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A fictitious or pseudo outward force on a particle rotating about an axis which by Newton's third law is equal and opposite to the centripetal force. Like all such action-reaction pairs of forces, they are equal and opposite but do not act on the same body and so do not cancel each other. Consider a mass M tied by a string of length R to a pin at the center of a smooth horizontal table and whirling around the pin with an angular velocity of ω radians per second. The mass rotates in a circular path because of the centripetal force FC = Mω2R which is exerted on the mass by the string. The reaction force exerted by the rotating mass M, the so-called centrifugal force, is Mω2R in a direction away from the center of rotation. See also Centripetal force.
From another point of view, consider an experimenter in a windowless, circular laboratory that is rotating smoothly about a centrally located vetical axis. No object remains at rest on a smooth surface; all such objects move outward toward the wall of the laboratory as though an outward, centrifugal force were acting. To the experimenter partaking in the rotation, in a rotating frame of reference, the centrifugal force is real. An outside observer would realize that the inward force which the experimenter in the rotating laboratory must exert to keep the object at rest does not keep it at rest, but furnishes the centripetal force required to keep the object moving in a circular path. The concept of an outward, centrifugal force explains the action of a centrifuge. See also Centrifugation.
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| Geography Dictionary: centrifugal forces |
In human geography, those forces which encourage a movement of people, business, and industry away from central urban areas. These forces include traffic congestion, restricted sites, high local taxes and rents, obsolete technology, and lack of amenity. see counter-urbanization, decentralization, urban sprawl.
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An outwardly directed force acting on a body rotating around a central point. It is a reaction force that is equal in magnitude, but opposite in direction to the centripetal force acting on the body.

| Science Dictionary: centrifugal force |
A force that tends to move objects away from the center in a system undergoing circular motion. Centrifugal force keeps the water in a whirling bucket from spilling or throws a rider in a car against the door when the car goes around a sharp curve. Centrifugal force is actually a form of inertia.
| Wikipedia: Centrifugal force (rotating reference frame) |
In classical mechanics, centrifugal force is an outward force associated with curved motion, that is, rotation about some (possibly not stationary) center. Centrifugal force is one of several so-called pseudo-forces (also known as inertial forces), so named because, unlike fundamental forces, they do not originate in interactions with other bodies situated in the environment of the particle upon which they act. Instead, centrifugal force originates in the curved motion of the frame of reference within which observations are made.[1][2][3][4][5][6]
Contents |
Next, a formal mathematical expression for centrifugal force is derived by comparing the form of Newton's second law in an inertial frame with its form in a frame rotating about a fixed axis.
Newton's law of motion for a particle of mass m can be written in vector form as

where F is the vector sum of the physical forces applied to the particle and a is the absolute acceleration[7] of the particle, given by:

where r is the position vector of the particle. The differentiations are performed in terms of an inertial reference frame.
In a rotating frame of reference, a body appears to move differently due to the rotation of the frame. Consequently, the time derivatives of its position will appear different as well. As shown in Rotating reference frame, for any vector Q depending upon time, its time derivative [dQ/dt] evaluated in terms of a reference frame with a coincident origin but rotating with the absolute angular velocity Ω[8] is related to the absolute derivative dQ/dt by:[9]
![\frac{d}{dt}\boldsymbol{Q} = \left[\frac{d}{dt}\boldsymbol{Q}\right] + \boldsymbol{\Omega} \times \boldsymbol{Q}\ ,](http://wpcontent.answers.com/math/b/d/5/bd57acae95a319bc40dc3452f563511a.png)
where × denotes the vector cross product and square brackets […] denote evaluation in the rotating frame of reference. As a particular example, it follows that the absolute acceleration of the particle can be written as (for more detail, see Rotating frame of reference):
![\begin{align}
\boldsymbol{a} &=\frac{d^2}{dt^2}\boldsymbol{r}\\
&= \left[ \frac{d^2 \boldsymbol{r}}{dt^2} \right] + \frac{d \boldsymbol{\Omega}}{dt}\times\boldsymbol{r} + 2 \boldsymbol{\Omega}\times \left[ \frac{d \boldsymbol{r}}{dt} \right] + \boldsymbol{\Omega}\times ( \boldsymbol{\Omega} \times \boldsymbol{r}) \ .
\end{align}](http://wpcontent.answers.com/math/f/5/b/f5bef7ffff976f7f476dbe702f48a239.png)
From the viewpoint of the rotating frame, where an observer sees merely the acceleration relative to the rotating frame, the first term on the right hand side appears to be the absolute acceleration. Of course, using this first term alone in Newton's law will lead to incorrect prediction of the trajectory and, to obtain agreement, the observer in the rotating frame is forced to add additional force terms on the force-side of Newton's law. When these forces are added, the equation of motion has the form:[3][10][11][12][13]
![= m\left[ \frac{d^2 \boldsymbol{r}}{dt^2} \right] \ ,](http://wpcontent.answers.com/math/2/1/f/21f56e25d332e2b594108efef194c513.png)
which, from a formal mathematical standpoint, is the same result as simply moving the extra acceleration terms to the left hand side (the force side) of the equation. From the viewpoint of the rotating frame, however, the terms on the force side all result from forces really experienced as forces.[14][15] The last term on the force side is commonly called the centrifugal force. It points directly away from the axis of rotation of the rotating reference frame, with magnitude mΩ2r.
Notice that for a non-rotating frame the centrifugal force is zero. The disappearance of centrifugal force in an inertial frame of reference is shared by all fictitious forces.[16]
Of course, living on Earth one is automatically in a rotating frame of reference, and it is not an option. However, even from the abstract stance of solving problems in mechanics, a rotating reference frame can have advantages over an inertial reference frame.[4][17] Sometimes the calculations are simpler (an example is inertial circles), and sometimes the intuitive picture coincides more closely with the rotational frame (an example is sedimentation in a centrifuge). By treating the extra acceleration terms due to the rotation of the frame as if they were forces, subtracting them from the physical forces, it's possible to treat the second time derivative of position (relative to the rotating frame) as absolute acceleration. Thus the analysis using Newton's law can proceed as if the reference frame was inertial, provided the fictitious force terms are included in the sum of forces. For example, centrifugal force is used in the FAA pilot's manual in describing turns.[18] Other examples are such systems as planets, centrifuges, carousels, turning cars, spinning buckets, and rotating space stations.[19][20][21] Regarding the advantages of rotating frames from the viewpoint of meteorology, Ryder says:[22]
A simple way of dealing with this problem is, of course, to transform all coordinates to an inertial system. This is, however, sometimes inconvenient. Suppose, for example, we wish to calculate the movement of air masses in the earth's atmosphere due to pressure gradients. We need the results relative to the rotating frame, the earth, so it is better to stay within this coordinate system if possible. This can be achieved by introducing fictitious (or "non-existent") forces which enable us to apply Newton's Laws of Motion in the same way as in an inertial frame.
– Peter Ryder: Classical Mechanics, pp. 78-79
The consideration of centrifugal force and absolute rotation is a topic of debate about relativity, cosmology, and the nature of physical laws.
Can absolute rotation be detected? In other words, can one decide whether an observed object is rotating or if it is you, the observer that is rotating? Newton suggested two experiments to resolve this problem. One is the effect of centrifugal force upon the shape of the surface of water rotating in a bucket. The second is the effect of centrifugal force upon the tension in a string joining two spheres rotating about their center of mass. A related third suggestion was that rotation of a sphere (such as a planet) could be detected from its shape (or "figure"), which is formed as a balance between containment by gravitational attraction and dispersal by centrifugal force.
Below several examples illustrate both the inertial and rotating frames of reference, and the role of centrifugal force and its relation to Coriolis force in rotating frameworks. For more examples see Fictitious force, rotating bucket and rotating spheres.
Figure 2 shows a simplified version of an apparatus for studying centrifugal force called the "whirling table".[23] The apparatus consists of a rod that can be whirled about an axis, causing a bead to slide on the rod under the influence of centrifugal force. A cord ties a weight to the sliding bead. By observing how the equilibrium balancing distance varies with the weight and the speed of rotation, the centrifugal force can be measured as a function of the rate of rotation and the distance of the bead from the center of rotation.
From the viewpoint of an inertial frame of reference, equilibrium results when the bead is positioned to select the particular circular orbit for which the weight provides the correct centripetal force.
As a lab experiment, it seems arbitrary whether to deal with centripetal force or centrifugal force. From the bead's standpoint, however, centrifugal force is real and is pushing the bead.
What is the viewpoint of an airplane pilot engaged in skywriting? The plane's path is the smoky trail left behind, and progress can be registered as the distance s from the start of the trail to the plane's present position. The speed of the plane is v = ds / dt and the curvature of the path is measured by the osculating circle of radius ρ that is tangent to the path. For the inertial observer watching from the ground, the plane at any instant is executing circular motion about its (instantaneous) center of curvature, and so is subject to a centripetal force v2 / ρ acting radially inward toward this center of curvature.[24] To maintain trajectory, this centripetal force is provided by banking the airplane, generating a lift that provides this centripetal force. According to the pilot, however, the plane is stationary, but subject to a centrifugal force outward from the instantaneous center of curvature with a magnitude v2 / ρ.[25] To maintain trajectory, this centrifugal force is combated by banking the airplane, generating a lift to counteract the centrifugal force, thereby maintaining the plane in its equilibrium motionless position.[26] For a detailed analysis, see Mechanics of planar particle motion.
Using the possible ω of the Universe as a fitting parameter one can deduce from the flatness of our galaxy whether the Universe is rotating.[27] The flatness of our galaxy depends on how fast it rotates. If we, the inhabitants of the galaxy, interpret its observed rotation as entirely its own, we ignore a possible centrifugal force due to the rotation of the Universe, and so predict an incorrect flatness. If we are smart enough to consider this possibility, we can include the rate of rotation of the Universe in our flatness calculation, include the centrifugal force as a force in our inventory of forces, and by adjusting this ω to fit the observed flatness we can deduce how fast the Universe is rotating. The inference is, in fact, that the Universe has rotated at most fifty times in the time since its birth.
In a similar fashion, if we did not know the Earth rotates about its axis, we could infer this rotation from the centrifugal force needed to account for its observed bulging at its equator.[28][29] The actual extent of oblateness in response to a centrifugal force requires an understanding of the make-up of the planet, not only today but during its formation.[30][31]
In general terms, the issue of whether one is in a rotating or an inertial frame can be determined by the need (or lack of need) for centrifugal force, as suggested by Newton in the rotating bucket problem and in the rotating spheres examples.[32]
Riding a car around a curve, we take a personal view that we are at rest in the car, and should be undisturbed in our seats. Nonetheless, we feel sideways force applied to us from the seats and doors and a need to lean to one side. To explain the situation, we propose a centrifugal force that is acting upon us and must be combated. Interestingly, we find this discomfort is reduced when the curve is banked, tipping the car inward toward the center of the curve.
A different point of view is that of the highway designer. The designer views the car as executing curved motion and therefore requiring an inward centripetal force to impel the car around the turn. By banking the curve, the force exerted upon the car in a direction normal to the road surface has a horizontal component that provides this centripetal force. That means the car tires no longer need to apply a sideways force to the car, but only a force perpendicular to the road. By choosing the angle of bank to match the car's speed around the curve, the car seat transmits only a perpendicular force to the passengers, and the passengers no longer feel a need to lean nor feel a sideways push by the car seats or doors.[33]
It has been mentioned that to deal with motion in a rotating frame of reference, one alternative to a solution based upon translating everything into an inertial frame instead is to apply Newton's laws of motion in the rotating frame by adding pseudo-forces, and then working directly in the rotating frame. Next is a simple example of this method. [34][35]
Figure 4 illustrates that a body that is stationary relative to the non-rotating inertial frame S' appears to be rotating when viewed from the rotating frame S, which is rotating at angular rate Ω. Therefore, application of Newton's laws to what looks like circular motion in the rotating frame S at a radius R, requires an inward centripetal force of −m Ω2 R to account for the apparent circular motion. According to observers in S, this centripetal force in the rotating frame is provided as a net force that is the sum of the radially outward centrifugal pseudo force m Ω2 R and the Coriolis force −2m Ω × vrot.[36] [37] To evaluate the Coriolis force, we need the velocity as seen in the rotating frame, vrot. According to the formulas in the Derivation section, this velocity is given by −Ω × R.[38] Hence, the Coriolis force (in this example) is inward, in the opposite direction to the centrifugal force, and has the value −2m Ω2 R. The combination of the centrifugal and Coriolis force is then m Ω2 R−2m Ω2 R = −m Ω2 R, exactly the centripetal force required by Newton's laws for circular motion.[39] [40][41]
For further examples and discussion, see Taylor.[42]
Figure 5 shows a ball dropping vertically (parallel to the axis of rotation Ω of the rotating frame). For simplicity, suppose it moves downward at a fixed speed in the inertial frame, occupying successively the vertically aligned positions numbered one, two, three. In the rotating frame it appears to spiral downward, and the right side of Figure 5 shows a top view of the circular trajectory of the ball in the rotating frame. Because it drops vertically at a constant speed, from this top view in the rotating frame the ball appears to move at a constant speed around its circular track. A description of the motion in the two frames is next.
In the inertial frame the ball drops vertically at constant speed. It does not change direction, so the inertial observer says the acceleration is zero and there is no force acting upon the ball.
In the rotating frame the ball drops vertically at a constant speed, so there is no vertical component of force upon the ball. However, in the horizontal plane perpendicular to the axis of rotation, the ball executes uniform circular motion as seen in the right panel of Figure 5. Applying Newton's law of motion, the rotating observer concludes that the ball must be subject to an inward force in order to follow a circular path. Therefore, the rotating observer believes the ball is subject to a force pointing radially inward toward the axis of rotation. According to the analysis of uniform circular motion

with
a unit vector in the outward radial direction, and where Ω is the angular rate of rotation, m is the mass of the ball, and R is the radius of the spiral in the horizontal plane. Because there is no apparent source for such a force (hence the label "fictitious"), the rotating observer concludes it is just "a fact of life" in the rotating world that there exists an inward force with this behavior. Inasmuch as the rotating observer already knows there is a ubiquitous outward centrifugal force in the rotating world, how can there be an inward force? The answer is the Coriolis force: the component of velocity tangential to the circular motion seen in the right panel of Figure 5 activates the Coriolis force, which cancels the centrifugal force and goes a step further to provide precisely the centripetal force demanded by the calculations of the rotating observer.
Some details of evaluation of the Coriolis force are shown in Figure 6. The Coriolis force is found to be (using the cross-product expansion):[43][44]

Combining this force with the centrifugal force:

as required for the necessary centripetal force to maintain circular motion.
Because the Coriolis force and centrifugal forces combine to provide the centripetal force the rotating observer requires for the observed circular motion, the rotating observer does not need to apply any additional force to the object, in complete agreement with the inertial observer, who also says there is no force needed. One way to express the result: the fictitious forces look after the "fictitious" situation, so the ball needs no help to travel the perceived trajectory: all observers agree that nothing needs to be done to make the ball follow its path.
To show a different frame of reference, let's revisit the dropping ball example in Figure 5 from the viewpoint of a parachutist falling at constant speed to Earth (the rotating platform). The parachutist aims to land upon the point on the rotating ground directly below the drop-off point. Figure 7 shows the vertical path of descent seen in the rotating frame. The parachutist drops at constant speed, occupying successively the vertically aligned positions one, two, three.
In the stationary frame, let us suppose the parachutist jumps from a helicopter hovering over the destination site on the rotating ground below, and therefore traveling at the same speed as the target below. The parachutist starts with the necessary speed tangential to his path (ωR) to track the destination site. If the parachutist is to land on target, the parachute must spiral downward on the path shown in Figure 7. The stationary observer sees a uniform circular motion of the parachutist when the motion is projected downward, as in the left panel of Figure 7. That is, in the horizontal plane, the stationary observer sees a centripetal force at work, -m ω2 R, as is necessary to achieve the circular path. The parachutist needs a thruster to provide this force. Without thrust, the parachutist follows the dashed vertical path in the left panel of Figure 7, obeying Newton's law of inertia.
The stationary observer and the observer on the rotating ground agree that there is no vertical force involved: the parachutist travels vertically at constant speed. However, the observer on the ground sees the parachutist simply drop vertically from the helicopter to the ground, following the vertically aligned positions one, two, three. There is no force necessary. So how come the parachutist needs a thruster?
The ground observer has this view: there is always a centrifugal force in the rotating world. Without a thruster, the parachutist would be carried away by this centrifugal force and land far off the mark. From the parachutist's viewpoint, trying to keep the target directly below, the same appears true: a steady thrust radially inward is necessary, just to hold a position directly above target. Unlike the dropping ball case, where the fictitious forces conspired to produce no need for external agency, in this case they require intervention to achieve the trajectory. The basic rule is: if the inertial observer says a situation demands action or does not, the fictitious forces of the rotational frame will lead the rotational observer to the same conclusions, albeit by a different sequence.
Notice that there is no Coriolis force in this discussion, because the parachutist has zero horizontal velocity from the viewpoint of the rotating ground observer.[45]
Centrifugal force arises in the analysis of orbital motion and, more generally, of motion in a central-force field – in the case of a two-body problem, it is easy to convert to an equivalent one-body problem with force directed to or from an origin, and motion in a plane,[46] so we consider only that.
The symmetry of a central force lends itself to a description in polar coordinates. The dynamics of a mass, m, expressed using Newton's second law of motion (F = ma), becomes in polar coordinates:[47][48]

where
is the force accelerating the object and the "hat" variables are unit direction vectors (
points in the centrifugal or outward direction, and
is orthogonal to it).
In the case of a central force, relative to the origin of the polar coordinate system,
can be replaced by
, meaning the entire force is the component in the radial direction. An inward force of gravity would therefore correspond to a negative-valued F(r).
The components of F = ma along the radial direction therefore reduce to

in which the term proportional to the square of the rate of rotation appears on the acceleration side as a "centripetal acceleration", that is, a negative acceleration term in the
direction.[47] In the special case of a planet in circular orbit around its star, for example, where
is zero, the centripetal acceleration alone is the entire acceleration of the planet, curving its path toward the sun under the force of gravity, the negative F(r).
As pointed out by Taylor,[49] for example, it is sometimes convenient to work in a co-rotating frame, that is, one rotating with the object so that the angular rate of the frame, Ω, equals the
of the object in the inertial frame. In such a frame, the observed
is zero and
alone is treated as the acceleration – so in the equation of motion, the
term is “reincarnated on the force side of the equation (with opposite signs, of course) as the centrifugal force mΩ2r in the radial equation”:[50]

where the
term is known as the centrifugal force. The centrifugal force term in this equation is called a "fictitious force", "apparent force", or "pseudo force", as its value varies with the rate of rotation of the frame of reference. When the centrifugal force term is expressed in terms of parameters of the rotating frame, replacing
with Ω, it can be seen that it is the same centrifugal force previously derived for rotating reference frames.
Because of the absence of a net force in the azimuthal direction, conservation of angular momentum allows the radial component of this equation to be expressed solely with respect to the radial coordinate, r, and the angular momentum
, yielding the radial equation (a "fictitious one-dimensional problem"[46] with only an r dimension):
.The L2 / mr3 term is again the centrifugal force, a force component induced by the rotating frame of reference. The equations of motion for r that result from this equation for the rotating 2D frame are the same that would arise from a particle in a fictitious one-dimensional scenario under the influence of the force in the equation above.[46] If F(r) represents gravity, it is a negative term proportional to 1/r2, so the net acceleration in r in the rotating frame depends on a difference of reciprocal square and reciprocal cube terms, which are in balance in a circular orbit but otherwise typically not. This equation of motion is similar to one originally proposed by Leibniz.[51] Given r, the rate of rotation is easy to infer from the constant angular momentum L, so a 2D solution can be easily reconstructed from a 1D solution of this equation.
When the angular velocity of this co-rotating frame is not constant, that is, for non-circular orbits, other fictitious forces – the Coriolis force and the Euler force – will arise, but can be ignored since they will cancel each other, yielding a net zero acceleration transverse to the moving radial vector, as required by the starting assumption that the
vector co-rotates with the planet.[52] In the special case of circular orbits, in order for the radial distance to remain constant the outward centrifugal force must cancel the inward force of gravity; for other orbit shapes, these forces will not cancel, so r will not be constant.
Concepts of centripetal and centrifugal force played a key early role in establishing the set of inertial frames of reference and the significance of fictitious forces, even aiding in the development of general relativity.
The operations of numerous common rotating mechanical systems are most easily conceptualized in terms of centrifugal force. For example:
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Nevertheless, all of these systems can also be described without requiring the concept of centrifugal force, in terms of motions and forces in an inertial frame, at the cost of taking somewhat more care in the consideration of forces and motions within the system.
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