Dictionary Definition
centripetal adj
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)]
User Contributed Dictionary
English
Etymology
From coined by Sir Isaac Newton, from L. centri (centrum) "center" + petere "to fall, rush out"Adjective
 directed or moving towards a centre
 of, relating to, or operated by centripetal force
 In the context of "neuroanatomyof a nerve impulse": directed towards the central nervous system; afferent
Antonyms
Derived terms
Translations
directed or moving towards a centre
 Czech: dostředivý
 Finnish: sentripetaalinen, keskihakuinen
 French: centripète
of, relating to, or operated by centripetal
force
directed towards the central nervous system
 Finnish: keskihakuinen
Extensive Definition
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 motion
The 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.
Common misunderstandings
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 cases
Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.Uniform circular motion
Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.Geometric derivation
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 vectors
Figure 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 righthand 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.
Consequently,
 \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 turn
Figure 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 socalled "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 \ ,
or,
 \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 motion
As 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 xaxis and the unit vector,
measured counterclockwise from the xaxis. 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
and
 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 nonzero 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
 Centrifugal force
 Circular motion
 Coriolis force
 Reactive centrifugal force a reaction force to the centripetal force required by Newton's third law
 Example: circular motion
 FrenetSerret formulas
 Orthogonal coordinates
 Statics
 Kinetics
Further reading
 Centripetal force vs. Centrifugal force, from an online Regents Exam physics tutorial by the Oswego City School District
Outside links
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
Synonyms, Antonyms and Related Words
approaching, asymptotic, centrolineal, concentrating, concurrent, confluent, confocal, connivent, consolidating, converging, focal, meeting, mutually approaching,
radial, radiating, tangent, tangential, unifying, uniting