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 \(\Delta t\):

Formula anchor$$ \begin{align} \bar{v} ~=~ \frac{\Delta s}{\Delta t} \end{align} $$

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

Formula anchor$$ \begin{align} v ~=~ \frac{\text{d}s}{\text{d}t} \end{align} $$

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

Formula anchor$$ \begin{align} v ~=~ v_1 ~=~ v_2 \end{align} $$

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:

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

Average acceleration

Formula anchor$$ \begin{align} \bar{\boldsymbol{a}} ~=~ \frac{\Delta \boldsymbol{v}}{\Delta t} \end{align} $$

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

Formula anchor$$ \begin{align} \boldsymbol{a} ~=~ \frac{\text{d}\boldsymbol{v}}{\text{d}t} \end{align} $$

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

Formula anchor$$ \begin{align} \sin(\Delta \varphi) ~=~ \frac{ \Delta v }{v} \end{align} $$

You can read off from the green triangle:

Sine of angle equals distance difference divided by radius

Formula anchor$$ \begin{align} \sin(\Delta \varphi) ~=~ \frac{ \Delta s }{r} \end{align} $$

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

Formula anchor$$ \begin{align} \frac{ \Delta v }{v} ~=~ \frac{ \Delta s }{r} \end{align} $$

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

Formula anchor$$ \begin{align} a_{\text z} \, \frac{1}{v} ~=~ v \, \frac{1}{r} \end{align} $$

Rearrange Eq. 11 with respect to the centripetal acceleration:

Formula anchor$$ \begin{align} a_{ \text z } ~=~ \frac{{\class{blue}{v}}^2}{ r } \end{align} $$

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:

Formula anchor$$ \begin{align} f ~=~ \frac{1}{T} \end{align} $$

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

Formula anchor$$ \begin{align} v ~=~ \frac{2\pi \, r}{T} ~=~ 2\pi \, r \, f \end{align} $$

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

Formula anchor$$ \begin{align} a_{ \text z } ~=~ {\class{red}{\omega}}^2 \, r \end{align} $$

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:

Formula anchor$$ \begin{align} F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r } \end{align} $$

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 \(\omega\):

Formula anchor$$ \begin{align} F_{\text z} ~=~ \class{brown}{m} \, {\class{red}{\omega}}^2 \, r \end{align} $$

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.

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