Derivation: Centripetal Acceleration and Force

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:

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:

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

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 \(\alpha\):

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

Substitute Eq. 5 into Eq. 4:

Rearranging Eq. 6 yields:

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:

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:

Set Eq. 8 and Eq. 9 equal to eliminate the angle \(\Delta \varphi\):

Divide both sides in Eq. 10 by \(\Delta t\):

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 \(v\):

Rearrange Eq. 12 with respect to the 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:

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:

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

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).