# Derivation Centripetal Acceleration and Force

**Level 2**requires school mathematics. Suitable for pupils.

In the following we want to derive the formula for the **centripetal acceleration** \( a_{\text{z}} \) (also called **radial force**). We are dealing here with a *uniform circular motion*, that is the magnitude \( v \) of the **orbital velocity** \( \boldsymbol{v} \) of the body is constant at every point on the circular path.

## Derivation of the centripetal acceleration

Consider a body moving with a constant magnitude of velocity \( v \) on a circular path with **radius** \( r \).

The velocity vector \( \boldsymbol{v} \) (shown here in bold) is a vector whose direction is *tangent* to the circular path at every point along the circular path. Although the magnitude \( v \) of the velocity is constant for uniform circular motion, its *direction* changes as the body circles.

The change of velocity results in an **acceleration** \( \boldsymbol{a}\) of the body. Acceleration is defined as the derivative of velocity \( \boldsymbol{v} \) with respect to **time** \( t \). To make the derivation 'illustrative', we do not consider the derivative, but its approximation:

**Acceleration is equal to velocity change per time**

If you consider a very small (infinitely small) time span \( \Delta t \), you get the exact value of the acceleration (called: instantaneous acceleration).

Let us assume that the body is at a **time** \(t_1\) at the **position** \(S_1\) on the circular path and has the **orbital velocity** \(\boldsymbol{v}_1\). At a **later time** \(t_2\) the body is at the **position** \(S_2\) on the circular path. It has moved a little further within the time \( \Delta t \) and has covered the **distance** \(\Delta s\).

At the position \(S_2\) the body has a different **velocity vector** \(\boldsymbol{v}_2\), because the direction of the velocity has changed (because the body moves on a *curved* path). The magnitude \(v\) naturally remains the same for a uniform circular motion:

**Magnitude of the velocity is constant**

The difference between the two velocity vectors indicates the change in velocity:

**Change of the velocity direction**

If we consider a very small **time span** \(\Delta t\), then the distance traveled \(\Delta s\) also becomes small. So small that \(\boldsymbol{v}_1\) and \(\boldsymbol{v}_2\) are approximately *parallel* to each other. (After an infinitely small time span, the velocity vectors are not approximately but exactly parallel). Then \(\delta \boldsymbol{v}\) is *perpendicular* to both \(\boldsymbol{v}_1\) and \(\boldsymbol{v}_2\) and consequently points towards the circle center \(C\).

But if the velocity change \(\delta \boldsymbol{v}\) points to the center of the circle, according to Eq. 1

the acceleration vector \(\boldsymbol{a}\) is also directed to the center of the circle. To indicate this, we rename \(\boldsymbol{a}\) to \(\boldsymbol{a}_{\text z}\). Index \(\text z\) because of centripetal acceleration in German (zentripetal = "directed to the center of the circle"). So we know immediately in which direction the acceleration vector always points.

We still have to find out the **magnitude** \(a_{\text z}\) of the acceleration vector \(\boldsymbol{a}_{\text z}\). In the following we consider only the magnitudes. The direction of the acceleration vector is already known.

Now we have two right triangles (see illustration 3):

A smaller triangle formed by \(\boldsymbol{v}_1\), \(\boldsymbol{v}_2\) and \(\Delta \boldsymbol{v}\).

And a larger triangle formed by the straight lines along the radius \(r\) and \(\Delta s\).

During the movement from \(S_1\) to \(S_2\) the body has covered the **angle** \(\Delta \varphi\). By a geometrical consideration it can be shown that the enclosed angle \(\delta \theta\) beween \(\boldsymbol{v}_1\) and \(\boldsymbol{v}_2\) is equal to the angle \(\delta \varphi\) (see illustration 3).

The angle between \(\boldsymbol{v}_1\) and the straight line \(C\,S_1\) is (approximately) 90 degrees. This angle is composed of \(\Delta \theta\) and \(\alpha\):

**Delta Theta plus Alpha equals 90 degrees**

The sum of angles in a triangle is 180 degrees. Then the angle \(\alpha\) is given by:

**Alpha is equal to 180 degrees minus 90 degrees - Delta Phi is equal to 90 degrees minus Delta Phi**

Substitute Eq. 5

into Eq. 4

:

**90 degrees equals Delta Theta plus 90 degrees minus Delta Phi**

Rearranging Eq. 6

yields:

**Delta Theta equals Delta Phi**

With the angles we are done. We can now exploit the equality of the angles in Eq. 7

! In the larger triangle, we can apply the trigonometric relationship for sine (opposite cathetus divided by hypotenuse). Here the opposite cathetus is \(\Delta s\) and the hypotenuse is the **radius** \(r\) of the circular path:

**Sine of the angle is equal to distance divided by radius**

To the smaller triangle we also apply the trigonometric relation for sine. Here the opposite cathetus is \(\Delta v\) and the hypotenuse is the magnitude \(v\) of the orbital velocity:

**Sine of angle equals change in velocity divided by speed**

Set Eq. 8

and Eq. 9

equal to eliminate the angle \(\Delta \varphi\):

**Change in distance divided by radius is equal to change in velocity divided by speed**

Divide both sides in Eq. 10

by \(\Delta t\):

**Velocity divided by radius is equal to acceleration divided by velocity**

This is how you bring the **centripetal acceleration** \(a_{\text z} = \Delta v / \Delta t\) into play. And \( \Delta s / \Delta t\) (distance per time) is the orbital velocity \(v\):

**Velocity divided by radius is equal to centripetal acceleration divided by velocity**

Rearrange Eq. 12

with respect to the centripetal acceleration:

**Magnitude of centripetal acceleration**

## Derivation of the centripetal force

To get the magnitude of the **centripetal force** \(F_{\text z}\), we have to multiply the centripetal acceleration given by Eq. 13

with the **mass** \(m\) of the body. The Second Law of Motion \(F = m \, a\) tells us how force is related to acceleration:

**Magnitude of the centripetal force**

Since mass \(m\) is just a number (not a vector), we can conclude from \(\boldsymbol{F} = m \, \boldsymbol{a}\) that the centripetal force \(\boldsymbol{F}_{\text z}\) like the centripetal acceleration \(\boldsymbol{a}_{\text z}\) points to the center of the circle.

We can alternatively express the centripetal force 14

using the **angular velocity** \(\omega\). For this purpose, we use the relationship between the orbital velocity \(v\) and the angular velocity \(\omega\), namely: \( v = \omega \, r \). We square the speed: \( v^2 = \omega^2 \, r^2 \) and use it in Eq. 14

:

**Circular motion (angular velocity, centripetal force, radius)**

We can cancel the radius \(r\) once:

**Centripetal force via angular velocity**

Since the angular velocity vector \(\boldsymbol{\omega}\) is perpendicular to the circular path, it is also perpendicular to the centripetal force \( \boldsymbol{F}_{\text z} \) (and of course to the centripetal acceleration).