To calculate the rotational inertia of a wheel, we will proceed step-by-step. Here’s the detailed solution:
1. Kinetic Energy Conversion:
The kinetic energy is given in kilojoules (kJ). To convert this to Joules (J):
[tex]\[
1 \text{ kJ} = 1000 \text{ J}
\][/tex]
Given kinetic energy is [tex]\( 15.6 \text{ kJ} \)[/tex]:
[tex]\[
15.6 \times 1000 = 15600 \text{ J}
\][/tex]
2. Angular Velocity Conversion:
The angular velocity is given in revolutions per minute (rev/min). We need to convert this to radians per second (rad/s).
Since:
[tex]\[
1 \text{ revolution} = 2\pi \text{ radians}
\][/tex]
and:
[tex]\[
1 \text{ minute} = 60 \text{ seconds}
\][/tex]
We convert:
[tex]\[
190 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{190 \times 2\pi}{60} \text{ rad/s}
\][/tex]
This simplifies to:
[tex]\[
\approx 19.89675347273536 \text{ rad/s}
\][/tex]
3. Rotational Inertia Calculation:
We know the expression for kinetic energy (K):
[tex]\[
K = \frac{1}{2} I \omega^2
\][/tex]
Where [tex]\( I \)[/tex] is the rotational inertia and [tex]\( \omega \)[/tex] is the angular velocity.
We rearrange this expression to solve for [tex]\( I \)[/tex]:
[tex]\[
I = \frac{2K}{\omega^2}
\][/tex]
Substituting the known values of [tex]\( K = 15600 \text{ J} \)[/tex] and [tex]\( \omega = 19.89675347273536 \text{ rad/s} \)[/tex]:
[tex]\[
I = \frac{2 \times 15600}{(19.89675347273536)^2}
\][/tex]
Simplifying this:
[tex]\[
I \approx 78.8116021240123 \, \text{kg·m}^2
\][/tex]
So, the rotational inertia of the wheel is approximately [tex]\( 78.8116 \, \text{kg·m}^2 \)[/tex].
