**– The radius of the earth is measured to be $6.37\times{10}^6m$. It completes one rotation around its orbit in $24$ hours.**

**– Part (a) – Calculate the angular speed of the earth.**

**– Part (b) – If the rotation of the earth is viewed from a location above the north pole, will the angular velocity have a positive notation or negative notation?**

**– Part (c) – Calculate the speed of a point on the equator of the earth.**

**– Part (d) – If a point lies halfway between the north pole and equator of the earth, calculate its speed.**

The aim of this question is to find the **angular speed of the earth**, its **direction**, and the **speed** of a point lying at certain **locations** on the earth.

The basic concept behind this article is the **Angular Speed** or **Angular Velocity** depending upon the **radius of rotation** and its relationship with **linear velocity**.

For any **object** moving in a **circle** or around its **orbit**, its **Angular** **Speed** $\omega$ is expressed as follows:

\[\omega=\frac{2\pi}{T}\]

Where:

$T=$ **Time period** taken to complete **one full rotation** around the **axis**.

The **Linear speed** of an object moving in **circular motion** is represented as follows:

\[v=r\omega\]

Where:

$r=$ **Distance** between the** axis of rotation** and the point at which **speed** is to be measured.

## Expert Answer

Given that:

The **Radius of Earth** $R=6.37\times{10}^6m$

**Time Period of Rotation** $T=24h$

\[T=24\times60\times60\ sec\]

\[T=86400s\]

**Part (a)**

**Angular Speed** $\omega$ is expressed as follows:

\[\omega=\frac{2\pi}{T}\]

\[\omega=\frac{2(3.14)}{86400s}\]

\[\omega=7.268\times{10}^{-5}s^{-1}\]

**Part (b)**

**Angular Speed** $\omega$ is considered **positive** if the **rotation** is **anti-clockwise** and it is considered **negative** if the **rotation** is **clockwise**.

If the **earth** is observed from a point directly above the **north pole**, the **rotation** is **anti-clockwise,** hence the **Angular Speed** $\omega$ is **positive**.

**Part (c)**

The **Linear speed** $v$ of an object that is in **rotation** is given by:

\[v=R\omega\]

At the **Equator**, the distance between the **axis of rotation** of the **earth** and the point at the **equator** is the **radius** $R$ of the **earth**. So, substituting the values in the above equation:

\[v=(6.37\times{10}^6m)(7.268\times{10}^{-5}s^{-1})\]

\[v=463\frac{m}{s}\]

**Part (d)**

For a point that lies **halfway** between the **north pole** and **equator** **of the earth**, the **radius** $r$ from the **rotational axis** is calculated from the following diagram:

Figure 1

\[r=Rsin\theta\]

\[r=(6.37\times{10}^6m)sin{45}^\circ\]

\[r=(6.37\times{10}^6m)(0.707)\]

\[r=4.504{\times10}^6m\]

And we know:

\[v=r\omega\]

\[v=(4.504{\times10}^6m)(7.268\times{10}^{-5}s^{-1})\]

\[v=327.35\frac{m}{s}\]

## Numerical Result

**Part (a)** – The **angular speed** $\omega$ of the **earth** is:

\[\omega=7.268\times{10}^{-5}s^{-1}\]

**Part (b)** –**Angular Speed** $\omega$ is **positive**.

**Part (c)** – The **speed** $v$ of a point on the **equator of earth** is:

\[v=463\frac{m}{s}\]

**Part (d)** – If a point lies **halfway** between the **north pole** and **equator of the earth**, its **speed** is:

\[v=327.35\frac{m}{s}\]

## Example

A car moving at $45\dfrac{km}{h}$ is taking a turn having a **radius** of $50m$. Calculate its **angular velocity**.

**Solution**

**Speed of the Car** $v=45\dfrac{km}{h}$

\[v=\frac{45\times1000}{60\times60}\frac{m}{s}\]

\[v=12.5\frac{m}{s}\]

**Radius of the turn** $r=50m$.

The **Linear speed** $v$ of an object that is in **rotation** is given by:

\[v=r\omega\]

So:

\[\omega=\frac{v}{r}\]

\[\omega=\frac{12.5\dfrac{m}{s}}{50m}\]

\[\omega=0.25s^{-1}\]

*Image/Mathematical drawings are created in Geogebra*