# 2.2: Applications of Radian Measure

**At Grade**Created by: CK-12

## Learning Objectives

- Solve problems involving angles of rotation using radian measure.
- Calculate the length of an arc and the area of a sector.
- Approximate the length of a chord given the central angle and radius.
- Solve problems about angular speed.

## Rotations

**Example 1:** The hands of a clock show 11:20. Express the obtuse angle formed by the hour and minute hands in radian measure.

**Solution:** The following diagram shows the location of the hands at the specified time.

Because there are 12 increments on a clock, the angle between each hour marking on the clock is \begin{align*}\frac{2\pi}{12}=\frac{\pi}{6}\end{align*}

## Length of Arc

The length of an arc on a circle depends on both the angle of rotation and the radius length of the circle. If you recall from the last lesson, the measure of an angle in radians is defined as the length of the arc cut off by one radius length. What if the radius is 4 cm? Then, the length of the half-circle arc would be \begin{align*}\pi\end{align*}

This results in a formula that can be used to calculate the length of any arc.

\begin{align*}s = r \theta,\end{align*}

where \begin{align*}s\end{align*} is the length of the arc, \begin{align*}r\end{align*} is the radius, and \begin{align*}\theta\end{align*} is the measure of the angle in radians.

Solving this equation for \begin{align*}\theta\end{align*} will give us a formula for finding the radian measure given the arc length and the radius length:

\begin{align*}\theta =\frac{s}{r}\end{align*}

**Example 2:** The free-throw line on an NCAA basketball court is 12 ft wide. In international competition, it is only about 11.81 ft. How much longer is the half circle above the free-throw line on the NCAA court?

**Solution:** Find both arc lengths.

\begin{align*}& \text{NCAA} && \text{INTERNATIONAL}\\ & s_1=r \theta && s_2 =r \theta\\ & s_1=\frac{12}{2}(\pi) && s_2 \approx \frac{11.81}{2}(\pi)\\ & s_1=6\pi && s_2 \approx 5.905\pi\end{align*}

So the answer is approximately \begin{align*}6 \pi - 5.905 \pi \approx 0.095\pi\end{align*}

This is approximately 0.3 ft, or about 3.6 inches longer.

**Example 3:** Two connected gears are rotating. The smaller gear has a radius of 4 inches and the larger gear’s radius is 7 inches. What is the angle through which the larger gear has rotated when the smaller gear has made one complete rotation?

**Solution:** Because the blue gear performs one complete rotation, the length of the arc traveled is:

\begin{align*}s &= r\theta\\ s&=4 \times 2 \pi\end{align*}

So, an \begin{align*}8\pi\end{align*} arc length on the larger circle would form an angle as follows:

\begin{align*}\theta &=\frac{s}{r}\\ \theta &=\frac{8\pi}{7}\\ \theta &\approx 3.6\end{align*}

So the angle is approximately 3.6 radians.

\begin{align*}3.6 \times \frac{180}{\pi} \approx 206^\circ\end{align*}

## Area of a Sector

One of the most common geometric formulas is the area of a circle:

\begin{align*}A = \pi r^2\end{align*}

In terms of angle rotation, this is the area created by \begin{align*}2 \pi\end{align*} radians.

\begin{align*}2\pi \ \text{rad} = \pi r^2 \ \text{area}\end{align*}

A half-circle, or \begin{align*}\pi\end{align*} radian rotation would create a section, or **sector** of the circle equal to half the area or:

\begin{align*}\frac{1}{2}\pi r^2\end{align*}

So an angle of 1 radian would define an area of a sector equal to:

\begin{align*}1=\frac{1}{2} r^2\end{align*}

From this we can determine the area of the sector created by any angle, \begin{align*}\theta\end{align*} radians, to be:

\begin{align*}A=\frac{1}{2}r^2 \theta\end{align*}

**Example 4:** Crops are often grown using a technique called center pivot irrigation that results in circular shaped fields.

Here is a satellite image taken over fields in Kansas that use this type of irrigation system.

If the irrigation pipe is 450 m in length, what is the area that can be irrigated after a rotation of \begin{align*}\frac{2\pi}{3}\end{align*} radians?

**Solution:** Using the formula:

\begin{align*}A&=\frac{1}{2}r^2 \theta\\ A&=\frac{1}{2}(450)^2 \left(\frac{2\pi}{3}\right)\end{align*}

The area is approximately 212,058 square meters.

## Length of a Chord

You may recall from your Geometry studies that a chord is a segment that begins and ends on a circle.

\begin{align*}\overline{AB}\end{align*} is a chord in the circle.

We can calculate the length of any chord if we know the angle measure and the length of the radius. Because each endpoint of the chord is on the circle, the distance from the center to \begin{align*}A\end{align*} and \begin{align*}B\end{align*} is the same as the radius length.

Next, if we bisect the angle, the angle bisector must be perpendicular to the chord and bisect it (we will leave the proof of this to your Geometry class). This forms a right triangle.

We can now use a simple sine ratio to find half the chord, called \begin{align*}c\end{align*} here, and double the result to find the length of the chord.

\begin{align*}& \sin \frac{\theta}{2}=\frac{c}{r}\\ & c=r \times \sin \frac{\theta}{2}\end{align*}

So the length of the chord is:

\begin{align*}2c=2r \sin \frac{\theta}{2}\end{align*}

**Example 5:** Find the length of the chord of a circle with radius 8 cm and a central angle of \begin{align*}110^\circ\end{align*}. Approximate your answer to the nearest mm.

**Solution:** We must first convert the angle measure to radians:

\begin{align*}110 \times \frac{\pi}{180}=\frac{11\pi}{18}\end{align*}

Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle.

\begin{align*}& \frac{11\pi}{18} \times \frac{1}{2}=\frac{11\pi}{36}\\ & 8 \times \sin \frac{11\pi}{36}\end{align*}

Multiply this result by 2.

So, the length of the arc is approximately 13.1 cm.

## Angular Velocity

What about objects that are traveling on a circular path? Do you remember playing on a merry-go-round when you were younger?

If two people are riding on the outer edge, their velocities should be the same. But, what if one person is close to the center and the other person is on the edge? They are on the same object, but their speed is actually not the same.

Look at the following drawing.

Imagine the point on the larger circle is the person on the edge of the merry-go-round and the point on the smaller circle is the person towards the middle. If the merry-go-round spins exactly once, then both individuals will also make one complete revolution in the same amount of time.

However, it is obvious that the person in the center did not travel nearly as far. The *circumference* (and of course the radius) of that circle is much smaller and therefore the person who traveled a greater distance in the same amount of time is actually traveling faster, even though they are on the same object. So the person on the edge has a greater *linear velocity* (recall that linear velocity is found using \begin{align*}\text{distance} = \text{rate} \cdot \text{time}\end{align*}). If you have ever actually ridden on a merry-go-round, you know this already because it is much more fun to be on the edge than in the center! But, there is something about the two individuals traveling around that is the same. They will both cover the same rotation in the same period of time. This type of speed, measuring the angle of rotation over a given amount of time is called the **angular velocity.**

The formula for angular velocity is:

\begin{align*}\omega =\frac{\theta}{t}\end{align*}

\begin{align*}\omega\end{align*} is the last letter in the Greek alphabet, omega, and is commonly used as the symbol for angular velocity. \begin{align*}\theta\end{align*} is the angle of rotation expressed in radian measure, and \begin{align*}t\end{align*} is the time to complete the rotation.

In this drawing, \begin{align*}\theta\end{align*} is exactly one radian, or the length of the radius bent around the circle. If it took point \begin{align*}A\end{align*} exactly 2 seconds to rotate through the angle, the *angular velocity* of \begin{align*}A\end{align*} would be:

\begin{align*}\omega &=\frac{\theta}{t}\\ \omega &=\frac{1}{2} \ \text{radians per second}\end{align*}

In order to know the *linear speed* of the particle, we would have to know the actual distance, that is, the length of the radius. Let’s say that the radius is 5 cm.

If linear velocity is \begin{align*}v=\frac{d}{t}\end{align*} then, \begin{align*}v=\frac{5}{2}\end{align*} or 2.5 cm per second.

If the angle were not exactly 1 radian, then the distance traveled by the point on the circle is the length of the arc, \begin{align*}s = r \theta\end{align*}, or, the radius length times the measure of the angle in radians.

Substituting into the formula for linear velocity gives: \begin{align*}v=\frac{r\theta}{t}\end{align*} or \begin{align*}v=r \cdot \frac{\theta}{t}\end{align*}.

Look back at the formula for angular velocity. Substituting \begin{align*}\omega\end{align*} gives the following relationship between linear and angular velocity, \begin{align*}v = r\omega\end{align*}. So, the *linear velocity* is equal to the radius times the *angular velocity.*

Remember in a unit circle, the radius is 1 unit, so in this case the linear velocity is the same as the angular velocity.

\begin{align*}v &= r\omega\\ v &= 1 \times \omega\\ v &= \omega\end{align*}

Here, the distance traveled around the circle is the same for a given unit of time as the angle of rotation, measured in radians.

**Example 6:** Lindsay and Megan are riding on a Merry-go-round. Megan is standing 2.5 feet from the center and Lindsay is riding on the outside edge 7 feet from the center. It takes them 6 seconds to complete a rotation. Calculate the linear *and* angular velocity of each girl.

**Solution:** We are told that it takes 6 seconds to complete a rotation. A complete rotation is the same as \begin{align*}2\pi\end{align*} radians. So the angular velocity is:

\begin{align*}\omega=\frac{\theta}{t}=\frac{2\pi}{6}=\frac{\pi}{3}\end{align*} radians per second, which is slightly more than 1 (about 1.05), radian per second. Because both girls cover the same angle of rotation in the same amount of time, their *angular speed* is the same. In this case they rotate through approximately 60 degrees of the circle every second.

As we discussed previously, their linear velocities are different. Using the formula, Megan’s linear velocity is:

\begin{align*}v=r\omega = (2.5)\left(\frac{\pi}{3}\right) \approx 2.6 \ \text{ft per sec}\end{align*}

Lindsay’s linear velocity is:

\begin{align*}v=r\omega=(7)\left(\frac{\pi}{3}\right) \approx 7.3 \ \text{ft per sec}\end{align*}

## Points to Consider

- What is the difference between finding arc length and the area of a sector?
- What is the difference between linear velocity and angular velocity?
- How are linear and angular velocity related?

## Review Questions

- The following image shows a 24-hour clock in Curitiba, Paraná, Brasil.
- What is the angle between each number of the clock expressed in:
- exact radian measure in terms of \begin{align*}\pi\end{align*} ?
- to the nearest tenth of a radian?
- in degree measure?

- Estimate the measure of the angle between the hands at the time shown in:
- to the nearest whole degree
- in radian measure in terms of \begin{align*}\pi\end{align*}

- What is the angle between each number of the clock expressed in:
- The following picture is a window of a building on the campus of Princeton University in Princeton, New Jersey.
(a) What is the exact radian measure in terms of \begin{align*}\pi\end{align*} between two consecutive circular dots on the small circle in the center of the window? (b) If the radius of this circle is about 0.5 m, what is the length of the arc between the centers of each consecutive dot? Round your answer to the nearest cm.

- Now look at the next larger circle in the window.
- Find the exact radian measure in terms of \begin{align*}\pi\end{align*} between two consecutive dots in this window.
- The radius of the glass portion of this window is approximately 1.20 m. Calculate an estimate of the length of the highlighted chord to the nearest cm. Explain the reasoning behind your solution.

- The state championship game is to be held at Ray Diaz Memorial Arena. The seating forms a perfect circle around the court. The principal of Archimedes High School is sent the following diagram showing the seating allotted to the students at her school. It is 55 ft from the center of the court to the beginning of the stands and 110 ft from the center to the end. Calculate the approximate number of square feet each of the following groups has been granted:
- the students from Archimedes.
- general admission.
- the press and officials.

- Doris and Lois go for a ride on a carousel. Doris rides on one of the outside horses and Lois rides on one of the smaller horses near the center. Lois’ horse is 3 m from the center of the carousel, and Doris’ horse is 7 m farther away from the center than Lois’. When the carousel starts, it takes them 12 seconds to complete a rotation.
- Calculate the linear velocity of each girl.
- Calculate the angular velocity of the horses on the carousel.

- The Large Hadron Collider near Geneva, Switerland began operation in 2008 and is designed to perform experiments that physicists hope will provide important information about the underlying structure of the universe. The LHC is circular with a circumference of approximately 27,000 m. Protons will be accelerated to a speed that is very close to the speed of light (\begin{align*}\approx 3 \times 10^8\end{align*} meters per second).
- How long does it take a proton to make a complete rotation around the collider?
- What is the approximate (to the nearest meter per second) angular speed of a proton traveling around the collider?
- Approximately how many times would a proton travel around the collider in one full second?

## Review Answers

1. a. i. \begin{align*}\frac{\pi}{12}\end{align*}

- ii.\begin{align*}\approx \ 0.3 \ \text{radians}\end{align*}
- iii.\begin{align*}15^\circ\end{align*}
- b. i.\begin{align*}20^\circ\end{align*}. Answers may vary, anything above \begin{align*}15^\circ\end{align*} and less than \begin{align*}25^\circ\end{align*} is reasonable.
- ii.\begin{align*}\frac{\pi}{9}\end{align*} Again, answers may vary

2. a. \begin{align*}\frac{\pi}{6}\end{align*}

- b. \begin{align*}\approx 26 \ cm\end{align*}

3. a. \begin{align*}\frac{\pi}{16}\end{align*}

- b. Let’s assume, to simplify, that the chord stretches to the center of each of the dots. We need to find the measure of the central angle of the circle that connects those two dots.
- Since there are 13 dots, this angle is \begin{align*}\frac{13\pi}{16}\end{align*}. The length of the chord then is:
- \begin{align*}&=2r \sin \frac{\theta}{2}\\ &=2 \times 1.2 \times \sin (\frac{1}{2}\times \frac{13 \pi}{16})\end{align*}
- The chord is approximately 2.30 m, or 230 cm.

4. Each section is \begin{align*}\frac{\pi}{6}\end{align*} radians. The area of one section of the stands is therefore the area of the outer sector minus the area of the inner sector:

- \begin{align*}A &= A_{outer}-A_{inner}\\ A &= \frac{1}{2}(r_{outer})^2 \times \frac{\pi}{6}-\frac{1}{2}(r_{inner})^2 \times \frac{\pi}{6}\\ A &= \frac{1}{2}(110)^2 \times \frac{\pi}{6}-\frac{1}{2}(55)^2 \times \frac{\pi}{6}\end{align*}
- The area of each section is approximately \begin{align*}2376 \ ft^2\end{align*}.
- a. The students have 4 sections or \begin{align*}\approx 9503 \ ft^2\end{align*}
- b. There are 3 general admission sections or \begin{align*}\approx 7127 \ ft^2\end{align*}
- c. There is only one press and officials section or \begin{align*}\approx 2376 \ ft^2\end{align*}

5. It is actually easier to calculate the angular velocity first. \begin{align*}\omega=\frac{2\pi}{12}=\frac{\pi}{6}\end{align*}, so the angular velocity is \begin{align*}\frac{\pi}{6} \ rad\end{align*}, or \begin{align*}0.524\end{align*}. Because the linear velocity depends on the radius,each girl has her own.

- Lois: \begin{align*}v=r\omega = 3 \cdot \frac{\pi}{6}=\frac{\pi}{2} \ \text{or} \ 1.57 \ m/sec\end{align*}
- Doris: \begin{align*}v=r\omega=10 \cdot \frac{\pi}{6}=\frac{5\pi}{3} \ \text{or} \ 5.24 \ m/sec\end{align*}}}

6. a. \begin{align*}v=\frac{d}{t}\rightarrow 3 \times 10^8=\frac{27,000}{t} \rightarrow t=\frac{2.7 \times 10^4}{3 \times 10^8}=0.9 \times 10^{-4} = 9 \times 10^{-5}\end{align*} or 0.00009 seconds.

- b. \begin{align*}\omega=\frac{\theta}{t}=\frac{2\pi}{0.00009}\approx 69,813 \ rad/sec\end{align*}
- c. The proton rotates around once in 0.00009 seconds. So, in one second it will rotate around the LHC \begin{align*}1 \div 0.00009 = 11,111.11\end{align*} times, or just over 11,111 rotations.

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