/*! 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 20 A compact disc (CD) stores music... [FREE SOLUTION] | 91Ó°ÊÓ

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A compact disc (CD) stores music in a coded pattern of tiny pits 10\(^-\)7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of 1.25 m/s. (a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.

Short Answer

Expert verified
(a) 50 rad/s and 21.55 rad/s; (b) 5550 m; (c) -0.00641 rad/s²

Step by step solution

01

Understanding the Problem

We are given the dimensions of a CD and its linear scanning speed. We need to find the angular speeds at the inner and outer radii, the length of the track for a maximum-duration CD, and the average angular acceleration during the maximum playing time.
02

Finding Angular Speed for Inner Radius

The angular speed \( \omega \) can be calculated using the linear speed \( v \) and radius \( r \) with the formula: \[\omega = \frac{v}{r}\]For the inner radius, where \( r = 25.0 \) mm = 0.025 m:\[\omega_{\text{inner}} = \frac{1.25 \text{ m/s}}{0.025 \text{ m}} = 50 \text{ rad/s}\]
03

Finding Angular Speed for Outer Radius

For the outer radius, where \( r = 58.0 \) mm = 0.058 m:\[\omega_{\text{outer}} = \frac{1.25 \text{ m/s}}{0.058 \text{ m}} \approx 21.55 \text{ rad/s}\]
04

Calculating Length of the Track

The length of the track \( L \) is found using the relation: \[L = v \times t\]where \( t \) is the total playing time in seconds. Converting 74.0 min to seconds, \( t = 74 \times 60 = 4440 \) seconds.\[L = 1.25 \text{ m/s} \times 4440 \text{ s} = 5550 \text{ m}\]
05

Calculating Average Angular Acceleration

Angular acceleration \( \alpha \) is calculated using:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where \( \Delta \omega = \omega_{\text{outer}} - \omega_{\text{inner}} = 21.55 \text{ rad/s} - 50 \text{ rad/s} = -28.45 \text{ rad/s}\) and \( \Delta t = 4440 \) s.\[\alpha = \frac{-28.45 \text{ rad/s}}{4440 \text{ s}} \approx -0.00641 \text{ rad/s}^2\]

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Key Concepts

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

Linear Speed
Linear speed is a measure of how fast an object moves along a path. In the case of a CD player, the linear speed refers to the speed at which the laser reads the data on the track. It is given by the formula:
  • \( v = \omega \cdot r \)
This represents the relationship between linear speed \( v \), angular speed \( \omega \), and radius \( r \) of the track being read. Here, we are dealing with a constant linear speed of 1.25 m/s. This means that regardless of the radial position of the laser on the CD, it maintains this speed. This constancy is crucial for the uniform playback of audio. Understanding the concept of linear speed allows you to see why the CD player adjusts the spinning rate as it reads different parts of the disc. This ensures seamless audio output.
Angular Speed
Angular speed describes the rate of rotation of an object around a central point or axis. For a CD spinning in a player, angular speed can be found using the formula:
  • \( \omega = \frac{v}{r} \)
where \( \omega \) is angular speed, \( v \) is linear speed, and \( r \) is the radius from the center of the disc to the point being measured. For a CD, the angular speed varies based on the track's position. When the laser reads the inner part of the track (25.0 mm = 0.025 meters), the angular speed is higher, calculated as 50 rad/s. As the laser moves outward to 58.0 mm (0.058 meters), the angular speed reduces to approximately 21.55 rad/s. This change in angular speed occurs because the laser must cover increased track length at greater radii to maintain the constant linear speed. Understanding this helps in appreciating how devices maintain consistent performances.
Angular Acceleration
Angular acceleration measures how quickly an object's angular speed changes over time. This is particularly relevant for CDs, which adjust speeds gradually from the outer edge to the inner tracks. It is calculated using:
  • \( \alpha = \frac{\Delta \omega}{\Delta t} \)
where \( \Delta \omega \) is the change in angular speed and \( \Delta t \) is the change in time. For a CD, it's the difference between angular speeds at the outer and inner tracks over the entire playing time. For our CD example, the change in angular speed \( \Delta \omega \) is -28.45 rad/s over 4440 seconds (which is 74.0 minutes). Therefore, the average angular acceleration is approximately -0.00641 rad/s². It's negative since the CD decelerates as it moves from the inner to the outer track. Consistent deceleration is pivotal for gradual track alignment and balance sound quality.
CD Track Length
CD track length refers to the total length of the spiral track if it were unraveled into a straight line. This length represents all the data stored on the CD, which is a crucial aspect of how CDs carry information. Calculating this involves multiplying the constant linear speed by the total playing time:
  • \( L = v \times t \)
where \( L \) is the track length, \( v \) the linear speed, and \( t \) the playing time in seconds. For a CD player, with a linear speed of 1.25 m/s and maximum playtime of 4440 seconds (74 minutes), the track length is computed as 5550 meters. This length ensures enough space to store data for lengthy recordings. Understanding this conversion reveals the efficiency of storage mediums like CDs, which use compact spirals that physically extend to significant lengths when analyzed in terms of data capacity.

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Most popular questions from this chapter

If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \\( \frac{1}{2} \\)\(MV^2\). However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13% of a person's mass, while the legs and feet together account for 37%. For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm30^\circ\) (a total of 60\(^\circ\)) from the vertical in approximately 1 second. Assume that they are held straight, rather than being bent, which is not quite true. Consider a 75-kg person walking at 5.0 km/h, having arms 70 cm long and legs 90 cm long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

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