Centripetal force

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Th centripetal force is the external force required to make a body follow a circular path at constant speed. The force is directed inward, toward the center of the circle. Hence it is a force requirement, not a particular kind of force. Any force (gravitational, electromagnetic, etc.) can act as a centripetal force. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards").

The centripetal force always acts perpendicular to the direction of motion of the body. In the case of an object that moves along a circular arc with a changing speed, the net force on the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal force), and a parallel, or tangential component, that changes the speed.

1 Basic formula
2 Examples
3 Common misunderstandings
4 Geometric derivation
5 Derivation using calculus
6 See also
7 References & External Links

[edit] Basic formula

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, 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 is the centripetal acceleration.

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

$\mathbf{a}_c = -\frac{v^2}{r} \hat{\mathbf{r}} = -\frac{v^2}{r} \frac{\mathbf{r}}{r} = -\omega^2 \mathbf{r}$

where ω = v / r is the angular velocity. The negative sign indicates that the direction of this acceleration is towards the center of the circle, i.e., opposite to the position vector r. (We assume that the origin of r is the center of the circle.)

By Newton's second law of motion F = ma, a physical 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:

$\mathbf{F}_c = -\frac{m v^2}{r} \hat{\mathbf{r}} = -\frac{m v^2}{r} \frac{\mathbf{r}}{r} = -m \omega^2 \mathbf{r} = m \boldsymbol\omega \times (\boldsymbol\omega \times \boldsymbol r )$

where the formula has been written in several equivalent ways; here, $\hat{\mathbf{r}}$ is the unit vector in the r direction and ω is the angular velocity vector. Again, the negative sign indicates that the direction of the force is inwards, towards the center of the circle and opposite to the direction of the radius vector r. If the applied force is less or more than F_c, the object will "slip outwards" or "slip inwards", moving on a larger or smaller circle, respectively.

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.

[edit] Examples

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.

[edit] Common misunderstandings

Centripetal force should not be confused with centrifugal force. The centrifugal force is a fictitious force that arises from being in a rotating reference frame. To eliminate all such fictitious forces, one needs to be in a non-accelerating reference frame, i.e., in an inertial reference frame. Only then can one safely use Newton's laws of motion, such as F = ma.

Centripetal force should not be confused with central force, either. Central forces are 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 centres of these two objects. Examples of central forces include the gravitational force between two masses and the electrostatic force between two charges. The centripetal force maintaining an object in circular motion is often a central force.

Lastly, centripetal force is often mispoken for "centriphical" force. There is no such thing as centriphical force.

[edit] Geometric derivation

Figure 1: The position and velocity vectors both move in a circle.

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 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 1, 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

$T = \frac{2\pi R}{v}$

and, by analogy,

$T = \frac{2\pi v}{a}$

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

$a = \frac{v^{2}}{R}$

Comparing the two circles in Figure 1 also shows that the acceleration points toward the center of the R circle. For example, in the left circle in Figure 1, 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.

[edit] Derivation using calculus

Another derivation strategy is to 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. This direction is described by θ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. In terms of cartesian unit vectors in the x and y directions (i and j respectively):

u_R = cos(θ)i + sin(θ)j

Note: unlike cartesian unit vectors, which are constant, in polar coordinates the direction of the unit vectors depend on θ, and so in general have non-zero time derivatives.

We differentiate to find velocity:

$\mathbf{v} = r \frac {d\mathbf{u_r}}{dt} \,$

$\mathbf{v} = r \frac{d\theta}{dt} \mathbf{u_\theta} \,$

$\mathbf{v} = r \omega \mathbf{u_\theta} \,$

where ω is the angular velocity dθ/dt, and u_θ is the unit vector that is perpendicular to u_R and points in the direction of increasing θ. In Cartesian terms, u_θ = −sin(θ)i + cos(θ)j.

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:

$\mathbf{a} = r \left( \frac {d\omega}{dt} \mathbf{u_\theta} - \omega^2 \mathbf{u_r} \right) \,$

Thus, the radial component of the acceleration is:

a_R = −ω²r

[edit] See also

[edit] References & External Links

Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers, 6th ed., Brooks/Cole. ISBN 0-534-40842-7.
Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics, 5th ed., W. H. Freeman. ISBN 0-7167-0809-4.
Centripetal force vs. Centrifugal force, from an online Regents Exam physics tutorial by the Oswego City School District