Question

A disk-shaped flywheel, of uniform density \(\rho\), outer radius \(R\), and thickness \(w\), rotates with an angular velocity \(\omega\), in \(\mathrm{rad} / \mathrm{s}\). (a) Show that the moment of inertia, \(I=\int_{\text {vol }} \rho r^{2} d V\), can be expressed as \(I=\pi \rho w R^{4} / 2\) and the kinetic energy can be expressed as \(\mathrm{KE}=I \omega^{2} / 2\). (b) For a steel flywheel rotating at 3000 RPM, determine the kinetic energy, in \(\mathrm{N} \cdot \mathrm{m}\), and the mass, in \(\mathrm{kg}\), if \(R=0.38 \mathrm{~m}\) and \(w=0.025 \mathrm{~m}\). (c) Determine the radius, in \(\mathrm{m}\), and the mass, in \(\mathrm{kg}\), of an aluminum flywheel having the same width, angular velocity, and kinetic energy as in part (b).

Short answer

Moment of Inertia of steel flywheel is \( \pi \rho w R^4 / 2 \) and KE is \( 88414.67 \, \text{Nm}\). Aluminum flywheel: radius 0.52 m, mass 57.43 kg.

Step-by-step solution

Moment of Inertia Calculation

To find the moment of inertia, use the formula for a disk: \[ I = \rho \int_{\text{vol}} r^2 dV \] Considering the volume element in cylindrical coordinates: \[ dV = r \, dr \, d\theta \, dz \] Therefore: \[ I = \rho \int_0^R \int_0^{2\pi} \int_0^w r^3 \, dz \, d\theta \, dr \] Solving the integrals sequentially: \[ I = \rho \int_0^R r^3 \, dr \int_0^w dz \int_0^{2\pi} d\theta \] First, solve the integration over \( \theta \): \[ \int_0^{2\pi} d\theta = 2\pi \] Next, solve the integration over \( z \): \[ \int_0^w dz = w \] Lastly, solve the remaining integral over \( r \): \[ I = \rho \, w \, 2\pi \int_0^R r^3 \, dr = 2\pi \, \rho \, w \, \left[\frac{r^4}{4}\right]_0^R = 2\pi \, \rho \, w \, \frac{R^4}{4} \] Finally: \[ I = \frac{\pi \rho w R^4}{2} \]

Kinetic Energy Calculation

The kinetic energy of a rotating object can be found with: \[ KE = \frac{1}{2} I \omega^2 \] Substitute the moment of inertia: \[ KE = \frac{1}{2} \left( \frac{\pi \rho w R^4}{2} \right) \omega^2 = \frac{\pi \rho w R^4 \omega^2}{4} \]

Kinetic Energy and Mass of Steel Flywheel

Given \( R = 0.38 \) meters and \( w = 0.025 \) meters for the steel flywheel. \( \omega \) is 3000 RPM, which must be converted to rad/s: \[ \omega = 3000 \frac{2\pi}{60} = 100\pi \, \text{rad/s} \]Find mass \( m \): \[ m = \rho V = \rho \pi R^2 w \] Given steel density \( \rho = 7850 \text{kg/m}^3 \), calculate the volume: \[ V = \pi (0.38)^2 (0.025) \] Thus, \[ m = 7850 \, \pi \cdot 0.38^2 \cdot 0.025 = 88.41 \, \text{kg} \] Finally, calculate kinetic energy: \[ KE = \frac{\pi \rho w R^4 \omega^2}{4} = \frac{\pi \, 7850 \, \cdot 0.025 \, \cdot 0.38^4 \, (100\pi)^2}{4} = 88414.67 \, \text{Nm} \]

Radius and Mass of Aluminum Flywheel

Given the same kinetic energy and angular velocity as the steel flywheel, calculate the mass \( m \) and radius \( R \) of the aluminum flywheel. Aluminum density \( \rho = 2700 \text{kg/m}^3 \): First, set up the kinetic energy equation: \[ KE = \frac{\pi \rho w R^4 \omega^2}{4} \rightarrow R = \left(\frac{4KE}{\pi \rho w \omega^2}\right)^{1/4} \] Substitute known values: \[ R = \left(\frac{4 \, \cdot \, 88414.67}{\pi \, \cdot 2700 \, \cdot 0.025 \, \cdot (100\pi)^2}\right)^{1/4} = 0.52 \, \, m \] Mass calculation for aluminum flywheel: \[ m = \pi \rho R^2 w = \pi \cdot 2700 \cdot 0.52^2 \cdot 0.025 = 57.43 \, \, kg \]

Key concepts

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by an object due to its rotation. It is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
where \( KE \) is the kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
Just like linear kinetic energy depends on mass and velocity, rotational kinetic energy depends on the moment of inertia and angular velocity.
If the moment of inertia is high, even a slow rotation can store significant energy. Conversely, with high angular velocities, even smaller moment of inertia can result in high kinetic energy.
This relationship is crucial in systems like flywheels where energy storage and retrieval are important.

Cylindrical Coordinates Integration

In complex geometrical shapes like disks, using cylindrical coordinates simplifies the integration process.
For a volume element \( dV \) in cylindrical coordinates:
\[ dV = r \, dr \, d\theta \, dz \]
helps break the integration into simpler components.
Consider the formula for moment of inertia \( I \) used in the exercise:
\[ I = \rho \int_{\text{vol}} r^2 dV = \rho \int_0^R \int_0^{2\pi} \int_0^w r^3 \, dz \, d\theta \, dr \]
This means integrating over the radial distance \( r \), the angular component \( \theta \), and the thickness \( w \).
Using these coordinates makes the triple integration feasible, especially for symmetric objects like disks.
Each integral affects one dimension, making it simpler to solve step by step.

Flywheel Dynamics

Flywheels are devices designed to store rotational energy. Their main role is to maintain constant energy flow.
When energy input varies, the flywheel can absorb excess energy during high input periods and release it during low input periods.
The dynamics of a flywheel depend on three major factors:

• Moment of inertia \( I \)
• Angular velocity \( \omega \)
• Material density \( \rho \)

High moment of inertia and angular velocity means higher energy storage capacity.
Materials like steel and aluminum are common in flywheels—steel for higher density, and aluminum for weight reduction.
Choosing the right material involves a balance between storing sufficient energy and keeping the flywheel manageable in terms of size and mass.

Density Effects in Rotational Systems

Density \( \rho \) plays a crucial role in determining the flywheel's mass and energy storage capabilities.
For example, steel has a higher density than aluminum (7850 kg/m³ vs. 2700 kg/m³).
This means for the same volume, a steel flywheel will be heavier, and can store more energy due to higher moment of inertia.
However, heavier does not always mean better.
The exercise shows that for the same energy storage, changing the density affects the required radius and mass.
An aluminum flywheel needs to be larger in radius than a steel one to store the same energy due to its lower density.
Density directly influences the design considerations such as dimensions, material selection, and energy efficiency of rotational systems.