Centrifugal forces(psuedo forces)

no centrifugal force or for that matter any psuedo force is reqd in inertial reference frame .

pseudo force on body=mass*accelration of frame ...........apply this force opp to direction of motion of body on FBD......

– Marie-Antoinette Tonnelat in The Principles of Electromagnetic Theory and Relativity, p. 113

[edit] Fictitious forces on Earth

See also: centrifugal force, reactive centrifugal force, and Coriolis force

The surface of the Earth is a rotating reference frame. To solve classical mechanics problems exactly in an Earth-bound reference frame, two fictitious forces must be introduced, the Coriolis force and the centrifugal force (described below). Both of these 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.

[edit] 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:[10]

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 = m a, where a is the acceleration of the system K

– V.I. Arnol'd Mathematical Methods of Classical Mechanics, p. 129

Other accelerations also give rise to fictitious forces, as described mathematically 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. [11]

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.

[edit] Examples of fictitious forces

[edit] Acceleration in a straight line
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.
Center panel: an exploded view in the inertial frame. The passenger is subject to the accelerating force ma. The seat (assumed of negligible mass) is compressed between the reaction force –ma and the applied force from the car ma. The car is subject to the net acceleration force Ma that is the difference between the applied force (m + M)a from the axle and the reaction from the seat −ma.
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 −Ma and the force (m + M) a from the axle. The difference between these forces ma is applied to the seat, which exerts a reaction −ma upon the car, so zero net force is applied to the car. The seat (assumed massless) transmits the force ma to the passenger, who is subject also to the fictitious force −ma, resulting in zero net force on the passenger. The passenger exerts a reaction force −ma 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 accelerates hard, the common human response is to feel "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:[12]

1. Figure 1, (center panel). From the viewpoint of an inertial reference frame with constant velocity matching the initial motion of the car, the car is accelerating. In order for the passenger to stay inside the car, a force must be exerted on the passenger. 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 passenger is accelerating in this frame due to the unbalanced force of the seat.
2. 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) because the seat is subject to the force of acceleration 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.

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.

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.)

[show]Animation: driving from block to block
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.

[edit] Circular motion

See also: Centrifugal force, Reactive centrifugal force, and 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 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 the turning car):

1. 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 motion. This force is exerted by the ground upon the wheels, in this case thanks to the friction between the wheels and the road.[13] The car is accelerating, due to the unbalanced force, which causes it to move in a circle. (See also banked turn.)
2. 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 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.

[show]Animation: object released from a carousel
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.[14]

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.

[edit] Fictitious forces and work

Fictitious forces can be considered to do work, provided that they move an object on a trajectory that changes its energy from potential to 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.

[edit] Gravity as a fictitious force

Main article: General relativity

General relativity is outside the scope of this article. However, the notion of "fictitious force" comes up in general relativity, so it is discussed briefly here.[15][16] Notice however, that this use of the term "fictitious force" is not covered by the analysis in this article, which all is based upon the traditional Euclidean geometry of space, not upon "curved space time".

All fictitious forces are proportional to the mass of the object upon which they act, which is also true for gravity.[17] This led Albert Einstein to wonder whether gravity was a fictitious force as well. He noted that a freefalling 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.

As part of the general theory, all reference frames are equivalent, even rotating frames. If we compare lengths in a rotating system in a radial direction with those in an azimuthal direction, the varying azimuthal velocity of points at smaller radius compared with that at larger radius causes the azimuthal Lorentz contraction to vary with radius in the rotating frame. That is, there exists a distortion of geometry dependent on the rate of rotation of the rotating frame and the radial distance to a point. This observation is just one indication that it is not possible in general relativity to use Cartesian space coordinates in accelerated systems. A need for general curvilinear coordinates arises.[18]

[show]Animation: ball that rolls off a cliff
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[19] 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.

[edit] Mathematical derivation of fictitious forces

Figure 2: An object located at xA in inertial frame A is located at location xB in accelerating frame B. The origin of frame B is located at XAB in frame A. The orientation of frame B is determined by the unit vectors along its coordinate directions, uj with j = 1, 2, 3. Using these axes, the coordinates of the object according to frame B are xB = ( x1, x2, x3 ).

[edit] General derivation

Many problems require use of noninertial reference frames, for example, those involving satellites[20][21] and particle accelerators.[22] Figure 2 shows a particle with mass m and position vector xA(t) in a particular inertial frame A. Consider a non-inertial frame B whose origin relative to the inertial one is given by XAB(t). Let the position of the particle in frame B be xB(t). What is the force on the particle as expressed in the coordinate system of frame B? [23][24]

To answer this question, let the coordinate axis in B be represented by unit vectors uj with j any of { 1, 2, 3 } for the three coordinate axes. Then

\mathbf{x}_{B} = \sum_{j=1}^3 x_j\ \mathbf{u}_j \ .

The interpretation of this equation is that xB 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:

\mathbf{x}_A =\mathbf{X}_{AB} + \sum_{j=1}^3 x_j\ \mathbf{u}_j \ .

As an aside, the unit vectors { uj } cannot change magnitude, so derivatives of these vectors express only rotation of the coordinate system B. On the other hand, vector XAB 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:

\frac {d \mathbf{x}_{A}}{dt} =\frac{d \mathbf{X}_{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} \ .

The second term summation is the velocity of the particle, say vB as measured in frame B. That is:

\frac {d \mathbf{x}_{A}}{dt} =\mathbf{v}_{AB}+ \mathbf{v}_B + \sum_{j=1}^3 x_j \ \frac{d \mathbf{u}_j}{dt} \ .

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 vB, 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 vAB. 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:

\frac {d^2 \mathbf{x}_{A}}{dt^2} = \mathbf{a}_{AB}+\frac {d\mathbf{v}_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}\ .

Using the same formula already used for the time derivative of xB, the velocity derivative on the right is:

\frac {d\mathbf{v}_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}_B + \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} \ .

Consequently,

\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} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ .    (Eq. 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 aB, 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 aAB, 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 dvB / dt 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 Σxj \stackrel{\frac{d}{dt}}{}uj due to change in x j ).

To put matters in terms of forces, the accelerations are multiplied by the particle mass:

\mathbf{F}_A = \mathbf{F}_B + m\ \mathbf{a}_{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}\ .

The force observed in frame B, FB = m aB is related to the actual force on the particle, FA, by:

\mathbf{F}_B = \mathbf{F}_A + \mathbf{F}_{\mbox{fictitious}} \ .

where:

\mathbf{F}_{\mbox{fictitious}} =-m\ \mathbf{a}_{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}\ .

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 Ffictitious as an additional force.[25][10][26]

Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on centrifugal force.

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