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Chapter 11: Problem 53

Calculate the angular momentum of the Earth about its own axis, due to its daily rotation. Assume that the Earth is a uniform sphere.

Short Answer

Expert verified
The angular momentum of the Earth is approximately 7.07 x 10^{33} kg m^2/s.

Step by step solution

01

Understand the Given

The problem asks us to calculate the angular momentum of the Earth about its own axis due to its daily rotation. We'll assume the Earth is a uniform sphere, with known radius and mass.
02

Identify Known Values

The Earth's average radius is approximately 6371 km, and its mass is about 5.97 x 10^24 kg. We'll need to use these values to find the moment of inertia.
03

Recall the Formula for Moment of Inertia

For a uniform sphere, the moment of inertia about an axis through its center is given by \[ I = \frac{2}{5} m r^2 \]where \( m \) is the mass of the Earth, and \( r \) is its radius.
04

Calculate the Moment of Inertia

Substitute the known values:\[ I = \frac{2}{5} \times 5.97 \times 10^{24} \times (6.371 \times 10^6)^2 \] Calculate to find \( I \approx 9.72 \times 10^{37} \text{ kg m}^2 \).
05

Recall the Formula for Angular Momentum

Angular momentum \( L \) can be calculated using:\[ L = I \omega \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
06

Determine Angular Velocity

The Earth completes one rotation in about 24 hours. Therefore, the angular velocity \( \omega \) is:\[ \omega = \frac{2\pi}{86400} \text{ rad/s} \]\( 86400 \) is the number of seconds in a day.
07

Calculate Angular Momentum

Now, substitute \( I \) and \( \omega \) into the formula for \( L \) to find:\[ L = 9.72 \times 10^{37} \times \frac{2\pi}{86400} \approx 7.07 \times 10^{33} \text{ kg m}^2/\text{s} \].

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Key Concepts

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

Moment of Inertia
The moment of inertia is a measure of an object's resistance to rotational motion. It is a critical part of calculating angular momentum. For a uniform sphere like the Earth, this calculation can be simplified.

To determine the moment of inertia (\( I \)) for a uniform sphere, we use the formula:\[ I = \frac{2}{5} m r^2 \]where:
  • \( m \) is the mass of the sphere.
  • \( r \) is the radius of the sphere.
The factor \( \frac{2}{5} \) comes from integrating \( r^2 \) over the volume of a sphere, which accounts for distribution of mass in a spherical object. For Earth's scenario, using its average mass and radius, the moment of inertia is a massive \( 9.72 \times 10^{37} \text{ kg m}^2 \). This reveals how much rotational resistance the Earth offers during its rotational motion.
Angular Velocity
Angular velocity is the rate at which an object rotates or revolves around a point or axis. For Earth's daily rotation, we are interested in knowing how fast it spins around its axis.

The formula for angular velocity (\( \omega \)) is:\[ \omega = \frac{\Delta \theta}{\Delta t} \]For a full rotation, \( \Delta \theta \) equals \( 2\pi \) (one complete revolution), and \( \Delta t \) is the time taken, which is a day or \( 86400 \) seconds.
  • Hence, the angular velocity \( \omega \) becomes approximately \( \frac{2\pi}{86400} \text{ rad/s} \).
Simply put, angular velocity measures the spin rate of the Earth as it completes its spins in a single day. This small figure reflects the Earth's slow rotational speed, yet at a massive scale due to its large size.
Uniform Sphere
Assuming Earth as a uniform sphere simplifies our calculation by potentially ignoring minor variations in density and irregularities.

A uniform sphere implies that its mass is evenly distributed throughout its volume. This assumption is pivotal in using the moment of inertia equation we earlier described.
  • For real-world applications, this means approximations and assumptions. Earth's reality might differ due to its equatorial bulge or underground mass distribution.
  • This simplification, however, is generally sufficient for most large-scale calculations, such as those involving celestial bodies or planetary rotations.
Thus, while not perfectly accurate, treating Earth as a uniform sphere allows scientists to perform large-scale calculations efficiently.

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Most popular questions from this chapter

A mass \(M\) is attached to a rope that passes over a diskshaped pulley of mass \(m\) and radius \(r .\) The mass hangs to the left side of the pulley. On the right side of the pulley, the rope is pulled downward with a force \(F .\) Find \((a)\) the acceleration of the mass, (b) the tension in the rope on the left side of the pulley, and \((c)\) the tension in the rope on the right side of the pulley. (d) Check your results in the limits \(m \rightarrow 0\) and \(m \rightarrow \infty\)

A student on a piano stool rotates freely with an angular speed of 2.95 rev /s. The student holds a 1.25 -kg mass in each outstretched arm, 0.759 \(\mathrm{m}\) from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is \(5.43 \mathrm{kg} \cdot \mathrm{m}^{2}\), a value that remains constant. (a) As the student pulls his arms inward, his angular speed increases to 3.54 rev \(/ \mathrm{s} .\) How far are the masses from the axis of rotation at this time, considering the masses to be points? (b) Calculate the initial and final kinetic energies of the system.

To loosen the lid on a jar of jam \(8.9 \mathrm{cm}\) in diameter, a torque of \(8.5 \mathrm{N} \cdot \mathrm{m}\) must be applied to the circumference of the lid. If a jar wrench whose handle extends \(15 \mathrm{cm}\) from the center of the jar is attached to the lid, what is the minimum force required to open the jar?

A string that passes over a pulley has a \(0.321-\mathrm{kg}\) mass attached to one end and a \(0.635-\mathrm{kg}\) mass attached to the other end. The pulley, which is a disk of radius \(9.40 \mathrm{cm}\), has friction in its axle. What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium?

Suppose partial melting of the polar ice caps increases the moment of inertia of the Earth from \(0.331 M_{\mathrm{E}} R_{\mathrm{E}}{ }^{2}\) to \(0.332 M_{\mathrm{E}} R_{\mathrm{E}}^{2}\) (a) Would the length of a day (the time required for the Earth to complete one revolution about its axis) increase or decrease? Explain. (b) Calculate the change in the length of a day. Give your answer in seconds.
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