/*! 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 44 Calculate the kinetic energy tha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(9.3 \times 10^{19} \mathrm{~J}\).

Short Answer

Expert verified
The Earth's rotational kinetic energy is approximately \(2.57 \times 10^{29} \, \text{J}\) and its orbital kinetic energy is approximately \(2.67 \times 10^{33} \, \text{J}\), both significantly larger than the US annual energy consumption.

Step by step solution

01

Define Kinetic Energy Formula

The kinetic energy of an object can be calculated using the formula \( KE = \frac{1}{2} I \omega^2 \) for rotational motion, where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For translational motion, it is \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity.
02

Calculate Earth's Rotational Kinetic Energy

For the Earth's rotation about its own axis: Assume Earth is a uniform sphere. The moment of inertia \( I \) for a sphere is \( \frac{2}{5} m r^2 \), where \( m \) is the mass of Earth (\(5.97 \times 10^{24} \text{ kg}\)) and \( r \) is the radius of Earth (\(6.371 \times 10^6 \text{ m}\)). The angular velocity \( \omega \) of Earth is \(2 \pi / T\), where \( T \) is the period of rotation (24 hours or 86400 seconds). Plug these into the formula \( KE = \frac{1}{2} I \omega^2 \) to find the kinetic energy.
03

Plug Values for Earth's Rotation

Substitute \( I = \frac{2}{5}(5.97 \times 10^{24} \text{ kg})(6.371 \times 10^6 \text{ m})^2 \) into the equation and \( \omega = \frac{2\pi}{86400 \text{ s}}\). Calculate for \( KE \).
04

Calculate Earth's Translational Kinetic Energy

For the Earth's motion around the Sun: Earth's orbit is roughly a circle with radius of \(1.496 \times 10^{11} \text{ m}\). The velocity \( v \) is determined by \( v = \frac{2 \pi r}{T} \), with \( T \) as 1 year in seconds. Use Earth's mass \( m = 5.97 \times 10^{24} \text{ kg} \) in the formula \( KE = \frac{1}{2} m v^2 \) to calculate the kinetic energy.
05

Plug Values for Earth's Orbit

Substitute the values \( m = 5.97 \times 10^{24} \text{ kg} \), \( r = 1.496 \times 10^{11} \text{ m} \), and \( T \approx 3.156 \times 10^7 \, \text{s} \) to find the velocity \( v \). Then substitute \( v \) in the translational kinetic energy formula to calculate \( KE \).
06

Compare with US Annual Energy Consumption

Compare the calculated kinetic energies to the US annual energy consumption value of \(9.3 \times 10^{19} \, \text{J}\).

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 91Ó°ÊÓ!

Key Concepts

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

Rotational Motion
Rotational motion refers to the movement of an object around a central axis. In the context of Earth's rotation, this axis is the line connecting the North and South Poles. Rotational motion is characterized by how an object spins or rotates without changing its position.

If you imagine the Earth spinning like a ballerina, you're visualizing rotational motion. Key elements of rotational motion include:
  • Rotation Axis: The fixed line around which the rotation happens (e.g., Earth's polar axis).
  • Angular Displacement: The angle covered by the object during rotation.
  • Angular Velocity: The rate at which the angular displacement changes over time.
Grasping these concepts helps in understanding how the Earth's rotational movement contributes to its overall kinetic energy.
Angular Velocity
Angular velocity is the rate of change of angular position of an object with respect to time. It shows how swiftly an object is rotating. Imagine tracing a circle. The faster you complete one loop, the greater your angular velocity.

For Earth, its angular velocity refers to how fast it completes a rotation around its own axis. Angular velocity is measured in radians per second and can be found using: \[\omega = \frac{2\pi}{T}\]where \(T\) is the period of rotation. Earth's angular velocity is approximately \(7.27 \times 10^{-5} \, \text{rad/s}\) because it completes a full rotation (360 degrees or \(2\pi\) radians) in around 24 hours. Understanding angular velocity is crucial for calculating the rotational kinetic energy of celestial bodies.
Moment of Inertia
The moment of inertia is a measure of how difficult it is to change the rotational motion of an object. It is the rotational equivalent of mass in linear motion and is often described as the angular mass or rotational inertia.

For Earth, considered as a uniform sphere, the moment of inertia \(I\) is calculated using:\[I = \frac{2}{5} m r^2\]where \(m\) is the mass and \(r\) is the radius of Earth. A higher moment of inertia means the object is harder to spin at the same angular velocity. Being familiar with the moment of inertia allows for deeper insights into how rotational bodies like Earth acquire and distribute kinetic energy.
Translational Motion
Translational motion refers to the movement of an object from one location to another, distinct from rotational motion. When Earth revolves around the Sun, it undergoes translational motion.

In translational motion, kinetic energy depends on both the mass and velocity of an object. For Earth orbiting the Sun, this kinetic energy is computed as:\[KE = \frac{1}{2} m v^2\]where \(m\) is Earth's mass and \(v\) is its orbital velocity. This velocity is derived from the formula:\[v = \frac{2 \pi r}{T}\]With \(r\) being the radius of Earth's orbit and \(T\) the time for one complete orbit (about a year). Understanding translational motion aids in comparing Earth's energy consumption with its vast motion-related energy.

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 solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s}\). Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s}\). The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of \(-2.4 \mathrm{rad} /\) s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

A square, \(0.40 \mathrm{~m}\) on a side, is mounted so that it can rotate about an axis that passes through the center of the square. The axis is perpendicular to the plane of the square. A force of \(15 \mathrm{~N}\) lies in this plane and is applied to the square. What is the magnitude of the maximum torque that such a force could produce?

Interactive Solution 9.35 at presents a method for modeling this problem. MultipleConcept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{~m}\), an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{~s}\). (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

See all solutions

Recommended explanations on Physics Textbooks

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.