A flywheel is attached to an engine to keep its rotational speed between 100 rad/s and 110 rad/s. If the energy fluctuation in the flywheel between these two speeds is 1.05 kJ then the moment of inertia of the flywheel is _______ kg.m² (round off to 2 decimal places).

The correct answer is 1.

Understanding the Concept of Flywheel Energy Fluctuation:
A flywheel is a mechanical device specifically designed to store rotational kinetic energy. The kinetic energy ($KE$) of a rotating body is given by the formula:
$KE=\frac{1}{2}I{\omega }^2$
where:
- $I$ is the moment of inertia of the flywheel (in kg.m²),
- $\omega$ is the angular velocity (in rad/s).

When the rotational speed fluctuates between a minimum speed (${\omega }_2$) and a maximum speed (${\omega }_1$), the energy stored in the flywheel fluctuates as well. The maximum fluctuation of energy ($\Delta KE$) is the difference between the kinetic energy at the maximum speed and the kinetic energy at the minimum speed:

Factoring out the common terms, we get:
$\Delta KE=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)$

Step-by-Step Derivation and Calculation:
Given the following data:
- Maximum angular speed, ${\omega }_1=110\text{rad/s}$
- Minimum angular speed, ${\omega }_2=100\text{rad/s}$
- Energy fluctuation, $\Delta KE=1.05\text{kJ}=1050\text{J}$

Substitute these values into the energy fluctuation formula:
$1050=\frac{1}{2}\times I\times ({110}^2-{100}^2)$

Calculate the squares of the angular speeds:
${110}^2=12100$
${100}^2=10000$

Find the difference between the squares:
${110}^2-{100}^2=12100-10000=2100$

Now, substitute this back into the equation:
$1050=\frac{1}{2}\times I\times 2100$
Simplify the equation:
$1050=1050\times I$
Solve for the moment of inertia ($I$):
$I=\frac{1050}{1050}=1{\text{kg.m}}^2$

Thus, the moment of inertia of the flywheel is 1 kg.m².