8. Applications of Radian Measure

by M. Bourne

Linear velocity applet

Don't miss the interactive angular/linear velocity animation on this page.

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Angular velocity applet

In this section, we see some of the common applications of radian measure, including arc length, area of a sector of a circle, and angular velocity.

Go back to the section on Radians if you are not sure what is going on.

Arc Length

The length, s, of an arc of a circle radius r subtended by θ (in radians) is given by:

s = r θ

If r is in meters, s will also be in meters. Likewise, if r is in cm, s will also be in cm.

Example 1

Find the length of the arc of a circle with radius `4\ "cm"` and central angle `5.1` radians.

Answer

We need to find length s.

s = r θ

= 4 × 5.1

= 20.4 cm

Easy to understand math videos:
MathTutorDVD.com

Area of a Sector

Area, A, of a sector of a circle.

The area of a sector with central angle θ (in radians) is given by:

`"Area"=(theta\ r^2)/2`

If r is measured in `"cm"`, the area will be in `"cm"`2. If r is in `"m"`, the area will be in `"m"`2.

Example 2

Find the area of the sector with radius `7\ "cm"` and central angle `2.5` radians.

Answer

This is the area we need to find:

Area, a sector of a circle, radius 7 and central angle 2.5 rad.

`"Area"=(theta\ r^2)/2=(2.5xx7^2)/2=61.25\ "cm"^2`

Angular Velocity

The time rate of change of angle θ by a rotating body is the angular velocity, written ω (omega). It is measured in radians/second.

If v is the linear velocity (in m/s) and r is the radius of the circle (in m), then

v =

Note: If r is in `"cm"`, v will be in `"cm/s"`.

Example 3

A bicycle with tyres `90\ "cm"` in diameter is travelling at `25` km/h. What is the angular velocity of the tyre in radians per second?

Answer

Bicycle wheel, radius 90 cm.
The arrows represent the linear speed of 25 km/h

We learned that linear velocity for a wheel rotating at ω rad/s is given by:

v = r ω

The units are a mix of cm and km. Let's present everything in meters.

We need to convert v to m/s first.

`25\ "km/h" = 25000\ "m/h"`

` = 25000/3600 "m/s"`

` = 6.94444\ "m/s"`

Also, we have

`r = (90\ "cm")/2 = 45\ "cm" = 0.45\ "m"`

So `ω = v/r = 6.94444/0.45 = 15.43\ "rad/s"`

Interactive linear velocity applet

Background

A car is going around a circular track of radius 0.5 km. The speedometer in the car shows the (magnitude) of the linear velocity.

At first, the car goes around the track once in just over 6 minutes. It's angular velocity is 1 rad/min or one complete revolution in 2π = 6.28 min.

The distance travelled in this time is the circumference of the circle, C = 2πr = 2π(0.5) = 3.14 km. So the car is travelling at `(3.14" km")/(6.28 min) = 0.5" km/min" = 30" km/h"`.

The linear velocity showing on the speedo is 30 km/h.

Things to do

In this applet, you can:

  • Vary the radius of the track
  • Vary the angular velocity of the car

Observe the change in linear speed as you do so.

Of course, angle measures are in radians in this applet.

r = 0 km ω = 0 rad/min

v = rω = 0 × 0 = 0 km/min = 0 km/h

Radius:

Angular velocity:

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Exercises:

1. A section of side walk is a circular sector of radius `1.25\ "m"` and central angle `50.6°`. What is the area of this section of sidewalk?

Answer

Circular sector, radius 1.25 m, central angle 56°.

First we must convert `50.6°` to radians:

`50.6° = 50.6 × π/180 = 0.8831\ "radians"`

`"Area"=(theta\ r^2)/2`

`=(0.8831xx1.25^2)/2`

`=0.690\ "m"^2`

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2. A cam is in the shape of a circular sector with radius `1.875\ "cm"` and central angle `165.58°`. What is the perimeter of the cam?

Answer

Circular sector, radius 1.875 cm, central angle 165.58°.

The length of the arc is given by s = rθ.

First we must convert `165.58°` into radians:

`165.58° = 165.58 × π/180 = 2.8899` radians.

So arc length is: `s = 1.875 × 2.8899 = 5.419` cm.

So the perimeter of the cam is:

`2 × 1.875 + 5.419 = 9.169` cm.

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3. The roller on a computer printer makes `2200` rev/min. What is its angular velocity?

Answer

Angular velocity is:

`ω = 2200 r/min × (2π) / 60 = 230.4\ "rad/s"`.

Easy to understand math videos:
MathTutorDVD.com

4. The propeller on a motorboat is rotating at `130` rad/s. What is the linear velocity of a point on the tip of a blade if the blade is `22.5` cm long?

Answer

Linear velocity = v = ωr

In this example, ω = 130 rad/s and r = 0.225 m

So the linear velocity is:

v = 130 × 0.225 = 29.3 ms-1.

Note 1: ms-1 is an equivalent way of writing m/s. This comes from the index laws where the rule is `s^-1= 1/s`.

Note 2: It is common in physics to write velocity using ms-1 and the units for acceleration as ms-2.

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5. The sweep second hand of a watch is `15.0` mm long. What is the linear velocity of the tip?

Answer

The second hand rotates `2pi` every minute, so per second we have:

`omega=(2pi)/60=pi/30"rad""/"s`

and `r= 0.015\ "m"`.

So

`v=omega r`

`=(pi/30)(0.015)`

` = 0.00157\ "m/s"`.

Pulley Problems

pulleys

You can investigate the linear velocity of a belt moving around two pulleys in this interactive example.

Go to Pulleys simulation.