/*! 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 91 On a compact disc (CD), music is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of \(v =\) 1.25m/s. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta) = r_0 + \beta\theta\), where \(r_0\) is the radius of the spiral at \(\theta =\) 0 and \(\beta\) is a constant. On a CD, \(r_0\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d\theta\), the distance scanned along the track is \(ds = rd\theta\). Using the above expression for \(r(\theta)\), integrate \(ds\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) Since the track is scanned at a constant linear speed \(v\), the distance s found in part (a) is equal to \(vt\). Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_z\) and the angular acceleration \(\alpha_z\) as functions of time. Is \(\alpha_z\) constant? (d) On a CD, the inner radius of the track is 25.0 mm, the track radius increases by 1.55 mm per revolution, and the playing time is 74.0 min. Find \(r_0, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_z\) (in rad/s) versus \(t\) and \(\alpha_z\) (in rad/s\(^2\)) versus \(t\) between \(t =\) 0 and \(t =\) 74.0 min. \(\textbf{The Spinning eel.}\) American eels (\(Anguilla\) \(rostrata\)) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could.

Short Answer

Expert verified
(a) Integral yields \( s = r_0\theta + \frac{1}{2}\beta\theta^2 \). (b) \( \theta = \frac{-r_0 + \sqrt{r_0^2 + 2\beta vt}}{\beta} \). (c) \( \alpha_z \) is not constant. (d) Compute \( r_0, \beta, \) and revolutions. (e) Graph \( \omega_z \) and \( \alpha_z \).

Step by step solution

01

Expression for Distance Scanned

The distance scanned along the track, \( ds \), when the disc rotates through a small angle \( d\theta \) is given by \( ds = r \, d\theta \). Substituting \( r(\theta) = r_0 + \beta\theta \), we have \( ds = (r_0 + \beta\theta) \, d\theta \). To find the total distance \( s \) as a function of \( \theta \), integrate \( ds \):\[ s = \int (r_0 + \beta\theta) \, d\theta = r_0 \theta + \frac{1}{2} \beta \theta^2. \]
02

Solve for Angle \( \theta \) as a Function of Time

Since the linear speed \( v \) is constant and given as \( v = 1.25 \, \text{m/s} \), the distance \( s \) is equal to \( vt \). Set the expression for \( s \) found in step 1 equal to \( vt \): \( r_0 \theta + \frac{1}{2}\beta \theta^2 = vt \). Rearrange this quadratic equation to find \( \theta \):\[ \frac{\beta}{2} \theta^2 + r_0 \theta - vt = 0. \]Solve for \( \theta \) using the quadratic formula: \( \theta = \frac{-r_0 \pm \sqrt{r_0^2 + 2\beta vt}}{\beta} \). Choose the positive root because it results in a positive and increasing \( \theta \) as the disc spins.
03

Determine Angular Velocity and Acceleration

Angular velocity \( \omega_z \) is given by \( \omega_z = \frac{d\theta}{dt} \). Differentiate the positive root of \( \theta(t) \):\[ \theta = \frac{-r_0 + \sqrt{r_0^2 + 2\beta vt}}{\beta},\]\[ \omega_z = \frac{v}{r_0 + \beta\theta}. \]Angular acceleration \( \alpha_z \) is the rate of change of angular velocity, \( \alpha_z = \frac{d\omega_z}{dt} \). Since \( \omega_z \) depends on \( \theta \) and \( \theta \) is a function of \( t \), \( \alpha_z \) is not constant.
04

Compute Constants and Total Revolutions

Given the inner radius \( r_0 = 25.0 \, \text{mm} = 0.025 \, \text{m} \) and track radius increases by \( 1.55 \, \text{mm} = 0.00155 \, \text{m} \) per revolution, solve for \( \beta \): \( \beta = \frac{0.00155 \, \text{m/rev}}{2\pi} \). The playing time is 74.0 minutes (4440 seconds). Using \( s = vt = 1.25 \, \text{m/s} \times 4440 \, \text{s} \approx 5550 \, \text{m} \), find the total angle \( \theta \) using \( r_0 \theta + \frac{1}{2}\beta \theta^2 = 5550 \, \text{m} \). Calculate total revolutions: \( \theta_{total} \div 2\pi \).
05

Graph \( \omega_z \) and \( \alpha_z \) vs. Time

Plot \( \omega_z(t) \) using \( \omega_z = \frac{v}{r_0 + \beta\theta(t)} \). For \( \alpha_z(t) \), differentiate \( \omega_z(t) \) over time to observe how \( \alpha_z \) changes. Graph both \( \omega_z \) and \( \alpha_z \) over the playing time, showing the decrease in \( \omega_z \) and the non-linear nature of \( \alpha_z \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Acceleration
Angular acceleration is a fundamental concept in physics, especially when dealing with rotating objects like a compact disc (CD). It describes how the angular velocity of an object changes over time.
When a CD spins in a player, the angular acceleration ensures that the disc adjusts its spinning speed according to the position of its spiral track. This is crucial because the linear scanning speed must remain constant to ensure an even reading of data. Yet, due to the outward spiraling track, the angular speed needs continuous adjustment.
Angular acceleration, denoted by \( \alpha_z \), is derived from the rate of change of angular velocity \( \omega_z \), given as \( \alpha_z = \frac{d\omega_z}{dt} \). During operation, because \( \omega_z \) is dependent on \( \theta(t) \) and thereby on time \( t \), \( \alpha_z \) is not constant; instead, it varies throughout the CD's playtime. This varying nature of angular acceleration makes it essential to apply integral calculus to explore its behavior over time.
Spiral Track Equation
The spiral track equation is a key part of understanding how discs like CDs are designed for consistent data readout. The spiral track equation is given by \( r(\theta) = r_0 + \beta \theta \), where \( r_0 \) represents the initial radius at \( \theta = 0 \), and \( \beta \) is a constant describing how the radius of the spiral increases as the angle \( \theta \) increases.
This mathematical equation demonstrates how a CD's spiral grows, making it essential to adjust the angular speed correspondingly to maintain a constant linear speed of 1.25 m/s. Substituting this equation into real scenarios, like the given CD problem, helps to understand the systematic increase in the radius as it spirals outward.
The equation \( ds = (r_0 + \beta\theta) \, d\theta \) translates these theoretical concepts into practical applications. Here, it plays a pivotal role in calculating the distance scanned along the track by integrating the designed spiral equation for \( r(\theta) \). This integration provides insights into the physical journey the CD's laser takes as it scans the data from start to finish.
Integral Calculus in Physics
Integral calculus is a powerful tool in physics to calculate quantities when dealing with change and accumulation, like those in motion and area calculations.
In this CD problem, integral calculus helps us solve for various quantities, like the total distance \( s \) scanned along the spiral track. By integrating the expression \( ds = (r_0 + \beta \theta) \, d\theta \), the total distance as a function of \( \theta \) is derived, \( s = r_0 \theta + \frac{1}{2} \beta \theta^2 \).
This is essential to link the linear distance with time, thereby facilitating calculations of \( \theta(t) \). Such integrals are not merely abstract exercises but are instrumental in practical applications like timing the play rate of CDs by enabling a constant linear speed. Through calculus, we transition from simple equations to more complex functions, allowing for a nuanced understanding of how systems evolve over time.
The application of integral calculus thus goes beyond academics by deeply embedding into technologies and mechanical functionalities encountered daily, exemplified by how we read and process data from CDs effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A compound disk of outside diameter 140.0 cm is made up of a uniform solid disk of radius 50.0 cm and area density 3.00 g/cm\(^2\) surrounded by a concentric ring of inner radius 50.0 cm, outer radius 70.0 cm, and area density 2.00 g/cm\(^2\). Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

The Crab Nebula is a cloud of glowing gas about 10 lightyears across, located about 6500 light-years from the earth (\(\textbf{Fig. P9.86}\)). It is the remnant of a star that underwent a \(supernova\) \(explosion\), seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about 5 \(\times\) 10\(^3$$^1\) W, about 10\(^5\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning \(neutron\) \(star\) at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22 \(\times 10^{-13}\) s for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock (3000 kg/m\(^3\)) and to the density of an atomic nucleus (about 10\(^{17}\) kg/m\(^3\)). Justify the statement that a neutron star is essentially a large atomic nucleus.

A uniform 2.00-m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of 53.0\(^\circ\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?

A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at \(t =\) 0, the wheel turns through 8.20 revolutions in 12.0 s. At \(t =\) 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.