# A bicycle with 0.80 m diameter.

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 $\omega$.

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.