**Calculate the angular speed of the earth.****Calculate the direction (positive or negative) of the angular velocity. Assume you are viewing from a point exactly above the north pole.****Calculate the tangential speed of a point on the earth’s surface located on the equator.****Calculate the tangential speed of a point on the earth’s surface located halfway between the pole and the equator.**

The aim of the question is to understand the concept of angular and tangential speeds of a rotating body and the points on its surface, respectively.

If $\omega$ is the angular speed and $T$ is the time period of rotation, the **angular speed** is defined by the following formula:

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

If the radius $r$ of the rotation of a point around the axis of rotation, then the** tangential speed $v$** is defined by the following formula:

\[v = r \omega\]

## Expert Answer

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

If $\omega$ is the **angular speed** and $T$ is the **time period** of rotation, then:

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

For our case:

\[T = 24 \times 60 \times 60 \ s\]

So:

\[\omega = \frac{2\pi}{24\times 60 \times 60 \ s} = 7.27 \times 10^{-5} \ rad/s\]

**Part (b): Calculate the direction (positive or negative) of the angular velocity. Assume you are viewing from a point exactly above the north pole.**

When viewed from a point exactly above the north pole, the earth rotates counter-clockwise, so the angular velocity is positive (following the right-hand convention).

**Part (c): Calculate the tangential speed of a point on the earth’s surface located on the equator.**

If the radius $r$ of the rigid body is known, then the** tangential speed $v$** can be calculated using the formula:

\[v = r \omega\]

For our case:

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

And:

\[ \omega = 7.27 \times 10^{-5} rad/s\]

So:

\[v = ( 6.37 \times 10^{6} m)(7.27 \times 10^{-5} rad/s)\]

\[v = 463.1 m/s\]

**Part(d): Calculate the tangential speed of a point on the earth’s surface located halfway between the pole and the equator.**

A point on the earth’s surface located halfway between the pole and the equator rotates in a circle of **radius given by** the following formula:

\[\boldsymbol{r’ = \sqrt{3} r }\]

\[r’ = \sqrt{3} (6.37 \times 10^{6} m) \]

Where $r$ is the radius of the earth. Using the **tangential speed formula**:

\[v = \sqrt{3} ( 6.37 \times 10^{6} m)(7.27 \times 10^{-5} rad/s)\]

\[v = 802.11 m/s\]

## Numerical Result

Part (a): $\omega = 7.27 \times 10^{-5} \ rad/s$

Part (b): Positive

Part (c): $v = 463.1 m/s$

Part (d): $v = 802.11 m/s$

## Example

The radius of the Moon is $1.73 \times 10^{6} m$

**– Calculate the angular speed of the moon.****– Calculate the tangential speed of a point on the moon’s surface located midway between the poles.**

Part (a): **One day on Moon** is equal to:

\[T = 27.3 \times 24 \times 60 \times 60 \ s\]

So:

\[\omega = \frac{2\pi}{T} = \frac{2\pi}{27.3 \times 24 \times 60 \times 60 \ s}\]

\[\boldsymbol{\omega = 2.7 \times 10^{-6} \ rad/s}\]

Part (b): **Tangential speed** on the given point is:

\[v = r \omega\]

\[v = ( 1.73 \times 10^{6} m)(2.7 \times 10^{-6} \ rad/s)\]

\[ \boldsymbol{v = 4.67 m/s}\]