# Centripetal force

The centripetal force is the force pulling an object toward the center of a circular path as the object goes around the circle. An object can travel in a circle only if there is a centripetal force on it.

In the case of an orbiting satellite the centripetal force is its weight and acts towards the satellite's primary; in the case of an object at the end of a rope, the centripetal force is the tension of the rope and acts towards whatever the rope is anchored to.

Centripetal force must not be confused 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 has zero velocity. In this case, the centripetal force appears to be exactly cancelled by a pseudo-force, the centrifugal force. Centripetal forces are true forces, appearing in inertial reference frames; centrifugal forces appear only in rotating frames.

Centripetal force must not be confused with central force either.

Objects moving in a straight line with constant speed also have constant velocity. However an object moving in an arc with constant speed has a changing direction of motion. As 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 direction of this acceleration towards the center of the circle.

[itex] \mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}[itex]

By Newton's second law of motion, as there is an acceleration there has to be a force in the direction of the acceleration. This is the centripetal force, and is equal to:

[itex] \mathbf{F} = - \frac{m v^2}{r} \frac{\mathbf{r}}{r} = - m \omega^2 \mathbf{r}[itex]

(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)

### Proof of law

Proving the formula is a trivial matter. Simply use a polar coordinate system, assume a constant radius, and take two derivatives.

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·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. In terms of Cartesian unit vectors:

[itex]u_r = cos(\theta)u_x + sin(\theta)u_y \, [itex]

Note well: 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 take the first derivative to find velocity:

[itex]v = R \frac {du_r}{dt} \, [itex]
[itex]v = R \frac{d\theta}{dt} u_\theta \, [itex]
[itex]v = R \omega u_\theta \, [itex]

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. Taking another derivative, we find that the acceleration, a is:

[itex]a = R \left( \frac {d\omega}{dt} u_\theta - \omega^2 u_r \right) \, [itex]

since the motion is uniform, the magnitude of v is constant, and thus there can be no dv/dt that points in the same direction as v. This fact implies ω is constant which simplifies the equation to:

[itex]a = -R \omega^2 u_r \, [itex]

A simple substitution brings us to the equation above.

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy