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Chapter 8: Problem 71

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

Short Answer

Expert verified
(a) The tangential speed is approximately 1.25 m/s. (b) The angular speed for radius 0.0249 m is approximately 8.00 rev/s.

Step by step solution

01

Understand the Given

We are provided with information that the CD spins with an angular speed of 3.50 revolutions per second when the music is at the outer edge with a radius \(r = 0.0568\) m. We need to find two things: (a) the constant tangential speed \(v_T\) and (b) the angular speed \(\omega\) when the radius is 0.0249 m.
02

Convert Angular Speed to Radians per Second

We calculate the angular speed in radians per second from revolutions per second. Since 1 revolution is \(2\pi\) radians, \[ \omega = 3.50 \times 2\pi \] So, \( \omega = 7.00\pi \) rad/s.
03

Calculate Tangential Speed

Using the relationship \( v_T = r \omega \), we substitute the known values for the outer edge of the CD.\[ v_T = 0.0568 \times 7.00\pi \] Calculate \( v_T \): \[ v_T = 0.3981 \pi \text{ m/s} \] Substitute \( \pi \approx 3.1416 \),\( v_T \approx 1.25 \text{ m/s} \).
04

Calculate New Angular Speed at Inner Radius

With the constant tangential speed \(v_T = 1.25\) m/s at the inner radius \( r = 0.0249 \) m, use the formula \(v_T = r \omega'\).Rearrange for \(\omega'\): \[ \omega' = \frac{v_T}{r} \] Substitute the known values: \[ \omega' = \frac{1.25}{0.0249} \approx 50.20 \] rad/s.
05

Convert Rad/s to Rev/s

Finally, convert \(\omega' = 50.20\) rad/s back to revolutions per second.\[ \text{Revolutions per second} = \frac{50.20}{2\pi} \approx 8.00 \text{ rev/s} \].

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Key Concepts

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

Angular Speed
In the context of rotational motion, angular speed refers to how quickly an object rotates or revolves around a specific point. It is typically measured in revolutions per second (rev/s) or radians per second (rad/s). For a compact disc (CD), this is crucial because it determines how fast the disc spins to read music when played. Angular speed varies depending on the location on the CD. For example, the outer edge and the inner part of the CD have different speeds. The reason is due to the mathematical relation between tangential speed and angular speed:
  • The formula is given by \( v_T = r \omega \), where \(v_T\) is the tangential speed, \(r\) is the radius, and \(\omega\) is the angular speed.
  • As the radius \(r\) increases, for a constant tangential speed, the angular speed \(\omega\) must decrease.
Therefore, to maintain a constant tangential speed across varying radial distance, a CD must adjust its angular speed.
Compact Disc
A compact disc (CD) is a simple yet ingenious storage medium for music and other forms of data. It consists of a spiral track that stores information which can be accessed by a laser in a CD player. The fascinating aspect of CDs is that the data is read at a constant tangential speed despite varying angular speeds.
  • During playback, the CD's motor adjusts the spinning speed to ensure the laser reads the data at this constant speed.
  • This ensures the music or data playback is smooth and without distortion.
The CD changes its angular speed depending on the radial distance from the center to ensure this constant tangential speed. This ability is vital for the consistent sound quality enjoyed by users.
Radial Distance
Radial distance is the measurement from the center of a circular object, like a compact disc, to a point on the edge or surface. It's vital in calculating both angular and tangential speeds. In physics, understanding radial distance helps describe how objects move in circular paths.
  • The larger the radial distance, the slower the angular speed needed to maintain a certain tangential speed, according to \( v_T = r \omega \).
  • On a CD, the radial distance changes as the laser moves from the center to the edge.
This variation allows CDs to adjust their angular speed to keep the tangential speed constant, thus playing music at the correct speed regardless of the track's location.
Revolutions
Revolutions are complete turns or cycles of a rotating object. When discussing CDs, revolutions indicate how many complete turns the disc makes per second. This value directly relates to angular speed and is essential in understanding the disc's rotational dynamics.
  • One revolution equals a complete 360-degree turn, or \(2\pi\) radians.
  • In the exercise, the CD initially spins at 3.50 revolutions per second at its outer edge.
The transition of revolutions per second is crucial for maintaining the desired tangential speed across the entire disc, ensuring that information is accessed correctly despite varying distances from the CD's center.

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Most popular questions from this chapter

A racing car, starting from rest, travels around a circular turn of radius 23.5 m. At a certain instant, the car is still accelerating, and its angular speed is 0.571 rad/s. At this time, the total acceleration (centripetal plus tangential) makes an angle of 35.0 with respect to the radius. (The situation is similar to that in Figure 8.12b.) What is the magnitude of the total acceleration?

mmh The drawing shows a chain-saw blade. The rotating sprocket tip at the end of the guide bar has a radius of \(4.0 \times 10^{-2} \mathrm{m} .\) The linear speed of a chain link at point A is 5.6 m/s. Find the angular speed of the sprocket tip in rev/s.

A thin rod (length 1.50 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of an object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.) (b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

A car is traveling along a road, and its engine is turning over with an angular velocity of 220 rad/s. The driver steps on the accelerator, and in a time of 10.0 s the angular velocity increases to 280 rad/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 rad/s during the entire 10.0-s interval? (b) What would have been the angular displacement if the angular velocity had been equal to its final value of 280 rad/s during the entire 10.0-s interval? (c) Determine the actual value of the angular displacement during the 10.0-s interval.

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates in a horizontal circle. A trainee riding in the chamber of a centrifuge rotating with a constant angular speed of 2.5 rad/s experiences a centripetal acceleration of 3.2 times the acceleration due to gravity. In a second training exercise, the centrifuge speeds up from rest with a constant angular acceleration. When the centrifuge reaches an angular speed of 2.5 rad/s, the trainee experiences a total acceleration equal to 4.8 times the acceleration due to gravity. (a) How long is the arm of the centrifuge? (b) What is the angular acceleration of the centrifuge in the second training exercise?
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