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**What is the angular speed of the tires?****What is the speed of the blue dot when it is $0.80\, m$ above the road?****What is the speed of the blue dot when it is $0.40\, m$ above the road?**

This question aims to find the angular speed of the tire of a bicycle.

The rate at which an object travels a given distance is said to be speed. Consequently, angular speed is the rate of rotation of an object. More generally, it is the change in an object’s angle per unit of time. As a result, the rotational motion’s speed can be calculated if its angular speed is known. The formula of angular speed computes the distance traveled by a body with regard to rotations/revolutions per unit of time. In other words, we can define angular speed as the rate of change of angular displacement having the mathematical form $\omega=\dfrac{\theta}{t}$, in which $\theta$ defines the angular displacement, $t$ defines the time and $\omega$ defines the angular speed. It is measured in radians which are known as circular measurements.

It is a scalar quantity describing how fast a body rotates. The term scalar refers to a quantity that does not have a direction but possesses a magnitude. On the other hand, angular velocity refers to a vector quantity. The angular velocity measures the rotation of an object in a particular direction and is also measured in radians per second. Angular velocity has the formula: $\omega=\dfrac{\Delta\theta}{\Delta t}$.There are two forms of angular velocity: orbital angular velocity and spin angular velocity.

## Expert Answer

Given that:

$d=0.80\,m$

$r=\dfrac{0.80}{2}\,m$

$r=0.4\,m$

Let $v_{cm}=5.6\,m/s$ be the linear velocity of the center of mass of the wheel then angular speed can be calculated as:

$\omega=\dfrac{v_{cm}}{r}$

$\omega=\dfrac{5.6}{0.4}$

$\omega=14\,rad/s$

Speed of the blue dot can be found as:

$v=v_{cm}+r\omega$

$v=5.6+(0.4)(14)$

$v=5.6+5.6$

$v=11.2\,m/s$

Finally, the speed of the blue dot, using the Pythagoras theorem, when it is $0.40\, m$ above the road is:

$v^2=(r\omega)^2+(v_{cm})^2$

$v=\sqrt{(r\omega)^2+(v_{cm})^2}$

$v=\sqrt{(0.4\cdot 14)^2+(5.6)^2}$

$v=\sqrt{31.36+31.36}$

$v=\sqrt{62.72}$

$v=7.9195\,m/s$

## Example 1

Determine the angular velocity of a particle traveling along the straight line denoted by $\theta(t)=4t^2+3t-1$ when $t=6\,s$.

### Solution

The formula for the angular velocity is:

$\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{d\theta}{dt}$

Now, $\dfrac{d\theta}{dt}=\dfrac{d}{dt}(4t^2+3t-1)$

$\omega=8t+3$

Now at $t=6\,$, we have:

$\omega=8(6)+3$

$\omega=48+3$

$\omega=51\,units/second$

## Example 2

On the road, an $18$-inch-radius car wheel rotates at $9$ revolutions per second. Find the tire’s angular speed.

### Solution

The angular speed is given by:

$\omega=\dfrac{\theta}{t}$

A full rotation is $360^\circ$ or $2\pi$ in radians, so multiply the $9$ revolutions by $2\pi$ and find the angular speed as:

$\omega=\dfrac{(9)(2\pi)}{1\,s}=18\pi\,rad/s$