/*! 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 63 A compact disc (CD) contains mus... [FREE SOLUTION] | 91Ó°ÊÓ

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A compact disc (CD) contains music on a spiral track. Music is put onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. Since \(\mathcal{V}_{\mathrm{T}}=r \omega,\) a CD rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music near the inner part of the disc. For music at the outer edge \((r=0.0568 \mathrm{~m})\), the angular speed is \(3.50 \mathrm{rev} / \mathrm{s}\). Find (a) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of \(0.0249 \mathrm{~m}\) from the center of a \(\mathrm{CD}\).

Short Answer

Expert verified
(a) 3.91 m/s (b) 7.96 rev/s

Step by step solution

01

Convert Units to Standard Form

First, convert the angular speed provided in revolutions per second to radians per second. We know that one revolution is equal to \(2\pi\) radians. Hence:\[\omega = 3.50 \frac{\text{rev}}{\text{s}} \times 2\pi \frac{\text{rad}}{\text{rev}} = 7\pi \frac{\text{rad}}{\text{s}}\]
02

Calculate Tangential Speed at the Outer Edge

Using the formula for tangential speed, \( v_{T} = r \omega \):\[v_{T} = 0.0568 \text{ m} \times 7\pi \frac{\text{rad}}{\text{s}} = 1.2448\pi \frac{\text{m}}{\text{s}} \approx 3.91 \frac{\text{m}}{\text{s}}\]
03

Calculate Angular Speed for Inner Radius

Using the same formula \( v_{T} = r \omega \), rearrange to solve for \( \omega \) for the music nearer the center:\[\omega = \frac{v_{T}}{r} = \frac{1.2448\pi \frac{\text{m}}{\text{s}}}{0.0249 \text{ m}} \approx 50 \frac{\text{rad}}{\text{s}}\]
04

Convert Angular Speed Back to Revolutions per Second

Convert \( \omega \) from radians per second back to revolutions per second:\[\omega = 50 \frac{\text{rad}}{\text{s}} \times \frac{1 \text{ rev}}{2\pi \text{ rad}} \approx 7.96 \text{ rev/s}\]

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Key Concepts

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

Tangential Speed
Tangential speed is a concept that often pops up in rotational motion discussions. It's the speed at which something moves along the edge of a rotating object. Imagine you're on a merry-go-round. Your tangential speed is how fast you're moving in a straight line, although you're following a circular path. To compute tangential speed, we use the formula \(v_{T} = r \omega\), where:
  • \(v_{T}\) is the tangential speed
  • \(r\) is the radius from the center of rotation
  • \(\omega\) is the angular speed
The tangential speed is constant across the whole CD surface, meaning the music flows smoothly regardless of position. However, since the radius \(r\) changes from inner to outer tracks, \(\omega\) must adjust to keep \(v_{T}\) steady. This relationship ensures perfect sound quality from start to finish.
CD Playback Mechanics
One fascinating aspect of CDs is their playback mechanics. A CD player reads music data off a continuously spiraled track. Unlike older formats that play evenly from start to finish by spinning at a constant speed, the trick with CDs is that the system must adjust the speed of the disc's rotation to maintain constant tangential speed.
When music is being read from the outer parts of the disk, the process slows down because the distance \(r\) from the spinning center is larger. Imagine a car in circular motion. The outer circles traced by its wheels would be slower than inner those because they cover greater distances while keeping the same time.
As the laser reads music moving towards the center, the disc spins faster. This compensates for the reduction in the radius. So, an innovative design cleverly balances rotational speed to ensure your favorite songs play flawlessly!
Unit Conversion
In physics, unit conversion is an essential skill that often saves the day when tackling problems. The problem specifies angular speed in revolutions per second. Yet, our calculations rely on radians per second, the standard angular unit in physics equations. Here's how we convert:
  • Recognize: One full revolution equals \(2\pi\) radians.
  • Apply: Multiply given revolutions by \(2\pi\) to gain radians.
For our problem, the initial speed of \(3.50\ \text{rev/s}\) transforms to \(7\pi\ \text{rad/s}\).
Unit conversion steps are vital, ensuring all components use compatible units for easier problem-solving. A solid understanding of these conversions smooths the path to correct results and builds a proper foundation for more complex topics.
Physics Problem Solving
Solving physics problems efficiently involves utilizing a set of strategies that bring clarity to complex scenarios. Let’s break it down into simple steps you can follow:
  • Identify known and unknown quantities. Here, we know the radius, and angular speed at the outer edge. We need to find tangential speed and the angular speed closer to the center.
  • Convert units where necessary. For example, turning revolutions into radians makes formulas easier to handle.
  • Use the correct formulas. The key to this problem is using \(v_{T} = r \omega\).
  • Rearrange formulas as needed. Solving \(\omega = \frac{v_{T}}{r}\) for different conditions lets us adapt to varying scenarios.
  • Stay organized and double-check your math. Even small errors can lead to major confusion.
By following these steps, students can approach any physics problem with structured confidence, paving their way to successful solutions and a deeper understanding of physical concepts.

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Most popular questions from this chapter

The front and rear sprockets on a bicycle have radii of 9.00 and \(5.10 \mathrm{~cm}\), respectively The angular speed of the front sprocket is \(9.40 \mathrm{rad} / \mathrm{s}\). Determine (a) the linear speed (in \(\mathrm{cm} / \mathrm{s}\) ) of the chain as it moves between the sprockets and \((\mathrm{b})\) the centripetal acceleration (in \(\mathrm{cm} / \mathrm{s}^{2}\) ) of the chain as it passes around the rear sprocket.

A baseball pitcher throws a baseball horizontally at a linear speed of \(42.5 \mathrm{~m} / \mathrm{s}\) (about 95 \(\mathrm{mi} / \mathrm{h}\) ). Before being caught, the baseball travels a horizontal distance of \(16.5 \mathrm{~m}\) and rotates through an angle of 49.0 rad. The baseball has a radius of \(3.67 \mathrm{~cm}\) and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

During a time \(t,\) a wheel has a constant angular acceleration, so its angular velocity increases from an initial value of \(\omega_{0}\) to a final value of \(\omega\). During this time, is the angular displacement less than, greater than, or equal to (a) \(\omega_{0} t\) or (b) \(\omega t\) ? In each case, justify your answer. (c) If you conclude that the angular displacement does not equal \(\omega_{0} t\) or \(\omega t,\) then how does it depend on \(\omega_{0}, \omega,\) and \(t ?\) Problem A car is traveling along a road, and the engine is turning over with an angular velocity of \(+220 \mathrm{rad} / \mathrm{s}\). The driver steps on the accelerator, and in a time of \(10.0 \mathrm{~s}\) the angular velocity increases to \(+280 \mathrm{rad} / \mathrm{s}\). (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of \(+220 \mathrm{rad} / \mathrm{s}\) during the \(10.0 \mathrm{~s} ?\) (b) What would have been the angular displacement if the angular velocity had been equal to its final value of \(+280 \mathrm{rad} / \mathrm{s}\) during the \(10.0 \mathrm{~s} ?\) (c) Determine the actual value of the angular displacement during this period. Check your answers to see that they are consistent with those to the Concept Questions.

A bicycle is rolling down a circular hill that has a radius of \(9.00 \mathrm{~m}\). As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. The radius of each wheel is \(0.400 \mathrm{~m}\). What is the angle (in radians) through which each tire rotates?

ssm An electric circular saw is designed to reach its final angular speed, starting from rest, in \(1.50 \mathrm{~s}\). Its average angular acceleration is \(328 \mathrm{rad} / \mathrm{s}^{2}\). Obtain its final angular speed.

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