Problem 77 - University Physics with Modern Physics

Question

It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3} )\) in the shape of a \(10.0-\mathrm{cm}\) -hick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajonles of kinetic energy when spinning at 90.0 \(\mathrm{rpm}\) about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

Step-by-Step Solution

Answer

Diameter required is approximately 11.4 meters. Centripetal acceleration is \( 9.87 \times 10^3 \mathrm{m/s^2} \).

Step 1: Understand the Given Problem

We need to determine the diameter of an iron flywheel disk that can store 10.0 megajoules of energy at 90.0 rpm. Also, we need to calculate the centripetal acceleration at the rim of the disk when spinning at this rate.

Step 2: Calculate the Moment of Inertia

The moment of inertia for a uniform disk rotating about an axis through its center perpendicular to its plane is given by the formula \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius.

Step 3: Relate Angular Speed and Moment of Inertia to Kinetic Energy

Kinetic energy of a rotating object is given by \( KE = \frac{1}{2} I \omega^2 \). Here, \( \omega = \frac{2\pi N}{60} \), where \( N \) is the rotational speed in rpm. For this problem, \( \omega = \frac{2\pi \times 90}{60} = 3\pi \) rad/s.

Step 4: Use Kinetic Energy to Find Mass and Radius

Plug the kinetic energy and angular speed into the equation: \( 10^7 = \frac{1}{2} \times \frac{1}{2} \times m \times r^2 \times (3\pi)^2 \). "Solve for \( m \times r^2 \): \( m \times r^2 = \frac{10^7 \times 8}{9\pi^2} \).

Step 5: Solve for Diameter with Mass-Volume Relationship

The mass of the disk can also be written in terms of density: \( m = \rho \times \text{Volume} = \rho \times \pi r^2 h \). Plug it back into the equation: \( \rho \times \pi r^4 h = \frac{10^7 \times 8}{9\pi^2} \). Solve for \( r \) and then the diameter: \( r^4 = \frac{10^7 \times 8}{9\pi \rho h} \) and calculate \( d = 2 \times r \).

Step 6: Calculate the Centripetal Acceleration

The centripetal acceleration is given by \( a_c = \omega^2 r \). Use \( \omega = 3\pi \) rad/s and the calculated radius \( r \) to find the centripetal acceleration.

Key Concepts

Moment of Inertia

Calculating the moment of inertia is a fundamental part of understanding how objects rotate. For a disk, this property measures how much torque is needed for a certain angular acceleration. The standard formula to express the moment of inertia for a uniform disk rotating around its center is:

• \( I = \frac{1}{2} m r^2 \)

Where:

• \( I \) is the moment of inertia
• \( m \) is the mass of the disk
• \( r \) is the radius of the disk

This equation reveals that both the mass and the square of the radius significantly impact how much rotational energy the disk can store. By understanding these relationships, we can solve problems related to rotational dynamics with more ease. When designing systems like flywheels for energy storage, calculating the moment of inertia is crucial to ensure efficiency and effectiveness.

Centripetal Acceleration

When an object rotates, every particle of the object moves in a circular path. This path leads to an acceleration directed towards the center of the circle, known as centripetal acceleration. It can be calculated using:

• \( a_c = \omega^2 r \)

Where:

• \( a_c \) is the centripetal acceleration
• \( \omega \) is the angular velocity
• \( r \) is the radius of rotation

Even when the rotating object, like our flywheel, is stable and spinning at a constant speed, the acceleration remains necessary to keep each point on the object moving in its curved path. Knowing how to determine this acceleration is crucial for ensuring the structural integrity of rotating systems, as large centripetal forces could lead to mechanical failures if not properly managed.

Mechanical Energy Storage

Mechanical energy storage systems like flywheels are powerful tools in energy management. These systems store energy in the form of kinetic energy, which can be released when needed. In the case of classic energy storage in flywheels:

• The amount of energy is determined by the rotational speed and the moment of inertia.
• Using kinetic energy equation: \( KE = \frac{1}{2} I \omega^2 \) allows us to calculate how much energy can be stored.

Flywheels can change the game in power management for applications requiring burst energy during peak demand. They can charge during low demand and then discharge when the energy is needed most.This principle is immensely beneficial for balancing energy grids and increasing efficiency. By understanding and utilizing kinetic energy equations, engineers can design systems with optimal energy storage capabilities.

Analytic Problem-Solving

Approaching complex physics problems can be made easier through a systematic analytic problem-solving method. This involves breaking down the problem into manageable parts and tackling each piece with a step-by-step strategy. Here's how it can be applied:

• Identify what is given and what needs to be solved.
• Use known formulas, like those for kinetic energy and centripetal acceleration, relevant to the problem.
• Substitute the known values, making sure units are consistent, and solve the intermediate steps step-by-step.
• Verify each calculation for accuracy before proceeding to the next.
• Consolidate the results into the final solution.

In our flywheel example, calculations were dissected into understanding the problem, applying formulas for moment of inertia, and determining centripetal acceleration. With practice, this method develops strong problem-solving skills and confidence in tackling similar types of physics challenges.