Question

A CD has a mass of \(15.0 \mathrm{~g}\), an inner diameter of \(1.50 \mathrm{~cm},\) and an outer diameter of \(11.9 \mathrm{~cm} .\) Suppose you toss it, causing it to spin at a rate of 4.30 revolutions per second. a) Determine the moment of inertia of the CD, approximating its density as uniform. b) If your fingers were in contact with the CD for 0.250 revolutions while it was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque?

Short answer

Given: - Mass of CD, m_CD = 17.0 g - Inner diameter = 1.50 cm - Outer diameter = 11.9 cm - Spinning rate = 4.30 revolutions per second - Fingers in contact for 0.250 revolutions a) The moment of inertia of the CD can be calculated using the steps mentioned above, which gives us: \(I_{CD} = I_{outer} - I_{inner}\) b) To calculate the torque applied by the fingers, we can use the formula: \(\tau = I_{CD} \cdot \alpha\) Please complete the calculations using the given values and steps mentioned in the solution.

Step-by-step solution

To find the moment of inertia of the CD, we can use the formula for a solid cylinder: \(I = \frac{1}{2}mR^2\), where \(m\) is the mass, and \(R\) is the radius. However, the CD has a hole in the middle, so we need to subtract the moment of inertia of the inner part. The moment of inertia of the CD can be written as \(I_{CD} = I_{outer} - I_{inner}\). Here, we're given inner diameter = 1.50 cm, which gives us an inner radius of \(r_{inner}=\frac{1.50}{2}\) cm. Similarly, we're given outer diameter = 11.9 cm, which gives us an outer radius of \(r_{outer}=\frac{11.9}{2}\) cm. To find the mass of the inner part \(m_{inner}\), we need to compare the area of the inner circle and the outer circle. We can then calculate \(m_{inner}\) by taking the ratio of the areas and multiplying it by the mass of the CD. Area ratio: \(\frac{A_{inner}}{A_{outer}} = \frac{\pi r_{inner}^2}{\pi r_{outer}^2} = \frac{r_{inner}^2}{r_{outer}^2}\) Mass of inner part: \(m_{inner} = m_{CD} \cdot \frac{r_{inner}^2}{r_{outer}^2}\) Now, we can calculate the moment of inertia of the inner part: \(I_{inner} = \frac{1}{2}m_{inner}r_{inner}^2\) Moment of inertia of outer part: \(I_{outer} = \frac{1}{2}m_{outer}r_{outer}^2\), where \(m_{outer} = m_{CD} - m_{inner}\) Finally, moment of inertia of the CD: \(I_{CD} = I_{outer} - I_{inner}\) #b) Torque

We are given that the fingers were in contact with the CD for 0.250 revolutions while it was acquiring its angular velocity. To find the angular acceleration, we can use the relation: \(\alpha = \frac{\omega^2 - \omega_0^2}{2\theta}\), where \(\alpha\) is the angular acceleration, \(\omega\) is the final angular velocity, \(\omega_0\) is the initial angular velocity (which is 0 since the CD starts from rest), and \(\theta\) is the angle through which the CD rotates while the fingers were in contact. We are given that the CD spins at a rate of 4.30 revolutions per second. We need to convert this to angular velocity in radians per second: \(\omega = 4.30 \times 2\pi\) radians/second Since the fingers are in contact for 0.250 revolutions, the angle through which the CD rotates can be calculated: \(\theta = 0.250 \times 2\pi\) radians Now, we can calculate the angular acceleration using the relation mentioned above. Finally, to determine the magnitude of the torque, we can use the formula: \(\tau = I_{CD} \cdot \alpha\).

Key concepts

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around a fixed axis. In the context of a spinning CD, it tells us how many revolutions the object makes per unit of time. It is generally denoted by the Greek letter \( \omega \). To calculate the angular velocity of a CD spinning 4.30 revolutions per second, as in our exercise, we convert revolutions to radians, because radians are the standard unit for angular measurements in physics. Since one revolution is equal to \( 2\pi \) radians, the angular velocity is \( \omega = 4.30 \times 2\pi \) radians per second.

The concept of angular velocity is crucial because it sets the foundation for understanding other aspects of rotational motion such as torque, angular acceleration, and moment of inertia, all of which depend upon how fast an object is spinning.

Torque

Torque, often denoted by \( \tau \), is a measure of the force that causes an object to rotate. The equation for torque is the cross product of the radius vector (from the axis of rotation to the point of force application) and the force vector, which can also be expressed as \( \tau = rF \sin(\theta) \), where \( r \) is the distance from the axis to the point where the force is applied, \( F \) is the magnitude of the force, and \( \theta \) is the angle between the force vector and the radius vector.

In our CD example, if the fingers apply a constant force over 0.250 revolutions to reach the angular velocity of 4.30 rev/s from rest, the torque is the product of the moment of inertia and the angular acceleration. Determining torque gives us a clearer picture of the forces at play during the CD's rotation and is also indicative of how much work is done to set the CD in motion.

Angular Acceleration

Angular acceleration, notated as \( \alpha \), describes how rapidly the angular velocity of an object changes with time. It is a vector quantity, pointing in the direction of the axis of rotation. Just like acceleration in linear motion, it's a measure of how quickly the velocity changes, but for rotational motion.

The step by step solution references the formula \( \alpha = \frac{\omega^2 - \omega_0^2}{2\theta} \) wherein \( \omega_0 \) is the initial angular velocity (0 in the case of the CD starting from rest), \( \omega \) is the final angular velocity, and \( \theta \) is the angle of rotation in radians. This equation gives us the rate of change of angular velocity, or angular acceleration. Understanding angular acceleration helps us describe how quickly an object can reach a certain rotational speed, which is directly related to the torque applied to it.

Physics of Rotational Motion

The physics of rotational motion encompasses the concepts, laws, and mathematical expressions that describe the rotation of objects. It parallels many aspects of linear motion; for example, just as mass is a measure of an object's resistance to changes in linear motion, moment of inertia (I) is a measure of an object's resistance to changes in rotational motion.

Rolational motion is defined by parameters such as torque (), which influences the rate of spin (angular acceleration ), and depends on the distribution of mass around the axis of rotation (moment of inertia ). The exercise with the spinning CD conveniently integrates these aspects of rotational motion, using equations and concepts to calculate physical quantities like moment of inertia and torque. Understanding these concepts is essential for solving problems in rotational dynamics, which is a fundamental part of physics and essential to comprehend machines, planets' rotations, and everyday objects like the discussed CD.