The circumference of a circle of radius $r$ is $2\pi r$. Thus for the unit circle, where $r=1$, the circumference is $2\pi$. If a point starts at $A$ and travels around the circle (Fig. 1), it will travel a distance of $2\pi$. If it travels halfway around the circle (Fig. 2), it will travel a distance of ${\displaystyle {\displaystyle \frac{1}{2}}}\cdot 2\pi$, or $\pi$.

If a point $C$ travels ${\displaystyle {\displaystyle \frac{1}{8}}}$ of the way around the circle (Fig. 3), it will travel a distance of ${\displaystyle {\displaystyle \frac{1}{8}}}\cdot 2\pi$, or $\pi /4$. Note that $C$ is ${\displaystyle {\displaystyle \frac{1}{4}}}$ of the way from $A$ to $B$. If a point $D$ travels${\displaystyle {\displaystyle \frac{1}{6}}}$ of the way around the circle (Fig. 4), it will travel a distance of ${\displaystyle {\displaystyle \frac{1}{6}}}\cdot 2\pi$, or $\pi /3$. Note that $D$ is ${\displaystyle {\displaystyle \frac{1}{3}}}$ of the way from $A$ to $B$.

Example 1

How far will a point travel if it goes (a) ${\displaystyle {\displaystyle \frac{1}{4}}}$, (b) ${\displaystyle {\displaystyle \frac{1}{12}}}$, (c) ${\displaystyle {\displaystyle \frac{3}{8}}}$, and (d) ${\displaystyle {\displaystyle \frac{5}{6}}}$ of the way around the unit circle?

Solution

1. a) ${\displaystyle {\displaystyle \frac{1}{4}}}$ of the total distance around the circle is ${\displaystyle {\displaystyle \frac{1}{4}}}\cdot 2\pi$, which is ${\displaystyle {\displaystyle \frac{1}{2}}}\pi$, or $\pi /2$.

2. b) The distance will be ${\displaystyle {\displaystyle \frac{1}{12}}}\cdot 2\pi$, which is ${\displaystyle {\displaystyle \frac{1}{6}}}\pi$, or $\pi /6$.

3. c) The distance will be ${\displaystyle {\displaystyle \frac{3}{8}}}\cdot 2\pi$, which is ${\displaystyle {\displaystyle \frac{3}{4}}}\pi$, or $3\pi /4$.

4. d) The distance will be ${\displaystyle {\displaystyle \frac{5}{6}}}\cdot 2\pi$, which is ${\displaystyle {\displaystyle \frac{5}{3}}}\pi$, or $5\pi /3$. Think of $5\pi /3$ as $\pi +{\displaystyle {\displaystyle \frac{2}{3}}}\pi$.

These distances are illustrated in the following figures.

A point may travel completely around the circle and then continue. For example, if it goes around once and then continues ${\displaystyle {\displaystyle \frac{1}{4}}}$ of the way around, it will have traveled a distance of $2\pi +{\displaystyle {\displaystyle \frac{1}{4}}}\cdot 2\pi$, or $5\pi /2$ (Fig. 5). Every real number determines a point on the unit circle. For the positive number 10, for example, we start at $A$ and travel counterclockwise a distance of 10. The point at which we stop is the point “determined” by the number 10. Note that $2\pi \approx 6.28$ and that $10\approx 1.6\left(2\pi \right)$. Thus the point for 10 travels around the unit circle about 1.6, or $1{\displaystyle {\displaystyle \frac{3}{5}}}$, times (Fig. 6).

For a negative number, we move clockwise around the circle. Points for $-\pi /4$ and $-3\pi /2$ are shown in the figures below. The number 0 determines the point $A$.

Degree measure is a common unit of angle measure in many everyday applications. But in many scientific fields and in mathematics (calculus, in particular), there is another commonly used unit of measure called the radian.

The word radian is derived from the word radius. Thus measuring 1 “radius” along the circumference of the circle determines an angle whose measure is 1 radian. One radian is about $57.3°$.

Angles that measure 2 radians, 3 radians, and 6 radians are shown below.

When we make a complete (counterclockwise) revolution, the terminal side coincides with the initial side on the positive x-axis. We then have an angle whose measure is $2\pi$ radians, or about 6.28 radians, which is the circumference of the circle:

Thus a rotation of $360°$ (1 revolution) has a measure of $2\pi$ radians. A half revolution is a rotation of $180°$, or $\pi$ radians. A quarter revolution is a rotation of $90°$, or $\pi /2$ radians, and so on.

To convert between degrees and radians, we first note that

It follows that

To make conversions, we multiply by 1, noting the following.

The radian–degree equivalents of the most commonly used angle measures are illustrated in the following figures.

When a rotation is given in radians, the word “radians” is optional and is most often omitted. Thus if no unit is given for a rotation, the rotation is understood to be in radians.

We can also find coterminal, complementary, and supplementary angles in radian measure just as we did for degree measure in Section 6.3.

Example 5

Find a positive angle and a negative angle that are coterminal with $2\pi \text{/}3$. Many answers are possible.

Solution

To find angles coterminal with a given angle, we add or subtract multiples of $2\pi$:

$$\begin{array}{ccc}\hfill {\displaystyle {\displaystyle \frac{2\pi }{3}}}+2\pi & =\hfill & {\displaystyle {\displaystyle \frac{2\pi }{3}}}+{\displaystyle {\displaystyle \frac{6\pi }{3}}}={\displaystyle {\displaystyle \frac{8\pi }{3}}},\hfill \\ \hfill {\displaystyle {\displaystyle \frac{2\pi }{3}}}-3\left(2\pi \right)& =\hfill & {\displaystyle {\displaystyle \frac{2\pi }{3}}}-{\displaystyle {\displaystyle \frac{18\pi }{3}}}=-{\displaystyle {\displaystyle \frac{16\pi }{3}}}.\hfill \end{array}$$

Thus, $8\pi /3$ and $-16\pi /3$ are two of the many angles coterminal with $2\pi /3$.

Now Try Exercise 51.

Radian measure can be determined using a circle other than a unit circle. In the figure at left, a unit circle (with radius 1) is shown along with another circle (with radius $r$, $r\ne 1$). The angle shown is a central angle of both circles.

From geometry, we know that the arcs that the angle subtends have their lengths in the same ratio as the radii of the circles. The radii of the circles are $r$ and 1. The corresponding arc lengths are $s$ and $s_1$. Thus we have the proportion

which also can be written as

Now $s_1$ is the radian measure of the rotation in question. It is common to use a Greek letter, such as $\theta$, for the measure of an angle or a rotation and the letter $s$ for arc length. Adopting this convention, we rewrite the proportion above as

In any circle, the measure (in radians) of a central angle, the arc length the angle subtends, and the length of the radius are related in this fashion. Or, in general, the following is true.

Example 8

Find the length of an arc of a circle of radius 5 cm associated with an angle of $2\pi /3$ radians.

Solution

We have

$$\theta ={\displaystyle {\displaystyle \frac{s}{r}}},\text{or}s=r\theta .$$

Thus, $s=5\mathrm{cm}\cdot 2\pi /3$, or about 10.47 cm.

Now Try Exercise 63.

Linear speed is defined to be distance traveled per unit of time. If we use $v$ for linear speed, $s$ for distance, and $t$ for time, then

Similarly, angular speed is defined to be amount of rotation per unit of time. For example, we might speak of the angular speed of a bicycle wheel as 150 revolutions per minute or the angular speed of the earth as $2\pi$ radians per day. The Greek letter $\omega$ (omega) is generally used for angular speed. Thus for a rotation $\theta$ and time $t$, angular speed is defined as

As an example of how these definitions can be applied, let’s consider the refurbished carousel at the Children’s Museum in Indianapolis, Indiana. It consists of three circular rows of animals. All animals, regardless of the row, travel at the same angular speed. But the animals in the outer row travel at a greater linear speed than those in the inner rows. What is the relationship between the linear speed $v$ and the angular speed $\omega \text{?}$

To develop the relationship we seek, recall that, for rotations measured in radians, $\theta =s/r$. This is equivalent to

We divide by time, $t$, to obtain

Now $s/t$ is linear speed $v$ and $\theta /t$ is angular speed $\omega$. Thus we have the relationship we seek,

Example 9 Linear Speed of an Earth Satellite.

An earth satellite in circular orbit 1200 km high makes one complete revolution every 90 min. What is its linear speed? Use 6400 km for the length of a radius of the earth.

Solution

To use the formula $v=r\omega$, we need to know $r$ and $\omega$:

$$\begin{array}{cccc}\hfill r& =\hfill & 6400\text{km}+1200\text{km}\hfill & \text{Radius}\text{of}\text{earth}\text{plus}\text{height}\text{of}\text{satellite}\hfill \\ \hfill & =\hfill & 7600\text{km};\hfill & \hfill \\ \omega \hfill & =\hfill & {\displaystyle {\displaystyle \frac{\theta }{t}}}={\displaystyle {\displaystyle \frac{2\pi }{90\text{min}}}}={\displaystyle {\displaystyle \frac{\pi }{45\text{min}}}}.\hfill & \text{We}\text{have,}\text{as}\text{usual,}\text{omitted}\text{the}\text{word}\text{radians}.\hfill \end{array}$$

Now, using $v=r\omega$, we have

$$v=7600\text{km}\cdot {\displaystyle {\displaystyle \frac{\pi }{45\text{min}}}}={\displaystyle {\displaystyle \frac{7600\pi }{45}}}\cdot {\displaystyle {\displaystyle \frac{\text{km}}{\text{min}}}}\approx 531{\displaystyle {\displaystyle \frac{\text{km}}{\text{min}}}}.$$

Thus the linear speed of the satellite is approximately 531 $\text{km/min}$.

Now Try Exercise 71.

Example 10 Angular Speed of a Capstan.

An anchor on a Navy vessel is hoisted at a rate of 2 $\text{ft/sec}$ as the line is wound around a capstan with a 1.4-yd diameter. What is the angular speed of the capstan?

Solution

We will use the formula $v=r\omega$ in the form $\omega =v/r$, taking care to use the proper units. Since $v$ is given in feet per second, we need to give $r$ in feet:

$$r={\displaystyle {\displaystyle \frac{d}{2}}}={\displaystyle {\displaystyle \frac{1.4}{2}}}\text{yd}\cdot {\displaystyle {\displaystyle \frac{3\text{ft}}{1\text{yd}}}}=2.1\text{ft}.$$

Then $\omega$ will be in radians per second:

$$\omega ={\displaystyle {\displaystyle \frac{v}{r}}}={\displaystyle {\displaystyle \frac{2\text{ft/sec}}{2.1\text{ft}}}}={\displaystyle {\displaystyle \frac{2\text{ft}}{\text{sec}}}}\cdot {\displaystyle {\displaystyle \frac{1}{2.1\text{ft}}}}\approx 0.952/\text{sec}.$$

Thus the angular speed is approximately 0.952 $\text{radian/sec}$.

Now Try Exercise 77.

The formulas $\theta =\omega t$ and $v=r\omega$ can be used in combination to find distances and angles in various situations involving rotational motion.

Example 11 Angle of Revolution.

A 2014 Toyota FJ Cruiser is traveling at a speed of 70 mph. Its tires have an outside diameter of 30.875 in. Find the angle through which a tire turns in 10 sec.

Solution

Recall that $\omega =\theta /t$, or $\theta =\omega t$. Thus we can find $\theta$ if we know $\omega$ and $t$. To find $\omega$, we use the formula $v=r\omega$. The linear speed $v$ of a point on the outside of the tire is the speed of the FJ Cruiser, 70 mph. For convenience, we first convert 70 mph to feet per second:

$$\begin{array}{c}v=70{\displaystyle {\displaystyle \frac{\text{mi}}{\text{hr}}}}\cdot {\displaystyle {\displaystyle \frac{1\text{hr}}{60\text{min}}}}\cdot {\displaystyle {\displaystyle \frac{1\text{min}}{60\text{sec}}}}\cdot {\displaystyle {\displaystyle \frac{5280\text{ft}}{1\text{mi}}}}\hfill \\ \phantom{v}\approx 102.667{\displaystyle {\displaystyle \frac{\text{ft}}{\text{sec}}}}.\hfill \end{array}$$

The radius of the tire is half the diameter. Now $r=d/2=30.875/2=15.4375\text{in}$. We will convert to feet, since $v$ is in feet per second:

$$\begin{array}{c}r=15.4375\text{in}.\cdot {\displaystyle {\displaystyle \frac{1\text{ft}}{12\text{in}.}}}\hfill \\ \phantom{r}={\displaystyle {\displaystyle \frac{15.4375}{12}}}\text{ft}\hfill \\ \phantom{r}\approx 1.29\text{ft}.\hfill \end{array}$$

Using $v=r\omega$, we have

$$102.667{\displaystyle {\displaystyle \frac{\text{ft}}{\text{sec}}}}=1.29\text{ft}\cdot \omega ,$$

so

$$\omega ={\displaystyle {\displaystyle \frac{102.667\text{ft/sec}}{1.29\text{ft}}}}\approx {\displaystyle {\displaystyle \frac{79.59}{\text{sec}}}}.$$

Then in 10 sec,

$$\theta =\omega t={\displaystyle {\displaystyle \frac{79.59}{\text{sec}}}}\cdot 10\text{sec}\approx 796.$$

Thus the angle, in radians, through which a tire turns in 10 sec is 796.

Now Try Exercise 79.

Example 12 Angular Speed of a Gear Wheel.

One gear wheel turns another, the teeth being on the rims. The wheels have 9-in. and 5-in. radii, and the smaller wheel rotates at 48 rpm. Find the angular speed of the larger wheel, in radians per second.

Solution

Let ${\omega }_1=\text{the angular speed of the smaller wheel}$ and ${\omega }_2=\text{the angular speed of the larger wheel}$. The wheels have the same linear speed, so we have

$$v=5{\omega }_1=9{\omega }_2.$$

We first convert the angular speed of the smaller wheel, 48 rpm (revolutions per minute), to radians per second:

$$\begin{array}{cccc}\hfill {\omega }_1& =\hfill & 48\text{rpm}={\displaystyle {\displaystyle \frac{48\cdot 2\pi }{1\text{min}}}}\hfill & \begin{array}{c}1\text{revolution}=2\pi ;\hfill \\ 48\text{revolution}=48\cdot 2\pi =96\pi \hfill \end{array}\hfill \\ \hfill & =\hfill & {\displaystyle {\displaystyle \frac{96\pi }{1\text{min}}}}\cdot {\displaystyle {\displaystyle \frac{1\text{min}}{60\text{sec}}}}\hfill & \hfill \\ \hfill & =\hfill & 1.6\pi /\text{sec}\hfill & \hfill \\ \hfill & \approx \hfill & 5.027/\text{sec}.\hfill & \hfill \end{array}$$

Next, we substitute $5.027/\text{sec}$ for ${\omega }_1$ and solve for ${\omega }_2$:

$$\begin{array}{ccc}\hfill 5{\omega }_1& =\hfill & 9{\omega }_2\hfill \\ \hfill 5\left(5.027/\text{sec}\right)& =\hfill & 9{\omega }_2\hfill \\ \hfill 25.135/\text{sec}& =\hfill & 9{\omega }_2\hfill \\ \hfill 2.793/\text{sec}& \approx \hfill & {\omega }_2.\hfill \end{array}$$

The angular speed of the larger wheel is about $2.793\text{radians/sec}$, or $2.793/\text{sec}$.

Now Try Exercise 81.