Problem 60 - University Physics with Modern Physics

Question

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 25.0 cm. Starting from rest at \(t =\) 0, the flywheel rotates with constant angular acceleration 3.00 rad/s\(^2\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m\(^3\), what thickness must it have to store 800 J of kinetic energy at \(t =\) 8.00 s?

Step-by-Step Solution

Answer

The required thickness of the flywheel is approximately 0.025 meters.

Step 1: Understanding the Physical Problem

We need to design a flywheel, a uniform disk, that stores 800 J of energy when rotating at an angular acceleration of 3.00 rad/s². We will find the required thickness of this disk, knowing its density and radius.

Step 2: Kinetic Energy of the Flywheel

The energy stored in a rotating flywheel is given by the rotational kinetic energy formula: \[ K = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. We need \( K = 800 \text{ J} \).

Step 3: Moment of Inertia for a Uniform Disk

For a uniform disk, the moment of inertia \( I \) is given by:\[ I = \frac{1}{2} m R^2 \]where \( m \) is the mass and \( R = 0.25 \text{ m} \) is the radius.

Step 4: Calculating Angular Velocity at t = 8s

Angular velocity \( \omega \) after 8 seconds can be found using:\[ \omega = \alpha t \]where \( \alpha = 3.00 \text{ rad/s}^2 \) and \( t = 8 \text{ s} \). Thus, \( \omega = 3.00 \times 8 = 24.00 \text{ rad/s} \).

Step 5: Solving for Mass Using Kinetic Energy Formula

Using the kinetic energy equation:\[ 800 = \frac{1}{2} \times \frac{1}{2} m \times 0.25^2 \times (24)^2 \]Solve for \( m \).

Step 6: Solve for Volume and Ultimately Thickness

Calculate the volume \( V \) of the disk using the density formula \( \rho = \frac{m}{V} \), with \( \rho = 8600 \text{ kg/m}^3 \). The volume \( V = \pi R^2 \times h \), solve for \( h \), the thickness.

Key Concepts

Rotational Kinetics

Rotational kinetics is a fascinating field that deals with the motion of rotating bodies. It is similar to linear kinetics, but instead of linear velocity, we deal with angular velocity, and instead of mass, we consider moment of inertia. When a rigid object spins around an axis, it experiences rotational motion, and this is where rotational kinetics comes in.

A key concept in rotational kinetics is **torque**, which is the rotational equivalent of linear force. Torque causes an angular acceleration, which changes the angular velocity of the rotating body. The relationship between torque (\( \tau \)) and angular acceleration (\( \alpha \)) is given by the equation:\[ \tau = I \alpha \]where \( I \) is the moment of inertia.

**Rotational kinetic energy** is another critical aspect of rotational kinetics. It's the energy an object possesses due to its rotation and is given by:\[ K = \frac{1}{2} I \omega^2 \]where \( \omega \) is the angular velocity. This equation tells us that a higher angular velocity or a larger moment of inertia will lead to more kinetic energy stored in a rotating object. Understanding rotational kinetics is essential for designing machines and engines, such as flywheels, which require precise control over rotational motion.

Moment of Inertia

The moment of inertia, often symbolized by \( I \), is a fundamental concept in understanding rotational motion. It is a measure of an object's resistance to changes in its angular motion. In simpler terms, it tells us how hard it is to change the speed at which an object is spinning.

For different shapes, the moment of inertia is calculated differently. Specifically for a uniform disk, like the flywheel in the exercise, the formula to find the moment of inertia is:\[ I = \frac{1}{2} m R^2 \]where \( m \) is the mass of the disk and \( R \) is the radius. This means that increasing either the mass or the radius of a disk will increase its moment of inertia, making it harder to start or stop its rotation.

The moment of inertia plays a crucial role in determining the rotational kinetic energy. A higher moment of inertia implies that more energy is needed to achieve a particular angular velocity, which is a key consideration in both designing and analyzing rotating systems like flywheels.

Angular Velocity

Angular velocity, denoted as \( \omega \), is a vector quantity that represents the rate of rotation of an object. It tells us how fast an object is rotating and in which direction. The unit of angular velocity is radian per second (rad/s).

Angular velocity can be calculated using the equation:\[ \omega = \alpha t \]where \( \alpha \) is the angular acceleration and \( t \) is time. In the given problem, the flywheel starts from rest and, given an angular acceleration of 3.00 rad/s², it reaches an angular velocity of 24.00 rad/s after 8 seconds.

This understanding of angular velocity is fundamental in rotational dynamics because it directly influences the rotational kinetic energy. When the angular velocity increases, the kinetic energy of the rotating body also increases, demonstrating the importance of precise control over rotational speeds in mechanical designs.

Energy Storage in Flywheels

Flywheels are remarkable devices that store energy in the form of rotational kinetic energy. They use their mass and speed of rotation to maintain a constant energy supply. This stored energy can be released to perform work when needed.

The energy stored in a flywheel is determined by its rotational kinetic energy, which is given by the formula:\[ K = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. A flywheel's ability to store energy efficiently makes it an important component in various applications, such as in energy recovery systems and mechanical batteries.

To maximize energy storage in a flywheel, you can either increase its moment of inertia or its angular velocity. Engineers often modify the flywheel's geometry, such as its thickness or radius, to optimize energy storage without making the device too heavy or occupying too much space. As in the exercise, calculating the required dimensions and materials is key to designing an effective flywheel.