/*! 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 24 A wheel rotates at \(720 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel rotates at \(720 \mathrm{rpm}\). Viewed from the edge, a point on the wheel appears to undergo simple harmonic motion. What are (a) the frequency in \(\mathrm{Hz}\) and (b) the angular frequency for this SHM?

Short Answer

Expert verified
The frequency is \(12 Hz\) and the angular frequency is \(24\pi rad/s\).

Step by step solution

01

Conversion of rotational speed from rpm to Hz

The given rotational speed is \(720\) revolutions per minute (rpm). 1 minute has 60 seconds, so the conversion factor from minute to seconds is needed. The frequency in Hertz (\(Hz\)), or cycles per second, can be calculated using the following formula: \(f = \frac{N}{T}\). Where \(N\) is the total number of revolutions (in this case, 720), and \(T\) is the time (in this case, 1 minute = 60 seconds). So, using the formula, the frequency in Hertz is: \(f = \frac{720}{60} = 12\) Hz.
02

Finding the angular frequency

Next, the angular frequency needs to be calculated. The angular frequency \(\omega\) is given in rad/s, and can be related to the frequency \(f\) by the equation: \(\omega = 2\pi f\). Substituting the value of the frequency found in the previous step, the angular frequency will be calculated as follows: \(\omega = 2\pi * 12 = 24\pi\) rad/s.

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Key Concepts

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

Angular Frequency
The concept of angular frequency is pivotal in understanding rotational motion and various types of oscillations, including simple harmonic motion (SHM). Angular frequency, denoted by the Greek letter \(\omega\), is a measure of how quickly an object rotates or oscillates, expressed in radians per second (rad/s). In the context of SHM, angular frequency tells us the speed at which an object moves back and forth through its equilibrium position.

An intuitive way to envision this is by picturing a point on the edge of a rotating wheel, as seen in our problem statement. While the wheel spins, that point oscillates along a diameter of the wheel, creating an oscillatory motion representative of SHM. The formula for angular frequency in terms of the regular frequency (\(f\)) is \(\omega = 2\pi f\). This formula ties together linear and rotational motion, enabling us to convert from cycles per second to radians per second, thus providing a more complete understanding of an object's motion in SHM.
Rotational Motion
Rotational motion comes into play when an object spins around a fixed axis. Think about a wheel turning on its axle or the Earth rotating on its axis; both are examples of rotational motion. There are several parameters we use to describe such motion, with angular frequency being one of the key variables, as it defines the speed of rotation.

For simple harmonic motion, specifically the type observed when looking at a point on the rim of a rotating wheel, it's fascinating to note how rotational motion translates into oscillatory motion. The path that the point traces out is a projection of the circular path onto a line, and this motion follows the principles of SHM. Conceptually, simple harmonic motion in this context is a shadow of the circular motion if we were to shine a light on the rotating wheel. Moreover, the understanding of rotational motion is not just academic; it has practical applications in mechanical engineering, astrophysics, and several other fields where rotational dynamics are essential.
Frequency Conversion
Frequency conversion is a necessary mathematical step when we are dealing with oscillatory and rotational motions. In our exercise, we converted rotations per minute (rpm) to Hertz (Hz), which is the unit for frequency. Frequency, in the simplest terms, is the number of complete cycles that occur every second. This conversion is crucial, as it allows us to switch between units that describe the same phenomenon, much like converting miles to kilometers.

The conversion from rpm to Hz is straightforward: divide the given revolutions per minute by 60, since there are 60 seconds in a minute (\(f = \frac{N}{T}\), with \(N\) being the number of cycles and \(T\) the time period in seconds). This step is fundamental for further calculations in physics, especially when determining the angular frequency (\(\omega\)) to describe SHM or when calculating centripetal forces in rotating systems. Comprehending frequency conversion is vital for students to bridge the gap between the observed rotational speed and the more abstract concept of angular velocity in radians per second.

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Most popular questions from this chapter

When in its cycle is the acceleration of an undamped simple harmonic oscillator zero? When is the velocity zero?

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A violin string playing the note A oscillates at \(440 \mathrm{~Hz}\). What's its scillation period?

This problem explores what would happen if a hole were drilled through Earth's center and out the other side, and an object were dropped into the hole. Approximating Earth as a uniform solid sphere, the gravitational acceleration within the planet (including inside the hypothetical hole) would be \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right)\), where \(g_{0}\) is the value at Earth's surface, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. This gravitational acceleration is directed toward Earth's center. (a) Write an expression for the force on a mass \(m\) at any point \(r\) in the hole, apply Newton's second law, and show that you get an equation analogous to Equation 13.3. Neglect air resistance. (b) Use your analogy to find an expression for the period of the simple harmonic motion that results when the mass is dropped into the hole. (c) Use appropriate values to find a numerical value for the period, and compare with the period for circular low- Earth orbit that we found in Chapter 8 .

Two mass-spring systems have the same mass and the same total energy. The amplitude of system 1 is 4 times that of system 2. How do (a) their frequencies and (b) their maximum accelerations compare?

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