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Chapter 10: Problem 25

(a) What is the angular speed \(\omega\) about the polar axis of a point on Earth's surface at latitude \(40^{\circ} \mathrm{N} ?\) (Earth rotates about that axis.) (b) What is the linear speed \(v\) of the point? What are (c) \(\omega\) and \((\mathrm{d}) v\) for a point at the equator?

Short Answer

Expert verified
(a) \( \omega = 7.27 \times 10^{-5} \text{ rad/s} \); (b) \( v = 354 \text{ m/s} \); (c) \( \omega = 7.27 \times 10^{-5} \text{ rad/s} \); (d) \( v = 465 \text{ m/s} \).

Step by step solution

01

Understand Earth鈥檚 Rotation

The Earth completes one full rotation around its axis in a period of 24 hours. This means the angular speed around that axis is constant for any point along the planet's surface. It is calculated using the formula: \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation. For Earth, \( T = 24 \) hours or 86400 seconds.
02

Calculate Angular Speed \(\omega\)

Using the formula from Step 1, compute \( \omega \):\[ \omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \text{ rad/s} \].This value remains the same for any latitude because angular speed is constant for rotations about the same axis.
03

Understand Linear Speed at Latitude

The linear speed \( v \) of a point on Earth鈥檚 surface depends on its distance from the axis of rotation. This distance changes with latitude. The formula for linear speed is \( v = r \omega \), where \( r \) is the radius of Earth at the latitude in question.
04

Calculate Distance From Axis

The Earth's radius decreases with latitude. At latitude \( \phi \), the effective radius \( r \) is calculated as \( r = R \cos \phi \), where \( R \approx 6371 \text{ km} \) is Earth's radius. For latitude \( 40^{\circ} N \), compute \( r \): \[ r = 6371 \times \cos 40^{\circ} \approx 4878 \text{ km} \].
05

Calculate Linear Speed \(v\) at \( 40^{\circ} \mathrm{N} \)

Substitute \( r \) and \( \omega \) into the formula for \( v \):\[ v = 4878 \times 10^3 \times 7.27 \times 10^{-5} \approx 354 \text{ m/s} \].
06

Calculate Angular Speed and Linear Speed at Equator

At the equator, the radius \( r \) is equal to Earth's radius \( R = 6371 \text{ km} \). The angular speed \( \omega = 7.27 \times 10^{-5} \text{ rad/s} \) is unchanged.The linear speed is then computed as:\[ v = 6371 \times 10^3 \times 7.27 \times 10^{-5} \approx 465 \text{ m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Speed
Angular speed, often denoted by the Greek letter omega (\( \omega \)), describes how fast an object rotates around an axis. For Earth, this refers to how quickly it spins around its own axis. This speed is constant because Earth completes a full rotation every 24 hours. To find the angular speed, use the formula: \( \omega = \frac{2\pi}{T} \), where \( T \) represents the time for one rotation, or in Earth's case, 24 hours (or 86400 seconds). When calculating this, you get \( \omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \text{ rad/s} \). This value remains consistent across all latitudes because it simply reflects how the entire planet rotates.
Linear Speed
Linear speed is a measure of how fast a point on the Earth's surface travels as the planet rotates. Unlike angular speed, linear speed varies depending on your position (latitude) on Earth's surface. It is given by the equation \( v = r \omega \), where \( r \) is the distance from that point to the Earth's axis of rotation. You can think of it this way:
  • At the equator, \( r \) is equal to Earth's max radius because it's furthest from the axis, which gives the highest linear speed.
  • At any other latitude, \( r \) can be calculated with \( r = R \cos \phi \), where \( R \approx 6371 \text{ km} \) is Earth鈥檚 radius, and \( \phi \) is the latitude.
As you move from the equator towards the poles, \( r \) decreases, resulting in a decrease in linear speed.
Latitude and Longitude
Latitude and longitude are coordinate systems used to determine any location on Earth's surface.
  • Latitude: Measures how far north or south a point is from the equator. It ranges from \( 0^{\circ} \) at the Equator to \( 90^{\circ} \) at the poles. Latitude impacts linear speed, as it affects the distance a point is from Earth's axis.
  • Longitude: Measures how far east or west a point is from the Prime Meridian, which runs through Greenwich, England. Longitude, unlike latitude, doesn't affect the Earth's linear speed because all points on a latitude circle cover the same rotational distance in 24 hours.
These coordinates help calculate how geographical location influences the characteristics of Earth's rotation.
Equator
The Equator is an imaginary line that runs around the middle of the Earth, dividing it into the Northern and Southern Hemispheres. It is situated at latitude \( 0^{\circ} \) and is the reference point for measuring latitude. The Equator marks the point on Earth where:
  • The linear speed of Earth's rotation is at its maximum because it experiences the largest radius from the axis.
  • Angular speed remains the same across all points on Earth's surface but exhibits the full effect of Earth's rotational speed.
at-G-at-one
Understanding the Equator is crucial for recognizing just how dynamic the Earth's rotation is, influencing global climates and weather patterns. Linear speed here is about \( 465 \text{ m/s} \), reflecting the fastest surface movement during Earth's rotation.

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Most popular questions from this chapter

(a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

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A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a \(1.25-\mathrm{m}\) -long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?
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