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A CD starts from rest and speeds up to the operating angular frequency of the CD player. Compare the angular velocity and acceleration of a point on the edge of the CD to those of a point halfway between the center and the edge of the CD. Do the same for the linear velocity and acceleration.

Short Answer

Expert verified
Answer: The angular velocity and angular acceleration are the same for both points. However, the linear velocity and linear acceleration are greater at the edge of the CD compared to the point halfway between the center and the edge of the CD.

Step by step solution

01

Understand the relationship between angular and linear quantities

Angular and linear quantities are related as follows: - Linear velocity (v) = Radius (r) × Angular velocity (ω) - Linear acceleration (a) = Radius (r) × Angular acceleration (α) Knowing these relationships, we can compare the angular and linear quantities between the two points.
02

Compare the angular velocity and acceleration

Since both points are on the same CD, the angular velocity and acceleration are the same for both points, regardless of their positions. Therefore, the angular velocity and acceleration of a point on the edge of the CD are equal to those of a point halfway between the center and the edge of the CD.
03

Compare the linear velocity and acceleration

Using the relationships between angular and linear quantities, we can compare the linear velocity and acceleration of both points. Linear velocity at the edge of the CD (v1) = Radius at the edge (r1) × Angular velocity (ω) Linear velocity halfway between the center and the edge (v2) = Radius halfway (r2) × Angular velocity (ω) Since r1 > r2, it means that v1 > v2. Therefore, the linear velocity of a point on the edge of the CD is greater than that of a point halfway between the center and the edge of the CD. Similarly, Linear acceleration at the edge of the CD (a1) = Radius at the edge (r1) × Angular acceleration (α) Linear acceleration halfway between the center and the edge (a2) = Radius halfway (r2) × Angular acceleration (α) Since r1 > r2, it means that a1 > a2. Therefore, the linear acceleration of a point on the edge of the CD is greater than that of a point halfway between the center and the edge of the CD.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a specific point or axis. It is denoted by the symbol \( \omega \) and is typically measured in radians per second (rad/s). In the case of a CD, as it starts from rest and speeds up to its operating frequency, every point on the CD increases its angular velocity at the same rate.

Because angular velocity is the same for all points on the disk, a point on the edge and a point halfway between the center and the edge will share the same angular velocity, despite being at different distances from the center. A critical aspect of angular motion to remember is that all points on a rigid rotating body experience the same angular velocity, making it a crucial parameter in describing rotational movement.
Angular Acceleration
Angular acceleration represents the rate of change of angular velocity over time and is denoted by \( \alpha \). It is a vector quantity, meaning it has both magnitude and direction, and is usually expressed in radians per second squared (rad/s²).

The angular acceleration experienced by the CD as it starts up is indeed the same for a point on its edge as it is for a point halfway to the edge. However, it's important to recognize that this uniformity in angular acceleration applies only to a rigid body in rotational motion where the angular acceleration doesn't vary across the object. So, no matter where a point is located on a CD, if the CD undergoes a change in rotation rate, every point will have the same angular acceleration.
Linear Velocity
Linear velocity is the rate at which a point on a rotating body moves in space and is dependent on its distance from the axis of rotation. Mathematically, it is given by the product of the radius (r) and angular velocity (\( \omega \)), as shown in the equation \( v = r \times \omega \).

The linear velocity of a point on the CD increases with its distance from the center. Thus, a point on the outer edge of the CD, having a larger radius, moves faster through space and has a higher linear velocity compared to a point halfway between the center and the edge. This concept is crucial as it illustrates how linear speed can vary on a rotating object, even though the angular velocity is shared by all points on the object.
Linear Acceleration
Linear acceleration describes how quickly the velocity of a point in a line is changing. This is entirely separate from angular acceleration, although it is related to it on a rotating body. The linear acceleration for a point on an object in circular motion is calculated by the equation \( a = r \times \alpha \), where \( r \) is the radius and \( \alpha \) is the angular acceleration.

In our CD example, the linear acceleration at the edge of the CD is higher than that at a point halfway to the edge. Even though angular acceleration is consistent throughout the CD, because the points on the edge are farther from the center, they experience a more significant change in linear velocity over time, and therefore, greater linear acceleration. It's a key concept when studying motion because it explains how different parts of a rotating object can move with different accelerations, contributing to the complexity of rotational dynamics.

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Most popular questions from this chapter

Unlike a ship, an airplane does not use its rudder to turn. It turns by banking its wings: The lift force, perpendicular to the wings, has a horizontal component, which provides the centripetal acceleration for the turn, and a vertical component, which supports the plane's weight. (The rudder counteracts yaw and thus it keeps the plane pointed in the direction it is moving.) A famous spy plane the SR-71 Blackbird, flying at \(4800 . \mathrm{km} / \mathrm{h}\), had a turning radius of \(290 . \mathrm{km}\). Find its banking angle.

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