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Compact Disc. 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 \(\mathrm{mm}\) 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 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 \(C D\) 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
Innermost angular speed: 50 rad/s; Outermost: 21.55 rad/s. Track length: 5550 m. Average angular acceleration: -0.0064 rad/s².

Step by step solution

01

Determine Angular Speed (Innermost part)

First, calculate the angular speed when the disc is scanned at the innermost radius. The linear speed is given by \( v = 1.25 \, \text{m/s} \). Use the relation \( v = r \omega \) to find \( \omega \), where \( r = 0.025 \, \text{m} \). \[ \omega = \frac{v}{r} = \frac{1.25}{0.025} = 50 \, \text{rad/s} \]
02

Determine Angular Speed (Outermost part)

Next, calculate the angular speed at the outermost radius using the same formula, with \( r = 0.058 \, \text{m} \). \[ \omega = \frac{v}{r} = \frac{1.25}{0.058} \approx 21.55 \, \text{rad/s} \]
03

Calculate Length of the Track

To find the total length of the track, multiply the maximum playing time by the linear speed. Convert 74.0 minutes to seconds. \[ 74.0 \, \text{min} = 74 \times 60 = 4440 \, \text{s} \] The length \( L \) is then \( v \times \text{time} \), so: \[ L = 1.25 \, \text{m/s} \times 4440 \, \text{s} = 5550 \, \text{m} \]
04

Calculate Average Angular Acceleration

To find the average angular acceleration, use the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \). The change in angular speed is \( \Delta \omega = 21.55 \, \text{rad/s} - 50 \, \text{rad/s} = -28.45 \, \text{rad/s} \). The total time is 4440 seconds: \[ \alpha = \frac{-28.45}{4440} \approx -0.0064 \, \text{rad/s}^2 \]
05

Summarize Results

The angular speeds at the innermost and outermost parts are 50 rad/s and 21.55 rad/s, respectively. The length of the track is 5550 meters. The average angular acceleration is approximately -0.0064 rad/s².

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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 a measure of how quickly an object rotates around a central point. In simpler terms, it describes how fast something spins. It is particularly important in understanding how objects like compact discs (CDs) function in devices such as CD players. The angular speed \(\omega\) is related to linear speed (\(v\)) and the radius \(r\) of the circular path the object follows.
To find angular speed, you use the formula:
  • \(\omega = \frac{v}{r}\)
This formula shows that angular speed depends on both the linear speed of something moving along the edge of a circle and the distance from the center (the radius).
In the case of a CD, when it spins, its angular speed is not constant because the radius of the spiral track changes as the laser moves from the innermost to the outermost part of the disc.
**Angular Speed in CDs**For example, a CD rotating inside a player has a constant linear speed of 1.25 m/s. When scanning the innermost part of a CD, which has a radius of 25 mm, the angular speed is calculated as 50 rad/s. At the outer edge, where the radius is 58 mm, the angular speed decreases to about 21.55 rad/s.
This demonstrates that angular speed decreases as you move outward along the track of a spinning disk.
Linear Speed
Linear speed refers to how fast something moves along a straight path. In circular motion, however, it is the speed of something moving tangentially along the edge of a circle. In simpler words, it's like the speed of a point on the rim of a spinning wheel.
Linear speed is given by the formula:
  • \(v = r \omega\)
Where \(v\) is linear speed, \(r\) is the radius, and \(\omega\) is angular speed. This formula highlights how linear speed is dependent on both the angular speed and the radius of the circular path.
**Constant Linear Speed in CD Players**In devices like CD players, the disc rotates at a constant linear speed to ensure the data is read consistently. For example, a CD player maintains a linear speed of 1.25 m/s. This consistency in linear speed ensures the seamless performance of audio playback despite changes in angular speed as the track spirals outward.
Linear speed being constant, often leads to adjustments in angular speed, especially when moving from the center to the edge of a disc or wheel.
Angular Acceleration
Angular acceleration refers to the rate at which angular speed changes with time. It signifies how quickly an object is speeding up or slowing down as it rotates around a point. Angular acceleration is crucial in understanding how the rotation of an object, like a CD, adjusts over time.
The formula for angular acceleration is:
  • \(\alpha = \frac{\Delta \omega}{\Delta t}\)
Where \(\alpha\) is angular acceleration, \(\Delta \omega\) is the change in angular speed, and \(\Delta t\) is the change in time.
**Understanding Angular Acceleration in CDs**For a CD with a maximum duration of 74 minutes, its track must be scanned entirely over this time. Given the outermost and innermost angular speeds (approximately 21.55 rad/s and 50 rad/s, respectively), we can calculate the average angular acceleration over the disc's playing time. Using the formula, we find the average angular acceleration as approximately -0.0064 rad/s².
It is important to note the negative sign of the angular acceleration indicates a decrease in the angular speed over time, as the CD spins from its inner track toward the outer edge during playback.

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Most popular questions from this chapter

Small blocks, each with mass \(m,\) are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2}\) . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through \(60.0^{\circ} ;(\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)

A uniform sphere with mass 28.0 \(\mathrm{kg}\) and radius 0.380 \(\mathrm{m}\) is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is \(176 \mathrm{J},\) what is the tangential velocity of a point on the rim of the sphere?

The Kinetic Energy of Walking. If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \(\frac{1}{2} M V^{2}\).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 \(\pm 30^{\circ}\) (a total of \(60^{\circ} )\) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 75 -kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) 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?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

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