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Chapter 8: Problem 45

A tornado has wind speed \(325 \mathrm{~km} / \mathrm{h}\) at a rotation radius of \(18 \mathrm{~m}\). What is the angular velocity at this point in the tornado?

Short Answer

Expert verified
The angular velocity is approximately \( 5.015 \text{ rad/s} \).

Step by step solution

01

Understanding the given parameters

We are given the wind speed, which is the linear velocity \( v = 325 \text{ km/h} \), and the rotation radius \( r = 18 \text{ m} \). We need to convert the wind speed from km/h to m/s to be consistent with the radius units.
02

Converting units

Convert the speed from kilometers per hour to meters per second. Use the conversion factor \( 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \). So, \( 325 \text{ km/h} = 325 \times \frac{1}{3.6} = 90.28 \text{ m/s} \).
03

Using the angular velocity formula

Angular velocity \( \omega \) can be calculated using the formula \( \omega = \frac{v}{r} \), where \( v \) is the linear velocity and \( r \) is the radius. Substitute \( v = 90.28 \text{ m/s} \) and \( r = 18 \text{ m} \).
04

Calculating angular velocity

Now calculate \( \omega = \frac{90.28 \text{ m/s}}{18 \text{ m}} \). This gives \( \omega = 5.015 \text{ rad/s} \). This is the angular velocity at the given point in the tornado.

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Key Concepts

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

Understanding Linear Velocity
Linear velocity refers to the rate of change of an object's position along a straight path. In the context of a tornado, linear velocity represents the wind speed at a particular point. It's quantified as a distance covered over time, such as kilometers per hour (km/h) or meters per second (m/s).
  • The formula for linear velocity is simply: \[ v = \frac{\text{distance}}{\text{time}} \]
  • For a tornado, the wind speed you feel is the linear velocity.
Linear velocity is crucial for understanding the intensity and potential impact of tornadoes, as higher velocities mean more severe weather conditions.
Importance of Unit Conversion
Unit conversion is an essential step when calculating angular velocity because it ensures that all measurements are in consistent and compatible units. This means converting the wind speed from kilometers per hour (km/h) to meters per second (m/s), which aligns with the radius measured in meters.To convert, you use the conversion factor:
  • \( 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \)
  • Simply multiply the speed in km/h by \( \frac{1}{3.6} \) to find the speed in m/s.
For example, a wind speed of \(325 \, \text{km/h}\) becomes approximately \(90.28 \, \text{m/s}\), ensuring precision in subsequent calculations.
Tornado Wind Speed Explained
Tornado wind speed describes the speed of the wind as it circulates within the tornado. These speeds vary widely and dictate the severity of the tornado's impact on its surroundings. High winds can cause severe damage:
  • They're fast enough to uproot trees and damage buildings.
  • Understanding wind speed helps in predicting potential damage.
Researchers track these speeds to classify tornadoes and issue safety advisories. The linear velocity of the tornado's wind provides vital data for calculating phenomena like angular velocity, which further quantifies the rotational energy of the storm.
Why Rotation Radius Matters
The rotation radius in a tornado is the distance from the center of rotation to the point where the wind speed is being measured. This radius influences the calculation of angular velocity:
  • Given as \( r \), it's measured in meters, helping in translating linear motion into rotational dynamics.
  • The formula for angular velocity \( \omega \) is: \[ \omega = \frac{v}{r} \] where \( v \) is the linear velocity.
Essentially, a smaller rotation radius leads to a greater angular velocity, showing the compact power of tornadoes. Understanding the rotation radius helps weather scientists and engineers design structures that can withstand high wind speeds.

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Most popular questions from this chapter

A solid cylinder has mass \(0.55 \mathrm{~kg}\) and radius \(3.5 \mathrm{~cm}\). It's rolling with center-of-mass speed \(0.75 \mathrm{~m} / \mathrm{s}\). What's its total kinetic energy? (a) \(0.07 \mathrm{~J}\)(b) \(0.16 \mathrm{~J} ;\) (c) \(0.23 \mathrm{~J}\)(d) \(0.30 \mathrm{~J}\).

A wheel has its rotation axis vertical, and turns counterclockwise as viewed from above. The direction of the wheel's angular momentum is (a) straight up; (b) straight down; (c) tangent to the wheel, in the direction of rotation; (d) tangent to the wheel, opposite the direction of rotation.

The Moon (mass \(7.35 \times 10^{22} \mathrm{~kg}\) ) takes 27.3 days to complete an essentially circular orbit of radius \(3.84 \times 10^{8} \mathrm{~m} .\) Estimate the magnitude of the Moon's orbital angular momentum. (a) \(7.33 \times 10^{25} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (b) \(2.81 \times 10^{34} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) (c) \(7.33 \times\) \(10^{41} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} ;\) (d) \(7.33 \times 10^{51} \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\)

A wheel consists of 20 thin spokes, each with mass \(0.055 \mathrm{~kg}\), attached to a rim of mass \(4.2 \mathrm{~kg}\) and radius \(0.75 \mathrm{~m}\). Find its rotational inertia.

A wheel rolls without slipping, with center-of-mass speed \(v_{\mathrm{cm}}\). What's the instantaneous velocity of the bottom of the wheel (in contact with the ground)? What's the instantaneous velocity of the top of the wheel?
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