/*! 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 26 The Earth rotates about its axis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 26

The Earth rotates about its axis with a period of \(24.0 \mathrm{h}\) Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be? (b) By what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

Short Answer

Expert verified
The new period of the earth's rotation for the object to have zero apparent weight is \(T = 2*Pi*sqrt{r/(2g)}\) and the speed of the object increases by a factor of 24/T.

Step by step solution

01

Understand the problem and find the gravitational force of the object

Under normal condition, the gravitational force \(F_g = m * g\), where m is the mass of the object and g is the acceleration due to gravity (鈮� 9.8 \(m/s^2\)).
02

Identify the conditions for zero apparent weight

The apparent weight of an object becomes zero when the normal force exerted on it (which is equal to the object's weight) is exactly balanced by the centrifugal force due to rotation. This is expressed as \(F_C = F_g\). And the centrifugal force \(F_C = \frac{m*r*\omega^2}{2}\), where r is radius of Earth and \(omega\) is angular velocity equal to \(2*Pi/T\), T is period of rotation.
03

Determine the new period of rotation

By setting the gravitational force equal to the centrifugal force, we get \(m * g = m * r * \omega^2 / 2\), the m's can be cancelled, simplifying the equation to \(g = r * (2*Pi/T)^2 / 2\). Solving for T, \(T = 2*Pi*sqrt{r/(2g)}\).
04

Find the factor of speed increase

Speed is given by the formula \(v = r * w\) (linear speed is radius times angular speed). So the factor by which speed increases is the same as the factor by which angular speed increases, i.e. \(w1/w2 = T2/T1 = 24/T\).

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.

Apparent Weight
Apparent weight is the force exerted by a body on a supporting surface, which reflects what we perceive as our weight. It's different from true weight due to additional forces like acceleration or centrifugal forces acting on the object.

At the Earth's equator, if the planet's rotational speed increases, the centrifugal force increases, reducing the apparent weight. To have zero apparent weight, the centrifugal force must equal gravitational force. Thus, at a specific rotational speed, the normal force (supporting force) becomes zero.
  • Apparent weight equals true weight minus centrifugal force.
  • When centrifugal force matches gravitational force, apparent weight becomes zero.
Gravitational Force
Gravitational force is the attractive force between any two masses. For an object on Earth, it is primarily the force exerted by Earth on the object.

Mathematically, it's given by the equation: \[ F_g = m \cdot g \] where:
  • m is the mass of the object
  • g is the acceleration due to Earth's gravity (approximately 9.8 \( m/s^2 \))
Gravity is what gives us our weight and keeps us grounded. No matter the rotation of the Earth, gravitational force remains approximately constant at the surface.
Angular Velocity
Angular velocity (\( \omega \)) refers to how fast an object rotates or revolves relative to another point. It is the rate of change of angular displacement, measured in radians per second.

For an object moving in a circular path, such as at Earth's equator:\[ \omega = \frac{2\pi}{T} \] where:
  • T is the period of rotation, the time it takes to complete one full rotation.
  • Angular velocity determines the centrifugal force acting on an object.
Higher angular velocity means higher centrifugal "pull" outward, affecting the object's apparent weight.
Period of Rotation
The period of rotation (\( T \)) is the time it takes for a planet or object to complete one full rotation around its axis. On Earth, this is approximately 24 hours.

When the Earth rotates faster:
  • The period decreases, meaning less time per rotation.
  • With a smaller period, angular velocity (\( \omega \)) increases.
In the exercise, to make an object's apparent weight zero, we want to find the new period. By setting centrifugal force equal to gravitational force, we see:\[ T = 2\pi\sqrt{\frac{r}{2g}} \] This shows that as rotational speed increases, the period must decrease accordingly to achieve the condition of zero apparent weight.

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 \(50.0-\mathrm{kg}\) parachutist jumps from an airplane and falls to Earth with a drag force proportional to the square of the speed, \(R=C v^{2} .\) Take \(C=0.200 \mathrm{kg} / \mathrm{m}\) (with the parachute closed) and \(C=20.0 \mathrm{kg} / \mathrm{m}\) (with the chute open). (a) Determine the terminal speed of the parachutist in both configurations, before and after the chute is opened. (b) Set up a numerical analysis of the motion and compute the speed and position as functions of time, assuming the jumper begins the descent at \(1000 \mathrm{m}\) above the ground and is in free fall for 10.0 s before opening the parachute. (Suggestion: When the parachute opens, a sudden large acceleration takes place; a smaller time step may be necessary in this region.)

If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time. The equation \(v=v_{i}+a t\) gives its velocity \(v\) as a function of time, where \(a\) is its constant acceleration. What if velocity is instead a linear function of position? Assume that as a particular object moves through a resistive medium, its speed decreases as described by the equation \(v=v_{i}-k x,\) where \(k\) is a constant coefficient and \(x\) is the position of the object. Find the law describing the total force acting on this object.

Consider a \(10.0-\mathrm{kg}\) projectile launched with an initial speed of \(100 \mathrm{m} / \mathrm{s},\) at an elevation angle of \(35.0^{\circ} .\) The resistive force is \(\mathbf{R}=-b \mathbf{v},\) where \(b=10.0 \mathrm{kg} / \mathrm{s} .\) (a) Use a numerical method to determine the horizontal and vertical coordinates of the projectile as functions of time. (b) What is the range of this projectile? (c) Determine the elevation angle that gives the maximum range for the projectile. (Suggestion: Adjust the elevation angle by trial and error to find the greatest range.)

A small piece of Styrofoam packing material is dropped from a height of \(2.00 \mathrm{m}\) above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by \(a=g-b v .\) After falling \(0.500 \mathrm{m},\) the Styrofoam effectively reaches terminal speed, and then takes \(5.00 \mathrm{s}\) more to reach the ground. (a) What is the value of the constant \(b ?\) (b) What is the acceleration at \(t=0 ?\) (c) What is the acceleration when the speed is \(0.150 \mathrm{m} / \mathrm{s} ?\)

A space station, in the form of a wheel 120 m in diameter, rotates to provide an "artificial gravity" of \(3.00 \mathrm{m} / \mathrm{s}^{2}\) for persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions per minute) that will produce this effect.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Measurements

Read Explanation

Electricity

Read Explanation

Waves Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Thermodynamics

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.