The earth’s radius is 6.37×10^6 m; it rotates once every 24 hours.

1. Calculate the angular speed of the earth?
2. Calculate the direction (positive or negative) of the angular velocity? Assume you are viewing from a point exactly above the north pole.
3. Calculate the tangential speed of a point on the earth’s surface located on the equator?
4. 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:

$v = r \omega$

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 anticlockwise, 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:

$v = r \omega$

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

$\boldsymbol{v = 4.67 m/s}$