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

The EarthS Radius Is 6.37×106M 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:

\[v = r \omega\]

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 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:

\[v = r \omega\]

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

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

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