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This article is about the fictitious force related to rotating reference frames. For the reaction force to the centripetal force, see Reactive centrifugal force. For the general subject of centrifugal force, see Centrifugal force (disambiguation). For general derivations and discussion of fictitious forces, see Fictitious force.
Dynamics


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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

[edit] Derivation

Main article: Rotating reference frame
See also: Fictitious force and Mechanics of planar particle motion

[edit] Velocity

In a rotating frame of reference bodies can have the same positions as in a nonrotating frame, but they will move differently due to the rotation of the frame. Consequently, the time derivatives of any position vector r depending on time (velocity dr/dt and acceleration dr2/dt2) will differ according to the rotation. When time derivative [dr/dt] is evaluated from a reference frame with a coincident origin at r = 0 but rotating with the absolute angular velocity Ω:[7]

\frac{d\boldsymbol{r}}{dt} = \left[\frac{d\boldsymbol{r}}{dt}\right] + \boldsymbol{\Omega} \times \boldsymbol{r}\ ,

where \times denotes the vector cross product and square brackets […] denote evaluation in the rotating frame of reference. In other words, the apparent velocity in the rotating frame is altered by the amount of the apparent rotation \boldsymbol{\Omega} \times \boldsymbol{r}, which is perpendicular to both the vector from the origin r and the axis of rotation Ω and directly proportional in magnitude to each of them. The vector Ω has magnitude Ω equal to the rate of rotation and is directed along the axis of rotation according to the right-hand rule. The angle must be measured in radians per unit of time.

[edit] Acceleration

Newton's law of motion for a particle of mass m can be written in vector form as

\boldsymbol{F} = m\boldsymbol{a}\ ,

where F is the vector sum of the physical forces applied to the particle and a is the absolute acceleration[8] of the particle, given by:

\boldsymbol{a}=\frac{d^2\boldsymbol{r}}{dt^2} \ ,

where r is the position vector of the particle. The differentiations are performed in the inertial frame.

By twice applying the transformation above from the inertial to the rotating frame, the absolute acceleration of the particle can be written as:

\begin{align} \boldsymbol{a} &=\frac{d^2\boldsymbol{r}}{dt^2} = \frac{d}{dt}\frac{d\boldsymbol{r}}{dt} = \frac{d}{dt} ( \left[\frac{d\boldsymbol{r}}{dt}\right] + \boldsymbol{\Omega} \times \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}

[edit] Force

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][9][10][11][12]

\boldsymbol{F} - m\frac{d \boldsymbol{\Omega}}{dt}\times\boldsymbol{r} - 2m \boldsymbol{\Omega}\times \left[ \frac{d \mathbf{r}}{dt} \right] - m\boldsymbol{\Omega}\times (\boldsymbol{\Omega}\times \boldsymbol{r})   = m\left[ \frac{d^2 \boldsymbol{r}}{dt^2} \right] \ ,

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.[13][14] The terms on the force side of the equation can be recognized as the Euler force m d\boldsymbol{\Omega}/dt \times\boldsymbol{r}, the Coriolis force 2m \boldsymbol{\Omega}\times \left[ d \mathbf{r}/dt \right], and the centrifugal force m\boldsymbol{\Omega}\times (\boldsymbol{\Omega}\times \boldsymbol{r}), respectively. The centrifugal force points directly away from the axis of rotation of the rotating reference frame, with magnitude mΩ2r.

Notice that for a non-rotating inertial frame of reference (\boldsymbol\Omega=0) the centrifugal force and all other fictitious forces disappear.[15]

[edit] Advantages of rotating frames

When living on Earth the rotating frame of reference is far more convenient to use than one in which our velocity changes daily by hundreds of kilometers per hour. But even from the abstract stance of solving problems in mechanics, a rotating reference frame can have advantages over an inertial reference frame.[4][16] 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.[17] Other examples are such systems as planets, centrifuges, carousels, turning cars, spinning buckets, and rotating space stations.[18][19][20] Regarding the advantages of rotating frames from the viewpoint of meteorology, Ryder says:[21]

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

[edit] Centrifugal force and absolute rotation

Main article: Centrifugal force and absolute rotation

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.[22]

[edit] Examples

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.

[edit] Earth

A calculation for Earth at the equator (Ω = 2π / 86164 seconds, r = 6378100 meters) shows that an object experiences a centrifugal force equal to approximately 1/289 of standard gravity.[23] Because centrifugal force increases according to the square of Ω, one would expect gravity to be cancelled for an object travelling 17 times faster than the Earth's rotation, and in fact satellites in low orbit at the equator complete 17 full orbits in one day.[24]

[edit] Planet shape

See also: Clairaut's theorem

Centrifugal force can be used to explain the shape of the earth, in particular the observed bulging at the equator.[25][26] 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.[27][28] Centrifugal force can also account for the difference in Earth's gravity between the poles and the equator, which effects escape velocity enough that space launch sites are best sited as close to the equator as possible.

This is seen in many other many other astronomical objects, in particular the planets of the solar system where both the rotation period and degree of flattening can be observed directly. In two cases, Jupiter and Saturn the flattening is pronounced enough to be seen with just a small telescope.[29]

[edit] Whirling table

Figure 2: The "whirling table". The rod is made to rotate about the axis and (from the bead's viewpoint) the centrifugal force acting on the sliding bead is balanced by the weight attached by a cord over two pulleys.

Figure 2 shows a simplified version of an apparatus for studying centrifugal force called the "whirling table".[30] 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.

[edit] Skywriter

See also: Centripetal force#Example: The banked turn
FIgure 3: Skywriter. (Image from NASA ASRS)

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.[31] 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.[32] 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.[33] For a detailed analysis, see Mechanics of planar particle motion.

[edit] The banked turn

Main article: Banked turn
See also: Centripetal force#Example: The banked turn and Reactive centrifugal force#Example: The turning car

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.[34]

[edit] Rotating frame

Main article: Fictitious force
Figure 4: Object stationary in inertial frame S' appears to circle clockwise in a counterclockwise rotating frame S. Top panel: In inertial (stationary) frame S' , frame S is rotating counterclockwise at angular rate Ω, and occupies successive positions at times t0, t1, and t2. Stationary object does not move, of course. Center panel: Positions of the stationary object as it appears in S at the times t0, t1, and t2. The object appears to move clockwise in S. Bottom panel: Assembly of the positions in center panel to construct the orbit of the stationary object as seen by S. Radius vectors from the origin of moving frame S to the object at times t0, t1, t2 are R0, R1, R2; these vectors all have magnitude equal to the radius of the circle R. At time t0, the object has a velocity v0 in frame S, but this velocity turns with motion of the object to remain tangential to its orbit at all times.

Though centrifugal force adequately describes the force on objects at rest relative to a steadily rotating frame of reference, the fictitious force on moving objects includes the Eötvös effect and Coriolis force. To deal with motion directly in a rotating frame of reference by applying Newton's laws, it is necessary to take these pseudo-forces into account. For example:[35][36]

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.[37] [38] 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.[39] 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.[40] [41][42]

For further examples and discussion, see Taylor.[43]

[edit] Dropping ball

Figure 5: A ball moving vertically along the axis of rotation in an inertial frame appears to spiral downward in the rotating frame. The right panel shows a downward view in the rotating frame. The rate of rotation |Ω| = ω is assumed constant in time.
Figure 6: Vector cross product used to determine the Coriolis force. The vector Ω represents the rotation of the frame at angular rate ω; the vector v shows the velocity tangential to the circular motion as seen in the rotating frame. The vector Ω × v is found using the right-hand rule for vector cross products. It is related to the negative of the Coriolis force (the Coriolis force is −2 m Ω × v).

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.

[edit] Inertial frame

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.

[edit] Uniformly rotating frame

See also: Coriolis force

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

\mathbf{F}_{\mathrm{fict}} = -m\Omega^2 R \boldsymbol { \hat r}\ ,

with \boldsymbol { \hat r} 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):[44][45]

\mathbf{F_{Cor}} \overset{\underset{\mathrm{def}}{}}{=}-2 m \mathbf{\Omega \times v} = -2m \mathbf{\Omega \times} \left(-\mathbf{\Omega \times} R\boldsymbol{\hat {r}}\right) =-2m\Omega^2R \boldsymbol { \hat r}\ ,

Combining this force with the centrifugal force:

\mathbf{F}_{\mathrm{fict}} = \mathbf{F_{Cor}} + \mathbf{F_{Cfgl}} = \left(-2m\Omega^2R + m\Omega^2R\right)\boldsymbol { \hat r} = -m\Omega^2 R \boldsymbol { \hat r}\ ,

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.

[edit] Parachutist

Figure 7: A parachutist moving vertically parallel to the axis of rotation in a rotating frame appears to spiral downward in the inertial frame. The parachutist begins the drop with a horizontal component of velocity the same as the target site. The left panel shows a downward view in the inertial frame. The rate of rotation |Ω| = ω is assumed constant in time.

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.[46]

[edit] Planetary motion

See also: Orbit, Two-body problem, and Kepler's laws of planetary motion

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,[47] 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:[48][49]

\boldsymbol{F} = m((\ddot r - r \dot \theta^2) \boldsymbol{\hat r} + (r \ddot\theta + 2 \dot r \dot\theta) \boldsymbol{\hat \theta})

where \boldsymbol{F} is the force accelerating the object and the "hat" variables are unit direction vectors (\boldsymbol{\hat r} points in the centrifugal or outward direction, and \boldsymbol{\hat \theta} is orthogonal to it).

In the case of a central force, relative to the origin of the polar coordinate system, \boldsymbol{F} can be replaced by F(r) \boldsymbol{\hat r}, 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

F(r) = m(\ddot r - r \dot\theta^2)

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 \boldsymbol{\hat r} direction.[48] In the special case of a planet in circular orbit around its star, for example, where \ddot r 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,[50] 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 \dot\theta of the object in the inertial frame. In such a frame, the observed \dot \theta is zero and \ddot r alone is treated as the acceleration – so in the equation of motion, the m r \dot\theta^2 term is “reincarnated on the force side of the equation (with opposite signs, of course) as the centrifugal force mΩ2r in the radial equation”:[51]

F(r) + m r \dot\theta^2 = m\ddot r

where the m r \dot\theta^2 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 \dot\theta 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 L=m \dot\theta r^2, yielding the radial equation (a "fictitious one-dimensional problem"[47] with only an r dimension):

F(r) + \frac{L^2}{mr^3} = m \ddot r .

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.[47] 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.[52] 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 \hat r vector co-rotates with the planet.[53] 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.

[edit] Development of the modern conception of centrifugal force

Main article: History of the concepts of centrifugal and centripetal forces

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.

[edit] Applications

The operations of numerous common rotating mechanical systems are most easily conceptualized in terms of centrifugal force. For example:

  • A centrifugal governor regulates the speed of an engine by using spinning masses that move radially, adjusting the throttle, as the engine changes speed. In the reference frame of the spinning masses, centrifugal force causes the radial movement.
  • A centrifugal clutch is used in small engine-powered devices such as chain saws, go-karts and model helicopters. It allows the engine to start and idle without driving the device but automatically and smoothly engages the drive as the engine speed rises. Inertial drum brake ascenders used in rock climbing and the inertia reels used in many automobile seat belts operate on the same principle.
  • Centrifugal forces can be used to generate artificial gravity, as in proposed designs for rotating space stations. The Mars Gravity Biosatellite will study the effects of Mars-level gravity on mice with gravity simulated in this way.
  • Spin casting and centrifugal casting are production methods that uses centrifugal force to disperse liquid metal or plastic throughout the negative space of a mold.
  • Centrifuges are used in science and industry to separate substances. In the reference frame spinning with the centrifuge, the centrifugal force induces a hydrostatic pressure gradient in fluid-filled tubes oriented perpendicular to the axis of rotation, giving rise to large buoyant forces which push low-density particles inward. Elements or particles denser than the fluid move outward under the influence of the centrifugal force. This is effectively Archimedes' principle as generated by centrifugal force as opposed to being generated by gravity.
  • Some amusement park rides make use of centrifugal forces. For instance, a Gravitron’s spin forces riders against a wall and allows riders to be elevated above the machine’s floor in defiance of Earth’s gravity.

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.

[edit] See also

Physics portal

[edit] Notes and references

  1. ^ Robert Resnick & David Halliday (1966). Physics. Wiley. p. 121. ISBN 0471345245. http://books.google.com/books?lr=&as_brr=0&q=%22cannot+associate+them+with+any+particular+body+in+the+environment+of+the+particle%22+inauthor%3ADavid+inauthor%3AHalliday&btnG=Search+Books. 
  2. ^ Jerrold E. Marsden, Tudor S. Ratiu (1999). Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems. Springer. p. 251. ISBN 038798643X. http://books.google.com/books?id=I2gH9ZIs-3AC&pg=PA251&vq=Euler+force&dq=isbn=038798643X&source=gbs_search_s&sig=ACfU3U0DkJL1h3lGMMbXyKY15GtPpspHuQ. 
  3. ^ a b John Robert Taylor (2005). Classical Mechanics. University Science Books. p. 343. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn=189138922X&lr=&as_brr=0&sig=JVfFlMT5TvXh1I64JAFBFq7pA6s#PPA343,M1. 
  4. ^ a b Stephen T. Thornton & Jerry B. Marion (2004). Classical Dynamics of Particles and Systems (5th ed.). Belmont CA: Brook/Cole. Chapter 10. ISBN 0534408966. http://worldcat.org/oclc/52806908&referer=brief_results. 
  5. ^ David McNaughton. "Centrifugal and Coriolis Effects". http://dlmcn.com/circle.html. Retrieved 2008-05-18. 
  6. ^ David P. Stern. "Frames of reference: The centrifugal force". http://www.phy6.org/stargaze/Lframes2.htm. Retrieved 2008-10-26. 
  7. ^ John L. Synge (2007). Principles of Mechanics (Reprint of Second Edition of 1942 ed.). Read Books. p. 347. ISBN 1406746703. http://books.google.com/books?id=YZjIg4Mo56UC&pg=PA347&dq=rotating+fictitious+force&lr=&as_brr=0&sig=ACfU3U3sRtWpje8_Y86_IxXT9N_0FnhzwQ. 
  8. ^ By "absolute" is meant as seen in any inertial frame of reference; for example "absolute acceleration" or "absolute derivative".
  9. ^ Vladimir Igorević Arnolʹd (1989). Mathematical Methods of Classical Mechanics (2nd ed.). Springer. p. 130. ISBN 978-0-387-96890-2. http://books.google.com/books?id=Pd8-s6rOt_cC&pg=PT149&dq=%22additional+terms+called+inertial+forces.+This+allows+us+to+detect+experimentally%22&lr=&as_brr=0&sig=ACfU3U1qRbkvn6x7FcBsHO8Bp4Ty95XbZw#PPT150,M1. 
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  12. ^ Louis N. Hand, Janet D. Finch (1998). Analytical Mechanics. Cambridge University Press. p. 267. ISBN 0521575729. http://books.google.com/books?id=1J2hzvX2Xh8C&pg=PA267&vq=fictitious+forces&dq=Hand+inauthor:Finch&lr=&as_brr=0&source=gbs_search_s&sig=ACfU3U33emV_6eJZihu3M6IZKurSt85_eg. 
  13. ^ Mark P Silverman (2002). A universe of atoms, an atom in the universe (2 ed.). Springer. p. 249. ISBN 0387954376. http://books.google.com/books?id=-Er5pIsYe_AC&pg=PA249. 
  14. ^ John Robert Taylor (2005). op. cit.. Sausalito, Calif.: Univ. Science Books. p. 329. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA329. 
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  16. ^ John Robert Taylor (2004). Classical Mechanics. Sausalito CA: University Science Books. Chapter 9, pp. 327 ff. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn=189138922X&lr=&as_brr=0&sig=JVfFlMT5TvXh1I64JAFBFq7pA6s#PPA327,M1. 
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  31. ^ Donald A. Danielson (2003). Vectors and Tensors in Engineering and Physics. Westview Press. p. 86. ISBN 0813340802. http://books.google.com/books?id=A9fiXTC3cxsC&pg=PA275&dq=circular+centripetal+centrifugal+osculating&lr=&as_brr=0#PPA86,M1. 
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  33. ^ As seen by the inertial observer, the plane also may accelerate along its path (dv / dt = d2s / dt2 > 0). For the pilot this acceleration introduces another fictitious force: the Euler force.
  34. ^ Lawrence S. Lerner (1996). Physics for Scientists and Engineers. Jones & Bartlett Publishers. p. 129. ISBN 0763702536. http://books.google.com/books?id=Bsruo5nz1eIC&pg=PA129. 
  35. ^ Louis N. Hand & Janet D. Finch (1998). Finch Analytical Mechanics. Cambridge UK: Cambridge University Press. p. 267. ISBN 0521575729. http://books.google.com/books?id=1J2hzvX2Xh8C&pg=PA267&vq=centrifugal&dq=Hand+inauthor:Finch&lr=&as_brr=0&source=gbs_search_s&sig=ACfU3U33emV_6eJZihu3M6IZKurSt85_eg#PPA267,M1 Finch. 
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  38. ^ John Robert Taylor (2004). Classical Mechanics. Sausalito CA: University Science Books. p. 348–349. ISBN 189138922X. http://books.google.com/books?lr=&as_brr=0&q=%22include+when+you+want+to+use+Newton%27s+second+law+in+a+rotating+frame.+This+is+the+Coriolis%22&btnG=Search+Books. 
  39. ^ The vector cross product of the two orthogonal vectors Ω and R is a vector of magnitude equal to the product of their magnitudes, namely Ω R = vrot, and with direction given by the right-hand rule, in this case found by aligning the thumb with Ω, the index finger with R, and the middle finger normal to these two fingers points in the direction of −vrot.
  40. ^ Louis Bevier Spinney (1911). A Text-book of Physics. Macmillan Co.. pp. 47–49. http://books.google.com/books?id=5zgFAAAAMAAJ&pg=PA47&dq=%22circular+motion%22&lr=&as_brr=0#PPA47,M1. 
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  42. ^ Burgel, B. (1967). "Centrifugal Force". American Journal of Physics 35: 649. doi:10.1119/1.1974204. 
  43. ^ John Robert Taylor (2004). pp. 349ff. Sausalito, Calif.: Univ. Science Books. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1&dq=isbn=189138922X&lr=&as_brr=0&sig=JVfFlMT5TvXh1I64JAFBFq7pA6s#PPA349,M1. 
  44. ^ John Robert Taylor (2005). Classical Mechanics. University Science Books. p. 349. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA348&dq=%22Coriolis+force%22+inauthor:Taylor&lr=&as_brr=0#PPA349,M1. 
  45. ^ Robin McIlveen (1991). Fundamentals of Weather and Climate. Routledge. p. 187. ISBN 0748740791. http://books.google.com/books?id=TmdlBqzl9WIC&pg=PA187&dq=centrifugal+turntable&lr=&as_brr=0#PPA187,M1. 
  46. ^ J. P. Den Hartog (1961). Mechanics (Corrected reprint of 1948 First ed.). Courier Dover Publications. p. 304. ISBN 0486607542. http://books.google.com/books?id=WRXrtu44W9UC&pg=PA309&dq=centrifugal+turntable&lr=&as_brr=0#PPA304,M1. 
  47. ^ a b c Herbert Goldstein (1950). Classical Mechanics. Addison-Wesley. p. 24–25, 61–64. http://books.google.com/books?id=A_RQAAAAMAAJ&q=%22fictitious+one-dimensional+problem%22+intitle:mechanics+inauthor:goldstein&dq=%22fictitious+one-dimensional+problem%22+intitle:mechanics+inauthor:goldstein&lr=&as_brr=0&as_pt=ALLTYPES&ei=is__Se67NIvOkwT9tMigBA&pgis=1. 
  48. ^ a b John Robert Taylor (2005). Classical mechanics. University Science Books. p. 358–359. ISBN 9781891389221. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA300&dq=centripetal+centrifugal+intitle:mechanics+inauthor:taylor&lr=&as_brr=0&as_pt=ALLTYPES&ei=D8D_SfGbJJzwkQTMq9jfBA#PPA359,M1. 
  49. ^ Henry M. Stommel and Dennis W. Moore (1989). An introduction to the Coriolis force. Columbia University Press. p. 28–40. ISBN 9780231066365. http://books.google.com/books?id=-JQx_t3yGB4C&pg=PA36&dq=centrifugal+inauthor:stommel+inauthor:moore&lr=&as_drrb_is=q&as_minm_is=0&as_miny_is=&as_maxm_is=0&as_maxy_is=&as_brr=0&as_pt=ALLTYPES&ei=N34CSs6mJJvIkAT2qKDHBA. 
  50. ^ John Robert Taylor (2005). op. cit.. Sausalito, Calif.: Univ. Science Books. p. 359. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA359. 
  51. ^ This quote is from Taylor (2005). op. cit.. Sausalito, Calif.: Univ. Science Books. p. 359. ISBN 189138922X. http://books.google.com/books?id=P1kCtNr-pJsC&pg=PA359. . The "reincarnation" on the force side of the equation is necessary because, without this force term, observers in the rotating frame would find they could not predict the motion correctly. They would have an incorrect radial equation.
  52. ^ Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson, and Victor Katz (1997). Learn from the masters!. Mathematical Association of America. p. 268–269. ISBN 9780883857038. http://books.google.com/books?id=gqGLoh-WYrEC&pg=PA269&dq=reaction+fictitious+rotating+frame+%22centrifugal+force%22&lr=&as_brr=3&as_pt=ALLTYPES&ei=JUH7SYr3GIzckQSSx4XVBA#PPA269,M1. 
  53. ^ Whiting, J.S.S. (November 1983). "Motion in a central-force field". Physics Education 18 (6): pp. 256–257. doi:10.1088/0031-9120/18/6/102. ISSN 0031-9120. http://www.iop.org/EJ/article/0031-9120/18/6/102/pev18i6p256.pdf. Retrieved May 7, 2009. 

[edit] Further reading

[edit] External links

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