Arcs and Sectors: Apply It

Lumen Learning, OpenStax

Arcs and Sectors: Apply It

Find the length of a circular arc.
Find the area of a sector of a circle.
Use linear and angular speed to describe motion on a circular path.

Circular motion is everywhere—from the hands of a clock to the rotation of a Ferris wheel, from irrigation systems watering crops to satellites orbiting Earth. Understanding arc length, sector area, and angular speed allows us to analyze and describe these circular motions precisely. In this page, you’ll apply these concepts to analyze a real Ferris wheel attraction.

The London Eye Ferris Wheel

The London Eye is one of the world’s largest observation wheels, located on the banks of the River Thames. This giant Ferris wheel allows passengers to view the entire city from enclosed capsules as it slowly rotates.

London Eye Specifications:

Radius: 60 meters (from center to passenger capsule)
One complete rotation: 30 minutes
Operates continuously during business hours

Arc length – The distance along the curved path of a circle between two points, calculated using [latex]s = r\theta[/latex] where [latex]r[/latex] is the radius and [latex]\theta[/latex] is the angle in radians.

Sector – A region of a circle bounded by two radii and the intercepted arc, like a slice of pie. The area is calculated using [latex]A = \frac{1}{2}\theta r^2[/latex] where [latex]\theta[/latex] is in radians.

A passenger boards the London Eye and stays on for a 10-minute segment of the ride.

What angle (in both degrees and radians) does the wheel rotate through in 10 minutes?
What distance does the passenger travel along the circular path during these 10 minutes?
If the wheel is divided into 32 passenger capsules equally spaced around the circle, what is the area of the sector between two adjacent capsules?
What is the angular speed of the wheel in radians per minute?
What is the linear speed of a passenger in meters per minute? Convert this to kilometers per hour.

The wheel completes one full rotation (360° or [latex]2\pi[/latex] radians) in 30 minutes.

In 10 minutes, the wheel completes: [latex]\frac{10 \text{ minutes}}{30 \text{ minutes}} = \frac{1}{3} \text{ of a rotation}[/latex]

In degrees: [latex]\frac{1}{3} \times 360° = 120°[/latex]

In radians: [latex]\frac{1}{3} \times 2\pi = \frac{2\pi}{3} \text{ radians}[/latex]

The wheel rotates through 120° or [latex]\frac{2\pi}{3}[/latex] radians in 10 minutes.
Using the arc length formula [latex]s = r\theta[/latex], where [latex]\theta[/latex] must be in radians:

[latex]\begin{aligned} s &= r\theta \\ &= 60 \text{ meters} \times \frac{2\pi}{3} \\ &= \frac{120\pi}{3} \\ &= 40\pi \\ &\approx 125.66 \text{ meters} \end{aligned}[/latex]

The passenger travels approximately 125.66 meters along the circular path.
With 32 equally spaced capsules, the angle between adjacent capsules is:

[latex]\theta = \frac{2\pi}{32} = \frac{\pi}{16} \text{ radians}[/latex]

Using the sector area formula [latex]A = \frac{1}{2}\theta r^2[/latex]:

[latex]\begin{aligned} A &= \frac{1}{2}\theta r^2 \\ &= \frac{1}{2} \times \frac{\pi}{16} \times (60)^2 \\ &= \frac{1}{2} \times \frac{\pi}{16} \times 3600 \\ &= \frac{3600\pi}{32} \\ &= \frac{225\pi}{2} \\ &\approx 353.43 \text{ square meters} \end{aligned}[/latex]

The sector area between two adjacent capsules is approximately 353.43 square meters.
Angular speed is the rate of angular rotation, calculated as [latex]\omega = \frac{\theta}{t}[/latex].

For one complete rotation:

[latex]\begin{aligned} \omega &= \frac{\theta}{t} \\ &= \frac{2\pi \text{ radians}}{30 \text{ minutes}} \\ &= \frac{2\pi}{30} \\ &= \frac{\pi}{15} \\ &\approx 0.209 \text{ radians per minute} \end{aligned}[/latex]

The angular speed of the London Eye is [latex]\frac{\pi}{15}[/latex] radians per minute or approximately 0.209 rad/min.
Linear speed is the actual distance traveled per unit time along the circular path. It relates to angular speed by [latex]v = r\omega[/latex].

[latex]\begin{aligned} v &= r\omega \ &= 60 \text{ meters} \times \frac{\pi}{15} \text{ rad/min} \ &= \frac{60\pi}{15} \ &= 4\pi \ &\approx 12.57 \text{ meters per minute} \end{aligned}[/latex]

Converting to kilometers per hour: [latex]\begin{aligned} v &= 12.57 \frac{\text{meters}}{\cancel{\text{minute}}} \times \frac{1 \text{ km}}{1000 \text{ meters}} \times \frac{60 \cancel{\text{minutes}}}{1 \text{ hour}} \ &= \frac{12.57 \times 60}{1000} \\ &\approx 0.754 \text{ km/h} \end{aligned}[/latex]

The linear speed of a passenger is approximately 12.57 meters per minute or 0.754 km/h.

This slow speed explains why passengers can easily board and exit while the wheel continues moving!

Sector area [latex]A = \frac{1}{2}\theta r^2[/latex] represents the fraction [latex]\frac{\theta}{2\pi}[/latex] of the total circle area [latex]\pi r^2[/latex]. This formula only works when [latex]\theta[/latex] is in radians.

A smaller Ferris wheel at a carnival has a radius of 8 meters and completes one full rotation every 2 minutes.

What angle does the wheel rotate through in 45 seconds? Express your answer in radians.
How far does a passenger travel in 45 seconds?
What is the angular speed of the wheel in radians per second?
What is the linear speed of a passenger in meters per second?