Centripetal Acceleration and Centripetal Force

Every four years, we witness the best athletes from around the world gather in one city to compete in the Olympics. The athletes have to be in peak shape. But did you know they also rely on science to optimize their performance? For example, athletes who compete in the hammer throw rely on physics to help them optimize their throwing distance as they swing a seven-kilogram iron ball attached to a steel wire before releasing it. Each competitor hopes their throw travels the farthest distance, enabling them to take the gold medal home. What these athletes do by letting go of the hammer at the precise right moment so that it zips in a straight line is a prime example of both centripetal acceleration and centripetal force. Why? The athletes enter a circle, begin to spin, and eventually release the hammer after about four or five rotations. Every rotation means the athlete is accelerating, thereby increasing velocity and optimizing the distance the hammer travels. Impressive right? Who would have thought athletes use physics?

Hammer Throw: example of centripetal acceleration and centripetal force

Now that we have an idea about how centripetal acceleration and force appear in everyday life, let us take a deeper look at the concepts. This article will introduce two components of rotational and circular motion: centripetal acceleration and centripetal force. We will define centripetal acceleration and centripetal force and provide a formula for each. Then, we will discuss how these concepts relate to one another before working on some examples.

Definition of Centripetal Acceleration and Centripetal Force

With these definitions in mind, we now look at the formulas to calculate the centripetal acceleration and centripetal force of an object undergoing circular motion.

Centripetal Acceleration and Centripetal Force Formulas

To calculate centripetal acceleration, we use the centripetal acceleration formula

$a_c=\frac{v^2}{r}$$a_c=\frac{v^2}{r}$

where$v$is velocity measured in$\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$and$r$is the radius of the circular path measured in$\mathrm{m}$.

We can obtain the centripetal force formula by following the formula for centripetal acceleration. Acceleration and force are related by Newton's Second Law of Motion, which is

$F=ma$.

Based on this law, we know that the force exerted on an object is the product of mass and acceleration. For centripetal force, we know its associated acceleration is also centripetal. Therefore, we must insert its formula into the above equation to obtain the formula for centripetal force. Plugging the centripetal acceleration formula into this expression, we get the centripetal force formula:

$$\begin{gathered} F_c=m{a}_c \\ =\frac{m{v}^2}{r}. \end{gathered}$$

Here$m$is the mass of the object measured in$\mathrm{kg}$,$v$is the velocity measured in $\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$, and $r$ is the radius measured in$\text{m}$.

The Relationship between Centripetal Force and Centripetal Acceleration

We just saw above how centripetal force is mathematically related to centripetal acceleration. Knowing the latter, one only has to multiply it by the object's mass to get the former. Let us now look in more detail at the conceptual relation between centripetal acceleration and centripetal force.

What Forces cause Centripetal Acceleration?

Centripetal forces cause centripetal acceleration. Tension, gravity, and friction are typical examples of centripetal forces responsible for circular motion. For example, if we attach an object to a string and swing it horizontally overhead, the object will travel in a circular path due to tension. The tension on the string keeps the object along its circular path and, therefore, provides the centripetal force.

Tension: example of centripetal force

Tension: example of centripetal force.

Another example is the orbit of the moon around the earth. The moon is kept in its orbit by gravity, and gravity provides the centripetal force applied to the moon.

Gravity: example of centripetal force

The same applies to friction. If a car is driving on a banked curve, friction acts as an external force causing the car to remain along its path.

From the above examples, we can appreciate that every time an object undergoes circular motion, it has a centripetal acceleration because the direction of its velocity at any point along the path constantly changes. This acceleration is caused by a centripetal force constraining the object's motion to a circle.

Derivation of Centripetal Acceleration and Force

In this section, we provide a geometric derivation of centripetal acceleration. The accurate derivation would require calculus to derive the equation for centripetal acceleration. However, it's possible to work around this using informal visual proof. Let us begin by considering a particle moving on a circle with constant speed and draw the position vector and the corresponding velocity vector at different times.

For objects undergoing circular motion, the velocity vector is always tangential to the position vector.

When an object is in a circular motion, its position vector and velocity vector constantly change. The position vector changes due to the velocity vector. Therefore, if the particle moves with a speed, $\mathrm{v}$, along a circle with radius, $\mathrm{r}$, we can determine the time it requires for the particle to make one complete revolution

For a circle, we know that the distance will be equal to $2\mathrm{\pi r}$, which is the circumference of a circle. Therefore, we find the time around the circumference of this first circle via:

${\mathrm{t}}_1=\frac{2\mathrm{\pi r}}{\mathrm{v}}$.

Now, we also know that the velocity direction is changing due to centripetal acceleration.

For objects undergoing circular motion, the acceleration vector is always pointing towards the center of the circle

As the particle moves along the green circle of radius,$\mathrm{r}$, the velocity vector changes due to the acceleration vector. Now, if we rearrange these velocity and acceleration vectors while keeping them perpendicular to each other, we get the following:

Recalling that acceleration is the rate of change of position, we can rearrange the vectors of the previous figure to form a circle analogous to the first one

The changes in the velocity vector describe a second circle. This circle will have a radius$\mathrm{v}$and requires the same amount of time that the particle takes to move along the circular path of radius$r$from the first image. However, for this second circle, we know that centripetal acceleration is responsible for the particle's speed. Therefore, if the circle has a radius$\mathrm{v}$and a speed${\mathrm{a}}_{\mathrm{c}}$we can determine the time it takes for the particle to make one complete revolution:

Both circles must complete a full revolution in the same amount of time because they are describing the same case of circular motion. Therefore, if we set both time equations equal to each other, we can solve for centripetal acceleration as follows:

$t_1=t_2$

$\frac{\cancel{2\mathrm{\pi }}\mathrm{r}}{\mathrm{v}}=\frac{\cancel{2\mathrm{\pi }}\mathrm{v}}{{\mathrm{a}}_{\mathrm{c}}}\to \frac{\mathrm{r}}{\mathrm{v}}=\frac{\mathrm{v}}{{\mathrm{a}}_{\mathrm{c}}}$.

Cross multiplying, one gets

${\mathrm{a}}_{\mathrm{c}}\mathrm{r}={\mathrm{v}}^2$

which, upon dividing both sides by $r$, yields

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{\mathrm{r}}$.

This is the equation for centripetal acceleration we were looking for.

Examples of Centripetal Acceleration and Centripetal Force

In this section, we'll work out two calculations involving examples of centripetal acceleration and centripetal force. Before beginning a problem, always remember first to read and identify all variables provided to you before applying the necessary formulas to answer the question.