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Why in every mechanics/dynamics book I look the centrifugal force due to Earth's rotation is measured from "the center of Earth"? For example in Marion's dynamics, the fixed inertial axes are placed at the center of Earth, and the ones rotating with the Earth somewhere on the ground. Then, the centrifugal force on an object is introduced as [ω cross (ω cross (r+R))] term, where r is the position vector in non-inertial frame and R the vector connecting two origins. I was thinking what I measure on the Earth as the centrifugal force wouldn't have anything to do with where I place the origin of the inertial frame, is it correct? Then what if I place the origin of the inertial fixed system somewhere in the space? Will the centrifugal force I measure in the non-inertial frame still be the same?


This is from the book, where the highlighted term is introduced as the centrifugal force:

enter image description here

On the other hand, when discussing motion relative to Earth, it takes something out of one of the terms and add it to aforesaid, and introduces the whole thing as the centrifugal term :/

enter image description here

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closed as unclear what you're asking by ACuriousMind, user36790, Gert, gigacyan, honeste_vivere Jun 6 '16 at 12:12

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The centrifugal force is a fictitious force which is why it does depend on the precise coordinate systems one uses to describe the mechanical phenomena.

Imagine that you sit on a spinning carousel that spins at frequency $\omega$ around its vertical axis. According to a (nearly) inertial system of the people who stand on the Earth away from the carousel, if you get shot out of the carousel, you will simply move along a straight line – which is tangent to the circular orbit on the carousel. (You also fall down vertically, but let's only focus on the horizontal motion.)

From the viewpoint of the people who are still sitting on the carousel, you had the radial coordinate $r=R$, the radius of the carousel, before you jumped out of it. So you were at rest relatively to this coordinate system. But once you lost your contact with the carousel, you were influenced by the centrifugal force that was pushing you away from the axis of the carousel at acceleration $r\omega^2$. This is why your distance from the axis of the carousel began to increase (with the acceleration $r\omega^2$).

The two descriptions are different. Straight lines according to inertial frames look like curved or accelerated lines in spinning frames. And the acceleration (or force) needed to get this bending or acceleration is nothing else than the fictitious forces. The centrifugal force is a major example of such a fictitious force and the centrifugal acceleration is indeed $$ a = \omega \times [\omega\times (R+r)] $$

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    $\begingroup$ What I'm trying to say is that the centrifugal force does depend on the object's position with respect to the non-inertial system, but not on its position with respect to the inertial one. I still don't get why there is a [ω cross (ω cross R)] term involved. If the position vector of an object in the non-inertial (rotating) system is r, then the centrifugal acceleration must be -[ω cross (ω cross r)]. $\endgroup$ – Simorq May 30 '16 at 10:26
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    $\begingroup$ If you used the symbol $r'$ for $r+R$, would it be clearer? The addition of $R$ is only to find out the distance from the actual axis of the rotation. This is really just about a difference between active and passive transformations, you shouldn't make a big deal out of the question whether the vector is written as $r+R$ or a single $R$ as long as you know how to apply the expressions correctly to a problem. $\endgroup$ – Luboš Motl May 30 '16 at 11:24
  • $\begingroup$ No :/ What if I put the origin of the inertial frame somewhere in the space? $\endgroup$ – Simorq May 30 '16 at 17:15
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    $\begingroup$ @Simorq, then you will be solving a different problem than the one you described in your question. In your question, you talk about the Earth that rotates around an axis through the center of the Earth. A nice inertial system has the origin at the center - on the axis of rotation. But you want to use a non-inertial system associated with a point on the Earth's surface whose coordinates are $\vec R$ relatively to the coordinate system I mentioned a sentence ago, and place the origin here, on the surface. $\endgroup$ – Luboš Motl May 30 '16 at 17:25
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    $\begingroup$ So a point near this point on the surface, just separated a mile, $\vec r$, from this point on the surface, really has coordinates $\vec R+\vec r$ relatively to the inertial origin of coordinates, the center of the Earth, and that's what matters for the centrifugal acceleration because centrifugal is one that repels you from the center of the Earth, right? So obviously you need $\vec R+\vec r$ there. ... The Earth is spinning around a particular axis, one that goes through the Earth center. You can't replace this axis by a shifted one somewhere in space because that's not how the Earth spins $\endgroup$ – Luboš Motl May 30 '16 at 17:26
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Motion equations are independent from origin and kind of the coordinate system.

What is the equations of motion?

They are those relations that we can determine the position, velocity, acceleration, etc of the particle by using them.

If the position vector of a particle in an inertial frame (coordinate system) be $\vec r(t)$ ($t$ is the time), we define a quantity $\vec v(t)$ as $\vec v(t)=\frac d{dt}\vec r(t)$ and we name it velocity. Then we define another quantity $\vec a(t)$ as $\vec a(t)=\frac d{dt}\vec v(t)$ and we name it acceleration.

On the other hand, we define force by first law of Newton. Then, we accept that "If the resultant force acting on a particle with mass $m$ in an inertial frame (coordinate system) be $\vec F$, then acceleration of that particle will be $\vec a=\large{\frac 1m}\vec F$".

As you are seeing, we never mention the origin of coordinate system. But, choosing a suitable kind of coordinate system with a suitable origin will make easy our calculations.

So, if you can express the position vector of the particle in an arbitrary kind of inertial coordinate system with an arbitrary origin, then you only need to derive it with respect time for calculating velocity and acceleration of that particle.

Will the centrifugal force I measure in the non-inertial frame still be the same?

Once you find the acceleration vector ($\vec a$) of the particle you can find the resultant force vector acting on that. Then you can break that force to any arbitrary set of vectors that those sum is $\vec F$. And you can call each of them by any name you like for example centrifugal force or Coriolis force or Lucas force :-)

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