/*! 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 113 Consider the polar axis of the e... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 113

Consider the polar axis of the earth to be fixed in space and compute the magnitudes of the velocity and acceleration of a point \(P\) on the earth's surface at latitude \(40^{\circ}\) north. The mean diameter of the earth is \(12742 \mathrm{km}\) and its angular velocity is \(0.7292\left(10^{-4}\right) \mathrm{rad} / \mathrm{s}\)

Short Answer

Expert verified
Velocity: 0.3555 km/s; Acceleration: 0.0000258 km/s虏.

Step by step solution

01

Understand the Problem

We aim to find the magnitudes of the velocity and acceleration of a point on Earth鈥檚 surface at a latitude of 40掳 north. We will use the Earth鈥檚 radius and its angular velocity to calculate these values.
02

Determine the Radius at Latitude

First, determine the effective radius at latitude 40掳 north. The radius at this latitude is calculated using: \[ r = R_e \cos(\theta) \]where \(R_e = 6371 \, \text{km}\) is the Earth's radius (half of the diameter), and \(\theta = 40^{\circ}\). So, \[ r = 6371 \times \cos(40^{\circ}) \approx 4878.65 \, \text{km} \]
03

Calculate the Magnitude of Velocity

Using the formula for linear velocity, which is \[ v = r \omega \]where \(\omega = 0.7292 \times 10^{-4} \, \text{rad/s}\) is the angular velocity, and \(r\) is the effective radius found in Step 2.So, \[ v = 4878.65 \times 0.7292 \times 10^{-4} \approx 0.3555 \, \text{km/s} \]
04

Calculate the Magnitude of Centripetal Acceleration

Use the formula for centripetal acceleration:\[ a = r \omega^2 \]Substitute the values from previous calculations:\[ a = 4878.65 \times (0.7292 \times 10^{-4})^2 \approx 0.0000258 \, \text{km/s}^2 \]
05

Convert Units (if necessary)

Although the results are in kilometers, if needed, convert to meters: - Velocity: \(0.3555 \, \text{km/s} = 355.5 \, \text{m/s}\) - Acceleration: \(0.0000258 \, \text{km/s}^2 = 0.0258 \, \text{m/s}^2\)

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Key Concepts

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

Angular Velocity
Angular velocity is a fundamental concept in understanding the motion of rotating bodies, such as Earth. It describes how fast an object is rotating around a specific axis. In our exercise's context, Earth's angular velocity is given as \(0.7292 \times 10^{-4} \, \text{rad/s}\). This tells us how swiftly Earth is spinning.
In simple terms, angular velocity measures the angle in radians through which a point or line has rotated in a given time period. For Earth, this measurement helps us understand the rotation rate across different latitudes.
Angular velocity is usually represented by the Greek letter \(\omega\) (omega) and is crucial when calculating other properties like linear velocity and centripetal acceleration for points on Earth's surface at various latitudes.
  • Angular velocity helps in determining linear velocity and centripetal acceleration.
  • It is expressed in radians per second \((\text{rad/s})\).
  • Change in rotation speed affects angular velocity.
Velocity Calculation
Velocity calculation in the context of Earth's rotation involves determining how fast a point on the Earth鈥檚 surface is moving due to its spinning motion. This calculation requires both the angular velocity and the effective radius at a specific latitude.
For instance, at 40掳 north, the Earth鈥檚 radius鈥攁 key component in this calculation鈥攊s affected by the latitude. We found the effective radius using \(r = R_e \cos(\theta)\), with Earth鈥檚 equatorial radius \(R_e\) being 6371 km.
Once we have the effective radius \(r\), the linear velocity \(v\) can be derived using the formula \(v = r \omega\), where \(\omega\) is the given angular velocity. This gives an indication of how fast a point is moving due to Earth's rotation.
  • Effective radius is critical, varying with latitude.
  • Linear velocity \(v\) is calculated using \(v = r \omega\).
  • At 40掳 north, the velocity was found to be approximately 0.3555 km/s.
Centripetal Acceleration
Centripetal acceleration is essential for understanding the forces that keep objects moving in circular paths. It is the acceleration directed towards the center of the circle along which the object is moving. For Earth, it keeps objects following the curving path of rotation.
To calculate centripetal acceleration \(a\) at a latitude of 40掳 north, we use the formula \(a = r \omega^2\). Here, \(r\) is the effective radius calculated earlier, and \(\omega\) is Earth's angular velocity.
In this exercise, we found that the centripetal acceleration was approximately \(0.0000258 \, \text{km/s}^2\). Understanding this reading helps us grasp how the Earth鈥檚 rotation impacts objects, such as fluid movements or atmospheric phenomena.
  • Centripetal acceleration is important for circular motion dynamics.
  • Calculating it involves \(a = r \omega^2\).
  • Ensures safety and stability of motion at different latitudes.

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Most popular questions from this chapter

At time \(t=0,\) the 1.8 -lb particle \(P\) is given an initial velocity \(v_{0}=1 \mathrm{ft} / \mathrm{sec}\) at the position \(\theta=0\) and subsequently slides along the circular path of radius \(r=1.5 \mathrm{ft}\). Because of the viscous fluid and the effect of gravitational acceleration, the tangential acceleration is \(a_{t}=g \cos \theta-\frac{k}{m} v,\) where the constant \(k=0.2 \mathrm{lb}\) -sec/ft is a drag parameter. Determine and plot both \(\theta\) and \(\dot{\theta}\) as functions of the time \(t\) over the range \(0 \leq t \leq 5\) sec. Determine the maximum values of \(\theta\) and \(\dot{\theta}\) and the corresponding values of \(t .\) Also determine the first time at which \(\theta=90^{\circ}\)

The aerodynamic resistance to motion of a car is nearly proportional to the square of its velocity. Additional frictional resistance is constant, so that the acceleration of the car when coasting may be written \(a=-C_{1}-C_{2} v^{2},\) where \(C_{1}\) and \(C_{2}\) are constants which depend on the mechanical configuration of the car. If the car has an initial velocity \(v_{0}\) when the engine is disengaged, derive an expression for the distance \(D\) required for the car to coast to a stop.

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Ball 1 is launched with an initial vertical velocity \(v_{1}=160 \mathrm{ft} / \mathrm{sec} .\) Three seconds later, ball 2 is launched with an initial vertical velocity \(v_{2}\). Determine \(v_{2}\) if the balls are to collide at an altitude of \(300 \mathrm{ft} .\) At the instant of collision, is ball 1 ascending or descending?
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