/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 What is the angular speed \(\ome... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 25

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) \(7.27 \times 10^{-5}\, \text{rad/s}\), (b) 356 m/s, (c) \(7.27 \times 10^{-5}\, \text{rad/s}\), (d) 465 m/s.

Step by step solution

01

Formula for Angular Speed

The angular speed \( \omega \) of a point on Earth's surface is the same as the angular speed of Earth's rotation because the Earth rotates solidly. The Earth makes one full rotation (\(2\pi\) radians) in 24 hours, so \( \omega = \frac{2\pi}{24 \times 3600} \) radians per second.
02

Calculate Angular Speed

Substitute 24 hours converted to seconds into the formula to find \( \omega \): \[ \omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \text{ radians per second.} \]
03

Formula for Linear Speed

To find the linear speed \( v \) at latitude \( 40^{\circ} \mathrm{N} \), use the formula \( v = r \cdot \omega \), where \( r \) is the radius of the circle traced out by the point, given by \( R \cos \phi \). Here \( R \) is Earth's radius (approximated as \(6.371 \times 10^6 \) m) and \( \phi = 40^{\circ} \).
04

Calculate Radius of Rotation

Determine \( r = R \cos \phi \) using Earth's radius \( R = 6.371 \times 10^6 \text{ m} \) and \( \phi = 40^{\circ} \): \[ r = 6.371 \times 10^6 \times \cos 40^{\circ} \approx 4.889 \times 10^6 \text{ m.} \]
05

Calculate Linear Speed

Using \( r \) from Step 4 and \( \omega \) from Step 2, calculate \( v = r \cdot \omega \): \[ v = 4.889 \times 10^6 \times 7.27 \times 10^{-5} \approx 356 \text{ m/s.} \]
06

Angular Speed at Equator

Since the angular speed \( \omega \) is the same no matter where you are on Earth, the angular speed at the equator is the same as calculated in Step 2: \( \omega \approx 7.27 \times 10^{-5} \text{ radians per second.} \)
07

Linear Speed at Equator

Calculate the linear speed at the equator using Earth's full radius: \( r = R = 6.371 \times 10^6 \text{ m} \). Thus, \[ v = R \cdot \omega = 6.371 \times 10^6 \times 7.27 \times 10^{-5} \approx 465 \text{ m/s.} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Earth's rotation
The Earth rotates around an imaginary line that extends from the North Pole to the South Pole, known as the polar axis. This rotational motion is what gives us day and night. The Earth completes one full rotation about its axis in approximately 24 hours, equivalent to an angular displacement of 360掳, or in radians, a complete circle of \(2\pi\) radians. This consistent rotational motion can be calculated as the angular speed \(\omega\), which for the Earth remains constant at about \(7.27 \times 10^{-5}\) radians per second, regardless of the point on Earth's surface. All points on Earth share this angular speed, but their linear speeds will differ based on their latitude.
linear speed
While angular speed remains constant, linear speed \(v\) varies depending on a point's position relative to the axis of rotation. Linear speed refers to the distance covered per unit of time. It measures how fast the point on Earth's surface moves physically through space as the Earth rotates. It can be calculated using the relation \(v = r \cdot \omega\), where \(r\) is the radius of rotation. The value of \(v\) will be larger at the equator, where the radius of rotation is maximum, and smaller at the poles.
latitude effect
Latitude affects both the radius of rotation \(r\) and the linear speed \(v\) because the Earth's surface curves and the radius of the circle traced by a point on the surface decreases as you move from the equator to the poles. At the equator, the latitude \(\phi\) is 0掳, making the radius \(r\) equal to the Earth's full radius \(R\). At higher latitudes, \(r\) is reduced to \(R \cdot \cos \phi\). For example, at 40掳 N latitude, the radius is \(R \cos 40掳\), resulting in a smaller linear speed compared to points on the equator.
radius of rotation
The radius of rotation \(r\) is crucial to calculate the linear speed of a point on Earth's surface. It is effectively the radius of the circle that the point traces due to Earth's rotation. To compute \(r\), consider the Earth's radius \(R\), approximately \(6.371 \times 10^6\) meters. The effective radius \(r\) changes based on latitude, represented in the formula \(r = R \cdot \cos \phi\), where \(\phi\) is the latitude. At the equator, \(\phi = 0掳\), so \(r = R\). At 40掳 N, as in this exercise, \(r\) becomes \(R \cdot \cos 40掳\). This concept illustrates why linear speed varies with latitude while angular speed remains constant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s} .\) Calculate \((\mathrm{a})\) the angular acceleration, \((\mathrm{b})\) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev \(/\) s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

An automobile crankshaft transfers energy from the engine to the axle at the rate of \(100 \mathrm{hp}(=74.6 \mathrm{~kW})\) when rotating at a speed of 1800 rev \(/\) min. What torque (in newton-meters) does the crankshaft deliver?

A good baseball pitcher can throw a baseball toward home plate at \(85 \mathrm{mi} / \mathrm{h}\) with a spin of \(1800 \mathrm{rev} / \mathrm{min} .\) How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the \(60 \mathrm{ft}\) path is a straight line.

A thin spherical shell has a radius of \(1.90 \mathrm{~m} .\) An applied torque of \(960 \mathrm{~N} \cdot \mathrm{m}\) gives the shell an angular acceleration of \(6.20 \mathrm{rad} / \mathrm{s}^{2}\) about an axis through the center of the shell. What are (a) the rotational inertia of the shell about that axis and (b) the mass of the shell?

In Fig. 10-61, four pulleys are connected by two belts. Pulley \(A\) (radius \(15 \mathrm{~cm}\) ) is the drive pulley, and it rotates at \(10 \mathrm{rad} / \mathrm{s}\). Pulley \(B\) (radius \(10 \mathrm{~cm})\) is connected by belt 1 to pulley \(A .\) Pulley \(B^{\prime}\) (radius \(5 \mathrm{~cm})\) is concentric with pulley \(B\) and is rigidly attached to it. Pulley \(C\) (radius \(25 \mathrm{~cm}\) ) is connected by belt 2 to pulley \(B^{\prime}\). Calculate (a) the linear speed of a point on belt \(1,\) (b) the angular speed of pulley \(B\), (c) the angular speed of pulley \(B^{\prime},(\mathrm{d})\) the linear speed of a point on belt \(2,\) and \((\mathrm{e})\) the angular speed of pulley \(C\). (Hint: If the belt between two pulleys does not slip, the linear speeds at the rims of the two pulleys must be equal.)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Measurements

Read Explanation

Electricity and Magnetism

Read Explanation

Mechanics and Materials

Read Explanation

Kinematics Physics

Read Explanation

Particle Model of Matter

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.