Jump to a key chapter

## Angular Displacement Definition

So what exactly is angular displacement? Let's start with a simple definition.

In circular motion, **angular displacement** is defined as the change in the angle of a body with respect to its initial angular position.

This is completely analogous to how linear displacement is defined as the distance a body moves from its initial position. Angular displacement is a vector quantity, i.e., it has both magnitude and direction.

The choice of direction for positive displacement is simply a convention. There is no physical reason for why it's like this.

Take a look at the example below. There is a circle with two points \(\mathrm{A}\) and \(\mathrm{B}\) marked on the graph. We want to calculate the angular displacement when moving from point \(\mathrm{A}\) to \(\mathrm{B}\). This can simply be done by measuring the angle of each point from the \(x\)-axis, as shown on the left. As you can see below, point \(\mathrm{A}\) is at \(20^{\circ}\) and point \(\mathrm{B}\) at \(80^{\circ}\). The angular displacement is\[\Delta\theta=\theta_{B}-\theta_{A} = 80^{\circ}-20^{\circ}=60^{\circ}\]

Fig.1 - Angular displacement can be calculated by finding the difference between two angular positions

If a dancer dancing takes a turn and makes one full rotation during dancing the angle subtended will be \(360^{\circ}\); hence we can say that the angular displacement is \(360^{\circ}\) whereas for half rotation the angle will be \(180^{\circ}\).

Fig. 2 - A dancer showing angular motion and defining an angular displacement.

### Radians vs Degrees

In the example given above, we used the unit of degrees \( (^\circ) \) when calculating the angular displacement. You may feel very familiar with this unit, having used it for most of your calculations so far. However, there is another unit used for angles and angular displacement, which is often much simpler and far more elegant. This unit is the **radian** \((\mathrm{rad}\)).

A **radian** is a unit of angular measurement such that an angle of one radian is subtended by a circular sector whose circular arc is equal to its radius.

This definition may seem a little wordy so let's break it down. Consider a circle, we can draw two distinct radii from the circle's center to any two points on the circumference. The section of the circle's circumference between the two points on the radii is called an arc. An angle of one radian produces an arc with a length equal to the circle's radius. The image below illustrates this idea clearly.

Using this definition, we can calculate some simple conversions between degrees and radians. Consider a circle of radius \(r\), we know that there are \( 360\; ^{\circ} \) in a circle. We also know that the circumference (total arc length) of a circle is \( 2\pi r \), which means that a complete circumference defines an angle of \( 2\pi\;\mathrm{rad} \). Thus \( 360\;^{\circ} \) degrees must be equal to \(2\pi\) radians. Similarly \(180\;^{\circ}=\pi \;\mathrm{rad}\). We can convert from degrees to angles using the following handy equations.

Degrees to radians | Radians to degrees |

\(\theta_\mathrm{radians}=\frac{2\pi\;\mathrm{rad}}{360^\circ}\theta_\mathrm{degrees}\) | \(\theta_\mathrm{degrees}=\frac{360^\circ}{2\pi\;\mathrm{rad}}\theta_\mathrm{radians}\) |

## Angular Displacement Formula

Radians are very useful for measuring angular displacements as it gives us a link to the linear displacement via the arc length. In fact, by measuring angular displacement in radians, we can immediately derive a formula for angular displacement in terms of the arc length of distance traveled.

\[\begin{aligned}\text{Angular displacement (radians)}&=\frac{\text{arc length (meters)}}{\text{radius of circular motion (meters)}}\\\Delta\theta&=\frac{s}{r}\end{aligned}\]

Here the arc length is the linear distance traveled by the object during it's circular motion, and the radius of circular motion is the shortest distance between the object and its axis of rotation.

It is important to keep in mind that a radian defines an angle measure, but it is a dimensionless unit! We can see this in the above formula, as both \( s \) and \( r \) have the same units of length. Then, the units cancel out when calculating their quotient.

Let's see how this applies in a real-world context.

Alice runs on a circular track that has a diameter of \(8\,\mathrm{m}\). If she runs around the complete track for a distance of \(80\,\mathrm{m}\), what is her angular displacement?

**Solution:**

According to the question, Alice's total linear distance is \(80\,\mathrm{m}\) so \(s=80\,\mathrm{m}\). The diameter of the track is \(d=8\,\mathrm{m}\), so the radius is \(r=\frac{d}{2}=4\,\mathrm{m}\). Plugging this into the equation for angular displacement, we have,

\[\Delta\theta=\frac{80\;\mathrm{m}}{4\;\mathrm{m}}=20\,\mathrm{rad}\]

If we want to find the angular displacement of Alice from her initial position, we only want to consider how far along the track she is rather than how many revolutions she has done. This means we want to find the remainder of \(20\,\mathrm{rad}\) after dividing by an integer multiple of \(2\pi\). To do this, first divide \(20\,\mathrm{rad}\) by \(2\pi\)

\[\frac{20}{2\pi}=3.18\]

Rounding down to 3, the nearest whole integer, tells us that Alice has done three full laps. The angular displacement of these three laps is \(3\cdot2\pi=6\pi\,\mathrm{rad}\). Taking this off the total angular displacement gives us the remainder or Alice's angular displacement from her initial position. \[20\;\mathrm{rad}-6\pi\;\mathrm{rad}=1.15\;\mathrm{rad}\]

## Relationship between Angular and Linear Displacement

As we can see from the previous formula, angular and linear displacement are closely related.

\[\Delta\theta=\frac{s}{r}\]

In fact, they are directly proportional. This means that as the angular displacement increases or decreases, so does the linear displacement. However, the formula also states that the angular displacement is inversely proportional to the radial distance from the axis of rotation. This means that if two objects rotate through the same angular displacement, the one rotating further from the axis of rotation will cover a larger linear distance than the one rotating closer. We can see this in the graph below.

It's important to remember that the equation for angular displacement includes the radial distance. Two objects can cover the same linear distance but have very different angular displacements and vice versa.

Two compasses are used to draw arcs of length \(20\,\mathrm{cm}\). If the first compass has a radius of \(5.0\,\mathrm{cm}\) and the second has a radius of \(3.0\,\mathrm{cm}\) what angle must each compass move through?

Using the angular displacement formula \(\Delta\theta=\frac{s}{r}\) we have,

\[\Delta\theta_1=\frac{20\,\mathrm{cm}}{5.0\;\mathrm{cm}}=4.0\;\mathrm{rad}\]

\[\Delta\theta_2=\frac{20\,\mathrm{cm}}{3.0\;\mathrm{cm}}=6.7\;\mathrm{rad}\]

The second compass has to turn through \(2.7\,\mathrm{rad} \) more to draw the same arc length!

## Angular Displacement Graph

Much like regular linear displacement, we can use motion graphs to find the angular displacement of objects moving in circular motion. By plotting an angular position vs time graph of an object, we can find its angular displacement between two specific times by calculating the vertical change between the points with the corresponding time coordinates.

We can use the graph above to find the **angular velocity **\(\omega\) of the object.

**Angular velocity** \(\omega\) is the rate of change of angular displacement with respect to time.

We measured angular velocity in \(\frac{\mathrm{rad}}{\mathrm{s}}\).

Therefore, we can find the angular velocity by calculating the slope of the graph.

The **slope** of a graph can be calculated using

\[ \text{slope} = \frac{\text{Change in vertical direction}}{\text{Change in horizontal direction}} \]

As we can see in the above graph, the angular displacement between \(1\,\mathrm{s}\) and \(4\,\mathrm{s}\) is \(\frac{3\pi}{2}\,\mathrm{rad}\). This implies that the slope of the graph is\[\text{Gradient}=\omega=\frac{\frac{3\pi}{2}\;\mathrm{rad}}{4\;\mathrm{s} -1\;\mathrm{s}}=\frac{\pi}{2}\,\frac{\mathrm{rad}}{s}.\]

Thus, the angular velocity for the object is \( \frac{\pi}{2}\,\frac{\mathrm{rad}}{s}. \)

## Angular Displacement Example

Let's dive into some more practical and application-oriented examples.

An object moves at a constant linear speed of \(20\,\frac{\mathrm{m}}{\mathrm{s}}\) around a circle of radius \(4\,\mathrm{m}\). How large of a central angle does it sweep out in \(3.5\,\mathrm{s}\)?

**Solution**:

Here we have the following quantities:

- time \(t=3.5\,\mathrm{s}\)
- velocity \(v=20.0\,\frac{\mathrm{m}}{\mathrm{s}}\)
- radius \(r=4.0\,\mathrm{m}\)

To use our equation for angular displacement, we need to find the distance traveled \(s\). We know from the definition of velocity that

\[\begin{aligned}\text{distance} &=\text{velocity}\times\text{time}\\ s &= vt\end{aligned}\]

Plugging this into our formula for angular displacement \(\Delta\theta=\frac{s}{r},\) gives

\[\begin{aligned}\Delta\theta &=\frac{vt}{r}\\[8pt] \Delta\theta &=\frac{20.0\,\frac{\bcancel{\mathrm{m}}}{\bcancel{\mathrm{s}}}\cdot 3.5\,\bcancel{\mathrm{s}}}{4.0\,\bcancel{\mathrm{m}}}\\[8pt] \Delta\theta &=18\,\mathrm{rad}\end{aligned}.\]

Make sure the units of the linear displacement and radius are the same when calculating the angular displacement from this formula.

**Angular Displacement of Earth**

As the Earth rotates through an angular displacement of \(0.5\,\mathrm{rad}\) a point on the equator sweeps out an arc length of approximately \(3000\,\text{km}\). Calculate the radius of the Earth.

**Solution:**

First, re-arrange the formula for angular displacement to isolate the radius.\[r=\frac{s}{\Delta\theta}\]

Then, we can plug in the values from the question and simplify it to find the Earth's radius.

\[r=\frac{s}{\Delta\theta}=\frac{3000\,\text{km}}{0.5\,\mathrm{rad}}=6000\,\text{km}\]

Note that the answer is \( 6000\,\text{km} \). The radians seemed to have disappeared, but remember that a radian is a dimensionless unit.

If a car wheel has a diameter of \(3\,\mathrm{m}\) and makes three full revolutions every \(2\,\mathrm{s}\) how far does the car move every second?

**Solution:**

The first thing to note is that the distance that the car moves will be the total arclength swept out by the wheel in one second. The second thing is to remember that three full revolutions correspond to \(\Delta\theta=3\cdot2\pi\,\mathrm{rad}=6\pi\,\mathrm{rad}\).

Finally don't forget that the radius is half of the diameter, so \(r=\frac{3}{2}\,\mathrm{m}\).

Using this information, we can find the linear displacement \(s\) using the formula that relates angular displacement and linear distance.

\[s=\Delta\theta r \]

Now, let's plug in all the values we have and simplify.

\[s=6\pi\,\mathrm{rad}\times\frac{3}{2}\,\mathrm{m}=9\,\mathrm{m}\]

Therefore, the car moves \(9\,\mathrm{m}\) every second.

## Angular Displacement - Key takeaways

- In circular motion, angular displacement is defined as the change in the angle of a body with respect to its initial angular position.
- Angular displacement can be calculated using the formula \(\Delta\theta=\theta_f-\theta_i\).
- Radian is a unit used for measuring angular displacement.
- A
**radian**is a unit of angular measurement defined by a circular sector whose circular arc is equal to its radius. - We can convert between radians and degrees using the formula: \( \theta\,\mathrm{rad}=\frac{360\cdot\theta}{2\pi}^{\circ} \)
- An angular displacement \(\Delta\theta\) in radians is related to a linear displacement by the formula: \( \Delta\theta=\frac{s}{r} \) where \(r\) is the radial distance of the object from its axis of rotation.
- We can find angular displacement and velocity using an angular position vs time graph. The angular displacement equals the vertical change, and the angular velocity equals the slope of the graph.

## References

- Fig. 1 - Sinulog Mardi Gras Dancer (https://commons.wikimedia.org/wiki/File:Sinulog_Mardi_Gras_Dancer.jpg) by Herbertkikoy (https://commons.wikimedia.org/wiki/User:Herbertkikoy) is licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/deed.en).

###### Learn with 10 Angular Displacement flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Angular Displacement

What is meant by angular displacement?

In circular motion, **angular displacement** is defined as the angle a body moves with respect to its initial angular position.

How do you find the angular displacement?

Angular displacement can be found by calculating the change in the angular position. Alternatively, it can be found using the linear displacement *s* and the radial distance from the axis of rotation *r* using the equation

*Angular displacement = s/r.*

What is an example of angular displacement?

A ballet dancer twirling around will undergo angular motion, defining an angular displacement as they spin. Similarly, when you go on a merry-go-round your angular position changes so you also define an angular displacement as you move.

What are the units for angular displacement?

Angular displacement is usually measured in radians but it can also be measured in degrees, or more rarely in arcseconds.

Is arc length angular displacement?

No, however the arc length is directly proportional to the angular displacement.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more