Understanding Flywheel Energy Storage

A flywheel is a mechanical device specifically designed to store rotational energy. It works by accumulating energy when its speed increases and releasing it when its speed decreases. This energy storage helps maintain a constant speed, especially when the energy source is intermittent or the load varies. The amount of energy stored in a flywheel depends on its mass, shape (which affects its moment of inertia), and its rotational speed.

Energy Stored in a Rim-Type Flywheel

The energy stored in a rotating body is its kinetic energy of rotation, given by the formula:

$\text{Kinetic Energy (KE)} = \frac{1}{2} I \omega^2$

Where:

• $I$ is the moment of inertia of the flywheel about its axis of rotation.
• $\omega$ is the angular speed of the flywheel.

For a rim-type flywheel, where the mass is concentrated primarily at the rim, the moment of inertia can be approximated as:

$I = MR^2$

Where:

• $M$ is the mass of the flywheel.
• $R$ is the mean radius of the rim.

So, the energy stored in a rim-type flywheel is:

$\text{Energy (E)} = \frac{1}{2} (MR^2) \omega^2$

Analyzing Flywheels X and Y

We are given two rim-type flywheels, X and Y, with the following properties:

• Mean radius of X: $R_X = R$
• Mean radius of Y: $R_Y = R/2$
• Both have the equivalent speed: $\omega_X = \omega_Y = \omega$ (let's call this speed $\omega$)
• Both have the equivalent rotating mass: $M_X = M_Y = M$ (let's call this mass $M$)
• Energy stored in X is $E_X = E$

Calculating Energy Stored in Flywheel X

Using the formula for energy stored in a rim-type flywheel:

$E_X = \frac{1}{2} M_X R_X^2 \omega_X^2$

Substitute the values for X ($M_X = M$, $R_X = R$, $\omega_X = \omega$):

$E = \frac{1}{2} M R^2 \omega^2$

Calculating Energy Stored in Flywheel Y

Now, let's calculate the energy stored in flywheel Y, denoted as $E_Y$:

$E_Y = \frac{1}{2} M_Y R_Y^2 \omega_Y^2$

Substitute the values for Y ($M_Y = M$, $R_Y = R/2$, $\omega_Y = \omega$):

$E_Y = \frac{1}{2} M \left(\frac{R}{2}\right)^2 \omega^2$

$E_Y = \frac{1}{2} M \left(\frac{R^2}{4}\right) \omega^2$

$E_Y = \frac{1}{4} \left(\frac{1}{2} M R^2 \omega^2\right)$

Comparing Energy Storage

From the calculation for $E_X$, we know that $E = \frac{1}{2} M R^2 \omega^2$. We can substitute this into the expression for $E_Y$:

$E_Y = \frac{1}{4} (E)$

So, the energy stored in flywheel Y is one-fourth of the energy stored in flywheel X.

Therefore, the energy collected in Y should be $E/4$.