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The fan blades on a jet engine make one thousand revolutions in a time of 50.0 \(\mathrm{ms}\) . Determine (a) the period (in seconds) and (b) the frequency (in \(\mathrm{Hz} )\) of the rotational motion. (c) What is the angular frequency of the blades?

Short Answer

Expert verified
(a) The period is \(5.0 \times 10^{-5}\) s; (b) the frequency is 20000 Hz; (c) the angular frequency is \(40000\pi\) rad/s.

Step by step solution

01

Understanding the Problem

First, let's understand what is being asked. We need to find the period, frequency, and angular frequency of the rotating fan blades. The fan makes 1000 revolutions in 50.0 ms. We'll need to convert time to seconds and relate period and frequency to revolutions.
02

Convert Time to Seconds

The given time is 50.0 milliseconds. To convert milliseconds to seconds, divide by 1000: \[ 50.0 \text{ ms} = \frac{50.0}{1000} \text{ s} = 0.050 \text{ s} \]
03

Calculate the Period

The period \( T \) is the time for one complete revolution. Since 1000 revolutions take 0.050 seconds, the period is \[ T = \frac{0.050 \text{ s}}{1000} = 5.0 \times 10^{-5} \text{ s} \] per revolution.
04

Calculate the Frequency

The frequency \( f \) is the number of revolutions per second. It is the inverse of the period:\[ f = \frac{1}{T} = \frac{1}{5.0 \times 10^{-5} \text{ s}} = 20000 \text{ Hz} \]
05

Calculate the Angular Frequency

The angular frequency \( \omega \) is related to the frequency by \( \omega = 2\pi f \). Using the calculated frequency,\[ \omega = 2\pi \times 20000 \text{ Hz} = 40000\pi \text{ rad/s} \]

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Key Concepts

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

Period Calculation
When discussing rotational motion, the concept of the period is a key starting point. The period, denoted as \( T \), is essentially the time it takes for one complete cycle of motion to occur. In rotational contexts, this cycle is one full revolution.

To illustrate this, consider the example where a fan blade completes 1000 revolutions in 50.0 milliseconds. To find the period, we first convert the millisecond measurement to seconds, because period calculations typically use this unit. Since 1 millisecond is \( 1/1000 \)th of a second, we convert 50.0 milliseconds to 0.050 seconds.
  • Conversion: \( 50.0 \, \text{ms} = 0.050 \, \text{s} \)
To obtain the period \( T \), we divide the total time by the number of revolutions:
  • \( T = \frac{0.050 \, \text{s}}{1000} = 5.0 \times 10^{-5} \, \text{s} \)
This result tells us that each revolution of the blades takes \( 5.0 \times 10^{-5} \) seconds.
Frequency of Rotation
The frequency of a rotation is the number of cycles, or revolutions, that happen in a unit of time, typically measured in seconds. It is represented by the symbol \( f \) and is expressed in hertz (Hz).

There is a simple yet crucial relationship between frequency and period: they are inverses of each other. Thus, if you know one, you can easily find the other. Using the above example of the fan blades:
  • Formula for frequency: \( f = \frac{1}{T} \)
If the period \( T \) is \( 5.0 \times 10^{-5} \) seconds, then the frequency is:
  • \( f = \frac{1}{5.0 \times 10^{-5} \, \text{s}} = 20000 \, \text{Hz} \)
This means the fan blades complete 20000 revolutions each second, a testament to their rapid movement. Such high frequencies are typical in mechanical and electromagnetic applications.
Angular Frequency
Angular frequency, represented by \( \omega \), is another fundamental concept in rotational motion. It relates to how quickly an object rotates or revolves. Measured in radians per second (rad/s), it gives a more pure measure of rotation by considering the angle in radians, a mathematical constant, each revolution makes.

The formula to find angular frequency is directly related to the linear frequency:
  • \( \omega = 2\pi f \)
Given the frequency \( f \) calculated previously as 20000 Hz, the angular frequency is:
  • \( \omega = 2\pi \times 20000 \; \text{Hz} = 40000\pi \, \text{rad/s} \)
This conversion highlights how angular frequency connects the temporal frequency of rotations with the physical concept of radians. This application is extremely useful, for example, in defining the speeds of wheels, gears, or fan blades

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Most popular questions from this chapter

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 \(\mathrm{N} / \mathrm{m}\) . The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2 . The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2 .

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