by M. Bourne

### Linear velocity applet

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

Go to

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=rθ

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 arcof 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

## Area of a Sector

Area, *A*, of a sector of a circle.

The area of a sector with central angle *θ* (in radians) isgiven 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 sectorwith 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 inradians/second.

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

v=rω

**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 angularvelocity 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:

Copyright © www.intmath.com Frame rate: 0

### 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`

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* =*r**θ*.

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.

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

Answer

Angular velocity is:

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

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

Answer

Linear velocity = *v* = *ω**r*

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}.

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

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

Go to Pulleys simulation.