Problem 67 - 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 kg/m\(^3\)) in the shape of a 10.0-cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 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

(a) Diameter is approximately 5.56 m. (b) Centripetal acceleration is approximately 49.5 m/s².

Step 1: Understand the Problem

The exercise involves calculating the dimensions and physical properties of a flywheel required to store a specific amount of kinetic energy (10.0 megajoules) when rotating at a specific speed (90.0 rpm). We need to find the diameter of the disk and the centripetal acceleration of a point on its rim.

Step 2: Find the Moment of Inertia

The kinetic energy stored in a rotating object 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. For a solid disk, \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass and \( r \) is the radius of the disk.

Step 3: Relate Mass to Volume and Density

The mass \( m \) of the disk can be expressed in terms of volume \( V \) and density \( \rho \): \( m = \rho V \). The volume of the disk is \( V = \pi r^2 t \), where \( t \) is the thickness of the disk (0.10 m).

Step 4: Calculate Angular Velocity

Convert the rotational speed from rpm to radians per second. \( \omega = 90 \cdot \frac{2\pi}{60} \approx 9.42 \) rad/s.

Step 5: Set Up Energy Equation

Substitute \( I \) and \( \omega \) into the kinetic energy equation: \( 10 \times 10^6 = \frac{1}{2} \times \frac{1}{2} \times \rho \times \pi \times r^2 \times t \times r^2 \times (9.42)^2 \). Solving this will give us the disk's radius \( r \).

Step 6: Solve for Radius

Rearrange and solve the equation for \( r \) to get \( r^4 = \frac{20 \times 10^6}{\rho \pi t \times (9.42)^2} \). Substitute the known values (\( \rho = 7800 \) kg/m³, \( t = 0.10 \) m) to find \( r \).

Step 7: Calculate Diameter

Once \( r \) is found, the diameter \( D \) is simply \( 2r \). Substitute \( r \) to find \( D \).

Step 8: Calculate Centripetal Acceleration

The centripetal acceleration \( a_c = \omega^2 \times r \). Using the value of \( \omega \) calculated earlier and \( r \), find \( a_c \).

Key Concepts

Kinetic Energy

Kinetic energy is a form of energy associated with motion. In the context of flywheel energy storage, it's important to understand how kinetic energy works. The energy stored in a rotating object is given by the equation \( K = \frac{1}{2} I \omega^2 \). Here, \( I \) represents the moment of inertia, which is an indicator of how much mass is distributed around the axis of rotation. The term \( \omega \) is the angular velocity, representing how fast the object is spinning.
The goal with flywheels is to maximize the stored kinetic energy, allowing a power plant to store excess energy until it's needed. This approach is useful during off-peak times, when less electricity is required. Flywheels achieve this by converting electrical energy into kinetic energy and storing it within their spinning motions.
Kept in motion by nearly frictionless bearings, these flywheels can retain energy efficiently over long durations, making them ideal for balancing supply and demand in energy grids.

Moment of Inertia

The moment of inertia \( I \) is critical when discussing rotational motion, including that of a flywheel. It's a measurement that reflects how the mass of an object is spread in relation to the axis of rotation. For a solid disk, which is a common shape for flywheels, the moment of inertia is given by \( I = \frac{1}{2} m r^2 \).
Here, \( m \) denotes the mass of the disk, and \( r \) is its radius. For energy storage purposes, a larger moment of inertia implies that more energy can be stored because the distribution of mass allows for more significant rotational kinetic energy.
Designing a flywheel involves balancing mass and size because a larger radius increases the moment of inertia but also requires more material, affecting cost and practicality. Engineers must consider both the energy needs and physical constraints to design an efficient system.

Angular Velocity

Angular velocity \( \omega \) is a measure of how quickly an object rotates or spins. It's expressed in radians per second (rad/s) and is a key factor in calculating a flywheel's kinetic energy.
To convert rotational speeds from revolutions per minute (rpm) to radians per second, a simple conversion is used: \( \omega = \, \text{rpm} \times \frac{2\pi}{60} \). For example, a flywheel that rotates at 90 rpm has an angular velocity of approximately 9.42 rad/s.
This conversion is crucial, as most equations dealing with rotational dynamics, including those for kinetic energy, require angular velocity to be in radians per second. The higher the angular velocity, the greater the kinetic energy stored, assuming the moment of inertia remains constant.

Centripetal Acceleration

Centripetal acceleration \( a_c \) refers to the acceleration directed towards the center of a circular path. It is an important concept when analyzing points on the rim of a spinning flywheel, as these points experience the highest levels of acceleration.
The formula to calculate centripetal acceleration is \( a_c = \omega^2 \times r \), where \( \omega \) is the angular velocity and \( r \) is the radius of the disk. For points on the rim, it illustrates the impact of rotational speed and radius on the acceleration experienced.
In practical terms, ensuring that \( a_c \) does not exceed material limits is crucial to avoid structural failure. High centripetal forces can lead to significant stress on the flywheel material, necessitating careful design to ensure these forces are manageable.