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[[Fisica_classica/DinamicaEnergia_e_lavoro| Argomento precedente: DinamicaEnergia e lavoro]]
Nei sistemi di riferimento non inerziali le leggi della dinamica sono modificate e si manifestano delle forze che vengono chiamate fittizie.
La forza '''F''' non proviene 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, mentre le leggi del molto pssono varie 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. Per la [[Fisica_classica/Dinamica#Seconda_legge_della_dinamica_.28detta_anche_II_legge_di_Newton.29|II legge della dinamica]] nella forma ''F'''&nbsp;= ''m'''''a''', le forze apparenti sono proporzionalì alla accelerazione ''a''.
Le forze derivano il loro nome dall'azione dei muscoli del corpo umano. Le forze possono essere moltiplicate o divise, cambiate di direzione mediante varie macchine inventate fin dagi albori della civiltà. Un esempio di macchina semplice è
la [[w:Leva_(fisica)|leva]] che permette di sollevare pesi che non sarebbe possibile sollevare con i muscoli umani. Esiste una altra grandezza fisica che introduciamo in questa parte del libro che ha un ruolo essenziale in tutta la fisica l''''energia''', tale grandezza scalare, connessa al concetto di forza, è possibile trasformare, ma non è possibile moltiplicarla. Viene affermato come legge generale della natura che l'energia si trasforma nelle sue varie forme, ma non è possibile nè crearla nè distruggerla.
 
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 variabile, 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]]);
=Lavoro di una forza=
4) sistema in rotazione velocità angolare variabile.
 
La più semplice forma di energia è il lavoro fatto dalle forze.
Una forza è detta fare un lavoro quando agendo su un corpo ne provoca uno spostamento del punto di applicazione nella direzione della forza.
Il termine lavoro fu introdotto da [[w:Gaspard_Gustave_de_Coriolis|Gustave di Coriolis]] descrivendo l'azione di innalzare un peso ad una certa altezza, che era in effetti il lavoro fatto dalle prime macchine a vapore per innalzare secchi di acqua nelle miniere.
 
La unità di misura del lavoro nel [[w:Sistema_internazionale_di_unità_di_misura|Sistema Internazionale]] è il newton x metro o [[w:Joule|joule]] (J). Questa unità di misura è anche l'unità di misura di tutte le forme di energia. Nelle grandezze elemantari MLT del sistema internazionale le dimensioni fisiche dell'energia sono <math>[M][L]^2[T]^{-2}</math> essendo le forze delle <math>[M][L][T]^{-2}</math>.
 
==Fictitious forces on Earth==
Il lavoro fatto da una forza costante di grandezza ''F'' su un punto che si muove compiendo uno spostamento ''d'' nella direzione della forza è semplicemente il prodotto:
{{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==
:<math>W = Fd.</math>
{{See also|Inertial frame of reference}}
Ad esempio, se una forza di 100 newton (''F'' = 100 N) agisce su un punto materiale che percorre
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
2 metri (''d'' = 2 m), nella direzione della forza, essa compie un lavoro
|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
''W'' = (100 N)x(2 m) = 200 N m = 200 J. Questo è approssimativamente il lavoro che si fa alzando
|author= Vladimir Igorevich Arnold |publisher=Springer |location=Berlin |isbn=0-387-96890-3 |year=1989 |pages=§27 pp. 129 ff.}}</ref>
una massa di 10&nbsp;kg da terra a sopra la testa di una persona.
{{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''.}}
Poichè le dimensioni fisiche del lavoro sono eguali a quelle del momento meccanico (che vedremo nel seguito) che ha un ben diverso significato, viene sconsigliato di misurare il lavoro in newton x metro ma si consiglia di usare il joule.
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>
[[File:Mehaaniline_töö.png|thumb|250px|Lavoro di una forza]]
 
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.
Nel caso più generale consideriamo una forza risultante, <math>\vec F\ </math>, che agendo su un punto materiale ne provochi uno spostamento <math>d\vec s\ </math>, il prodotto [[w:Prodotto_scalare|prodotto scalare]]:
:<math>dW=\vec F \cdot d\vec s=F \cos \alpha ds= F_T ds</math>
viene definito lavoro infinitesimo delle forza risultante. Avendo indicato con <math>\alpha\ </math> l'angolo tra la forza e lo spostamento e con <math>F_T\ </math> la componente tangenziale della forza lungo la traiettoria. I casi in cui il lavoro è nullo sono quelli dove non agisce nessuna forza, oppure la risultante delle forze è perpendicolare alla traiettoria, così che <math>\cos \alpha = 0 \,\!</math>. Se invece vi è una componente della forza nella direzione dello spostamento, il lavoro fatto è diverso da zero. Il lavoro è positivo se provoca un aumento della velocità, mentre è negativo se provoca una diminuzione della velocità (come è il caso dell'attrito dinamico e di quello viscoso).
Nel caso più generale di un punto materiale che si muove su di una traiettoria curvilinea, il lavoro è dato dall'integrale di linea del lavoro infinitesimale e quindi se il punto si sposta dal punto A al punto B possiamo scrivere:
:<math>W=\int_A^B \vec F \cdot d\vec s</math>
 
==Examples of fictitious forces==
Dal punto di vista lessicale il lavoro fatto da una forza non è posseduto, ma scambiato tra sistemi.
===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.
 
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.
== [[w:Potenza_(fisica)|Potenza]] di una forza==
La potenza istantanea corrisponde al lavoro per unità di tempo:
:<math>P=\frac{dW}{dt}=\vec F \cdot \frac {d\vec s}{dt}=\vec F \cdot \vec v=F_T v</math>
Cioè esprime in qualche maniera quanto velocemente viene erogato il lavoro. Tale grandezza serve a quantificare le prestazioni sia nel lavoro umano o animale che nelle macchine. La potenza ha le dimensioni di una energia diviso un tempo. La sua unità di misura è il watt che ha come simbolo '''W'''. Il concetto di potenza è ben noto dagli albori della civiltà e veniva quantizzato dalla potenza dei cavalli da cui deriva l'unità di misura ora obsoleta il [[w:Cavallo_vapore|cavallo vapore]] (simbolo hp) che corrisponde a 735 W.
 
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.
Estendendo l'esempio precedente di una massa di 10 kg alzata da terra fino a sopra la testa.
Se tale azione viene svolta in 10 s, la potenza necessaria per eseguirla è di 20 W: facile
per una persona anche non allenata. Se invece tale lavoro viene eseguito in 1 s, la potenza
necessaria diviene 200 W: richiede una persona giovane e ben allenata.
 
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.)
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
= Energia Cinetica =
!''Animation:'' driving from block to block
L'energia cinetica di un corpo materiale di massa <math>m</math> con velocità <math>v</math> è dato da:
|-
:<math>E_k = \frac 12 mv^2.</math>
|[[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.]]
Le dimensioni fisiche di tale quantità sono quelle di una energia: <math>[M][L]^2[T]^{-2}</math> (J). La giustificazione dell'espressìone della energia cinetica si ricava dall'analisi di quello che succede se una forza agendo su un corpo di massa <math>m</math> con velocità iniziale <math>v_o</math> ne varia la velocità portandola fino <math>v_f</math> lungo una traiettoria descritta dal tratto infinitesimo <math>d\vec s</math>:
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.
:<math>dW= F_T ds=ma_Tds=m\frac {dv}{dt}ds=m\frac {ds}{dt}dv</math>
|}
Integrando tale differenziale si ha che il collegamento tra lavoro e variazione di energia cinetica:
:<math>W= \int_o^fmvdv=\frac 12mv_f^2-\frac 12mv_o^2=\Delta E_k</math>
Il simbolo <math>\Delta</math> indica la differenza tra l'energia cinetica finale e quella iniziale prodotta dal lavoro fatto dalle forze esterne, qualsiasi sia la loro natura.
Se il lavoro è positivo l'energia cinetica aumenta, se il lavoro è negativo l'energia cinetica diminuisce. Notiamo che se le forze agiscono in direzione perpendicolare alla traiettoria (forze centripete) il lavoro fatto è nullo e l'energia cinetica non varia.
 
===Circular motion===
La relazione tra il lavoro fatto dalla risultante delle forze agenti su un corpo e la variazione di energia cinetica prende il nome di [[w:Teorema_dell'energia_cinetica|teorema del lavoro]], tale teorema vale per qulasiasi tipo di forze anche quelle variabili con il tempo o con la posizione, ma solo per sistemi a massa costante.
{{See also|Centrifugal force|Reactive centrifugal force|Coriolis force}}
 
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]]):
Vi è una relazione tra l'energia cinetica e la [[Fisica_classica/Dinamica#Quantità di Moto|quantità di moto]] ricordando che <math>\vec p = m\vec v.</math> :
:<math>E_k = \frac {p^2}{2m}\qquad p = \sqrt {2mE_k}</math>
 
# 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]].)
L'energia cinetica al contrario del lavoro è una proprietà che viene posseduta dal punto materiale, ma che possiamo associare anche ai sistemi.
# 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.
=Energia potenziale=
== Lavoro della forza peso==
[[File:Work_of_gravity_F_dot_d_equals_mgh.JPG|thumb|250px|Lavoro di una forza costante verticale: ad esempio la forza peso]]
 
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.
Nel caso di forze costanti il calcolo del lavoro
totale è semplicemente dato da (riferendosi alla figura a fianco):
:<math>W=\int_o^f \vec F \cdot d\vec s=\vec F \cdot \int_o^f d\vec s=\vec F\cdot \vec d =Fh</math>
essendo <math>\vec F</math> dietto lungo la verticale.
[[File:PotentialEnergy_Gravitational2.png|thumb|350px|Forza peso e sua energia potenziale]]
Nel caso specifico della [[Fisica_classica/Dinamica#Forza peso|forza peso]] in cui <math>F=-mg</math> e quindi il lavoro fatto è:
:<math>W=-mgh</math>
Notiamo come il lavoro non dipenda dalla traiettoria, ma solo dalla differenza di quota (positivo se si diminuisce la quota e negativo nel caso opposto9.
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
Definendo <math>h_o</math> e <math>h_f</math> le quote iniziali e finali il teorema del lavoro nel caso della sola forza peso diviene:
!''Animation:'' object released from a carousel
:<math>W= -mg(h_f-h_o)=\frac 12mv_f^2-\frac 12mv_o^2</math>
|-
:<math>mgh_o+\frac 12mv_o^2=mgh_f+\frac 12mv_f^2</math>
|[[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.]]
Se definiamo <math>E_p=mgz</math> (energia potenziale gravitazionale) possiamo riscrivere:
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.
:<math>E_{po}+\frac 12mv_o^2=E_{pf}+\frac 12mv_f^2</math>
|}
Che esprime la conservazione dell'energia meccanica se agiscono solo forze gravitazionali, in quanto la somma dell'energia cinetica e di quella potenziale gravitazionale sono costanti nel tempo.
 
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>
Ritornando alla equazione esplicita si ha che:
:<math>v_f=\sqrt{2g(h_o-h_f)+v_o^2}</math>
Tale equazione viene scritta senza interessarsi della cinematica dell'oggetto, nell'ipotesi che la sola forza agente che compie lavoro meccanico sia la forza di gravità.
 
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.
== Lavoro di una forza elastica==
[[File:PotentialEnergy_Elastic.png|thumb|350px|Grafico della forza elastica e della sua energia potenziale nel caso unidimensionale]]
Consideriamo la [[Fisica_classica/Dinamica#Forza elastica|forza elastica]] nel caso unidimensionale assunta come origine la posizione di equilibrio:
:<math>F =- k x</math>
Il lavoro per andare dalla posizione iniziale <math>x_o</math> a quella finale <math>x_f</math> è:
:<math>W=\int_o^f F dx=-k\int_o^f xdx=\frac 12kx_o^2-\frac 12kx_f^2</math>
Per il teorema del lavoro:
:<math>W= \frac 12kx_o^2-\frac 12kx_f^2=\frac 12mv_f^2-\frac 12mv_o^2</math>
Quindi:
:<math>\frac 12kx_o^2+\frac 12mv_o^2=\frac 12kx_f^2+\frac 12mv_f^2</math>
Se definiamo <math>E_p=\frac 12kx^2</math> (energia potenziale elastica):
:<math>E_{po}+\frac 12mv_o^2=E_{pf}+\frac 12mv_f^2</math>
Anche in questo caso la energia meccanica si conserva. Se consideriamo [[Fisica_classica/Cinematica#Moto_armonico_semplice|moto armonico]], quando l'elongazione è massima l'energia potenziale è massima ed l'energia cinetica è nulla, mentre quando il sistema passa per la posizione di equilbrio la energia potenziale è nulla e tutta l'energia diviene cinetica.
 
=== Fictitious forces and work ===
== Lavoro della forza di attrito dinamico==
La [[Fisica_classica/Dinamica#Forza_di_attrito_dinamico|forza di attrito dinamico]] è diretta nella direzione opposta alla velocità e quindi dello spostamento <math>d\vec s</math>. Quindi il lavoro che si compie andando dalla posizione iniziale a quella finale è pari a:
:<math>W=\int_o^f \vec F \cdot d\vec s=-\mu_d N \int_o^f d\vec s=-\mu_d N \ell</math>
dove <math>\ell</math> è il cammino totale percorso. A differenza dei casi precedenti la posizione finale ed iniziale non bastano per caratterizzare il lavoro svolto, anzi per stesse posizioni iniziali e finali il lavoro, sempre negativo, può assumere valori molto diversi.
In questo caso l'energia meccanica non si conserva in quanto via via che viene percorsa la traiettoria l'energia cinetica diminiusce senza aumentare una qualche forma di energia potenziale. In realtà l'energia meccanica viene trasformata in calore che è un'altra forma di energia (non meccanica). Quindi il punto materiale ha una energia cinetica iniziale e via la perde per attrito, fino a fermarsi quando inizia a dominare la forza di attrito statico.
 
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.
Il teorema del lavoro diviene in questo caso:
:<math>W= -\mu_d N\ell =\frac 12mv_f^2-\frac 12mv_o^2</math>
:<math>\frac 12mv_f^2=\frac 12mv_o^2-\mu_d N\ell</math>
 
==Gravity as a fictitious force==
==Forze conservative==
{{Main|General relativity}}
[[File:Konservative_Kraft_Wege.png|thumb|250px|Due cammini diversi tra il punto 1 ed il punto 2]]
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
I tre casi esaminati descrivono dei casi che si possono generalizzare, infatti mentre il lavoro della forza peso e di quella elastica non dipendono dal percorso seguito per andare dalla posizione iniziale a quella finale, nel terzo caso, quello relativo alla forza di attrito, il lavoro dipende dal percorso. Le forze in cui il lavoro non dipende dal percorso seguito vengono dette '''forze conservative'''. Nelle figura a fianco vengono mostrati due qualsiasi cammini diversi che collegano i punti 1 e 2. Se:
|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
:<math>W=\int_{1}^2(\vec F \cdot ds)_{S1}= \int_{1}^2(\vec F \cdot ds)_{S2}</math>
|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]].
la forza è conservativa.
 
:{| class="toccolours collapsible collapsed" width="60%" style="text-align:left"
Inoltre poichè:
!''Animation:'' ball that rolls off a cliff
:<math>-\int_{1}^2(\vec F \cdot ds)_{S2}=\int_2^1(\vec F \cdot ds)_{S2}</math>
|-
segue che:
|[[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.
:<math>\oint \vec F \cdot ds=\int_{1}^2(\vec F \cdot ds)_{S1}+\int_2^1(\vec F \cdot ds)_{S2}=0
|}
</math>
Cioè nelle forze conservative l'integrale su una qualsiasi linea chiusa è nullo.
 
==Mathematical derivation of fictitious forces {{anchor|coordinates}}==
Per una forza conservativa, il lavoro si può scrivere come differenza di una funzione della coordinata spaziale calcolata nei punti di partenza e di arrivo detta energia potenziale:
[[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> ).]]
:<math>W=\int_o^f\vec F \cdot ds=E(r_f)-E(r_o)</math>
L'energia potenziale esprime la capacità di compiere lavoro, al diminuire dell'energia potenziale viene compiuto del lavoro che può aumentare l'energia cinetica. Se invece l'energia potenziale aumenta significa che bisogna fornire o lavoro dall'esterno o consumare energia cinetica.
L'energia potenziale è definita a meno di una costante additiva, che viene scelta per convenienza ad esempio nel caso della forza peso si è assunto che l'energia potenziale sia nulla a livello del mare e quindi l'espressione della energia potenziale è:
:<math>-W=\int_0^zmgdz'=E_p(z)</math>
:<math>E_p=mgz</math>
Cioè l'energia potenziale è eguale al lavoro combiato di segno per andare da quota 0 a quota
z. Ricordando che la forza peso è diretta verso il basso.
 
=== General derivation ===
Analogamente per la forza elastica assunto come costante additiva quella che annulla la energia potenziale nella posizione di riposo:
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
:<math>-W=\int_0^xkxdx=E_p(x)</math>
|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>
:<math>E_p=\frac 12kx^2</math>
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>
In forma differenziale:
:<math>dE_p=-dW=-\vec F\cdot d\vec s</math>
la forma differenziale è indipendente dal punto di riferimento esseno una differenza di energia potenziale.
 
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
Tutte le forze che dipendono dalla velocità, dal tempo o dalla lunghezza del percorso non sono conservative e per esse non è possibile definire una energia potenziale.
 
:<math> \mathbf{x}_\mathrm{B} = \sum_{j=1}^3 x_j \mathbf{u}_j \ . </math>
Se agiscono solo forze conservative l'energia meccanica totale determinata dalla energia cinetica e quella potenziale si conserva, cioè:
:<math>E_k+E_p=costante</math>
 
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>
==Dall'energia potenziale alla forza==
Se conosciamo in una regione di spazio l'energia potenziale di un punto materiale possiamo ricavare le forze che agiscono sul punto materiale con una operazione che l'inverso di quella che abbiamo usata per definire l'energia potenziale. Tale relazione si ottiene
esplicitando la relazione differenziale che collega l'energia potenziale al lavoro in coordinate cartesiane:
:<math>dE_p=\frac{\partial E_p}{\partial x}dx+\frac{\partial E_p}{\partial y}dy+\frac{\partial E_p}{\partial z}dz=-F_xdx-F_ydy-F_zdz</math>
Da cui:
:<math>F_x=-\frac{\partial E_p}{\partial x}\quad ,F_y=-\frac{\partial E_p}{\partial y},\quad F_z=-\frac{\partial E_p}{\partial z}</math>
 
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.
Un modo più compatto fa uso di un [[w:Operatore_(fisica)|operatore]] cioè una operazione matematica chiamata '''gradiente''' ed indicato con <math>\nabla=\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}</math>.
Questo operatore applicato ad uno scalare genera un vettore e quindi, in questo caso particolare, dal gradiente della funzione energia potenziale si ottengono le componenti cartesiane della forza in questione.
 
Taking a time derivative, the velocity of the particle is:
:<math>\vec F =- \vec{grad} E_p=-\vec \nabla E_p</math>
 
:<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>
= Momenti =
 
The second term summation is the velocity of the particle, say '''v'''<sub>B</sub> as measured in frame B. That is:
Introduciamo ora il concetto di momento di un vettore.
Definiamo come '''momento del vettore''' <math>\vec v</math> applicato in un punto '''P ''' ad una certa distanza da un punto '''O''' il vettore
:<math>\vec M=\vec r \times \vec v</math>
definendo <math>\vec r=\vec {OP}</math>. Il modulo è dato da :
:<math>M=rv \sin \theta = v d \,\!</math>
 
:<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>
dove <math>\theta \,\!</math> è l'angolo formato dalla direzione del vettore <math>\vec v</math> con la direzione di <math>\vec r</math> e quindi '''d''' non è altro che la distanza tra il punto '''O''' e la retta (direttrice) su cui giace il vettore <math>\vec v</math> e viene chiamato '''braccio'''.
 
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.
Facciamo notare come il modulo, essendo dipendente da '''d''' e non da '''r''', non dipende dal punto in cui viene applicato il vettore <math>\vec v</math> lungo la sua direttrice.
I momenti hanno una importanza fondamentale nella dinamica dei sistemi, ma si possono definire anche nella dinamica del punto materiale per quanto riguarda sia il '''momento angolare''' ed il '''momento di una forza'''
 
To find the acceleration, another time differentiation provides:
== Momento angolare ==
[[File:Torque animation.gif|frame|right|Relazione tra forza (F), momento della forza (τ), quantità di moto (p), e momento angolare (L) in un sistema ruotante]]
Il momento angolare <math>\vec L</math> è definito come il prodotto vettoriale tra il vettore posizione <math>\bar r</math> (rispetto alla stessa origine) e il vettore quantità di moto <math>\vec p</math>:
 
:<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>
:<math> \vec L = \vec r \times \vec p </math>
 
Using the same formula already used for the time derivative of '''x'''<sub>B</sub>, the velocity derivative on the right is:
Il [[w:modulo_(algebra)|modulo]] di <math>\vec L</math> è quindi definito da:
 
:<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>
:<math>{L} = {{r} {mv}}\sin{\theta}=mvd\, </math>
 
{{anchor|Eq. 1}}Consequently,
La direzione di <math>\vec L</math> è perpendicolare al piano definito da <math>\vec v</math> e da <math>\vec r</math>; il verso è quello di un osservatore che vede ruotare <math>\vec v</math> in senso antiorario. La grandezza <math>d={r}\sin{\theta}</math> è il ''braccio'' di <math>\vec p</math>.
:{{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.
Se <math>\vec v</math> e <math>\vec r</math> sono tra loro perpendicolari il momento angolare è massimo e questo avviene quando <math>\sin \theta = 1</math>. Il momento angolare è nullo invece se la quantità di moto o il braccio sono nulli, oppure se <math>\vec v</math> è parallelo ad <math>\vec r</math>, in tal caso infatti <math>\sin \theta = 0</math>.
 
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>'' ).
Ricordando come la velocità istantanea abbia una componente radiale <math>\vec v_r</math> ed una angolare <math>\vec v_\theta</math> in [[Fisica_classica/Cinematica#Velocit.C3.A0_in_coordinate_polari|coordinate polari]]
Quindi essendo
:<math>\vec L=\vec r \times m \vec v</math>
sostituendo a <math>\vec v=v_r\hat u_r+v_{\theta}\hat u_{\theta}</math> si ha che:
:<math>\vec L=\vec r \times m(v_r\hat u_r+v_{\theta}\hat u_{\theta})=mrv_\theta\hat u_r \times \hat u_{\theta}</math>
In modulo:
:<math>L=mr^2\frac {d\theta}{dt}</math>
 
To put matters in terms of forces, the accelerations are multiplied by the particle mass:
Si definisce '''momento angolare assiale''' il momento angolare proiettato su un asse passante per il polo.
 
:<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>
La dimensione fisica del momento angolare è [M}[L}<sup>2</sup>[T]<sup>-1</sup> e quindi nel [[w:Sistema internazionale di unità di misura|SI]] si misura in kg·m²/s. La sue dimensione coincide con quellea dell'[[w:Azione (fisica)|azione]] (ovvero di un'energia per un tempo), ma il significato fisico di azione e momento angolare sono completamenti differenti. Inoltre il momento angolare è una grandezza vettoriale mentre l'azione è uno scalare.
 
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
 
:<math>\mathbf{F}_\mathrm{B} = \mathbf{F}_\mathrm{A} + \mathbf{F}_{\mbox{fictitious}},</math>
== Momento della forza ==
 
where:
Il momento di una forza ha l'espressione:
:<math>\vec {\tau}= \vec r \times \vec F</math>
e possiamo notare che se vi sono più forze applicate la cui risultante è <math>\vec R</math> in un punto, si ha che:
:<math>\vec {\tau} = \vec r \times \vec R</math>
 
:<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>
Se consideriamo la variazione del momento angolare nel tempo allora possiamo scrivere:
:<math>\frac{d \vec L}{dt} = \frac {d \vec r}{dt} \times m \vec v + \vec r \times m \frac{d \vec v}{dt}</math>
 
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>
Nel caso che il polo '''O''' sia fermo il primo termine è nullo.
Il secondo termine coincide con la forza applicata moltiplicata vettorialmente per la distanza dal punto '''O'''. Ricaviamo così che
:<math>\frac{d \vec L}{dt}= \vec {\tau}</math>.
 
Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on [[centrifugal force]].
Inoltre è importante notare il caso in cui la forza sia applicata lungo la stessa direttice di <math>\vec r</math> allora <math>\vec {\tau}=0</math> e di conseguenza
:<math>\frac{d \vec L}{dt}=0</math>
e quindi
:<math>\vec L = costante</math>.
 
=== Rotating coordinate systems ===
Le dimensione fisica del momento di una forza sono [M}[L}<sup>2</sup>[T]<sup>-2</sup> e quindi nel [[w:Sistema internazionale di unità di misura|SI]] si misura in kg·m²/s². La dimensione coincide con quella dell'energia, ma qui è evidente la differenza. Inoltre il momento di una forza è una grandezza vettoriale mentre l'energia è uno scalare.
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
==Lavoro del momento di una forza==
Nel moto circolare l'espressione del lavoro:
:<math>W=\int_A^B \vec F \cdot d\vec s</math>
essendo <math>ds=Rd\theta</math> si riduce a:
:<math>W=\int_{\theta_A}^{\theta_B} rF_rd\theta=\int_{\theta_A}^{\theta_B} \tau d\theta</math>
 
:<math> |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t), </math>
[[File:ArealVelocity.svg|thumb|left|300px|alt=|L'area percorsa in giallo è eguale al triangolo formato dai lati <math>r(t)</math> e <math>r(t+dt)=rd\theta </math> .]]
 
==Forze centrali==
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>
Una forza è detta centrale di centro O se in ogni punto escluso il centro la direzione della forza passa per il centro medesimo ed Il modulo è unicamente funzione della distanza tra il punto e il centro:
 
:<math> \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t), </math>
 
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>
 
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>
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>
 
The [[proper acceleration|physical acceleration]] '''a'''<sub>A</sub> due to what observers in the inertial frame A call ''real external forces'' on the object is, therefore, not simply the acceleration '''a'''<sub>B</sub> seen by observers in the rotational frame B, but has several additional geometric acceleration terms associated with the rotation of B. As seen in the rotational frame, the acceleration '''a'''<sub>B</sub> of the particle is given by rearrangement of the above equation as:
:<math>
\mathbf{a}_\mathrm{B} = \mathbf{a}_\mathrm{A} - 2\boldsymbol{\Omega} \times \mathbf{v}_\mathrm{B} - \boldsymbol{\Omega} \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B}) - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}.
\vec F = F(r)\hat{\mathbf{r}}
</math>
Una forza centrale on dipende nè dal tempo nè da altre proprietà locali.
[[File:Keplero_legge_delle_aree.svg|thumb|left|300px|alt=|Esempio dell'applicazione della legge sulla costanza della velocità areolare al moto dei pianeti attorno al sole]]
Essendo <math>\vec r</math> e <math>\vec F</math> paralleli:
:<math>\vec {\tau}= 0\rightarrow \frac{d \vec L}{dt}=0</math>
Quindi il momento angolare <math>\vec L</math> è costante, inoltre per come è definito, è <math>\vec L</math> sia costante determina tra l'altro che il verso di percorrenza sia fissato.
Inoltre essendo:
:<math>\frac {dL}{dt}=0</math>
segue che:
:<math>r^2\frac {d^2\theta}{dt^2}=0</math>
Ma come si vede dalla figura subito sopra tale grandezza nulla corrisponde alla derivata della [[w:Velocità_areolare|velocità areolare]]. Si chiama velocità areolare una quantità pari alla derivata rispetto al tempo dell'area spazzata dal vettore posizione (raggio vettore) <math>\vec r</math>. In definitiva il fatto che il modulo di L sia costante comporta che nella dinamica in presenza di forze centrali la velocità areolare rimane costante.
 
The net force upon the object according to observers in the rotating frame is '''F'''<sub>B</sub> = ''m'''''a'''<sub>B</sub>. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force '''F'''<sub>fict</sub> is present, so the end result is '''F'''<sub>B</sub> = '''F'''<sub>A</sub> + '''F'''<sub>fict</sub>. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:
Se la traiettoria è chiusa essendo la derivata costante si ha che il rapporto tra l'area totale <math>A_t</math> (area all'interno della traiettoria) ed il periodo <math>T</math> vale:
 
:<math>\frac {A_t}T=\frac L2</math>
:<math>
\mathbf{F}_{\mathrm{fict}} = - 2 m \boldsymbol\Omega \times \mathbf{v}_\mathrm{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B}) - m \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}.
</math>
 
Here, the first term is the ''[[Coriolis force]]'',<ref name=Joos>{{cite book |title=Theoretical Physics|author= Georg Joos & Ira M. Freeman|url=http://books.google.com/books?id=vIw5m2XuvpIC&pg=PA233&dq=%22Coriolis+force%22+intitle:Theoretical+intitle:physics#PPA233,M1
|page=233 |isbn=0-486-65227-0 |publisher=Courier Dover Publications |location=New York|year=1986}}</ref> the second term is the ''[[centrifugal force (fictitious)|centrifugal force]]'',<ref name=Smith>{{cite book |title=Theoretical Mechanics|author=Percey Franklyn Smith & William Raymond Longley|url=http://books.google.com/books?id=qL4EAAAAYAAJ&pg=PA118&dq=%22centrifugal+force%22+intitle:theoretical|page=118 |year=1910|publisher=Gin|location=Boston}}</ref> and the third term is the ''[[Euler force]]''.<ref name=Lanczos>{{cite book |author=Cornelius Lanczos |title=The Variational Principles of Mechanics |url=http://books.google.com/books?id=ZWoYYr8wk2IC&pg=PA103&dq=%22Euler+force%22#PPA103,M1
|page=103|year=1986|publisher=Courier Dover Publications|isbn=0-486-65067-7|location=New York}}</ref><ref name=Marsden>{{cite book |author=Jerold E. Marsden & Tudor.S. Ratiu |year= 1999 |edition=2nd Edition |title= Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems: Texts in applied mathematics, 17 |publisher= Springer-Verlag|location = NY|url=http://books.google.com/books?id=I2gH9ZIs-3AC&printsec=frontcover&dq=intitle:Introduction+intitle:to+intitle:mechanics+intitle:and+intitle:symmetry#PPA251,M1 |isbn=0-387-98643-X|page= 251}}</ref> When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.
 
=== Orbiting coordinate systems ===
{{Unreferenced section|date=July 2008}}
[[Image:Orbiter.PNG|thumb|250px|Figure 3: An orbiting but fixed orientation coordinate system ''B'', shown at three different times. The unit vectors '''u'''<sub>j</sub>, j = 1, 2, 3 do ''not'' rotate, but maintain a fixed orientation, while the origin of the coordinate system ''B'' moves at constant angular rate ω about the fixed axis '''Ω'''. Axis '''Ω''' passes through the origin of inertial frame ''A'', so the origin of frame ''B'' is a fixed distance ''R'' from the origin of inertial frame ''A''.]]
As a related example, suppose the moving coordinate system ''B'' rotates in a circle of radius ''R'' about the fixed origin of inertial frame ''A'', but maintains its coordinate axes fixed in orientation, as in Figure 3. The acceleration of an observed body is now (see [[#Eq. 1|Eq. 1]]):
:<math> \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_{AB}+\mathbf{a}_{B} + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} </math>&ensp;<math>+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ . </math>
::<math>=\mathbf{a}_{AB}\ +\mathbf{a}_B\ , </math>
where the summations are zero inasmuch as the unit vectors have no time dependence. The origin of system ''B'' is located according to frame ''A'' at:
:<math>\mathbf{X}_{AB} = R \left( \cos ( \omega t) , \ \sin (\omega t) \right) \ ,</math>
leading to a velocity of the origin of frame ''B'' as:
:<math>\mathbf{v}_{AB} = \frac{d}{dt} \mathbf{X}_{AB} = \mathbf{\Omega \times X}_{AB} \ , </math>
 
leading to an acceleration of the origin of ''B'' given by:
:<math>\mathbf{a}_{AB} = \frac{d^2}{dt^2} \mathbf{X}_{AB} </math>&ensp;<math>= \mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) </math>&ensp;<math>= - \omega^2 \mathbf{X}_{AB} \ .</math>
Because the first term, which is
::::<math>\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right)\ , </math>
is of the same form as the normal centrifugal force expression:
::::<math>\boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ ,</math>
it is a natural extension of standard terminology (although there is no standard terminology for this case) to call this term a "centrifugal force". Whatever terminology is adopted, the observers in frame ''B'' must introduce a fictitious force, this time due to the acceleration from the orbital motion of their entire coordinate frame, that is radially outward away from the center of rotation of the origin of their coordinate system:
:<math>\mathbf{F}_{\mathrm{fict}} = m \omega^2 \mathbf{X}_{AB} \ , </math>
and of magnitude:
:<math>|\mathbf{F}_{\mathrm{fict}}| = m \omega^2 R \ . </math>
 
Notice that this "centrifugal force" has differences from the case of a rotating frame. In the rotating frame the centrifugal force is related to the distance of the object from the origin of frame ''B'', while in the case of an orbiting frame, the centrifugal force is independent of the distance of the object from the origin of frame ''B'', but instead depends upon the distance of the origin of frame ''B'' from ''its'' center of rotation, resulting in the ''same'' centrifugal fictitious force for ''all'' objects observed in frame ''B''.
 
===Orbiting and rotating===
[[Image:Center-facing orbiting coordinate system.PNG|thumb|250px|Figure 4: An orbiting coordinate system ''B'' similar to Figure 3, but in which unit vectors '''u'''<sub>j</sub>, j = 1, 2, 3 rotate to face the rotational axis, while the origin of the coordinate system ''B'' moves at constant angular rate ω about the fixed axis '''Ω'''.]]
As a combination example, Figure 4 shows a coordinate system ''B'' that orbits inertial frame ''A'' as in Figure 3, but the coordinate axes in frame ''B'' turn so unit vector '''u'''<sub>1</sub> always points toward the center of rotation. This example might apply to a test tube in a centrifuge, where vector '''u'''<sub>1</sub> points along the axis of the tube toward its opening at its top. It also resembles the Earth-Moon system, where the Moon always presents the same face to the Earth.<ref name=Newcomb>However, the Earth-Moon system rotates about its [[Barycentric coordinates (astronomy)|barycenter]], not the Earth's center; see {{cite book |author= Simon Newcomb |title=Popular Astronomy |page=307 |url=http://books.google.com/books?id=VS7aS8QS91oC&pg=PA307&dq=centrifugal+revolution+and+rotation+date:1970-2009|isbn=1-4067-4574-X |year=2007 |publisher=Read Books}}</ref> In this example, unit vector '''u'''<sub>3</sub> retains a fixed orientation, while vectors '''u'''<sub>1</sub>, '''u'''<sub>2</sub> rotate at the same rate as the origin of coordinates. That is,
:<math>\mathbf{u}_1 = (-\cos \omega t ,\ -\sin \omega t )\ ;\ </math>&ensp;<math>\mathbf{u}_2 = (\sin \omega t ,\ -\cos \omega t ) \ . </math>
:<math>\frac{d}{dt}\mathbf{u}_1 = \mathbf{\Omega \times u_1}= \omega\mathbf{u}_2\ ;</math>&ensp;<math> \ \frac{d}{dt}\mathbf{u}_2 = \mathbf{\Omega \times u_2} = -\omega\mathbf{u}_1\ \ .</math>
Hence, the acceleration of a moving object is expressed as (see [[#Eq. 1|Eq. 1]]):
:<math> \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_{AB}+\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} </math>&ensp;<math>+\ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ </math>
::<math>=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) +\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j\ \mathbf{\Omega \times u_j}</math>&ensp;<math> \ +\ \sum_{j=1}^3 x_j\ \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{u}_j \right)\ </math>
::<math>=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) + \mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ </math>&ensp;<math> \ +\ \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ </math>
::<math>=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times} (\mathbf{ X}_{AB}+\mathbf{x}_B) \right) + \mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ \ ,</math>
where the angular acceleration term is zero for constant rate of rotation.
Because the first term, which is
::::<math>\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times} (\mathbf{ X}_{AB}+\mathbf{x}_B) \right)\ , </math>
is of the same form as the normal centrifugal force expression:
::::<math>\boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ ,</math>
it is a natural extension of standard terminology (although there is no standard terminology for this case) to call this term the "centrifugal force". Applying this terminology to the example of a tube in a centrifuge, if the tube is far enough from the center of rotation, |'''X'''<sub>AB</sub>| = ''R'' >> |'''x'''<sub>B</sub>|, all the matter in the test tube sees the same acceleration (the same centrifugal force). Thus, in this case, the fictitious force is primarily a uniform centrifugal force along the axis of the tube, away from the center of rotation, with a value |'''F'''<sub>Fict</sub>| = ω<sup>2</sup> ''R'', where ''R'' is the distance of the matter in the tube from the center of the centrifuge. It is standard specification of a centrifuge to use the "effective" radius of the centrifuge to estimate its ability to provided centrifugal force. Thus, a first estimate of centrifugal force in a centrifuge can be based upon the distance of the tubes from the center of rotation, and corrections applied if needed.<ref name=Singh>{{cite book |title=Constitutive and Centrifuge Modelling: Two Extremes |author=Bea K Lalmahomed, Sarah Springman, Bhawani Singh |isbn=90-5809-361-1 |year=2002 |publisher=Taylor and Francis |page=82 |url=http://books.google.com/books?id=MJkz_IBZZS0C&printsec=frontcover&dq=centrifuge#PPT102,M1 }}</ref><ref name=Nen>{{cite book |title=Consolidation of Soils: Testing and Evaluation: a Symposium |author=Raymond Nen |isbn=0-8031-0446-4 |year=1986 |publisher=ASTM International |page=590 |url=http://books.google.com/books?id=a-BKqGTXA6kC&pg=PA590&dq=radius+centrifuge+effective}}</ref>
 
Also, the test tube confines motion to the direction down the length of the tube, so '''v'''<sub>B</sub> is opposite to '''u'''<sub>1</sub> and the Coriolis force is opposite to '''u'''<sub>2</sub>, that is, against the wall of the tube. If the tube is spun for a long enough time, the velocity '''v'''<sub>B</sub> drops to zero as the matter comes to an equilibrium distribution. For more details, see the articles on [[sedimentation]] and the [[Lamm equation]].
 
A related problem is that of centrifugal forces for the Earth-Moon-Sun system, where three rotations appear: the daily rotation of the Earth about its axis, the lunar-month rotation of the Earth-Moon system about their center of mass, and the annual revolution of the Earth-Moon system about the Sun. These three motions influence the [[tides]].<ref name=Appleton>{{cite book |title=The Popular Science Monthly |year=1877 |author=D Appleton |page=276 |url=http://books.google.com/books?id=YO0KAAAAYAAJ&pg=PA276&dq=rotation+revolution+%22centrifugal+force%22}}</ref>
 
===Crossing a carousel===
{{See also|Coriolis effect#Cannon on turntable|Coriolis effect#Tossed ball on a rotating carousel}}
[[Image:Carousel walk.PNG|thumb |430px |Figure 5: Crossing a rotating carousel walking at constant speed from the center of the carousel to its edge, a spiral is traced out in the inertial frame, while a simple straight radial path is seen in the frame of the carousel.]]
Figure 5 shows another example comparing the observations of an inertial observer with those of an observer on a rotating [[carousel]].<ref name= Giancoli>For a similar example, see {{cite book |title=A Handbook for Wireless/ RF, EMC, and High-Speed Electronics, Part of the EDN Series for Design Engineers |author=Ron Schmitt |year=2002 |publisher=Newnes |isbn=0-7506-7403-2 |url=http://books.google.com/books?id=fUBPN8T9bwUC&pg=PA61&dq=spheres+rotating++Coriolis#PPA60,M1
|pages=60–61 }}, and {{cite book |title=Physics for Scientists And Engineers With Modern Physics |author=Douglas C. Giancoli |page=301 |isbn=0-13-149508-9 |year=2007 |publisher=Pearson Prentice-Hall |url=http://books.google.com/books?id=xz-UEdtRmzkC&pg=PA301&dq=spheres+rotating++Coriolis#PPA301,M1 }}</ref> The carousel rotates at a constant angular velocity represented by the vector '''Ω''' with magnitude ω, pointing upward according to the [[right-hand rule]]. A rider on the carousel walks radially across it at constant speed, in what appears to the walker to be the straight line path inclined at 45° in Figure 5 . To the stationary observer, however, the walker travels a spiral path. The points identified on both paths in Figure 5 correspond to the same times spaced at equal time intervals. We ask how two observers, one on the carousel and one in an inertial frame, formulate what they see using Newton's laws.
 
====Inertial observer====
The observer at rest describes the path followed by the walker as a spiral. Adopting the coordinate system shown in Figure 5, the trajectory is described by '''r'''(''t''):
:<math>\mathbf{r}(t) =R(t)\mathbf{u}_R = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} R(t)\cos (\omega t + \pi/4) \\ R(t)\sin (\omega t + \pi/4) \end{bmatrix}, </math>
where the added π/4 sets the path angle at 45° to start with (just an arbitrary choice of direction), '''u'''<sub>''R''</sub> is a unit vector in the radial direction pointing from the center of the carousel to the walker at time ''t''. The radial distance ''R''(''t'') increases steadily with time according to:
:<math>R(t) = s t,</math>
with ''s'' the speed of walking. According to simple kinematics, the velocity is then the first derivative of the trajectory:
:<math>\mathbf{v}(t) = \frac{dR}{dt} \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} + \omega R(t) \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix}</math>
::<math>=\frac{dR}{dt} \mathbf{u}_R + \omega R(t) \mathbf{u}_{\theta}, </math>
with '''u'''<sub>θ</sub> a unit vector perpendicular to '''u'''<sub>R</sub> at time ''t'' (as can be verified by noticing that the vector [[dot product]] with the radial vector is zero) and pointing in the direction of travel.
The acceleration is the first derivative of the velocity:
:<math>\mathbf{a}(t) = \frac{d^2 R}{dt^2} \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} + 2 \frac {dR}{dt} \omega \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix} - \omega^2 R(t) \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix}</math>
::<math>=2s\omega \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix} -\omega^2 R(t) \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix}</math>
::<math>=2s\ \omega \ \mathbf{u}_{\theta}-\omega^2 R(t)\ \mathbf{u}_R \ . </math>
The last term in the acceleration is radially inward of magnitude ω<sup>2</sup> ''R'', which is therefore the instantaneous [[centripetal force|centripetal acceleration]] of [[circular motion]].<ref>{{Anchor|Note1}}'''Note''': There is a subtlety here: the distance ''R'' is the instantaneous distance from the rotational axis ''of the carousel''. However, it is not the [[osculating circle|radius of curvature]] ''of the walker's trajectory'' as seen by the inertial observer, and the unit vector '''u'''<sub>R</sub> is not perpendicular to the path. Thus, the designation "centripetal acceleration" is an approximate use of this term. See, for example, {{cite book |title=Orbital Mechanics for Engineering Students |author=Howard D. Curtis |isbn=0-7506-6169-0 |publisher=Butterworth-Heinemann |year=2005 |page=5 |url=http://books.google.com/books?id=6aO9aGNBAgIC&pg=PA5&vq=curvature&dq=orbit+%22coordinate+system%22}} and
{{cite book |title=Accelerator physics |author=S. Y. Lee |page= 37 |url=http://books.google.com/books?id=VTc8Sdld5S8C&pg=PA37&vq=curvature&dq=orbit+%22coordinate+system%22|isbn=981-256-182-X |publisher=World Scientific |location=Hackensack NJ |edition=2nd Edition |year=2004 }}</ref> The first term is perpendicular to the radial direction, and pointing in the direction of travel. Its magnitude is 2''s''ω, and it represents the acceleration of the walker as the edge of the carousel is neared, and the arc of circle traveled in a fixed time increases, as can be seen by the increased spacing between points for equal time steps on the spiral in Figure 5 as the outer edge of the carousel is approached.
 
Applying Newton's laws, multiplying the acceleration by the mass of the walker, the inertial observer concludes that the walker is subject to two forces: the inward, radially directed centripetal force, and another force perpendicular to the radial direction that is proportional to the speed of the walker.
 
====Rotating observer====
The rotating observer sees the walker travel a straight line from the center of the carousel to the periphery, as shown in Figure 5. Moreover, the rotating observer sees that the walker moves at a constant speed in the same direction, so applying Newton's law of inertia, there is ''zero'' force upon the walker. These conclusions do not agree with the inertial observer. To obtain agreement, the rotating observer has to introduce fictitious forces that appear to exist in the rotating world, even though there is no apparent reason for them, no apparent gravitational mass, electric charge or what have you, that could account for these fictitious forces.
 
To agree with the inertial observer, the forces applied to the walker must be exactly those found above. They can be related to the general formulas already derived, namely:
:<math>
\mathbf{F}_{\mathrm{fict}} =
- 2 m \boldsymbol\Omega \times \mathbf{v}_\mathrm{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B} ) - m \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}.
</math>
In this example, the velocity seen in the rotating frame is:
:<math>\mathbf{v}_\mathrm{B} = s \mathbf{u}_R, </math>
with '''u'''<sub>R</sub> a unit vector in the radial direction. The position of the walker as seen on the carousel is:
:<math>\mathbf{x}_\mathrm{B} = R(t)\mathbf{u}_R, </math>
and the time derivative of '''Ω''' is zero for uniform angular rotation. Noticing that
:<math>\boldsymbol\Omega \times \mathbf{u}_R =\omega \mathbf{u}_{\theta} \ </math>
and
:<math>\boldsymbol\Omega \times \mathbf{u}_{\theta} =-\omega \mathbf{u}_R \ ,</math>
we find:
:<math>
\mathbf{F}_{\mathrm{fict}} = - 2 m \omega s \mathbf{u}_{\theta} + m \omega^2 R(t) \mathbf{u}_R.
</math>
To obtain a [[straight-line motion]] in the rotating world, a force exactly opposite in sign to the fictitious force must be applied to reduce the net force on the walker to zero, so Newton's law of inertia will predict a straight line motion, in agreement with what the rotating observer sees. The fictitious forces that must be combated are the [[Coriolis force]] (first term) and the [[centrifugal force]] (second term). (These terms are approximate.<ref>A circle about the axis of rotation is not the [[osculating circle]] of the walker's trajectory, so "centrifugal" and "Coriolis" are approximate uses for these terms. [[#Note1|See note]].</ref>) By applying forces to counter these two fictitious forces, the rotating observer ends up applying exactly the same forces upon the walker that the inertial observer predicted were needed.
 
Because they differ only by the constant walking velocity, the walker and the rotational observer see the same accelerations. From the walker's perspective, the fictitious force is experienced as real, and combating this force is necessary to stay on a straight line radial path holding constant speed. It's like battling a crosswind while being thrown to the edge of the carousel.
 
===Observation===
Notice that this [[kinematics|kinematical]] discussion does not delve into the mechanism by which the required forces are generated. That is the subject of [[kinetics (physics)|kinetics]]. In the case of the carousel, the kinetic discussion would involve perhaps a study of the walker's shoes and the friction they need to generate against the floor of the carousel, or perhaps the dynamics of skateboarding, if the walker switched to travel by skateboard. Whatever the means of travel across the carousel, the forces calculated above must be realized. A very rough analogy is heating your house: you must have a certain temperature to be comfortable, but whether you heat by burning gas or by burning coal is another problem. Kinematics sets the thermostat, kinetics fires the furnace.
 
== See also ==
{{Col-begin}}
{{Col-1-of-3}}
* [[Newton's laws of motion]]
* [[inertial reference frame]]
* [[non-inertial reference frame]]
* [[rotating reference frame]]
* [[Coriolis force]]
* [[centrifugal force (fictitious)|Centrifugal force]]
* [[Gravity]]
* [[General relativity]]
* [[d'Alembert's principle]] of inertial forces
{{Col-2-of-3}}
* [[Centripetal force]]
* [[Circular motion]]
* [[Uniform circular motion]]
*[[Statics]]
*[[Kinetics (physics)]]
*[[Kinematics]]
*[[Applied mechanics]]
*[[Analytical mechanics]]
*[[Dynamics (physics)]]
{{Col-3-of-3}}
*[[Classical mechanics]]
*[[Generalized force]]
*[[Free motion equation]]
*[[Orthogonal coordinates]]
*[[Curvilinear coordinates]]
*[[Generalized coordinates]]
*[[Frenet-Serret formulas]]
{{Col-end}}
{{Portal|Physics}}
 
==Notes==
{{Reflist|2}}
 
==Further reading==
* {{cite book |author=[[Lev D. Landau]] and E. M. Lifshitz|isbn=0-7506-2896-0|edition=3rd Edition |year=1976|title=Mechanics |volume=Vol. 1 |series=[[Course of Theoretical Physics]] |url=http://books.google.com/books?id=LmAV8q_OOOgC&printsec=frontcover&dq=inauthor:lifshitz|publisher= Butterworth-Heinenan|pages= 128–130}}
* {{cite book |author=Keith Symon|year=1971|title=Mechanics|publisher= Addison-Wesley|edition=3rd Edition|isbn=0-201-07392-7}}
* {{cite book |author=Jerry B. Marion |year=1970|title=Classical Dynamics of Particles and Systems|publisher= Academic Press |isbn=0-12-472252-0 }}
* {{cite book|author=Marcel J. Sidi |title=Spacecraft Dynamics and Control: A Practical Engineering Approach |isbn=0-521-78780-7 |year=1997 |publisher=Cambridge University Press |url=http://books.google.com/books?id=xQpZJMtDehQC&pg=RA1-PA101&dq=orbit+%22coordinate+system%22#PRA1-PA88,M1 |nopp=true |pages=Chapter 4.8 }}
<!-- The text referencing these must have been deleted. Kept as a comment in case somebody knows where they are relevant:
# {{note|kandk2}} Kleppner, pages 355-360.
# {{note|fetter}} Alexander Fetter and John Walecka, ''Theoretical Mechanics of particles and continua'', McGraw-Hill, pages 33-39.
# {{note|landau}} [[Lev D. Landau]] and E. M. Lifshitz, (1976) ''Mechanics'', Butterworth-Heinenan, pages 128-130.
#Jerry B. Marion, (1970), ''Classical Dynamics of Particles and Systems'', Academic Press.
#Keith Symon, (1971), ''Mechanics'', Addison-Wesley
-->
 
==External links==
Tali proprietà sono particolarmente importanti nella dinamica celeste, la costanza della velocità areolare va infatti sotto il nome di
* [http://www.hcc.hawaii.edu/~rickb/SciColumns/FictForce.04Feb96.html Q and A from Richard C. Brill, Honolulu Community College]
[[w:Leggi_di_Keplero#Seconda_legge_.281609.29_-_Legge_delle_aree|II legge di Keplero]].
* [http://www-istp.gsfc.nasa.gov/stargaze/Sframes2.htm NASA's David Stern: Lesson Plans for Teachers #23 on ''Inertial Forces'']
* [http://scienceworld.wolfram.com/physics/CoriolisForce.html Coriolis Force]
* [http://mensch.org/physlets/merry.html Motion over a flat surface] Java physlet by Brian Fiedler illustrating fictitious forces. The physlet shows both the perspective as seen from a rotating and from a non-rotating point of view.
* [http://mensch.org/physlets/inosc.html Motion over a parabolic surface] Java physlet by Brian Fiedler illustrating fictitious forces. The physlet shows both the perspective as seen from a rotating and as seen from a non-rotating point of view.
 
{{DEFAULTSORT:Fictitious Force}}
[[Category:Classical mechanics]]
[[Category:Force]]
[[Category:Fictitious forces| ]]
 
=Bibliografia=
* {{cita libro | cognome= Mazzoldi | nome= Paolo | coautori= Massimo Nigro, Cesare Voci | titolo= Fisica | volume= 1 | editore= Edises | città= | ed= 2 | anno= 2000 | id= ISBN 8879591371 | cid= Mazzoldi | url= http://books.google.it/books?id=O6s7AAAACAAJ}}
 
[[Fisica_classica/Moti relativi| Argomento seguente: Moti relativi]]