# Uniform Circular Motion and its Important Quantities

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

The movement of the moon around the earth,

a circulating electric charge in the magnetic field

or a looping.

These are all nearly *uniform* circular motions. Let's get to know this simplest form of circular motion a little better.

Consider a body that has **mass** \(m\) and is moving in a circle. We say: *The body moves on a circular path*. The circular path has the **radius** \(r\). This is the distance of the body from the center of the circular path.

Let's look at a snapshot at a particular **time** \(t_1\). The body is then located, for example, at the point \(S_1\) on the circular path. At a later time \(t_2\) the body is at \(S_2\). The time it took the body to get from \(S_1\) to \(S_2\) is the **time span** \( \Delta t ~=~ t_2 - t_1 \).

The body has traveled a distance \(\Delta s\), called **arc length**, as it moved from \(S_1\) to \(S_2\). The average speed \(\bar{v}\) is distance traveled \(\Delta s\) per time \(\Delta t\):

**Average velocity**

If we choose \(\Delta t\) to be infinitely small (we note an infinitely small time span as \(\text{d}t\)), then the distance traveled \(\Delta s\) also becomes infinitely small: \(\text{d}s\)). The average velocity \(\bar{v}\) is then also the **instantaneous velocity** \(v\):

**Instantaneous velocity**

The instantaneous velocity \(v\) is the derivative of the position \(s\) with respect to the time \(t\). Eq. 2

is only the *magnitude* of the velocity. We can assign a direction to the magnitude and thus transform the velocity \(v\) into a vector quantity \(\boldsymbol{v}\). We represent vectors **in bold** to distinguish the vector \(\boldsymbol{v}\) from its magnitude \(v\).

The velocity vector touches the circular path at the point \(S_1\). We say: The velocity vector \(\boldsymbol{v}\) is *tangential* to the circular path. We also call this tangential instantaneous velocity **trajectory velocity**. We can therefore assign an orbital velocity to the body at each point of the circular path. At the point \(S_1\) the body has the orbital velocity \(\boldsymbol{v}_1\) and at the location \(S_2\) the orbital velocity is \(\boldsymbol{v}_2\).

A *uniform* circular motion is characterized by the fact that the magnitude \(v\) of the orbital velocity is *constant* at every point:

**Constant velocity**

The direction of the velocity is *different* at each point of the circular path. At the point \(S_1\) the orbital velocity \(\boldsymbol{v}_1\) points in a different direction than the orbital velocity \(\boldsymbol{v}_2\) at the point \(S_2\). Also, if the magnitude of the orbital velocity does not change as the body moves, the direction (due to the curved path) and thus the velocity vector \(\boldsymbol{v}_1\) changes to \(\boldsymbol{v}_2\).

The change \(\delta \boldsymbol{v} \) of the velocity vector is given by:

**Difference of velocity vector**

The **average acceleration** \( \boldsymbol{a} \) is defined as the change \(\delta \boldsymbol{v}\) per time span \(\delta t\):

**Average acceleration**

Here in Eq. 5

, if we consider an infinitesimally small time span \(\text{d} t\), the velocity change \(\text{d} \boldsymbol{v}\) also becomes infinitesimally small. The average acceleration becomes the **instantaneous acceleration** \(\boldsymbol{a}\) (exact acceleration at a point of the circular path). It corresponds to the derivative of the orbital velocity with respect to time:

**Instantaneous acceleration**

Since the velocity *changes* during a circular motion, the body experiences an acceleration according to Eq. 6

.

Here, we do not consider an infinitely small time span, but a *finitely* small time span \(\Delta t\). So we are dealing here with the average velocity 1

. However, we choose the time span so small that the body covers only a small distance \( \Delta s \). Then \( \boldsymbol{v}_1 \) and \( \boldsymbol{v}_2\) are approximately *parallel* to each other. Thus \( \delta \boldsymbol{v} \) is perpendicular to the two velocity vectors.

If \(\Delta \boldsymbol{v}\) is perpendicular to \(\boldsymbol{v}_1\), then \(\Delta \boldsymbol{v}\) points to the center of the circle. \(\delta \boldsymbol{v}\) is parallel to the radius \(r\).

According to Eq. 5

, \(\delta \boldsymbol{v}\) determines the direction of acceleration \( \boldsymbol{a} \). Consequently, \( \boldsymbol{a} \) also points to the center of the circle. The acceleration vector \( \boldsymbol{a} \) is *perpendicular* to the velocity vector \( \boldsymbol{v} \) at every point of the circular path. Let's add \( \boldsymbol{a} \) with the small index \(\text z\): \( \boldsymbol{a}_{\text z} \) to indicate that it is **centripetal acceleration** ("centripetal" = "acting towards the center of the circle"). \( \boldsymbol{a}_{\text z} \) is sometimes also called **radial acceleration** ("radial" = "along the radius").

We have figured out the direction of the centripetal acceleration. Now we have to find out its amount \( a_{\text z} \).

When the body moved from \(S_1\) to \(S_2\), it covered the angle \(\Delta \varphi\). \(\Delta \varphi\) is the angle between the straight lines \(C\,S_1\) and \(C\,S_2\) (see illustration 1). It can also be easily seen that the angle \(\delta \theta\) between \( \boldsymbol{v}_1 \) and \( \boldsymbol{v}_2 \) is equal to the angle \(\delta \varphi \): \(\Delta \theta ~=~ \Delta \varphi \) (see illustration 2).

We now exploit the trigonometric relationship for sines. The sine of an angle is equal to the opposite cathetus divided by the hypotenuse. From the blue triangle (see illustration 2) you can read the following:

**Sine of angle equals velocity difference divided by velocity**

You can read off from the green triangle:

**Sine of angle equals distance difference divided by radius**

If we just equate Eq. 7

and Eq. 8

, we eliminate the angle:

**Velocity difference divided by velocity is equal to distance difference divided by radius**

Divide both sides in Eq. 9

by \(\Delta t\):

**Speed difference by velocity is equal to distance difference divided by radius all divided by time**

In this way you bring the centripetal acceleration \(a_{\text z} = \Delta v / \Delta t\) into play. Let's take advantage of the fact that \( \Delta s / \Delta t\) (distance per time) is the orbital velocity \(v\):

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

Rearrange Eq. 11

with respect to the centripetal acceleration:

**Magnitude of centripetal acceleration**

So e can say: If a body moves on a circular path with radius \(r\), with a constant velocity \(v\), then the body is accelerated towards the center of the circle.

We can also express the velocity \(v\) with the **period** \(T\) and the **frequency** \(f\) of the oscillation. Period \(T\) is the *duration* of one revolution and frequency \(f\) is the number of revolutions per second. The two quantities are related as follows:

**Frequency is the reciprocal of the period**

The circumference is given by \(2\pi r\). This distance is covered within the period \(T\) because \(T\) is the duration of one revolution. The velocity \(v\) is then given as distance per time (here circumference per time):

**Velocity is equal to circumference divided by period**

The centripetal acceleration can also be expressed with the frequency \(f\):

**Circular Motion (Centripetal Acceleration, Frequency)**

Or alternatively with the **circular frequency** \(\omega = 2\pi \, f \):

**Circular Motion (Centripetal Acceleration, Angular Velocity)**

## Centripetal force

If we still bring the **mass** \(m\) of the body into play, we can get a force using the acceleration. According to Newton's second law of motion, force \(F\) is equal to mass \(m\) multiplied by acceleration \(a\): \( F = m \, a \). In our case, we multiply the mass of the body by the centripetal acceleration \(a_{\text z}\) and thus obtain the centripetal force:

**Magnitude of the centripetal force**

Centripetal force points in the same direction as centripetal acceleration, i.e. always to the center of the circle. Or in other words: perpendicular to the orbital velocity \(v\).

With Eq. 16

the centripetal force 17

can also be expressed by means of angular frequency \(\omega\):

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

In order for a body of mass \(m\) with velocity \(v\) to be kept on a circular path of radius \(r\), the centripetal force must act on this body given by Eq. 17

.

If the force acting towards the center of the circle is not sufficient, the body performs a circular motion with a larger radius. For example, when an electrically charged particle moves in a magnetic field, it experiences a magnetic force (Lorentz force) that forces the particle to move in a circle. The magnetic force IS the centripetal force in this case, because it acts towards the center of the circle and keeps the body on the circular path.