# centripetal force

centripetal force
The centripetal force is transmitted through the runners.

The centripetal force (also radial force ) is the external force that has to act on a body so that it moves on a curved path in the inertial system. The centripetal force is directed to the center of the circle of curvature and is perpendicular to the velocity vector in the inertial system. The centripetal force satisfies the principle of action and reaction , since there is a counterforce to it on another body. The centrifugal force , on the other hand, is an apparent force of the same magnitude as the centripetal force, but in the opposite direction.

Without this force, the body would move uniformly in the direction of the instantaneous velocity vector (the tangential vector of the path) according to the law of inertia , as is the case e.g. B. is observed in sparks that come off a grinding wheel.

The movement on a given path, e.g. B. in roller coasters or in road traffic, requires a centripetal acceleration (also radial acceleration ), which results from the current values ​​for the radius of curvature of the path and the speed. The centripetal force required for this is the product of this centripetal acceleration and the mass of the body.

Deviating from the modern definition given here, centripetal force is often the term for the force with which a fixed center of force attracts the body in older texts. This is known today as the central force .

## Etymology and conceptual history

The term centripetal force is derived from petere ( Latin for strive for, go ). It was introduced as vis centripeta by Isaac Newton . Newton coined the name as a contrast to the centrifugal force previously introduced by Christian Huygens . Newton understood this to mean what is now called central force . This means a difference if the paths are not exactly circular.

## Difference between centripetal force and central force

Central force

While a central force is always directed towards the same point (or away from it), the centripetal force points towards the center of the momentary circle of curvature. The centripetal force is only a central force with a pure circular movement . In an elliptical planetary orbit z. B. the central force is directed at each point on the fixed center of force, which is in a focal point of the ellipse. A central force can always be broken down into the two right-angled components centripetal force and tangential force at the location of the body. The centripetal force is directed to the momentary center of the curvature of the orbit and only changes the direction at the speed of the body. The tangential component only changes the amount at the speed, which z. B. the reason for this in planets is that they move faster near the sun than at greater distances.

## Examples

• When a car drives through a curve, this is only possible because a centripetal force is acting towards the inside of the curve. It results from the sum of the lateral forces that arise between the tire and the road surface and act on the vehicle. If this force is missing (e.g. on black ice), the car continues to move in a straight line and is therefore carried out of the curve. The vehicle occupant moves on the same circular path as the car because the seat exerts a centripetal force on them.
• The earth moves (approximately) on a circular orbit around the sun . This circular motion is caused by the gravitational force exerted on the earth by the sun , which in this approximation is both a central force and a centripetal force. More precisely, the orbit of the earth, like the orbits of all planets, is not a circular orbit, but an elliptical orbit (if one disregards the small disturbances caused by the gravitation of the moon and the other planets). The gravity points as a central force on the sun, which is located in one of the elliptical focal points . This central force deviates slightly from the centripetal force that points to the current center of the curvature of the path. The difference between central force and centripetal force is a tangential component that ensures that the earth moves faster near the sun (in the perihelion ) than at a distance from the sun.
• If electrons move perpendicular to a homogeneous magnetic field , they are deflected into a circular path by the Lorentz force perpendicular to the direction of movement and the magnetic field. In this example, the Lorentz force is the centripetal force.
• In air vortices, the centripetal force is the pressure gradient , i. H. there is negative pressure in the vortex core.

## Mathematical derivation

A point moves on a circular path. For the times and the point is in or (snapshots). The speed vectors and illustrate the change in the direction of movement.${\ displaystyle t_ {1}}$${\ displaystyle t_ {2}}$${\ displaystyle P_ {1}}$${\ displaystyle P_ {2}}$${\ displaystyle v_ {1}}$${\ displaystyle v_ {2}}$

### Simple derivation

If a point moves with constant path speed on a circular path, the speed is directed perpendicular to the radius of the circle at any moment . The adjacent drawing illustrates these relationships for the times and${\ displaystyle v}$ ${\ displaystyle r_ {1} = r_ {2} = r}$${\ displaystyle t_ {1}}$${\ displaystyle t_ {2}.}$

First of all, the relationships can be viewed purely geometrically: The arrow shown in blue in the sketch is created by parallel displacement of the arrow Its lengths correspond to the length of the arrow The following applies for the lengths of these three arrows: ${\ displaystyle v '_ {1}}$${\ displaystyle v_ {1}.}$${\ displaystyle v_ {2}.}$

${\ displaystyle v_ {1} = v '_ {1} = v_ {2} = v}$

In addition, the triangles and are similar in the geometric sense, because: ${\ displaystyle Q_ {2} P_ {2} Q_ {1}}$${\ displaystyle P_ {2} MP_ {1}}$

• Both and and and are each sides of an isosceles triangle .${\ displaystyle v_ {2}}$${\ displaystyle v '_ {1}}$${\ displaystyle r_ {2}}$${\ displaystyle r_ {1}}$
• The angles enclosed by the above mentioned sides are equal because the legs of the angles are pairwise orthogonal : is orthogonal to , and due to the parallelism of and are also and orthogonal.${\ displaystyle \ alpha}$${\ displaystyle v_ {2}}$${\ displaystyle r_ {2}}$${\ displaystyle v_ {1}}$${\ displaystyle v '_ {1}}$${\ displaystyle v '_ {1}}$${\ displaystyle r_ {1}}$

From the similarity of the triangles and it follows: ${\ displaystyle Q_ {2} P_ {2} Q_ {1}}$${\ displaystyle P_ {2} MP_ {1}}$

${\ displaystyle {\ frac {\ Delta v} {v}} = {\ frac {\ Delta s} {r}}}$

Multiplied by we get: ${\ displaystyle v}$

${\ displaystyle \ Delta v = \ Delta s \, {\ frac {v} {r}}}$

Division by the time period gives: ${\ displaystyle \ Delta t: = t_ {2} -t_ {1}}$

${\ displaystyle {\ frac {\ Delta v} {\ Delta t}} = {\ frac {v} {r}} {\ frac {\ Delta s} {\ Delta t}}}$,

If now sufficiently small is chosen, the following applies: ${\ displaystyle \ Delta t}$

• The path covered by the object corresponds to a section on the circular path and is the path speed of the object.${\ displaystyle \ Delta s}$${\ displaystyle {\ tfrac {\ Delta s} {\ Delta t}} = v}$
• The centripetal acceleration is the acceleration that the object experiences towards the center of the circle.${\ displaystyle a _ {\ mathrm {Z}} = {\ tfrac {\ Delta v} {\ Delta t}}}$

Then the equation tends towards:

${\ displaystyle a _ {\ mathrm {Z}} = {\ frac {v ^ {2}} {r}} = v ^ {2} \ cdot \ kappa}$,

with the curvature of the path. If the circling object is not just a geometric point, but an object with mass , there must be a force that keeps the object on its path. The force must be directed towards the center of the circle and is called "centripetal force". According to Newton's 2nd law, the following applies to the amount of the centripetal force : ${\ displaystyle \ kappa}$ ${\ displaystyle m}$${\ displaystyle F _ {\ mathrm {Z}}}$

${\ displaystyle F _ {\ mathrm {Z}} = m \, a _ {\ mathrm {Z}}}$

This centripetal force acts on every body with the mass that moves with the speed on a path with the local radius of curvature . ${\ displaystyle m}$${\ displaystyle v}$ ${\ displaystyle r}$

### Vector illustration

For a point that moves on any (smooth) curve in space, there is a clearly defined oscillating ball for every point on the path , so that the path follows the surface of the sphere up to the third spatial derivative. The center of the sphere is the center of curvature. Together with the path tangent, which also indicates the direction of the velocity vector, it determines the current path plane. This intersects the oscillating ball in a great circle , on which the point is in the state of a circular movement around the center of curvature at the observed moment. The axis of this circular movement is at its center perpendicular to the plane of the path. The velocity vector and the vector from the center of curvature to the location of the point under consideration are perpendicular to one another and, together with the angular velocity vector of the circular motion, satisfy the equation ${\ displaystyle {\ vec {v}}}$${\ displaystyle {\ vec {r}}}$${\ displaystyle {\ vec {\ omega}}}$

${\ displaystyle {\ vec {v}} = {\ vec {\ omega}} \ times {\ vec {r}}}$.

If the point is not accelerated in a tangential direction, the first derivative of vanishes . The acceleration then points to the center of curvature and indicates the centripetal acceleration . ${\ displaystyle {\ vec {\ omega}}}$${\ displaystyle {\ vec {a}} = {\ frac {\ mathrm {d} {\ vec {v}}} {\ mathrm {d} t}}}$${\ displaystyle {\ vec {a}} _ {\ mathrm {Z}}}$

${\ displaystyle {\ vec {a}} _ {\ mathrm {Z}} = {\ vec {\ omega}} \ times {\ frac {\ mathrm {d} {\ vec {r}}} {\ mathrm { d} t}} = {\ vec {\ omega}} \ times {\ vec {v}}}$.

Since the vectors and are perpendicular to each other, the amounts can be used. With the same equation results for the amount of centripetal acceleration as above: ${\ displaystyle {\ vec {v}}}$${\ displaystyle {\ vec {\ omega}}}$${\ displaystyle \ omega = v / r}$

${\ displaystyle a _ {\ mathrm {Z}} = \ omega \, v = {\ frac {v ^ {2}} {r}}}$.

### Derivation in the Cartesian coordinate system

First for a uniform circular movement of a point at speed on a circular path with a radius : In an xy coordinate system in the circular plane with the origin in the center of the circle, the point has the coordinates (with a suitable choice of the time zero point and ) ${\ displaystyle v}$${\ displaystyle r}$${\ displaystyle \ omega = {\ tfrac {v} {r}}}$

${\ displaystyle {\ vec {r}} = {\ begin {pmatrix} r \, \ cos \ omega t \\ r \, \ sin \ omega t \ end {pmatrix}}}$.

Its acceleration is the second derivative

${\ displaystyle {\ vec {a}} = {\ begin {pmatrix} - \ omega ^ {2} r \; \ cos \ omega t \\ - \ omega ^ {2} r \; \ sin \ omega t \ end {pmatrix}}}$.

thats why

${\ displaystyle {\ vec {a}} _ {Zp} = - \ omega ^ {2} \; {\ vec {r}}}$,

or according to the amount

${\ displaystyle a_ {Zp} = \ omega ^ {2} \; r = {\ frac {v ^ {2}} {r}} = v ^ {2} \ cdot \ kappa}$.

This derivation uses a specific coordinate system to represent the simplest possible way. The result, however, is an equation between coordinate-independent quantities and is therefore valid in every coordinate system. The derivation is also spatially and temporally local and therefore applies to any curved movement and variable path speed, if the local radius of curvature is used for r and the current path speed for v.

## Applications

When moving in everyday life, the centripetal force is often transmitted through static friction . With sliding friction, the frictional force is directed against the sliding speed and does not allow a controlled movement. The centripetal acceleration must have the following condition:

${\ displaystyle a_ {Zp} \ leqq \ mu \ cdot g}$

with the coefficient of static friction and the acceleration due to gravity . Studies show that a centripetal acceleration of 4 m / s 2 is seldom exceeded when driving a car normally . In the case of a motorcycle, this corresponds to an incline of around 20 degrees. This is still far from the physical limits on a dry road, but it shows that humans are able to adjust their speed so that the product of the driving speed squared and the curvature remains within the specified limits. ${\ displaystyle \ mu}$${\ displaystyle g}$

For many problems, determining the radius of curvature can be simplified. If the external forces are known, the equation of motion provides acceleration, speed and position of the center of mass. The train, e.g. B. the movement of the center of gravity of a vehicle is viewed in the projection onto a reference surface. In this, the component of the acceleration perpendicular to the velocity is the centripetal acceleration sought. In the simplest case, the reference surface is the xy plane of the inertial reference system. ${\ displaystyle {\ vec {a}} = {\ vec {F}} / m}$

In the experiment, the acceleration is usually measured in components of a vehicle-fixed coordinate system. In order to get the acceleration parallel to the reference plane, the portion of the acceleration due to gravity that is measured in the transverse direction based on the roll angle must be corrected.

## Individual evidence

1. M. Alonso, EJ Finn: Physics , 3rd edition
2. Bruno Assmann, Peter Selke: Technical Mechanics . 13th edition. Oldenbourg Wissenschaftsverlag GmbH, 2004, ISBN 3-486-27294-2 , p. 79 ( limited preview in Google Book search).
3. Principia , Definition 5 at the beginning of the work
4. ^ I. Bernard Cohen: Newton's Third Law and Universal Gravity. In: Paul B. Scheurer, G. Debrock: Newtons Scientific and Philosophical Legacy. Kluwer, Dordrecht 1988, p. 47. ISBN 90-247-3723-0
5. ^ I. Bernard Cohen: Introduction to Newton's Principia. London 1971, pp. 53, 296.
6. Klaus Becker (Ed.): Making subjective driving impressions visible . expert verlag, 2000, ISBN 3-8169-1776-3 , p. 44 ( limited preview in Google Book search).
7. Bernt Spiegel: The upper half of the motorcycle . 5th edition. Motorbuch Verlag, 2006, ISBN 3-613-02268-0 , p. 43-44 .