on other sites

Here's a little experiment you can try...

First of all get a rotating round surface (I did it with a rotating stool...), then get a plastic stick (like a sipping straw but longer and wider, the main point is that it must be hole inside), and cut it seccionally. Then place it in a radial postion (it must stick out of the surface) and finally place a little ball in the straw, in the center of the surface (it must fit in the straw's cavity)

Now you are ready for this brief experiment.

Gently start spinning the rotating surface, and watch.

Let me know if it worked.

Ms. M

P.S.: My english isn't very good, I live in a spanish-speaking country, I hope you understood the procedure, if you have any doubts please let me know

##### Share on other sites

If you're in an accelerating frame, you need a term to counter your acceleration so that F=ma will hold. That's why there would be a centrifugal term in that case, and why you also have a Coriolis term when treating the earth as an inertial frame of reference. If the rotation changes with time, you have an Euler force term.

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

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

##### Share on other sites

Imagine this situation:

Frame 1: stationary

Frame 2: moving at speed u w.r.t F1 (frame 1)

A particle is moving at v in F1

v' = velocity of particle in F2

= v - u

Differentiating:

a' = acc of particle in F2

= dv/dt - du/dt

dv/dt = a = acc of particle in F1

du/dt = acc of F2 w.r.t F1

If F2 is inertial then du/dt=0, and in this case:

F1 gives: F=ma

F2 gives: F'=ma'

If F2 is non-inertial then du/dt=aframe

F1 still gives: F = ma

But F2 gives: F' = ma' - maf

and so Newton's Laws are not valid in F2, when F2 is non-inertial (i.e. is accelerating).

To "fix" this we add a pseudo-force.

Let this force be: Fp=-maf

So F'tot=F'+Fp

And therefore: F'tot=ma'

To use Newton's 2nd Law in non-inertial frames you need to include an extra pseudo-force.

I can give a nice little example if you would like?

##### Share on other sites

Wanna show me?
Assume two reference frames, one inertial and one rotating. Denote the angular velocity and angular acceleration of the rotating frame with respect to inertial as $\mathbf \omega$ and $\dot{\mathbf \omega}$.

Now assume two observers of some vector quantity $\mathbf q$, one observer fixed in the inertial frame and the other fixed in the rotating frame. The two observers will see substantially different time derivatives of the vector quantity. The derivatives are related via

$\dot{\mathbf q}_I = \dot{\mathbf q}_R + \mathbf \omega \times \mathbf q$

If you want me to derive this result (called the transport theorem in several aeronautics engineering texts), I will. For now take it as a given.

Applying this transport theorem twice to the position vector,

$\ddot{\mathbf r}_I = \ddot{\mathbf r}_R + \dot{\mathbf \omega} \times \mathbf r + 2 \mathbf \omega \times \dot {\mathbf r}_R + \mathbf \omega \times (\mathbf \omega \times \mathbf r)$

The left hand side is the subject of Newton's second law, $\mathbf F = m \ddot{\mathbf r}_I$. Making those extra terms on the right-hand side look like forces will make $\ddot{\mathbf r}_R$ act in a manner similar to $\ddot{\mathbf r}_I$. To make an acceleration look like a force, simply multiply by mass. To put it on the other side of the equal sign, simply negate.

The additive inverse final term on the right-hand side, $-\,\mathbf \omega \times (\mathbf \omega \times \mathbf r)$ is the centrifugal acceleration. The penultimate term is the Coriolis effect.

##### Share on other sites

• 1 month later...

My science teacher once said that centrifugal force doesn't in fact exist, because centrifugal force is essentially just inertia

##### Share on other sites

the centrifugal force does exist, but only in rotating reference frames.

when most people say centrifugal, they mean centripetal force anyway.

##### Share on other sites

Ah okay, thanks very much for clearing that up for me, it makes much more sense now.

##### Share on other sites

the centrifugal force does exist, but only in rotating reference frames.

when most people say centrifugal, they mean centripetal force anyway.

So if I'm swinging a cat around by the tail, the force I feel at the end of my arm is centripetal, but the force the cat feels is centrifugal?
##### Share on other sites

The centralpetal and fugal forces are essentially the same thing but with slightly different coordinate settups, it drops out when you solve all the forces in a rotating frame, they are described as pseudo-forces.

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

##### Share on other sites

Antimatter: your science teacher was right. That rotating frame doesn't really "exist", it's just a way of describing an environment subject to an ongoing circular motion. That means the centripetal force doesn't really "exist" either. But the inertia definitely does. Hence people call it an "inertial force". Check this out:

http://findarticles.com/p/articles/mi_m1511/is_n7_v16/ai_17040768

All sorts of interesting things pop up:

It's all to do with geometry.

##### Share on other sites

So if I'm swinging a cat around by the tail, the force I feel at the end of my arm is centripetal, but the force the cat feels is centrifugal?

Your motion is the result of forces exerted on you (action forces), not by the forces you exert (reaction forces), but the forces we feel are often the ones we exert. The cat's motion is circular, thus it must feel a centripetal force. In that case, your motion isn't circular (or easily identifiable as such) so it isn't the best example.

If you are in a car going around a circle, you feel like you are thrown out and pressing against the side of the car — that's a force you exert. But your motion is dictated by the force on you, which is pointed inward. In that frame, the centripetal and centrifugal forces are an action-reaction force pair, from Newton's third law, and the identification of the centrifugal force is the mixing of recognizing forces exerted by and forces exerted on.

##### Share on other sites

Antimatter: your science teacher was right. That rotating frame doesn't really "exist", it's just a way of describing an environment subject to an ongoing circular motion.

Why does that mean the reference frame doesn't exist? It's a valid perspective to use, isn't it?

##### Share on other sites

Why does that mean the reference frame doesn't exist? It's a valid perspective to use, isn't it?
Sure it's a valid perspective. But that's all it is. A perspective. It's how you view the world, because of your motion through it.

The thing is that you can't touch a reference frame, or smell it, weigh it, et cetera, because it's an abstraction. It isn't a quale like vision, which is emphatically "real" to you. It's just a term associated with what you measure and observe because of your motion. People say "in your reference frame" as if it's a place, or something real, but all it really means is "in your state of motion".

It's simple to understand this when you think of constant linear motion and and an inertial reference frame. It gets a little more complicated when you introduce acceleration, and more complicated again when you introduce gravity. But the same ontological principle applies. See:

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

##### Share on other sites

And why does that mean that centrifugal force "doesn't exist"?

##### Share on other sites

It's not a real force, it's just inertia...so it is a force, but it's just inertia...

now I'm just confusing myself...

##### Share on other sites

It's inertia if you look at it one way, but it's centrifugal force if you look at it from another frame. Who's to say which frame is the "right" one?

##### Share on other sites

Wouldn't that also apply to centripetal force then too?

Yes.

##### Share on other sites

It has always been to my teaching, that the centrifugal force does not exist.

##### Share on other sites

/me votes that it's a psydoforce again.

##### Share on other sites

It has always been to my teaching, that the centrifugal force does not exist.

"A laughable clame, graviphoton, perpetuated by overzealous teachers of science"

no?

## Create an account

Register a new account