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[[Fisica_classica/Energia_e_lavoro| Argomento precedente: Energia e lavoro]]
Nei sistemi di riferimento non inerziali le leggi della dinamica sono modificate e si manifestano delle forze che vengono chiamate fittizie.
LaTali forza '''F'''forze non provieneprovengono da nessuna interazione tra oggetti, ma piuttosto dalla accelerazione '''a''' propria del [[w:Sistema_di_riferimento_non_inerziale|sistema di riferimento non inerziale]]. Si noti che un cambiamento di sistema di riferimento ad esempio da Cartesiano a polare non comporta l'insorgere di forze apparenti, mentreanche se le leggi del molto pssonopossono varievariare da un tipo all'altro di sistema di riferimento.
 
Le forze dovute al moto relativo non uniforme tra i due sistemi di riferimento sono chiamate forza apparenti. PerLa la [[Fisica_classica/Dinamica#Seconda_legge_della_dinamica_.28detta_anche_II_legge_di_Newton.29|II legge della dinamica]] nellacontinua formaad ''F''' =essere ''m'''''a''',valida lee forzequindi apparentil'accelerazione sonosi proporzionalìmantiene proporzionale alla accelerazione ''a''forza.
 
Una forza apparente compare su un oggetto quando il sistema di riferimento usato per descrivere il movimento dell'oggetto stesso viene accelerato rispetto a un sistema di riferimento inerziale. Le accelerazioni possono avvenire in maniera molto variabilediverso, ma per spiegare i fenomeni si considerano solo 4 forze apparenti: 1) quella causata da accelerazioni in linea retta; 2) quella riguardante un sistema in rotazione ([[w:Forza_centrifuga|forza centrifuga]]); 3) moto in un sistema in rotazione ([[w:Forza_di_Coriolis|la forza di Coriolis]]);
4) sistema in rotazione velocità angolare variabile.
 
=Esempi di forze apparenti=
==Accelerazione in linea retta==
 
[[Image:Accelerating car.PNG|thumb |250px|Figura ''in alto'': una macchina in accelerazione di massa ''M'' con un passeggero di massa ''m''. Figura ''di mezzo'' cosa avviene dal punto di vista di un riferimento inerziale. Figura ''in basso'' cosa avviene nel sistema di riferimento solidale alla macchina]]
 
Un esempio facilmente comprensibile è quello mostrato nella figura a fianco nella porzione in alto. Una macchina di massa ''M'' con un passeggero di massa ''m'' in fase di accelerazione.
==Fictitious forces on Earth==
Il passaggero si sente schiacciare contro il sedile. Nel sistema di riferimento inerziale non vi è nessuna forza che spinge il passaggero contro il sedile. Mentre nel sistema di riferimento non inerziale della macchina vi è una forza fittizia -m'''a''' che spinge il passeggero contro il sedile. Vi sono due modi possibili di analizzare il problema.
{{See also|centrifugal force|Coriolis force|Euler force}}
The surface of the Earth is a [[rotating reference frame]]. To solve [[classical mechanics]] problems exactly in an Earth-bound reference frame, three fictitious forces must be introduced, the [[Coriolis force]], the [[centrifugal force (fictitious)|centrifugal force]] (described below) and the [[Euler force]]. The Euler force is typically ignored because the variations in the angular velocity of the rotating Earth surface are usually insignificant. Both of the other fictitious forces are weak compared to most typical forces in everyday life, but they can be detected under careful conditions. For example, [[Léon Foucault]] was able to show the [[Coriolis force]] that results from the Earth's rotation using the [[Foucault pendulum]]. If the Earth were to rotate a thousand times faster (making each day only ~86 seconds long), people could easily get the impression that such fictitious forces are pulling on them, as on a spinning carousel.
 
==Detection of non-inertial reference frame==
{{See also|Inertial frame of reference}}
Observers inside a closed box that is moving with a constant [[velocity]] cannot detect their own motion; however, observers within an accelerating reference frame can detect that they are in a non-inertial reference frame from the fictitious forces that arise. For example, for straight-line acceleration [[Vladimir Arnold]] presents the following theorem:<ref name="Arnolʹd">{{cite book |title=Mathematical Methods of Classical Mechanics
|url=http://books.google.com/books?id=Pd8-s6rOt_cC&printsec=frontcover&dq=intitle:Mathematical+intitle:Methods+intitle:of+intitle:Classical+intitle:Mechanics#PPA129,M1
|author= Vladimir Igorevich Arnold |publisher=Springer |location=Berlin |isbn=0-387-96890-3 |year=1989 |pages=§27 pp. 129 ff.}}</ref>
{{blockquote|In a coordinate system ''K'' which moves by translation relative to an inertial system ''k'', the motion of a mechanical system takes place as if the coordinate system were inertial, but on every point of mass ''m'' an additional "inertial force" acted: '''F'''&nbsp;{{=}} −''m'''''a''', where '''a''' is the acceleration of the system ''K''.}}
Other accelerations also give rise to fictitious forces, as described mathematically [[#Mathematical_derivation_of_fictitious_forces|below]]. The physical explanation of motions in an inertial frames is the simplest possible, requiring no fictitious forces: fictitious forces are zero, providing a means to distinguish inertial frames from others.<ref name=transformations>As part of the requirement of simplicity, to be an inertial frame, in all other frames that differ only by a uniform rate of translation, the description should be of the same form. However, in the Newtonian system the [[Galilean transformation]] connects these frames and in the special theory of relativity the [[Lorentz transformation]] connects them. The two transformations agree for speeds of translation much less than the [[speed of light]].</ref>
 
# Figura centrale. Punto di vista di un sistema di riferimento inerziale la macchina sta accelerando. Per mantenere il passeggero sul sedile dellla macchina deve essere esercitata una forza sul passeggero. La forza è la reazione vincolare del sedile, che ha incominciato a muoversi in avanti quando la macchina ha accelerato e ha compresso il sedile verso il passeggero fino a raggiungere una situazione di equilibrio dinamico in maniera che la macchina e il passeggero si muovono insieme. Il passeggero quindi viene accelerato esattamente come la macchina con la forza della spinta del sedile.
An example of the detection of a non-inertial, rotating reference frame is the precession of a [[Foucault pendulum]]. In the non-inertial frame of the Earth, the fictitious [[Coriolis force]] is necessary to explain observations. In an inertial frame outside the Earth, no such fictitious force is necessary.
# Figura in basso. Dal punto di vista dell'interno della macchina, un sistema di riferimento non inerziale, vi è una forza fittizia che spinge il passeggero contro il sedile, con valore pari alla massa del passegero moltiplicata per la accellerazione della macchina. Questa forza spinge il passegero contro il sedile, fino a quando il sedile compresso non compensa esattamente la forza fittizia. In questo sistema di riferimento il passeggero è in quiete.
Mentre nel sistema inerziale vi è solo la forza propulsiva della macchina che accelera la vettura e con essa il sedile rigidamente connesso. La spinta del sedile accelera il passeggero che si muove con la vettura. Nel sistema di riferimento non inerziale lo stato di quiete è giustificato da due forze equali ed opposte quella fittizia del passeggero contro il sedile e quella del sedile contro il passeggero. La spiegazione fisica è più semplice nel sistema inerziale, ma la formulazione matematica potrebbe essere più complicata nel sistema inerziale. In questo caso specifico non vi è differenza di calcolo nei due sistemi di riferimento.
 
L'esempio illustra come le forze apparenti vengono osservando i fatti in un sistema di riferimento non inerziale. Vi sono dei casi in cui è più semplice studiare i fenomeni dal punto di vista di un sistema non inerziale.
==Examples of fictitious forces==
===Acceleration in a straight line===
[[Image:Accelerating car.PNG|thumb |350px|Figure 1: ''Top panel'': accelerating car of mass ''M'' with passenger of mass ''m''. The force from the axle is (''m'' + ''M'')'''a'''. In the inertial frame, this is the only force on the car and passenger.<br /> ''Center panel'': an exploded view in the inertial frame. The passenger is subject to the accelerating force ''m'''''a'''. The seat (assumed of negligible mass) is compressed between the reaction force –''m'''''a''' and the applied force from the car ''m'''''a'''. The car is subject to the net acceleration force ''M'''''a''' that is the difference between the applied force (''m'' + ''M'')'''a''' from the axle and the reaction from the seat −''m'''''a'''. <br />''Bottom panel'': an exploded view in the non-inertial frame. In the non-inertial frame where the car is not accelerating, the force from the axle is balanced by a fictitious backward force −(''m'' + ''M'')'''a''', a portion −''M'''''a''' applied to the car, and −''m'''''a''' to the passenger. The car is subject to the fictitious force −''M'''''a''' and the force (''m'' + ''M'')'''a''' from the axle. The difference between these forces ''m'''''a''' is applied to the seat, which exerts a reaction −''m'''''a''' upon the car, so zero net force is applied to the car. The seat (assumed massless) transmits the force ''m'''''a''' to the passenger, who is subject also to the fictitious force −''m'''''a''', resulting in zero net force on the passenger. The passenger exerts a reaction force −''m'''''a''' upon the seat, which is therefore compressed. In all frames the compression of the seat is the same, and the force delivered by the axle is the same.]]
Figure 1 (top) shows an accelerating car. When a car [[acceleration|accelerates]], a passenger feels like they're being pushed back into the seat. In an inertial frame of reference attached to the road, there is no physical force moving the rider backward. However, in the rider's non-inertial reference frame attached to the accelerating car, there ''is'' a backward fictitious force. We mention two possible ways of analyzing the problem:{{clarify|reason=How is this a problem?|date=June 2011}}<ref name=Motz>{{cite book |title=The Concepts of Science: From Newton to Einstein |author=Lloyd Motz & Jefferson Hane Weaver |url=http://books.google.com/books?id=SnufjnGnqD8C&pg=PA95&dq=history+velocity+%22law+of+inertia%22#PPA101,M1
|page=101 |isbn=0-7382-0834-5 |publisher=Basic Books |year=2002}}</ref>
# Figure 1, (center panel). From an [[inertial reference frame]], with a constant velocity matching the initial motion{{clarify|reason=What initial motion?|date=June 2011}} of the car, the car is accelerating. In order for the passenger to stay inside the car, a force must be exerted on them. This force is exerted by the seat, which has started to move forward with the car and is compressed against the passenger until it transmits the full force to keep the passenger moving with the car. Thus, the forces of the seat are unbalanced,{{clarify|reason=On or created by?|date=June 2011}} so the passenger is accelerating in this frame.
# Figure 1, (bottom panel). From the point of view of the interior of the car, an accelerating reference frame, there is a fictitious force pushing the passenger backwards, with magnitude equal to the [[mass]] of the passenger times the acceleration of the car. This force pushes the passenger back into the seat, until the seat compresses and provides an equal and opposite force. Thereafter, the passenger is stationary in this frame, because the fictitious force and the real force of the seat are balanced.
 
== Forza centrifuga==
How can the accelerating frame be discovered to be non-inertial? In the accelerating frame, everything appears to be subject to zero net force, and nothing moves. Nonetheless, compression of the seat is observed and is explained in the accelerating frame (and in an inertial frame) by the force of acceleration on the seat from the car on one side, and the opposing force of reaction to acceleration by the passenger on the other. Identification of the accelerating frame as non-inertial cannot be based simply on the compression of the seat, which all observers can explain; rather it is based on the ''simplicity'' of the physical explanation for this compression.
Un effetto simile si ha nel moto lungo un circuito circolare di una macchina. Dal punto di vista di un sistema di riferimento sulla strada si osserva un moto circolare della macchina. Quando lo stesso fenomeno è osservato in un sistema sulla macchina appare una forza apparente detta forza centrifuga. Se la macchina si muove a velocità costante lungo il tratto di strada circolare, gli occupanti della macchina si sentono spinti in fuori dal centro di rotazione dalla forza centrifuga. Anche in questo caso il fenomeno può essere visto dal punto di vista del sistema inerziale e da quello non inerziale solidale con la macchina.
 
# Dal punto di vista del sistena di riferimento inerziale stazionario rispetto alla strada, la macchina accelera verso il centro del cerchio. L'accelerazione è necessaria in quanto la direzione della velocità cambia, anche se in modulo rimane costante. La accelerazione verso il centro è chiamata accelerazione centripeta e richiede una forza centripeta per permettere il moto circolare. Nel caso di una macchina questa forza è fornita in genere dall'attrito statico tra le ruote e la strada. Questa forza causa il movimento lungo una circoferenza. In questo caso a causa dell'attrito statico tra passeggero e sedile, anche il passegero si muove di moto circolare uniforme come la macchina (se non vi fosse attrito si muoverebbe di moto rettilineo uniforme quindi non in direzione del centro).
The explanation of the seat compression in the accelerating frame requires not only the thrust from the axle of the car, but additional (fictitious) forces. In an inertial frame, only the thrust from the axle is necessary. Therefore, the inertial frame has a ''simpler'' physical explanation (not necessarily a simpler mathematical formulation, however), indicating the accelerating frame is a non-inertial frame of reference. In other words, in the inertial frame, fictitious forces are zero. See [[inertial frame]] for more detail.
# Dal punto di vista del sistema del sistema di riferimento ruotante, che si muove con la macchina, vi è una forza fittizia centrifuga che spinge gli occupanti della macchina verso l'esterno a cui ci oppone l'attrito del sedile che impedisce che le persone vadano a urtare la portiera. In questo riferimento il passeggero è fermo.
 
== Forza di Coriolis ==
This example illustrates how fictitious forces arise from switching from an inertial to a non-inertial reference frame. Calculations of physical quantities (compression of the seat, required force from the axle) made in any frame give the same answers, but in some cases calculations are easier to make in a non-inertial frame. (In this simple example, the calculations are equally complex for the two frames described.)
Se da una torre alta 50 m viene lanciato un oggetto, se l'attrito dell'aria è trascurabile, cadrà in un punto a 7.7 mm ad est dalla verticale. Dal punto di vista di un osservatore sulla terra, sistema ruotante non inerziale, in quanto vi è la forza apparente di Coriolis che causa un moto non rettilineo, dal punto di vista di un osservatore inerziale durante il tempo di caduta la terra ruota e quindi l'oggetto cade in linea retta e la terra gli si sposta di sotto.
 
== Forza di trascinamento angolare==
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
Se durante il moto di una macchina lungo una circonferenza aumenta la velocità angolare si ha un qualcosa di simile a quello che si è descritto nel moto accelerato lineare. Dal punto di vista del sistema non inerziale vi è oltre alla forza centrifuga una forza apparente che spinge il passeggero verso il sedile. Se invece è in fase di decelerazione lo spinge verso il parabrezza. Dal punto di vista di un osservatore inerziale vi sono solo la forza centrifuga e la accelerazione angolare con la relativa forza.
!''Animation:'' driving from block to block
|-
|[[Image:Carframe.gif|thumb|360px|center|Map and car frame perspectives of physical (red) and fictitious (blue) forces for a car driving from one stop sign to the next.]]
In this illustration the car accelerates after a stop sign until midway up the block, at which point the driver is immediately off the accelerator and onto the brake so as to make the next stop.
|}
 
= Formulazione analitica=
===Circular motion===
[[Image:Moving coordinate system.PNG|thumb|400px|| Un oggetto in '''x'''<sub>A</sub> nel sistema di riferimento inerziale ''A'' ha una posizione '''x'''<sub>B</sub> nel sistema accelerato ''B''. L'origine di ''B'' è in '''X'''<sub>AB</sub> nel sistema ''A''. Le orientazioni del sistema ''B'' è determinato dai versori lungo i suoi assi delle coordinate '''u'''<sub>j</sub> con ''j'' = 1, 2, 3. Usando questi assi, le coordinate dell'oggetto nel sistema ''B'' sono '''x'''<sub>B</sub> = ( '''''x'''''<sub>1</sub>, '''''x'''''<sub>2</sub>, '''''x'''''<sub>3</sub></font> )]]
{{See also|Centrifugal force|Reactive centrifugal force|Coriolis force}}
 
La figura mostra un esempio di formulazione dei moti relativi. Un punto materiale di massa ''m'' e posizione '''x'''<sub>A</sub>(''t'') in un sistema di riferimento inerziale A. Consideriamo
A similar effect occurs in [[circular motion]], circular from the standpoint of an inertial frame of reference attached to the road. When seen from a non-inertial frame of reference attached to the car, the fictitious force called the [[centrifugal force (fictitious)|centrifugal force]] appears. If the car is moving at constant speed around a circular section of road, the occupants will feel pushed outside by this centrifugal force, away from the center of the turn. Again the situation can be viewed from inertial or non-inertial frames (for free body diagrams, see [[Reactive_centrifugal_force#Example:_The_turning_car|the turning car]]):
un sistema di riferimento non inerziale B la cui origine è in '''X'''<sub>AB</sub> nel sistema ''A''. Nel sistema B la posizione del punto materiale è '''x'''<sub>B</sub>(''t''). Vogliamo determinare le forze agenti nel sistema di riferimento B sul punto materiali
 
Gli assi delle coordinate nel riferimento B sono identificati dai versori '''u'''<sub>j</sub> con ''j'' {&thinsp;1,&thinsp;2,&thinsp;3&thinsp;} per i tre assi delle coordinate.
# From the viewpoint of an inertial reference frame stationary with respect to the road, the car is accelerating toward the center of the circle. This acceleration is necessary because the ''direction'' of the velocity is changing, despite a constant speed. This inward acceleration is called [[centripetal acceleration]] and requires a [[centripetal force]] to maintain the circular motion. This force is exerted by the ground upon the wheels, in this case thanks to the [[friction]] between the wheels and the road.<ref>The force in this example is known as ''[[Ground reaction force|ground reaction]],'' and it could exist even without friction, for example, a sled running down a curve of a bobsled track.</ref> The car is accelerating, due to the unbalanced force, which causes it to move in a circle. (See also [[banked turn]].)
# From the viewpoint of a rotating frame, moving with the car, there is a fictitious centrifugal force that tends to push the car toward the outside of the road (and to push the occupants toward the outside of the car). The centrifugal force balances the friction between wheels and road, making the car stationary in this non-inertial frame.
 
A classic example of fictitious force in circular motion is the experiment of [[Centrifugal_force#Rotating_identical_spheres|rotating spheres]] tied by a cord and spinning around their center of mass. In this case, as with the linearly accelerating car example, the identification of a rotating, non-inertial frame of reference can be based upon the vanishing of fictitious forces. In an inertial frame, fictitious forces are not necessary to explain the tension in the string joining the spheres. In a rotating frame, Coriolis and centrifugal forces must be introduced to predict the observed tension.
 
To consider another example, where a rotating reference frame is very natural to us, namely the surface of the rotating Earth, centrifugal force reduces the apparent force of gravity by about one part in a thousand, depending on latitude. This reduction is zero at the poles, maximum at the equator.
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!''Animation:'' object released from a carousel
|-
|[[Image:Spinframe.gif|thumb|360px|center|Map and spin frame perspectives of physical (red) and fictitious (blue) forces for an object released from a carousel.]]
From the map frame perspective, what's dangerous on losing centripetal acceleration may be your speed. From the spin frame perspective, the danger instead may lie with the geometric acceleration which gives rise to that fictitious force.''Note:'' With some browsers, you can hit [Esc] to freeze the motion for more detailed analysis. However you may have to reload the page to restart.
|}
 
The fictitious [[Coriolis force]], which is observed in rotational frames, is ordinarily visible only in very large-scale motion like the projectile motion of long-range guns or the circulation of the Earth's atmosphere (see [[Rossby number]]). Neglecting air resistance, an object dropped from a 50-meter-high tower at the [[equator]] will fall 7.7 millimeters eastward of the spot below where it is dropped because of the Coriolis force.<ref name=Kleppner>{{cite book |author=Daniel Kleppner and Robert J. Kolenkow |year=1973 |title= An Introduction to Mechanics |publisher=McGraw-Hill |page= 363 |isbn=0-07-035048-5}}</ref>
 
In the case of distant objects and a rotating reference frame, what must be taken into account is the resultant force of centrifugal and Coriolis force. Consider a distant star observed from a rotating spacecraft. In the reference frame co-rotating with the spacecraft, the distant star appears to move along a circular trajectory around the spacecraft. The apparent motion of the star is an apparent centripetal acceleration. Just like in the example above of the car in circular motion, the centrifugal force has the same magnitude as the fictitious centripetal force, but is directed in the opposite, centrifugal direction. In this case the Coriolis force is twice the magnitude of the centrifugal force, and it points in centripetal direction. The vector sum of the centrifugal force and the Coriolis force is the total fictitious force, which in this case points in centripetal direction.
 
=== Fictitious forces and work ===
 
Fictitious forces can be considered to do [[Mechanical work|work]], provided that they move an object on a [[trajectory]] that changes its [[energy]] from [[potential energy|potential]] to [[kinetic energy|kinetic]]. For example, consider a person in a rotating chair holding a weight in his outstretched arm. If he pulls his arm inward, from the perspective of his rotating reference frame he has done work against centrifugal force. If he now lets go of the weight, from his perspective it spontaneously flies outward, because centrifugal force has done work on the object, converting its potential energy into kinetic. From an inertial viewpoint, of course, the object flies away from him because it is suddenly allowed to move in a straight line. This illustrates that the work done, like the total potential and kinetic energy of an object, can be different in a non-inertial frame than an inertial one.
 
==Gravity as a fictitious force==
{{Main|General relativity}}
The notion of "fictitious force" comes up in general relativity.<ref name=Rohrlich>{{cite book |title=Classical charged particles |author=Fritz Rohrlich |page=40 |url=http://books.google.com/books?id=lJrPrP6L6tQC&pg=PA39&dq=acceleration+transformation+formula+%22Coriolis+force%22#PPA40,M1
|isbn=981-270-004-8 |year=2007 |publisher=World Scientific |location =Singapore }}</ref><ref name=Stephani>{{cite book |title=Relativity: An Introduction to Special and General Relativity |author=Hans Stephani |url=http://books.google.com/books?id=WAW-4nd-OeIC&pg=PA104&dq=acceleration+transformation+formula+%22Coriolis+force%22#PPA105,M1
|page=105 |isbn=0-521-01069-1 |year=2004 |publisher=Cambridge University Press |location=Cambridge UK }}</ref> All fictitious forces are proportional to the mass of the object upon which they act, which is also true for [[gravity]].<ref name=Eötvös>The gravitational mass and the inertial mass are found experimentally to be equal to within experimental error.</ref> This led [[Albert Einstein]] to wonder whether gravity was a fictitious force as well. He noted that a [[freefall]]ing observer in a closed box would not be able to detect the force of gravity; hence, freefalling reference frames are equivalent to an inertial reference frame (the [[equivalence principle]]). Following up on this insight, Einstein was able to formulate a theory with gravity as a fictitious force; attributing the apparent acceleration of gravity to the [[curvature]] of [[spacetime]]. This idea underlies Einstein's theory of [[general relativity]]. See [[Eötvös experiment]].
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
!''Animation:'' ball that rolls off a cliff
|-
|[[Image:Shellframe.gif|thumb|360px|center|Rain and shell frame perspectives of physical (red) and fictitious (blue) forces for an object that rolls off a cliff.]] ''Note:'' The rain frame perspective here, rather than being that of a raindrop, is more like that of a trampoline jumper whose trajectory tops out just as the ball reaches the edge of the cliff. The shell frame perspective<ref>Edwin F. Taylor and John Archibald Wheeler (2000) ''Exploring black holes'' (Addison Wesley Longman, NY) ISBN 0-201-38423-X</ref> may be familiar to planet dwellers who rely minute by minute on upward physical forces from their environment, to protect them from the geometric acceleration due to curved spacetime.
|}
 
==Mathematical derivation of fictitious forces {{anchor|coordinates}}==
[[Image:Moving coordinate system.PNG|thumb|400px|Figure 2: An object located at '''x'''<sub>A</sub> in inertial frame ''A'' is located at location '''x'''<sub>B</sub> in accelerating frame ''B''. The origin of frame ''B'' is located at '''X'''<sub>AB</sub> in frame ''A''. The orientation of frame ''B'' is determined by the unit vectors along its coordinate directions, '''u'''<sub>j</sub> with ''j'' = 1, 2, 3. Using these axes, the coordinates of the object according to frame ''B'' are '''x'''<sub>B</sub> = ( '''''x'''''<sub>1</sub>, '''''x'''''<sub>2</sub>, '''''x'''''<sub>3</sub></font> ).]]
 
=== General derivation ===
Many problems require use of noninertial reference frames, for example, those involving satellites<ref name=Isidon>{{cite book |title=Robust Autonomous Guidance: An Internal Model Approach |author=Alberto Isidori, Lorenzo Marconi & Andrea Serrani |page= 61 |url=http://books.google.com/books?id=Mo0YhjLeTA0C&pg=RA2-PA60&dq=orbit+%22coordinate+system%22#PRA2-PA61,M1
|isbn=1-85233-695-1 |publisher=Springer |year=2003 }}</ref><ref name=Ying>{{cite book |title=Advanced Dynamics |author=Shuh-Jing Ying |url=http://books.google.com/books?id=juO1Hj8ocx8C&pg=PA172&dq=orbit+%22coordinate+system%22|page=172 |isbn=1-56347-224-4 |publisher=American Institute of Aeronautics, and Astronautics|location=Reston VA |year=1997 }}</ref> and particle accelerators.<ref name=Bryant>{{cite book |title=The Principles of Circular Accelerators and Storage Rings |author=Philip J. Bryant & Kjell Johnsen |page=xvii |url=http://books.google.com/books?id=6P-UWLmfD4wC&pg=PR17&dq=orbit+%22coordinate+system%22|isbn=0-521-35578-8 |publisher=Cambridge University Press|location=Cambridge UK |year=1993 }}</ref>
Figure 2 shows a particle with [[mass]] ''m'' and [[position (vector)|position]] [[vector (geometric)|vector]] '''x'''<sub>A</sub>(''t'') in a particular [[inertial frame]] A. Consider a non-inertial frame B whose origin relative to the inertial one is given by '''X'''<sub>AB</sub>(''t''). Let the position of the particle in frame B be '''x'''<sub>B</sub>(''t''). What is the force on the particle as expressed in the coordinate system of frame B? <ref name=Fetter>{{cite book |author=Alexander L Fetter & John D Walecka|title=Theoretical Mechanics of Particles and Continua |publisher= Courier Dover Publications |url=http://books.google.com/books?id=olMpStYOlnoC&printsec=frontcover&dq=intitle:theoretical+intitle:mechanics+inauthor:fetter#PPA32,M1 |year=2003 |isbn=0-486-43261-0 |pages= 33–39}}</ref><ref name=Lim>{{cite book |title=Problems and Solutions on Mechanics: Major American Universities Ph.D. Qualifying Questions and Solutions |author=Yung-kuo Lim & Yuan-qi Qiang |isbn=981-02-1298-4 |page=183|url=http://books.google.com/books?id=93b3cjVJ2l4C&pg=PA183&dq=%22Coriolis+force%22+-snippet+date:1990-2008|publisher=World Scientific |location=Singapore|year=2001}}</ref>
 
To answer this question, let the coordinate axis in B be represented by unit vectors '''u'''<sub>j</sub> with ''j'' any of {&thinsp;1,&thinsp;2,&thinsp;3&thinsp;} for the three coordinate axes. Then
 
Quindi la posizione del punto materiale secondo il sistema B è:
:<math> \mathbf{x}_\mathrm{B} = \sum_{j=1}^3 x_j \mathbf{u}_j \ . </math>
Mentre nel sistema A la posizione è:
 
The interpretation of this equation is that '''x'''<sub>B</sub> is the vector displacement of the particle as expressed in terms of the coordinates in frame B at time ''t''. From frame A the particle is located at:
 
:<math>\mathbf{x}_\mathrm{A} =\mathbf{X}_\mathrm{AB} + \sum_{j=1}^3 x_j \mathbf{u}_j \ . </math>
I versori {&thinsp;'''u'''<sub>''j''</sub>&thinsp;} non variano di ampiezza nel tempo, quindi la loro derivata corripsonde solo ad una rotazione del sistema di coordinate B. Inoltre il vettore '''X'''<sub>AB</sub> fornisce la posizione dell'origine di B rispetto ad A, e non include la rotazione del sistema B.
== Velocità relativa==
 
Quindi facendo la derivate temporale, la velocità del punto materiale diviene:
As an aside, the unit vectors {&thinsp;'''u'''<sub>''j''</sub>&thinsp;} cannot change magnitude, so derivatives of these vectors express only rotation of the coordinate system B. On the other hand, vector '''X'''<sub>AB</sub> simply locates the origin of frame B relative to frame A, and so cannot include rotation of frame B.
 
Taking a time derivative, the velocity of the particle is:
 
:<math> \frac {d \mathbf{x}_\mathrm{A}}{dt} =\frac{d \mathbf{X}_\mathrm{AB}}{dt} + \sum_{j=1}^3 \frac{dx_j}{dt} \mathbf{u}_j + \sum_{j=1}^3 x_j \frac{d \mathbf{u}_j}{dt} \ . </math>
Il primo termine è la velocità con cui si sposta l'origine di B ('''v'''<sub>AB</sub>). Il secondo termine è la velocità del punto materiale, cioè '''v'''<sub>B</sub> nel sistema di riferimento B, quindi possiamo scrivere:
 
The second term summation is the velocity of the particle, say '''v'''<sub>B</sub> as measured in frame B. That is:
 
:<math> \frac {d \mathbf{x}_\mathrm{A}}{dt} =\mathbf{v}_\mathrm{AB}+ \mathbf{v}_\mathrm{B} + \sum_{j=1}^3 x_j \frac{d \mathbf{u}_j}{dt}. </math>
L'interpretazione di questa equazione è che la velocità del punto materiale vista dall'osservatore in A considte di quella che l'osservatore in B chiama velocità, cioè '''v'''<sub>B</sub>, più due termini aggiuntivi uno dovuto alla velocità dell'origine
e l'altro dovuto alla rotazione del sistema di riferimento, l'effetto di quest'ultimo termine è tanto più grande quanto il punto materiale è lontano dall'origine in B.
== Accelerazione relativa ==
Per ottenere l'accelerazione bisogna fare una ulteriore derivata nel tempo:
:<math> \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2} = \mathbf{a}_\mathrm{AB}+\frac {d\mathbf{v}_\mathrm{B}}{dt} + \sum_{j=1}^3 \frac {dx_j}{dt} \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}=\mathbf{a}_\mathrm{AB}+\frac {d\mathbf{v}_\mathrm{B}}{dt} + \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}. </math>
 
Usando la stessa formula già usata per la derivata temporale di '''x'''<sub>B</sub>, le derivata della velocità ('''v'''<sub>B</sub>) in forma esplicita diviene:
The interpretation of this equation is that the velocity of the particle seen by observers in frame A consists of what observers in frame B call the velocity, namely '''v'''<sub>B</sub>, plus two extra terms related to the rate of change of the frame-B coordinate axes. One of these is simply the velocity of the moving origin '''v'''<sub>AB</sub>. The other is a contribution to velocity due to the fact that different locations in the non-inertial frame have different apparent velocities due to rotation of the frame; a point seen from a rotating frame has a rotational component of velocity that is greater the further the point is from the origin.
 
To find the acceleration, another time differentiation provides:
 
:<math> \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2} = \mathbf{a}_\mathrm{AB}+\frac {d\mathbf{v}_\mathrm{B}}{dt} + \sum_{j=1}^3 \frac {dx_j}{dt} \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}. </math>
 
Using the same formula already used for the time derivative of '''x'''<sub>B</sub>, the velocity derivative on the right is:
 
:<math>\frac {d\mathbf{v}_\mathrm{B}}{dt} =\sum_{j=1}^3 \frac{d v_j}{dt} \mathbf{u}_j+ \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} =\mathbf{a}_\mathrm{B} + \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt}. </math>
Di conseguenza
 
:<math> \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2}=\mathbf{a}_\mathrm{AB}+\mathbf{a}_\mathrm{B} + 2\ \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}.</math>
{{anchor|Eq. 1}}Consequently,
Moltiplicando per la massa si ha che:
:{{NumBlk|:|<math> \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2}=\mathbf{a}_\mathrm{AB}+\mathbf{a}_\mathrm{B} + 2\ \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}.</math>|{{EquationRef|1}}}}
 
The interpretation of this equation is as follows: the acceleration of the particle in frame A consists of what observers in frame B call the particle acceleration '''a'''<sub>B</sub>, but in addition there are three acceleration terms related to the movement of the frame-B coordinate axes: one term related to the acceleration of the origin of frame B, namely '''a'''<sub>AB</sub>, and two terms related to rotation of frame B. Consequently, observers in B will see the particle motion as possessing "extra" acceleration, which they will attribute to "forces" acting on the particle, but which observers in A say are "fictitious" forces arising simply because observers in B do not recognize the non-inertial nature of frame B.
 
The factor of two in the Coriolis force arises from two equal contributions: (i) the apparent change of an inertially constant velocity with time because rotation makes the direction of the velocity seem to change (a ''d'''''v'''<sub>B</sub>/d''t'' term) and (ii) an apparent change in the velocity of an object when its position changes, putting it nearer to or further from the axis of rotation (the change in <math>\sum x_j\, d\mathbf{u}_j/dt</math> due to change in ''x <sub>j</sub>'' ).
 
To put matters in terms of forces, the accelerations are multiplied by the particle mass:
 
:<math>\mathbf{F}_\mathrm{A} = \mathbf{F}_\mathrm{B} + m\mathbf{a}_\mathrm{AB}+ 2m \sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} + m \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}\ . </math>
== Forze apparenti==
La forza osservata nel riferimento B, '''F'''<sub>B</sub> = ''m'''''a'''<sub>B</sub> è dovuta alla forza reale , '''F'''<sub>A</sub>, da:
 
:<math>\mathbf{F}_\mathrm{B} = \mathbf{F}_\mathrm{A} + \mathbf{F}_{\mbox{apparenti}},</math>
The force observed in frame B, '''F'''<sub>B</sub> = ''m'''''a'''<sub>B</sub> is related to the actual force on the particle, '''F'''<sub>A</sub>, by
 
dove:
:<math>\mathbf{F}_\mathrm{B} = \mathbf{F}_\mathrm{A} + \mathbf{F}_{\mbox{fictitious}},</math>
 
:<math> \mathbf{F}_{\mbox{apparenti}} = -m\mathbf{a}_\mathrm{AB} - 2m\sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} - m \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}\ . </math>
where:
La prima forza apparente è dovuta all'accelerazione dell'origine di B: <math>-ma_{AB}</math>.
Il secondo termine è la cosidetta accelerazione di Coriolis (il fattore due deriva da due contributi diversi come ricavato con la derivazione analitica)
Il terzo termine contiene sia la accelerazione centrifuga che l'eventuale accelerazione angolare.
 
La seconda legge della dinamica vale anche per le forze apparenti, che possono essere considerate forze a tutti gli effetti.
:<math> \mathbf{F}_{\mbox{fictitious}} = -m\mathbf{a}_\mathrm{AB} - 2m\sum_{j=1}^3 v_j \frac{d \mathbf{u}_j}{dt} - m \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}\ . </math>
 
Alcuni casi particolari permettono di esplicitare meglio le cose.
Thus, we can solve problems in frame B by assuming that Newton's second law holds (with respect to quantities in that frame) and treating '''F'''<sub>fictitious</sub> as an additional force.<ref name="Arnolʹd"/><ref name=Taylor>{{cite book |title=Classical Mechanics |author=John Robert Taylor |location=Sausalito CA |isbn=1-891389-22-X |year=2004 |url=http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn=1-891389-22-X#PPA343,M1 |publisher=University Science Books|pages=343–344}}</ref><ref>Kleppner, pages 62-63</ref>
==Sistema di riferimento ruotante==
 
Una situazione comune è quando il sistema di riferimento ruota. A causa di tale rotazione il sistema di riferimento B non è inerziale, dovuto al fatto che per avere rotazione è necesaria una accelerazione, quindi in questo caso se ci si mette nel riferimento in rotazione sono sempre presenti forze apparenti.
Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on [[centrifugal force]].
 
=== Rotating coordinate systems ===
A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Because such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved.
 
To derive expressions for the fictitious forces, derivatives are needed for the apparent time rate of change of vectors that take into account time-variation of the coordinate axes. If the rotation of frame ''B'' is represented by a vector '''Ω''' pointed along the axis of rotation with orientation given by the [[right-hand rule]], and with magnitude given by
 
Per derivare l'espressione delle forze apparenti, è necessario esplicitare le derivate dei versori delle coordinate del sistema in rotazione. Se la rotazione del sistema ''B'' è rappresentata da un vettore '''Ω''' che punta lungo l'asse di rotazione con direzione determinata dalla [[w:Regola_della_mano_destra|regola della mano destra]] e con ampiezza data da:
:<math> |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t), </math>
Allora la derivata prima temporale dei tre versori che descrivono il sistema ''B'' è:
 
then the time derivative of any of the three unit vectors describing frame ''B'' is<ref name=Taylor/><ref>See for example, {{cite book |author=JL Synge & BA Griffith |title=Principles of Mechanics |publisher=McGraw-Hill |edition=2nd Edition |year=1949 |pages=348–349}}</ref>
 
:<math> \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t), </math>
La derivata seconda temporale è:
 
and
 
:<math>\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j +\boldsymbol{\Omega} \times \frac{d \mathbf{u}_j (t)}{dt} = \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \left[ \boldsymbol{\Omega} \times \mathbf{u}_j (t) \right], </math>
abbiamo usato le regole del prodotto vettoriale. Queste derivate sono sostituite
 
nella espressione finale dell'ultima sezione ponendo '''a'''<sub>AB</sub> = 0 (escludendo traslazione dell'origine e ponendo l'accento sulla sola rotazione):
as is verified using the properties of the [[vector cross product]]. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω(''t''). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting '''a'''<sub>AB</sub> = 0 to remove any translational acceleration, and focusing on only rotational properties (see [[#Eq. 1|Eq. 1]]):
 
:<math> \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2}=\mathbf{a}_\mathrm{B} + 2\sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2},</math>
:<math>\mathbf{a}_\mathrm{A} = \mathbf{a}_\mathrm{B} +\ 2\sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t) + \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j \ + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left[ \boldsymbol{\Omega} \times \mathbf{u}_j (t) \right]</math>
::<math>=\mathbf{a}_\mathrm{B} + 2 \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) + \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j + \boldsymbol{\Omega} \times \left[\boldsymbol{\Omega} \times \sum_{j=1}^3 x_j \mathbf{u}_j (t) \right].</math>
Riunendo i termini, risulta:
Collecting terms, the result is the so-called ''acceleration transformation formula'':<ref name=Gregory>{{cite book |title=Classical Mechanics: An Undergraduate Text |url=http://books.google.com/books?id=uAfUQmQbzOkC&printsec=frontcover&dq=%22rigid+body+kinematics%22#PRA1-PA475,M1
|author=R. Douglas Gregory |year=2006 |isbn=0-521-82678-0 |publisher=Cambridge University Press |location=Cambridge UK |pages=Eq. (17.16), p. 475}}</ref>
 
:<math>\mathbf{a}_A=\mathbf{a}_B + 2\boldsymbol{\Omega} \times\mathbf{v}_\mathrm{B} + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_\mathrm{B} + \boldsymbol{\Omega} \times \left(\boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .</math>