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Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{~m}\) 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 \(25.0 \mathrm{~mm}\) and \(58.0 \mathrm{~mm}\), respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). (a) What is the angular speed of the \(\mathrm{CD}\) 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 \(\mathrm{CD}\) 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
The angular speeds for innermost and outermost tracks are roughly \(3.27 \times 10^1 \, \text{rad/sec}\) and \(2.16 \times 10^1 \, \text{rad/sec}\) respectively. The track length is approximately \(5.55 \times 10^4 \, \text{m}\) and the average angular acceleration is roughly \(2.38 \times 10^{-4} \, \text{rad/sec}^2\).

Step by step solution

01

Part A - Angular Speed

Using the relation between linear and angular speed \(v = r \omega\), where \(v\) is the linear speed, \(r\) is the radius and \(\omega\) is the angular speed, we can find the angular speed at both inner and outer radii. For the innermost track, \(\omega = \frac{v}{r_{\text{in}}} = \frac{1.25 \, \text{m/s}}{0.025 \, \text{m}}\). For the outermost track, \(\omega = \frac{v}{r_{\text{out}}} = \frac{1.25 \, \text{m/s}}{0.058 \, \text{m}}\)
02

Part B - Track Length

The total playing time of the CD is 74.0 minutes and the linear speed is 1.25 m/s. The length of the track is equal to the product of time and linear speed. So, length = time x speed = \(74.0 \times 60 \, \text{sec} \times 1.25 \, \text{m/s}\)
03

Part C - Average Angular Acceleration

Average angular acceleration can be found using the formula \(\alpha_{\text{avg}} = \frac{\Delta \omega}{\Delta t}\). Here, \(\Delta \omega\) is the change in angular speed which would be difference between the angular speeds calculated in Part A. And \(\Delta t\) is the duration of time which is 74 minutes. So, \(\alpha_{\text{avg}} = \frac{\omega_{\text{outer}} - \omega_{\text{inner}}}{74.0 \times 60 \, s}\)

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Key Concepts

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

Angular Speed
Angular speed is an important concept when studying rotating objects like a compact disc (CD). It describes how quickly an object rotates around a fixed axis and is usually measured in radians per second (rad/s). When a CD is spinning inside a player, different parts of the disc spin at different angular speeds. This happens because the CD spirals outward from the center. The equation to find angular speed, \(\omega\), is given by the formula \(v = r \omega\), where \(v\) is the linear speed and \(r\) is the radius from the center of rotation. - **For the innermost radius**, the distance to the center is smaller. With the CD's constant linear speed of \(1.25 \, \text{m/s}\), the angular speed is calculated as \(\omega = \frac{v}{r_{\text{in}}} = \frac{1.25 \, \text{m/s}}{0.025 \, \text{m}}\).- **For the outermost radius**, the distance is larger, giving a different angular speed: \(\omega = \frac{v}{r_{\text{out}}} = \frac{1.25 \, \text{m/s}}{0.058 \, \text{m}}\).This variation ensures that the laser reads the CD at a constant linear speed, despite changes in angular speed.
Linear Speed
Linear speed refers to the rate at which an object travels along a path, which in the context of a compact disc is the speed at which the track is scanned by the laser. For a CD, this speed is maintained as constant at \(1.25 \, \text{m/s}\) throughout the duration of its playtime.In calculating the track's **total length**, we multiply the linear speed by the total playing time. The formula used is:\[ \text{Length} = \text{time} \times \text{speed} = 74.0 \times 60 \, \text{sec} \times 1.25 \, \text{m/s} \]This calculation demonstrates that the length of the track, if it were to be unrolled in a straight line, is directly proportional to both the speed and the time the CD can play, which showcases how uniformly the disc is read at a constant speed.
Average Angular Acceleration
Average angular acceleration provides insight into how the rate of rotation changes over time for the CD. It is a measure of how quickly the angular velocity of an object varies, expressed in radians per second squared (rad/s²).The formula to calculate average angular acceleration, \(\alpha_{\text{avg}}\), is:\[ \alpha_{\text{avg}} = \frac{\Delta \omega}{\Delta t} \]where \(\Delta \omega\) is the change in angular speed, and \(\Delta t\) is the time duration. For the given situation of a maximum-duration CD:- **\(\Delta \omega\)** is the difference in angular speeds at the inner and outermost parts of the CD, calculated previously.- **\(\Delta t\)** is the total playing time of \(74.0 \, \text{minutes}\) converted to seconds, which is \(74.0 \times 60 \, \text{seconds}\).Average angular acceleration helps us understand how the CD player maintains the necessary speed to ensure that the music plays effectively throughout its duration.

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Most popular questions from this chapter

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

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The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova \(e x\) plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{~W},\) 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 \(0.0331 \mathrm{~s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{~s}\) 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 \(\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

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