This question aims to find the **angular speed** of the tires of the bicycle and the **speed** of the **blue dot** painted on the tires of **0.8m diameter**.

A bicycle is coasting on a level road at a speed of **5.6 m/s**. The tires of this bicycle have a diameter of **0.80 m** and a blue dot is painted on the tread of the rear tire of this bicycle. We have to find the angular speed of the tires. The **Angular speed** is defined as the speed of the rotating body with its **central angle**. The speed of the rotating body changes with **time**.

The blue dot rotates as the tire rotates with some speed. We have to find the speed of the blue dot when it is **0.80 m** **above the ground** and the speed of the blue dot when it is **0.40 m** above the ground.

The **diameter** of the tire is represented by **d**, the **radius** is represented by **r**, the **speed** of the bicycle is represented as **v** and the **angular speed** of the tire is represented by $ \omega $.

## Expert Answer

The values are given as:

\[ d = 0 . 8 0 m \]

\[ r = \frac { d } { 2 } \]

\[ r = \frac { 0 . 8 0 } { 2 } \]

\[ r = 0 . 4 0 \]

The speed of the bicycle is given as:

\[ v = r \omega \]

\[ 5 . 6 = ( 0 . 4 0 ) \omega \]

\[ \omega = \frac { 5 . 6 } { 0 . 4 0 } \]

\[ \omega = 14 rad/s \]

The speed of the blue dot is given by:

\[ v’ = v + r \omega \]

\[ v’ = 5 . 6 + ( 0 . 4 0 ) \times 14 \]

[ v’ = 11 . 2 m/s \]

The angle between the speed and angular speed of tires is** 90°**. Using the **Pythagoras theorem**, we get:

\[ v ^ 2 = ( r \omega ) ^ 2 + ( v ) ^ 2 \]

Taking square-root on both sides:

\[ v = \sqrt { ( r \omega ) ^ 2 + ( v ) ^ 2 } \]

\[ v = \sqrt { ( 0.40 \times 14 ) ^ 2 + ( 5.6 ) ^ 2 } \]

\[ v = 7 . 9 1 9 m/s \]

## Numerical Solution

**The angular speed $ \omega $ of the tires is 14 rad/s. The speed of the blue dot rotating with the tires is 11.2 m/s when it is 0.80 m above the ground. The speed changes to 7.919 m/s when it is 0.40 m above the ground.**

## Example

Find the **angular speed** of the tire of a car moving with a velocity of **6.5 m/s.** The diameter of the tires is **0.60 m.**

The values are given as:

\[ d = 0 . 6 0 m \]

\[ r = \frac { d } { 2 } \]

\[ r = \frac { 0 . 6 0 } { 2 } \]

\[ r = 0 . 3 0 \]

The speed of the bicycle is given as:

\[ v = r \omega \]

\[ 6 . 5 = ( 0 . 3 0 ) \omega \]

\[ \omega = \frac { 6 . 5 } { 0 . 3 0 } \]

\[ \omega = 21.6 rad/s \]

The angular speed of the tires is **21.6 rad/s.**

*Image/Mathematical drawings are created in Geogebra**.*