A fly wheel rotating about a fixed axis has a kinetic energy of `360 J`. When its angular speed is `30 rad s^(-1)`. The moment of inertia of the wheel about the axis of rotation is

Options

A. `0.6kgm^(-2)`
B. `0.15kgm^(-2)`
C. `0.8kgm^(-2)`
D. `0.75kgm^(-2)`

Explanation

To find the moment of inertia of the flywheel, we can use the formula for rotational kinetic energy:

\[ KE = \frac{1}{2} I \omega^2 \]

where:
- \( KE \) is the kinetic energy,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular speed.

**Step 1: Write down the known values.**
- Kinetic energy \( KE = 360 \, J \)
- Angular speed \( \omega = 30 \, rad/s \)

**Step 2: Substitute the known values into the kinetic energy formula.**
\[ 360 = \frac{1}{2} I (30)^2 \]

**Step 3: Calculate \( (30)^2 \).**
\[ (30)^2 = 900 \]

**Step 4: Substitute \( 900 \) back into the equation.**
\[ 360 = \frac{1}{2} I \cdot 900 \]

**Step 5: Simplify the equation.**
Multiply both sides by 2 to eliminate the fraction:
\[ 720 = I \cdot 900 \]

**Step 6: Solve for \( I \).**
\[ I = \frac{720}{900} \]

**Step 7: Simplify \( \frac{720}{900} \).**
\[ I = \frac{8}{10} = 0.8 \, kg \cdot m^2 \]

Thus, the moment of inertia of the flywheel about the axis of rotation is \( 0.8 \, kg \cdot m^2 \).