Opening the book…
- Note to the Instructor (click to expand)
- There is no subtlety to what is in this section or why I placed it here — you need radian measure to discuss either of the topics in this section, so it could not occur until this chapter. I suppose we could have covered this material directly after introducing radian measure (and prior to the Unit Circle definition of the trigonometric functions); however, I happened to place it here before thinking about that order. Feel free to make the switch if you feel so inclined (it actually might make the connection between the Unit Circle definition of the trigonometric functions and the area of the sector subtended by \(\theta\) more digestible).
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
-
- Calculate the area of a sector formed by a central angle.
- Solve an application involving sector area.
- Calculate the linear velocity of a point moving with uniform circular motion.
- Calculate the angular velocity of a point moving with uniform circular motion.
- Convert from linear velocity to angular velocity and vice versa.
- Solve application problems involving linear and angular velocity.
Areas of Sectors
Now that we have paired angles with real numbers via radian measure, a world of applications awaits us.
We recently discovered the relationship between the length, \(s\), of the arc subtended by an angle \(\theta\) (in radians) in a circle of radius \(r\) to be \(s = r \theta\). We now investigate the area of the sector formed by this central angle \(\theta\).
To find the area of a sector, recall that the total area of a circle of radius \(r\) is \(\pi r^2\); however, the shaded region in Figure \(\PageIndex{1}\) is not the entire circle. In fact, it is a specific portion of this circle. In terms of angular measure, we have only shaded the region covered by an angle rotated through \(\theta\) out of \(2\pi\) radians of a full circle. That is, the portion of the circle we have shaded is\[A = \dfrac{\theta}{2\pi} \, \text{of the area of the full circle} = \dfrac{\theta}{2 \cancel{\pi}} \cdot \cancel{\pi} r^2 = \dfrac{1}{2} r^2 \theta.\nonumber\]This proves the following theorem.
Finding the areas of sectors becomes increasingly essential as you move into Calculus (specifically, Integral Calculus). Be sure to spend the time now to master this relatively straightforward concept.
The following Interactive Element might help you visualize how a sector's area depends on its central angle and radius.
Interact: Drag the large red point on the rim. Move it around to change the central angle \(\theta\), and pull it in or out to change the radius \(r\).
Observation: Sweeping \(\theta\) fills a fraction \(\theta/2\pi\) of the disk — at \(\theta = 2\pi\) the sector is the whole circle. The area grows in proportion to \(\theta\), but pulling the point outward shows it grows with the square of \(r\): doubling \(r\) quadruples the area, matching \(A = \tfrac{1}{2}r^2\theta\).
Find the area of the sector subtended by a central angle of \(120^{\circ}\) in a circle of radius \(4\).
- Solution
- The most common mistake when using the formula for the area of a sector is not working in radian measure. Since \(120^{\circ} = \frac{2\pi}{3}\), we find the area of the sector to be\[A = \dfrac{1}{2}\cdot (4)^2 \left( \dfrac{2\pi}{3} \right) = \dfrac{16\pi}{3}.\nonumber\]
If the area of a sector of a circle having radius \(12\) meters is \(100\) square meters, find the measure of the central angle (in radians).
- Answer
-
\(\frac{25}{18}\) radians
In certain parts of the world near the Arctic, such as Norway, the sun doesn't set during part of the summer. This phenomenon is known as the Midnight Sun. A golf course in Norway takes advantage of this by hosting a special event: The Midnight Sun Golf Challenge. During this event, golfers can play as many holes as they want throughout the night.
The golf course has a unique challenge on their signature par 3 hole, which features a circular green with a radius of 15 meters. If a golfer lands their ball within a particular marked sector of the green, where the central angle is 30 degrees, they win a prize. To determine the likelihood of winning, calculate the area of the sector where golfers must land their ball.
- Solution
- We are given \(\theta = 30^{\circ} = \frac{\pi}{6}\) radians and \(r = 15\) meters. Therefore, the area of the sector is\[A = \dfrac{1}{2} \cdot (15)^2 \dfrac{\pi}{6} = \dfrac{225\pi}{12} \approx 59 \, \text{square meters}.\nonumber\]As an aside, the probability that the golfer's ball would land in this sector (given the ball lands on the green and that all areas are equally likely for the ball to land in) would be the ratio of this sector area to the area of the entire green, which is\[\dfrac{A_{\text{sector}}}{A_{\text{green}}} = \dfrac{225\pi/12}{\pi(15)^2} = \frac{1}{12} \approx 0.083 = 8.3\%.\nonumber\]This should make sense because \(30^{\circ}\) is one-twelfth of a full rotation of a circle.
A circular park has a sector fenced off for dogs. The sign posted by Sacramento County states that dogs can enjoy 6,000 square feet of play area; however, Paige finds this claim rather dubious. Unafraid about how foolish she might look, Paige takes a very large tape measure to the park and measures the radius of the park to be 120 feet. She then measures the length of fencing along the arc of the dog park area to be roughly 84 feet. Is the sign claiming that dogs have 6,000 square feet of play area correct?
- Answer
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No. The given measurements yield a central angle of \(0.7\) radians. Thus, the area is actually\[A = \dfrac{1}{2} (120)^2 (0.7) = 5,040 \, \text{square feet}.\nonumber\]
Linear Motion
We have already used the variable \(s\) to denote the arc length in a circle of radius \(r\) subtended by a central angle \(\theta\). We might not have mentioned that this same variable is used throughout Mathematics and Physics to denote spatial position (hence, the letter \(s\)). For example, in Algebra and Calculus, you might represent the height of a projectile (relative to the ground) at any time \(t\) by the function\[s(t) = -16t^2 + 20t + 5.\nonumber\]Note that we have used the words spatial position and height to represent the same thing. In truth, "height" only applies if an object rises or falls relative to some point in space. "Spatial position," on the other hand, is a bit more flexible and allows us to consider the distance an object is away from a centralized point. Another word commonly used in place of spatial position is displacement.
Once we have a function for an object's spatial position (or displacement) at a given time \(t\), we can expand our investigation of the object's motion to include its velocity. In Physics, the average velocity (sometimes called the average linear velocity) of an object, denoted \(\overline{v}\) and read as "\(v\)-bar," is defined as the average rate of change of the spatial position of the object with respect to time. As a result, we have\[\overline{v} = \dfrac{\text{displacement}}{\text{time}} = \dfrac{s}{t}.\nonumber\]The quantity \(\overline{v}\) has units of \(\frac{\text{length}}{\text{time}}\) and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. For example, if we define movement to the right as positive and movement to the left as negative, an object moving 10 feet to the right in 5 seconds has an average velocity over that period of\[\overline{v} = \dfrac{+10 \, \text{feet}}{5 \, \text{seconds}} = 2 \, \text{feet per second}.\nonumber\]In this situation, we have written a plus sign on the 10 feet because the object is moving to the right by 10 feet over this period. On the other hand, if the object moving to the left at 10 feet for every 5 seconds, its average velocity over that period would be\[\overline{v} = \dfrac{-10 \, \text{feet}}{5 \, \text{seconds}} = -2 \, \text{feet per second}.\nonumber\]If we could somehow measure the velocity of an object at a specific moment in time (rather than over an interval of time), we could call that the instantaneous velocity, \(v\). The instantaneous velocity of an object is often just called the velocity of the object.
It is a common mistake for people to confuse velocity with speed. The speed of an object is the magnitude of its instantaneous velocity. That is, \(\text{speed} = \left| v \right|\). Note that speed is never negative, while velocity (and average velocity) can be negative (denoting both speed and direction).
Ants are incredibly fast, especially when you consider their size! The Saharan silver ant (Cataglyphis bombycina) holds the record for the fastest speed ever measured in an ant. It can move at a breathtaking 168.3 inches in five seconds. Find the average velocity of this ant.
- Solution
- Let \(s = 168.3\) inches and \(t = 5\) seconds. Then the average velocity of the Saharan silver ant is\[\overline{v} = \dfrac{168.3 \, \text{inches}}{5 \, \text{seconds}} = 33.66 \, \text{inches per second}.\nonumber\]
Kenyan athlete Kelvin Kiptum set a men's world record time of 2:00:35 on October 8, 2023, at the 2023 Chicago Marathon (26.2 miles). Find his average velocity for the race in miles per hour. Round your answer to the nearest hundredth.
- Answer
-
\(13.04\) miles per hour.
What if we wanted to know how fast the Kenyan athlete in Checkpoint \(\PageIndex{3}\) was running at exactly 1 hour and 12 minutes into the marathon? That is, what if we wanted to know his instantaneous velocity at precisely 1 hour and 12 minutes into the marathon? We know that he had an average linear velocity of approximately 13.04 miles per hour; however, this does not mean he ran at 13.04 miles per hour the entire race. There were many moments when he ran faster than this and others when he ran slower. Since his velocity was not constant throughout the marathon, we would be hard-pressed to compute his instantaneous velocity at 1 hour and 12 minutes into the race—the answer to this type of question is the purview of Calculus.
Since we do not have Calculus at our disposal, if we ever want to answer a question about instantaneous velocity (again, which we call velocity), we must assume that the object's velocity is constant throughout the motion. Then, and only then, it can be shown that the average linear velocity is the same as the instantaneous velocity, \(v\). For the remainder of our discussion, we will make such an assumption. That is, from this sentence forward in this section, we will assume that the object, person, car, particle, or thing we are considering is moving at a constant speed. Hence, its average velocity is equivalent to its instantaneous velocity at any given moment. Again, we will simplify the wording to "velocity."
Circular Motion - Linear Velocity
Our next excursion into the realm of applications comes through circular motion. Suppose an object is moving as pictured below along a circular path of radius \(r\) from the point \(P\) to the point \(Q\) in an amount of time \(t\).
Like before, \(s\) represents a displacement so that \(s > 0\) means the object is traveling in a counterclockwise direction, and \(s < 0\) indicates movement in a clockwise direction. Note that with this convention, the formula we used to define radian measure, namely \(s = r\theta\), still holds since a negative value of \(s\) incurred from a clockwise displacement matches the negative we assign to \(\theta\) for a clockwise rotation.
Again, just as before, the quantity \(\frac{s}{t}\) is called the average linear velocity. This is despite the fact that \(s\) is a length along the curved edge of a circle. This is because if the path of the object were "uncurled" from a circle to form a line segment, then the average velocity of the object on that line segment would be the same as the average velocity on the circle.
As of 2024, the world's largest Ferris wheel is the Dubai Eye. It opened in 2021 on tiny Bluewaters Island in the Arabian Gulf. The diameter of the wheel is 820 feet, and a single rotation lasts 38 minutes. Assuming the wheel never stops turning, what is a rider's (instantaneous) linear velocity on the Dubai Eye?
- Solution
- We need to compute how far along the edge of the circle formed by the Dubai Eye that a rider moves through in a certain amount of time. Luckily, we were told a rider completes a full circuit in 38 minutes. Therefore, we let \(t = 38\) minutes. Moreover, the distance traveled in that time is the same as the circumference of the circle. Here, we can use the faithful formula,\[C = 2\pi r.\nonumber\]The diameter of the Dubai Eye is 820 feet, so the radius is 410 feet. Thus, a rider travels a distance of\[s = C = 2\pi(410 \, \text{feet}) = 820\pi \, \text{feet}.\nonumber\]We now use our velocity formula,\[v = \dfrac{s}{t} = \dfrac{820\pi \, \text{feet}}{38 \, \text{minutes}} \approx 67.8 \, \text{feet per minute}.\nonumber\]For reference, this is a glacial 0.8 miles per hour.
Example \(\PageIndex{4}\) showcases that computing the linear velocity of an object along a circular path is very much the same as computing the velocity along a line, we just compute \(\frac{\text{distance}}{\text{time}} = \frac{s}{t}\).
Gum gets stuck to the outer edge of your bike tire. Suppose the tire rotates so that the gum travels along the circular path of its edge 5 inches in 2 seconds. What is the linear velocity of the gum? (By the way, this is also the linear speed of the tire.)
- Answer
-
\(v = 2.5\) inches per second
Circular Motion - Angular Velocity
Dividing both sides of \(s = r\theta\) by \(t\), we get a very powerful relation:\[\dfrac{s}{t} = \dfrac{r\theta}{t} \implies v = r \cdot \dfrac{\theta}{t}\nonumber\]
A point on a circle rotates through \(38^{\circ}\) in 3 hours. What is the angular velocity of the point?
- Solution
- Remember that the formula, \(s = r\theta\), requires \(\theta\) to be in radians. This is the same \(\theta\) as the angle in \(\omega = \frac{\theta}{t}\). Therefore, we need to convert the given angle to radians.\[\omega = \dfrac{38^{\circ}}{3\, \text{hours}} = \dfrac{38\pi \, \text{radians}}{540\,\text{hours}} \approx 0.2211 \, \text{radians per hour}.\nonumber\]
Suppose a point on a circle rotates through \(300^{\circ}\) in two hours. Find the exact value of the angular velocity of the point.
- Answer
-
\(\frac{5\pi}{6}\) radians per hour
Circular Motion - Relating Linear and Angular Velocities
We now formalize our recent statement.
We need to talk about units here. The units of \(v\) are \(\frac{\text{length}}{\text{time}}\), the units of \(r\) are length only, and the units of \(\omega\) are \(\frac{\text{radians}}{\text{time}}\). Thus the left hand side of the equation \(v = r\omega\) has units \(\frac{\text{length}}{\text{time}}\), whereas the right hand side has units \(\text{length} \cdot \frac{\text{radians}}{\text{time}} = \frac{\text{length} \cdot \text{radians}}{\text{time}}\). The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identified with real numbers so that the units \(\frac{\text{length} \cdot \text{radians}}{\text{time}}\) reduce to the units \(\frac{\text{length}}{\text{time}}\). We are long overdue for an example.
Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during these 24 hours is the latitude of that point. Cosumnes River College is at \(38^{\circ} \, 27^{\prime} \, 14.07^{\prime\prime}\) north latitude, and it can be shown that the radius of the Earth at this latitude is approximately \(3958\) miles. Find the linear velocity, in miles per hour, of Cosumnes River College as the world turns.
- Solution
- To use the formula \(v = r\omega\), we first need to compute the angular velocity \(\omega\). The Earth makes one revolution in 24 hours. One revolution is \(2\pi\) radians, so\[\omega = \dfrac{2\pi \, \text{radians}}{24 \, \text{hours}} = \dfrac{\pi}{12 \, \text{hours}},\nonumber\]where, once again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity's sake, we are also assuming that we are viewing the rotation of the Earth as counterclockwise, so \(\omega > 0\).) Hence, the linear velocity is\[v = 3958 \, \text{miles} \cdot \dfrac{\pi}{12 \, \text{hours}} \approx 1036 \, \dfrac{\text{miles}}{\text{hour}}.\nonumber\]
How fast is a bike moving, in miles per hour, if its rear tire has a radius of 15 inches and the angular velocity of the rear tire is 20 radians per second?
- Answer
-
\(\approx 17\) miles per hour
It is worth noting that the quantity \(\frac{1 \, \text{revolution}}{24 \, \text{hours}}\) in Example \(\PageIndex{6}\) is called the ordinary frequency of the motion.
The higher the ordinary frequency, the faster the motion. This is because the circular motion must go through more cycles per unit time. An ordinary frequency that you are familiar with is a radio station's frequency. For example, if you listen to 93.7 on the FM dial, that means you are listening to a signal broadcasting at a frequency of 93.7 MHz. This means 93.7 million Hertz. A Hertz is defined to be 1 cycle per second. Therefore, 93.7 MHz means you are listening to a station broadcasting at 93.7 million cycles per second.
The angular frequency, on the other hand, tells us how many radians an object turns through in a given time.
Since there are \(2\pi\) radians per cycle and \(f\) cycles per unit time, the angular frequency is \(2\pi f\). However, angular frequency is radians per unit time, and so is angular velocity. Thus, angular frequency and angular velocity mean the same thing! That is, \(\omega = 2\pi f\).
One final "measuring stick" for circular motion is very important—the period of the circular motion.
In the scenario of Example \(\PageIndex{6}\), the period of the motion is 24 hours, or one day.
The concepts of frequency and period help frame the equation \(v = r\omega\) in a new light. That is, if \(\omega\) is fixed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one revolution in one period's time. The distance of the object to the center of rotation is the radius of the circle, \(r\), and is the "magnification factor," which relates \(\omega\) and \(v\). We will have more to say about frequencies and periods later in this text.
The following Interactive Element might help you visualize why points farther from the center of a spinning object move faster even though every point turns through the same angle in the same time.
Interact: Drag the large red point. Move it around the center to sweep the angle (both points turn together at the same angular velocity), and pull it inward or outward to change its radius.
Observation: Both points sweep the same angle in the same time, so they share one angular velocity \(\omega\). Yet the outer (red) arc is always longer than the inner (blue) arc — the farther point covers more distance in the same time, so its linear velocity is larger by exactly the factor \(r\). This is \(v = r\omega\).
We close out this section with a beautiful application requiring great attention to detail.
A bicycle with a 27-inch-diameter rear wheel has its gears set so that the chainring has a 4-inch diameter and the rear sprocket has a 3-inch diameter (see image below). The cyclist pedals at 250 rpm. What is the speed of the bicycle in miles per hour?
- Solution
- In a complex problem like this, it's best to collect the given data by object. That is, we will have data for the chainring, the sprocket, and the rear wheel. We might not use all of these objects in our computations, but just listing them out is often helpful.\[\begin{array}{rccc} & \mathbf{Chainring} & \mathbf{Sprocket} & \mathbf{Wheel} \\[6pt] \hline \mathbf{Radius} & r_c = 2 \, \text{inches} & r_s = 1.5 \, \text{inches} & r_w = 13.5 \, \text{inches} \\[6pt] \mathbf{Linear Velocity} & v_c = ? & v_s = ? & v_w = ? \\[6pt] \mathbf{Angular Velocity} & \omega_c = ? & \omega_s = ? & \omega_w = ? \\[6pt] \end{array}\nonumber\]Before we get into computations, a little bit of logic is helpful. First, how quickly the sprocket is spinning should match how quickly the rear wheel is spinning (in angles per unit time). Therefore, \(\omega_s = \omega_w\). Additionally, we were told the cyclist is pedaling at 250 rpm (revolutions per minute). Thus, the chainring (which the pedal is attached to) is spinning at 250 rpm. We need to convert this to "radians per minute" using the fact that there are \(2\pi\) radians per revolution. Therefore,\[250 \, \text{rpm} = \dfrac{250 \, \text{revolutions}}{\text{minute}} = \dfrac{250 \, \cancel{\text{revolutions}}}{\text{minute}} \cdot \dfrac{2\pi \, \text{radians}}{\cancel{\text{revolution}}} = \dfrac{500\pi \, \text{radians}}{\text{minute}}.\nonumber\]Notice that we kept the word "radians" in this computation. It's not necessary to do this, but it allows us to easily interpret the meaning of the quantity we are working with. At this point, let's update our table.\[\begin{array}{rccc} & \mathbf{Chainring} & \mathbf{Sprocket} & \mathbf{Wheel} \\[6pt] \hline \mathbf{Radius} & r_c = 2 \, \text{inches} & r_s = 1.5 \, \text{inches} & r_w = 13.5 \, \text{inches} \\[6pt] \mathbf{Linear Velocity} & v_c = ? & v_s = ? & v_w = ? \\[6pt] \mathbf{Angular Velocity} & \omega_c = 500\pi \, \dfrac{\text{radians}}{\text{minute}} & \omega_s = \omega_w & \omega_w = \omega_s \\[6pt] \end{array}\nonumber\]We now use the completed information about the chainring (the radius and angular velocity) to find the linear velocity of a point on the edge of the chainring.\[v_c = r_c \omega_c = (2 \, \text{inches})(500\pi \, \text{radians/minute}) = 1000\pi \, \dfrac{\text{inches}}{\text{minute}}\nonumber\]Remember, radians is a unitless measure. Since a point along the edge of the chainring is just a link in the chain connecting the chainring to the sprocket, this will also be the linear velocity of a link moving along the edge of the sprocket. Thus, our table gets updated to the following.\[\begin{array}{rccc} & \mathbf{Chainring} & \mathbf{Sprocket} & \mathbf{Wheel} \\[6pt] \hline \mathbf{Radius} & r_c = 2 \, \text{inches} & r_s = 1.5 \, \text{inches} & r_w = 13.5 \, \text{inches} \\[6pt] \mathbf{Linear Velocity} & v_c = 1000\pi \, \dfrac{\text{inches}}{\text{minute}} & v_s = 1000\pi \, \dfrac{\text{inches}}{\text{minute}} & v_w = ? \\[6pt] \mathbf{Angular Velocity} & \omega_c = 500\pi \, \dfrac{\text{radians}}{\text{minute}} & \omega_s = \omega_w & \omega_w = \omega_s \\[6pt] \end{array}\nonumber\]We now can find the angular velocity of the sprocket.\[v_s = r_s \omega_s \implies \omega_s = \dfrac{v_s}{r_s} = \dfrac{1000\pi \, \cancel{\text{inches}}\text{/minute}}{1.5 \, \cancel{\text{inches}}} = \dfrac{2000\pi}{3} \, \dfrac{\text{radians}}{\text{minute}}\nonumber\]Again, this angular velocity is exactly the same as the angular velocity of the wheel. Thus, our table becomes\[\begin{array}{rccc} & \mathbf{Chainring} & \mathbf{Sprocket} & \mathbf{Wheel} \\[6pt] \hline \mathbf{Radius} & r_c = 2 \, \text{inches} & r_s = 1.5 \, \text{inches} & r_w = 13.5 \, \text{inches} \\[6pt] \mathbf{Linear Velocity} & v_c = 1000\pi \, \dfrac{\text{inches}}{\text{minute}} & v_s = 1000\pi \, \dfrac{\text{inches}}{\text{minute}} & v_w = ? \\[6pt] \mathbf{Angular Velocity} & \omega_c = 500\pi \, \dfrac{\text{radians}}{\text{minute}} & \omega_s = \dfrac{2000\pi}{3} \, \dfrac{\text{radians}}{\text{minute}} & \omega_w = \dfrac{2000\pi}{3} \, \dfrac{\text{radians}}{\text{minute}} \\[6pt] \end{array}\nonumber\]We finally arrive at the penultimate computation—the linear velocity of the wheel.\[v_w = r_w \omega_w = (13.5 \, \text{inches}) \left(\dfrac{2000\pi}{3} \, \dfrac{\text{radians}}{\text{minute}}\right) = 9000\pi \, \dfrac{\text{inches}}{\text{minute}}\nonumber\]Did you know the word "penultimate" means "second-to-last?" This means we have one more step (the ultimate step?). First, the linear velocity of a point on the wheel gives the velocity of the bicycle; however, this is in inches per minute and we were asked to find the answer in miles per hour. This is a nice unit analysis problem. Let's convert the distance units first.\[9000\pi \, \dfrac{\cancel{\text{inches}}}{\text{minute}} \cdot \dfrac{\cancel{\text{foot}}}{12 \, \cancel{\text{inches}}} \cdot \dfrac{\text{mile}}{5280 \, \cancel{\text{feet}}} = \dfrac{9000\pi}{(12)(5280)} \, \dfrac{\text{miles}}{\text{minute}}\nonumber\]Now let's convert the time units.\[\dfrac{9000\pi}{(12)(5280)} \, \dfrac{\text{miles}}{\cancel{\text{minute}}} \cdot \dfrac{60 \, \cancel{\text{minutes}}}{\text{hour}} = \dfrac{(9000\pi)(60)}{(12)(5280)} \, \text{mph} \approx 26.8 \, \text{mph}\nonumber\]
If the chainring of a bicycle is 150 mm in diameter, the sprocket is 80 mm in diameter, and the wheel is 700 mm in diameter, how fast would one need to pedal, in revolutions per minute, in order to maintain a speed of 24 kilometers per hour?
- Answer
-
Approximately 97 rpm.
This page titled 7.4: Velocities and Areas of Sectors was last modified on Tue, 08 Jul 2025 17:44:30 GMT and is shared under a CC BY-SA license and was authored, remixed, and/or curated by Roy Simpson.
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