Question

How much work is done by the motor in a CD player to make a CD spin, starting from rest? The CD has a diameter of \(12.0 \mathrm{cm}\) and a mass of \(15.8 \mathrm{g} .\) The laser scans at a constant tangential velocity of \(1.20 \mathrm{m} / \mathrm{s} .\) Assume that the music is first detected at a radius of \(20.0 \mathrm{mm}\) from the center of the disk. Ignore the small circular hole at the CD's center.

Short answer

The work done by the motor is 0.05118 J.

Step-by-step solution

Convert Units

First, we need to convert the given units to standard SI units. The diameter of the CD is given as \(12.0 \text{ cm}\), so the radius \(r\) of the CD is \(6.0 \text{ cm} = 0.06 \text{ m}\). The mass \(m\) of the CD is given as \(15.8 \text{ g}\), which is \(0.0158 \text{ kg}\). The distance from the center to where the music is detected is given as \(20.0 \text{ mm}\), which is \(0.020 \text{ m}\).

Calculate the Moment of Inertia

The CD is a thin disk, so we use the formula for moment of inertia for a solid disk: \( I = \frac{1}{2} m r^2 \). Substitute the values: \( I = \frac{1}{2} \times 0.0158 \text{ kg} \times (0.06 \text{ m})^2 = 2.844 \times 10^{-5} \text{ kg m}^2\).

Find Angular Velocity

The tangential speed \( v \) is given as \(1.20 \text{ m/s}\). The initial radius from which we detect music is \(0.020 \text{ m}\). The angular velocity \( \omega \) is \( \omega = \frac{v}{r} = \frac{1.20 \text{ m/s}}{0.020 \text{ m}} = 60 \text{ rad/s}\).

Calculate Rotational Kinetic Energy

The rotational kinetic energy \( KE \) of a spinning object is given by \( KE = \frac{1}{2} I \omega^2 \). Substitute the values: \( KE = \frac{1}{2} \times 2.844 \times 10^{-5} \text{ kg m}^2 \times (60 \text{ rad/s})^2 = 0.05118 \text{ J}\).

Determine the Work Done

The work done by the motor is equal to the change in rotational kinetic energy of the CD from rest to the final kinetic energy. Since the CD starts from rest, \( KE_{initial} = 0 \). Therefore, the work done \( W \) is simply \( 0.05118 \text{ J}\).

Key concepts

Understanding Moment of Inertia

The moment of inertia is a crucial concept in rotational dynamics. It is the rotational equivalent of mass in linear motion. Essentially, it measures how much resistance an object offers to changes in its rotational state. For a solid disk, the formula used is:

• \( I = \frac{1}{2} m r^2 \)

where \( m \) is the mass and \( r \) is the radius of the disk (or CD in this case). The thinner the disk, the less resistance it poses to rotation, affecting how much torque is needed to spin it. In this exercise, we calculated the moment of inertia of the CD, which was used to determine how much work the motor had to do. This calculation provides insight into understanding how different shapes and mass distributions can affect rotational motion.

Getting to Know Angular Velocity

Angular velocity describes how fast something is rotating. It tells us how much angle, in radians, an object covers per unit of time. We used the relationship:

• \( \omega = \frac{v}{r} \)

where \( v \) is the tangential velocity, and \( r \) is the radius. It helps us translate linear motion into rotational motion. In the case of the CD, knowing the radius where the laser scans and the speed at which it does so allowed us to find the angular velocity. This value is essential for later calculations, such as finding kinetic energy.

The Role of Kinetic Energy

In physics, kinetic energy is the energy possessed by an object in motion, and is crucial for understanding energy changes in systems. For rotating objects, we use the rotational kinetic energy formula:

• \( KE = \frac{1}{2} I \omega^2 \)

This mirrors the linear kinetic energy formula but includes moment of inertia and angular velocity instead of mass and speed. By calculating the kinetic energy of the spinning CD, we understand how much energy is stored due to its rotation. This was particularly important to determine how much work the motor had done in bringing the CD from a state of rest to its scanning speed.

Exploring the Work-Energy Principle

The work-energy principle is a key concept that connects work and changes in energy within a system. Simply put, it states that the work done on an object results in a change in its kinetic energy. In rotational dynamics, this is reflected in how much energy is required to increase the rotational speed of an object. In our exercise, we calculated the work done by the motor to spin the CD using the formula:

• Work done = \( \Delta KE = KE_{final} - KE_{initial} \)

Given the CD started from rest, the initial kinetic energy was zero. Thus, the entire rotational kinetic energy at the end translated into work done. This principle helps in understanding how energy transfer in rotating systems work.