1 tending to move toward a center; "centripetal force" [ant: centrifugal]
2 tending to unify [syn: unifying(a)]
3 of a nerve fiber or impulse originating outside and passing toward the central nervous system; "sensory neurons" [syn: receptive, sensory(a)]
EtymologyFrom coined by Sir Isaac Newton, from L. centri- (centrum) "center" + petere "to fall, rush out"
directed or moving towards a centre
of, relating to, or operated by centripetal force
directed towards the central nervous system
- Finnish: keskihakuinen
The centripetal force is the external force required to make a body follow a curved path. Hence centripetal force is a force requirement, not a particular kind of force, like gravity or electromagnetic force. Any force (gravitational, electromagnetic, etc.) (or combination of forces) can act to provide a centripetal force. An example for the case of uniform circular motion is shown in Figure 1. Centripetal force is directed inward, toward the center of curvature of the path. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards", "aim at."), and can also be derived from Isaac Newton's original definitions described in Philosophiae Naturalis Principia Mathematica.
Centripetal force should not be confused with centrifugal force (see the Common misunderstandings section below).
Simple example: uniform circular motionThe velocity vector is defined by the speed and also by the direction of motion. Objects experiencing no net force do not accelerate and, hence, move in a straight line with constant speed: they have a constant velocity. However, even an object moving in a circle at constant speed has a changing direction of motion. The rate of change of the object's velocity vector in this case is the centripetal acceleration. See Figure 1.
The centripetal acceleration varies with the radius R of the path and speed v of the object, becoming larger for greater speed (at constant radius) and smaller radius (at constant speed). If an object is traveling in a circle with a varying speed, its acceleration can be divided into two components, a radial acceleration (the centripetal acceleration that changes the direction of the velocity) and a tangential acceleration that changes the magnitude of the velocity.
Below a number of examples of increasing complexity are discussed, and formulas for the motion are derived.
How is centripetal force provided?Centripetal force is inferred from the trajectory of the object, without regard for how the path was arrived at (regardless of the origin of the forces involved). The studies of trajectories and the forces they imply is kinematics, while the study of which motions result from given physical forces is kinetics, the other branch of dynamics.
Supposing the analysis of a trajectory (kinematics) has concluded that for an object to follow the observed path a centripetal force is required, one might reasonably ask the (kinetic) question: "Where is the centripetal force coming from?"
For a satellite in orbit around a planet, the centripetal force is supplied by the gravitational attraction between the satellite and the planet, and acts toward the center of mass of the two objects. For an object at the end of a rope rotating about a vertical axis, the centripetal force is the horizontal component of the tension of the rope, which acts towards the center of mass between the axis of rotation and the rotating object. For a spinning object, internal tensile stress is the centripetal force that holds the object together in one piece.
Centripetal force should not be confused with centrifugal force. Centripetal force is a force requirement deduced from an observed trajectory, not a kinetic force like gravity or electrical forces. Centripetal force requirements may be deduced from a trajectory in any frame of reference (although the trajectory of an object and the deduced centripetal force will vary from one frame to another). Because centripetal force is inferred from an established trajectory, and is not used to deduce a trajectory from a physical situation, centripetal force is not included in the inventory of forces that are used in applying Newton's laws F = m a to calculate a trajectory.
Centrifugal force, on the other hand, is a fictitious force that arises only when motion is described or experienced in a rotating reference frame, and it does not exist in an inertial frame of reference. In both inertial frames and rotating frames of reference one uses Newton's laws of motion, such as F = ma, but inertial frames never use fictitious forces, while rotating frames must include fictitious forces that express the effects of rotation, in particular, the centrifugal force and the Coriolis force. Examples are provided in the article on centrifugal force.
Centripetal force should not be confused with central force, either. To reiterate, centripetal force is a force requirement necessary for a curved trajectory to be possible: it is not a type of force, such as a nuclear or gravitational force. In contrast, central forces does refer to a type of force, more exactly to a class of physical forces between two objects that meet two conditions: (1) their magnitude depends only on the distance between the two objects and (2) their direction points along the line connecting the centers of these two objects. Examples of central forces include the gravitational force between two masses and the electrostatic force between two charges.
As an example relating these terms, the centripetal force implied by the circular motion of an object often is provided by a central force.
Analysis of several casesBelow are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.
Uniform circular motionUniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.
The circle on the left in Figure 2 shows an object moving on a circle at constant speed at two different times in its orbit. Its position is given by R and its velocity is v.
The velocity vector v is always perpendicular to the position vector (since the velocity vector is always tangent to the R circle); thus, since R moves in a circle, so does v. The circular motion of the velocity is shown in the circle on the right of Figure 2, along with its acceleration a. Just as velocity is the rate of change of position, acceleration is the rate of change of velocity.
Since the position and velocity vectors move in tandem, they go around their circles in the same time T. That time equals the distance traveled divided by the velocity
T = \frac
and, by analogy,
T = \frac
Setting these two equations equal and solving for a, we get
a = \frac
The angular rate of rotation in radians / s is:
- \omega = \frac \ .
Comparing the two circles in Figure 2 also shows that the acceleration points toward the center of the R circle. For example, in the left circle in Figure 2, the position vector R pointing at 12 o'clock has a velocity vector v pointing at 9 o'clock, which (switching to the circle on the right) has an acceleration vector a pointing at 6 o'clock. So the acceleration vector is opposite to R and toward the center of the R circle.
Derivation using vectorsFigure 3 shows the vector relationships for uniform circular motion. The rotation itself is represented by the vector Ω, which is normal to the plane of the orbit (using the right-hand rule) and has magnitude given by:
- \left| \boldsymbol \right| = \frac = \omega \ ,
with θ the angular position at time t. In this subsection, dθ / dt is assumed constant, independent of time t. The displacement of the particle in time dt along the circular path is
- d\boldsymbol = \boldsymbol \times \mathbf ( t ) dt \ ,
which, by properties of the vector cross product, has magnitude r d θ and is in the direction tangent to the circular path.
- \frac = \frac = \frac \ .
In other words,
- \mathbf\ \stackrel\ \frac = \frac = \boldsymbol \times \mathbf ( t )\ .
Differentiating with respect to time,
- \mathbf\ \stackrel\ \frac = \boldsymbol \times \frac = \boldsymbol \times \left[ \boldsymbol \times \mathbf (t)\right] \ .
Lagrange's formula states:
- a × (b × c) = b(a · c) − c(a · b).
Applying Lagrange's formula with the observation that Ω • r (t) = 0 at all times,
- \mathbf = - ^2 \mathbf (t) \ .
In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:
- |\mathbf| = |\mathbf| \left ( \frac \right) ^2 = R ^2\
where vertical bars |…| denote the vector magnitude, which in the case of r(t) is simply the radius R of the path. This result agrees with the previous section if the substitution is made for rate of rotation in terms of the period of rotation T:
- \frac = \omega = \frac = \frac \ .
Not surprisingly, this result also agrees with that of the nonuniform circular motion when the rate of rotation is made constant.
A merit of the vector approach is that it is manifestly independent of any coordinate system.
Example: The banked turnFigure 4 shows a ball in circular motion on a banked curve. The curve is banked at an angle θ from the horizontal, and the surface of the road is considered to be slippery. The object is to find what angle θ the bank must have so the ball does not slide off the road (the so-called "angle of bank"). Intuition tells us that on a flat curve with no banking at all, the ball will simply slide off the road; while with a very steep banking, the ball will slide to the center unless it travels the curve rapidly.
The right side of Figure 4 indicates the forces on the ball. There are two forces; one is the force of gravity vertically downward through the center of mass of the ball m g where m is the mass of the ball and g is the gravitational acceleration; the second is the upward normal force exerted by the road perpendicular to the road surface m an. The horizontal net force on the ball is the horizontal component of the force from the road, which has magnitude Fh = m an sin θ. The vertical component of the force from the road must counteract the gravitational force, that is Fv = m an cos θ = m g, so m an = m g / cos θ. Accordingly one finds the net horizontal force to be:
- F_h = \frac \mathrm \ \theta = mg \mathrm\ \theta \ .
On the other hand, at velocity v on a circular path of radius R, kinematics says that the force needed to turn the ball continuously into the turn is the radially inward centripetal force Fc of magnitude:
- F_c = m a _c = m\frac \ .
Consequently the ball is in a stable path when the angle of the road is set to satisfy the condition:
- mg \mathrm\ \theta = m\frac \ ,
- \mathrm\ \theta = \frac \ .
As the angle of bank θ approaches 90°, the tangent function approaches infinity, allowing larger values for v2 / R. In words, this equation states that for faster speeds (bigger v) the road must be banked more steeply (a larger value for θ), and for sharper turns (smaller R) the road also must be banked more steeply, which accords with intuition. When the angle θ does not satisfy the above condition, the horizontal component of force exerted by the road does not provide the correct centripetal force, and an additional frictional force tangential to the road surface is called upon to provide the difference. If friction cannot do this (that is, the coefficient of friction is exceeded), the ball slides to a different radius where the balance can be realized.
These ideas apply to air flight as well. See the FAA pilot's manual.
Nonuniform circular motionAs a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown in Figure 5. This case is used to demonstrate a derivation strategy based upon a polar coordinate system.
Let r(t) be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let r(t) = R·ur, where R is a constant (the radius of the circle) and ur is the unit vector pointing from the origin to the point mass. The direction of ur is described by θ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, uθ is perpendicular to ur and points in the direction of increasing θ. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted i and j respectively:
- ur = cos(θ) i + sin(θ) j
- uθ = −sin(θ) i + cos(θ) j.
Note: unlike the Cartesian unit vectors i and j, which are constant, in polar coordinates the direction of the unit vectors ur and uθ depend on θ, and so in general have non-zero time derivatives.
We differentiate to find velocity:
- \mathbf = R \frac = R \frac \left( \mathrm\ \theta \ \mathbf + \mathrm\ \theta \ \mathbf\right)
- = R \frac \left( -\mathrm\ \theta \ \mathbf + \mathrm\ \theta \ \mathbf\right)\,
- = R \frac \mathbf \,
- = R \omega \mathbf \,
where ω is the angular velocity dθ/dt.
This result for the velocity matches expectations that the velocity should be directed tangential to the circle, and that the magnitude of the velocity should be ωR. Differentiating again, and noting that
- = -\frac \mathbf = - \omega \mathbf \ ,
we find that the acceleration, a is:
- \mathbf = R \left( \frac \mathbf - \omega^2 \mathbf \right) \ .
Thus, the radial and tangential components of the acceleration are:
- \mathbf_ = - \omega^ R \ \mathbf=- \frac \ \mathbf \ and \ \mathbf_=R \ \frac \ \mathbf = \frac \ \mathbf \ ,
where |v| = Rω is the magnitude of the velocity (the speed).
These equations express mathematically that, in the case of an object that moves along a circular path with a changing speed, the acceleration of the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, or tangential component, that changes the speed.
General planar motion
- Centripetal force vs. Centrifugal force, from an online Regents Exam physics tutorial by the Oswego City School District
centripetal in Czech: Dostředivá síla
centripetal in Danish: Centripetalkraft
centripetal in German: Zentripetalkraft
centripetal in Spanish: Fuerza centrípeta
centripetal in French: Force centripète
centripetal in Korean: 구심력
centripetal in Indonesian: Gaya sentripetal
centripetal in Italian: Forza centripeta
centripetal in Hebrew: כוח צנטריפטלי
centripetal in Hungarian: Centripetális gyorsulás
centripetal in Japanese: 回転運動
centripetal in Norwegian: Sentripetalkraft
centripetal in Polish: Siła dośrodkowa
centripetal in Russian: Центростремительная сила
centripetal in Simple English: Centripetal force
centripetal in Slovenian: Centripetalna sila
centripetal in Finnish: Sentripetaalivoima
centripetal in Swedish: Centripetalkraft
centripetal in Urdu: مرکز مائل قوت
centripetal in Chinese: 向心力
centripetal in Slovak: Dostredivá sila