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its own right. Any physical force ( Gravity , Electrostatics , Friction , etc.) can be used to supply the centripetal force. The term
''centripetal force'' comes from the Latin words ''centrum'' ("center")
and ''petere'' ("tend towards").


BASIC IDEA


Objects moving in a straight line with constant speed have constant Velocity and require no force to do so, since they experience
no acceleration. However, an object moving in a circle at constant
speed has a changing direction of motion. Since velocity is a vector
of speed and direction, a changing direction implies a changing velocity. The rate of this change in velocity is the centripetal acceleration.
Differentiating the velocity vector gives .

The centripetal acceleration varies with the radius r of the circle and speed v, becoming larger for higher speed and smaller radius. More precisely, the centripetal acceleration is given by

: \mathbf{a}_c = - rac{v^2}{r} \hat{\mathbf{r}} = - rac{v^2}{r} rac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}

The direction of this acceleration is towards the center of the circle.

By Newton's Second Law Of Motion F=ma, a force
F must be applied to a mass m to
produce this acceleration. The amount of force needed to move at
speed v on a circle of radius r is exactly

: \mathbf{F}_c = - rac{m v^2}{r} \hat{\mathbf{R}} = - rac{m v^2}{r} rac{\mathbf{r}}{r} = - m \omega^2 \mathbf{r}

''(where m is mass, v is velocity, r is radius of the circle, and the minus sign denotes that the vector points to the center of the circle and ω = v / r is the Angular Velocity )''. If the applied force is less or more than
F_{c}, the object will move on a larger or smaller circle.

In vector notation we can write:

: \boldsymbol F_c = m \boldsymbol\omega imes (\boldsymbol\omega imes \boldsymbol r ),

where \boldsymbol\omega is the angular velocity vector of the rotation and \boldsymbol r is a vector from an arbitrary point on the rotation axis to the body (with mass m).

An object that moves in a Circular Path undergoes a continuous Acceleration towards the center of the circle. The Net Force that causes this acceleration is called a
centripetal force (from Latin ''centrum'' "center" and ''petere'' "tend towards"). This term refers to the ''effect'' of the force (namely, to maintain the circular motion of the object); the ''origin'' of the centripetal force can be anything that causes a Force to exist. An object can travel in a circle with a constant Speed only if the Net Force acting on it is a centripetal force. (And if the object is traveling in a circle with a varying Speed , the component of the Net Force along the Radius is the centripetal force.)


EXAMPLES


For an orbiting Satellite , the centripetal force is supplied by the gravitational attraction between the satellite and its primary, and acts toward the center of mass which lies in the satellite's primary. 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 Axis Of Rotation . For a spinning object, internal Tensile Stress gives the centripetal force that holds the object together in one piece.


COMMON MISUNDERSTANDINGS


Centripetal force should not be mixed up with Centrifugal Force . In an Inertial Reference Frame (not rotating or accelerating), the centripetal force accelerates a particle in such a way that it moves along a circular path. In a corotating Reference Frame , a particle in circular motion appears to have zero velocity, if the rotation is not accounted for. The centripetal force is exactly cancelled by a centrifugal force that in this approach appears as a Fictitious Force . Centripetal forces are according to Newtonian mechanics true forces, while centrifugal forces only appear relative to rotating frames.

Centripetal force should not be confused with Central Force , either.


DERIVATION

Simply use a Polar Coordinate System , assume a constant radius, and differentiate twice.

Let r(t) be a vector that describes the position of a Point Mass as a function of time. Since we are assuming uniform Circular Motion , let r(t) = R·'''u'''r, where R is a constant (the radius of the circle) and '''u'''r is the Unit Vector pointing from the origin to the point mass. In terms of Cartesian unit vectors:

:u_r = cos( heta)u_x + sin( heta)u_y \,

''Note: unlike in Cartesian Coordinates where the unit vectors are constants, in Polar Coordinates the direction of the unit vectors depend on the angle between the x_axis and the point being described; the angle θ.''

So we differentiate to find velocity:

:v = R rac {du_r}{dt} \,

:v = R rac{d heta}{dt} u_ heta \,

:v = R \omega u_ heta \,

where ω is the angular velocity (just a short way of writing dθ/dt), uθ is the unit vector that is perpendicular to ur that points in the direction of increasing θ. In cartesian terms: uθ = -sin(θ) ux + cos(θ) uy

This result for the velocity is good because it matches our expectation that the velocity should be directed around the circle, and that the magnitude of the velocity should be ωR. Differentiating again, we find that the acceleration, a is:

:a = R \left( rac {d\omega}{dt} u_ heta - \omega^2 u_r ight) \,

Thus, the radial component of the acceleration is:

:a_r = -\omega^2 R \,


GEOMETRIC DERIVATION (NO CALCULUS!)


The circle on the left in Figure 1 shows an object moving on a circle at
constant speed at four different times in its orbit. Its position is
given by \mathbf{R} and its velocity is \mathbf{v}.

The velocity vector is always perpendicular to the position vector; thus,
since \mathbf{R} moves in a circle, so does \mathbf{v}. The circular motion of the velocity is
shown in the circle on the right of Figure 1, along with its
accelaration \mathbf{a}. Just as velocity is the
rate of change of position, acceleration is the rate of change of
velocity.

The position and velocity vectors move in tandem, so they go around their
circles in the same time T. That time equals the
distance traveled divided by the velocity

:
T = rac{2\pi R}{v}


and, by analogy,

:
T = rac{2\pi v}{a}


Setting these two equations equal and solving for a, we get

:
a = rac{v^{2}}{R}


Comparing the two circles in Figure 1 also shows that the acceleration
points towards the center of the \mathbf{R} circle.


SEE ALSO



REFERENCES