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Chapter 11: Problem 90

For an \(84 \mathrm{~kg}\) person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation?

Short Answer

Expert verified
The magnitude of the angular momentum is approximately 7.06 x 10^33 kg m^2/s.

Step by step solution

01

Identify Known Quantities

We know the mass of the person, which is given as 84 kg. We also know that the radius of Earth is approximately 6,371,000 meters (or 6.371 x 10^6 meters). The angular velocity of Earth due to its rotation is approximately 7.27 x 10^-5 rad/s.
02

Understand Angular Momentum Formula

The angular momentum (L) of a point mass rotating around an axis is given by the formula \( L = m imes v imes r \), where \( m \) is the mass, \( v \) is the tangential velocity, and \( r \) is the radius. Alternatively, when using angular velocity (\( \omega \)), the formula is \( L = I \times \omega \), where \( I \) is the moment of inertia.
03

Find Tangential Velocity

The tangential velocity (v) can be found using the equation \( v = \omega \times r \). Here, \( \omega = 7.27 \times 10^{-5} \text{ rad/s} \) and \( r = 6.371 \times 10^6 \text{ m} \). So, \( v = 7.27 \times 10^{-5} \times 6.371 \times 10^6 \).
04

Calculate Moment of Inertia

Since the person is effectively a point mass at the Earth's surface, the moment of inertia \( I \) is calculated as \( I = m \times r^2 = 84 \times (6.371 \times 10^6)^2 \).
05

Calculate Angular Momentum

Use the alternate formula for angular momentum: \( L = I \times \omega \). Substitute the values obtained: \( L = 84 \times (6.371 \times 10^6)^2 \times 7.27 \times 10^{-5} \).
06

Compute the Magnitude of Angular Momentum

Compute the result from Step 5 to find the angular momentum. This will involve careful calculation to ensure correct use of scientific notation and significant figures.

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Key Concepts

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

Earth Rotation
Earth's rotation is a fundamental concept in physics, which underlines the daily turning of the Earth around its axis. This axis runs from the North Pole to the South Pole, creating the characteristic day-night cycle.
Understanding Earth's rotation helps to explain various physical phenomena, including the setting and rising of the sun, tides, and the Coriolis effect.
  • **Angular velocity**: Earth's rotation involves an angular velocity, denoted as \( \omega \). For Earth, this is approximately \( 7.27 \times 10^{-5} \text{ rad/s} \). This value is obtained based on Earth's complete rotation over a day (24 hours).
  • **Implications**: The rotation affects everything on Earth, grounding our local time system and influencing weather patterns and climates.
The Earth's rotation has significant importance when considering physics problems because it introduces concepts like angular momentum, which can have profound effects on motion and forces.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotation. It plays a crucial role in determining how rotational motion occurs.
In simple terms, dealing with rotational physics, the moment of inertia could be thought of as the rotational analogue to mass in linear motion.
  • **Formula**: The moment of inertia \( I \) for a point mass is given by the formula \( I = m \times r^2 \), where \( m \) is mass and \( r \) is the radius (distance from the axis of rotation).
  • **Dependence**: Unlike mass, the moment of inertia depends not only on how much mass there is, but also on how this mass is distributed with respect to the axis of rotation. For instance, in the problem provided, this simplifies to the point mass at a distance equivalent to Earth's radius.
The concept directly influences the angular momentum calculations as it connects with the speed of rotation in physical systems.
Tangential Velocity
Tangential velocity refers to the linear speed of something moving along a path. This speed is along the edge (or tangent) of the object's circular path.
In the context of Earth, tangential velocity can be calculated for a person standing on its surface.
  • **Calculation**: Using the formula \( v = \omega \times r \), where \( \omega \) is the angular velocity and \( r \) is the radius of the circular path (Earth's radius), tangential velocity gives a measure of how fast the equator is moving as Earth rotates.
  • **Example**: As in the exercise, substitution provides \( v = 7.27 \times 10^{-5} \times 6.371 \times 10^6 \). This result is essential for analyzing the contributions to angular momentum.
Understanding tangential velocity is key to comprehend how objects move in a curved path, combining both rotational and linear motion aspects.
Physics Calculations
Physics is replete with calculations that explain natural phenomena, translating concepts into numerical values that represent measurable quantities.
Using calculations, like those above, physicists can predict behaviors and quantify observables, ensuring a pragmatic understanding of otherwise abstract concepts.
  • **Step-by-step approach**: Clear steps such as identifying known quantities, employing formulas, and executing computations are typical strategies used to solve physics problems. For example, deriving angular momentum from the given physical data requires systematic application of defined formulas and careful handling of units and scientific notation.
  • **Analytical Importance**: Each calculation has a specific objective, whether to determine velocity, momentum, or other physical characteristics, and is integral in finding solutions to complex real-world scenarios.
Engaging with these calculations not only sharpens analytical skills but increasingly aligns theoretical understanding with mathematical precision, critical to mastering physics and related fields.

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Most popular questions from this chapter

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. \(11-48\) ). A toy train of mass \(m\) is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches speed \(0.15\) \(\mathrm{m} / \mathrm{s}\) with respect to the track. What is the angular speed of the wheel if its mass is \(1.1 \mathrm{~m}\) and its radius is \(0.43 \mathrm{~m}\) ? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.)

Force \(\vec{F}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}-(3.0 \mathrm{~N}) \hat{\mathrm{k}}\) acts on a pebble with posi- tion vector \(\vec{r}=(0.50 \mathrm{~m}) \hat{\mathrm{j}}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unitvector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) ?

A solid brass ball of mass \(0.280 \mathrm{~g}\) will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius \(R=14.0 \mathrm{~cm}\), and the ball has radius \(r \ll R\). (a) What is \(h\) if the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height \(h=\) \(6.00 R\), what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point \(Q\) ?

Shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800\) \(\mathrm{m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2} .\) By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1} ?\)

A top spins at \(30 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.0 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?
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