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Chapter 4: Problem 111

(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
(a) The centripetal acceleration at the equator is about 0.034 m/s虏. (b) Earth's rotation period would need to be about 84.4 minutes.

Step by step solution

01

Determine Earth's Radius

First, we need to identify the radius of Earth. Earth has an average radius of approximately 6,371 kilometers (or 6,371,000 meters). This will be used in further calculations.
02

Calculate Angular Velocity of Earth

Earth rotates once every 24 hours. To find the angular velocity, we use: \( \omega = \frac{2 \pi}{T} \) where \( T \) is the period in seconds. \( T = 24 \times 3600 \) seconds. Thus, \( \omega = \frac{2 \pi}{24 \times 3600} \).
03

Calculate Centripetal Acceleration at Earth's Equator

Using the formula for centripetal acceleration \( a_c = \omega^2 r \), where \( \omega \) is the angular velocity and \( r \) is the Earth's radius, we compute \( a_c \). Substitute \( \omega \) and \( r = 6,371,000 \) meters to find \( a_c \).
04

Set New Centripetal Acceleration

For part (b), we set the centripetal acceleration \( a_c = 9.8 \ \mathrm{m/s^2} \). We use the formula \( a_c = \omega'^2 r \) and solve for the new angular velocity \( \omega' \).
05

Find New Rotation Period

Using the new angular velocity \( \omega' \) found from the previous step, use the formula \( \omega' = \frac{2 \pi}{T'} \) and solve for the new period \( T' \) in seconds. Convert this result into hours for a more comprehensible answer.

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Key Concepts

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

Earth's Rotation Period
The Earth's rotation period refers to how long it takes the planet to complete one full spin around its axis. This is crucial because it's directly involved in calculating other dynamics like angular velocity and centripetal acceleration.
- The Earth completes one full rotation in about 24 hours, leading to the cycle of day and night. - In terms of seconds, this rotation period is exactly 24 hours multiplied by 3600 seconds, which equals 86,400 seconds.
Understanding the Earth's rotation period is an essential determinant in exercises involving centripetal acceleration. This period is a key component when using the formula for angular velocity, enabling us to delve deeper into Earth's dynamic characteristics.
Angular Velocity
Angular velocity is a measure of how fast something rotates or spins. For Earth, we are interested in how fast it rotates around its axis, which influences the centripetal acceleration at the equator.
- Angular velocity (\( \omega \)) is calculated using the formula \( \omega = \frac{2 \pi}{T} \), where \( T \) is the rotation period of Earth in seconds.
- For Earth, \( T \) is 86,400 seconds, leading to an angular velocity of approximately \( 7.27 \times 10^{-5} \) radians per second.
  • This velocity is crucial to understand how Earth's rotation influences objects at the equator.
  • Angular velocity helps us to compute the centripetal force needed to keep objects moving in a circular path as Earth spins.
Earth's Radius
The Earth's radius is fundamental in calculating dynamics like centripetal acceleration for an object at the equator.
- Earth's average radius is about 6,371 kilometers or 6,371,000 meters.
- The radius serves as the distance from the center of the Earth to the equator, which is essential when considering rotation-related calculations.
When we compute centripetal acceleration (\( a_c \)), we use the formula \( a_c = \omega^2 r \). Here, \( r \) represents Earth's radius.
- Having an accurate value for Earth's radius ensures precise calculations. Small changes in radius can lead to different insights when calculating phenomena related to rotation and acceleration.

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Most popular questions from this chapter

You are kidnapped by political-science majors (who are upset because you told them political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min},\) turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s},\) turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s},\) turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min},\) turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction

At one instant a bicyclist is \(40.0 \mathrm{~m}\) due east of a park's flag pole, going due south with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). Then \(30.0 \mathrm{~s}\) later, the cyclist is \(40.0 \mathrm{~m}\) due north of the flagpole, going due east with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). For the cyclist in this \(30.0 \mathrm{~s}\) interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magnitude and (d) direction of the average velocity, and the (e) magnitude and (f) direction of the average acceleration?

A wooden boxcar is moving along a straight railroad track at speed \(v_{1}\). A sniper fires a bullet (initial speed \(v_{2}\) ) at it from a high-powered rifle. The bullet passes through both lengthwise walls of the car, its entrance and exit holes being exactly opposite each other as viewed from within the car. From what direction, relative to the track, is the bullet fired? Assume that the bullet is not deflected upon entering the car, but that its speed decreases by \(20 \%\). Take \(v_{1}=85 \mathrm{~km} / \mathrm{h}\) and \(v_{2}=650 \mathrm{~m} / \mathrm{s}\). (Why don't you need to know the width of the boxcar?)

A woman who can row a boat at \(6.4 \mathrm{~km} / \mathrm{h}\) in still water faces a long, straight river with a width of \(6.4 \mathrm{~km}\) and a current of \(3.2 \mathrm{~km} / \mathrm{h}\). Let i point directly across the river and j point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to 矛 must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rows \(3.2 \mathrm{~km}\) down the river and then back to her starting point? (d) How long if she rows \(3.2 \mathrm{~km}\) up the river and then back to her starting point? (e) At what angle to i should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

A boat is traveling upstream in the positive direction of an \(x\) axis at \(14 \mathrm{~km} / \mathrm{h}\) with respect to the water of a river. The water is flowing at \(9.0 \mathrm{~km} / \mathrm{h}\) with respect to the ground. What are the (a) magnitude and (b) direction of the boat's velocity with respect to the ground? A child on the boat walks from front to rear at \(6.0 \mathrm{~km} / \mathrm{h}\) with respect to the boat. What are the (c) magnitude and (d) direction of the child's velocity with respect to the ground?
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