A flywheel rotating about a fixed axis has a kinetic energy of 225 J when its angular speed is 30 rad/s. What is the moment of inertia of the flywheel about its axis of rotation?

Options

A. `"0.5 kg m"^(2)`
B. `"0.6 kg m"^(2)`
C. `"0.8 kg m"^(2)`
D. `"0.3 kg m"^(2)`

Explanation

To find the moment of inertia of the flywheel about its axis of rotation, we can use the formula for rotational kinetic energy. The formula is:

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

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

Given:
- \( KE = 225 \, \text{J} \)
- \( \omega = 30 \, \text{rad/s} \)

**Step 1: Rearranging the formula to solve for moment of inertia \( I \)**

We can rearrange the formula to find \( I \):

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

**Step 2: Substitute the values into the equation**

Now, we will substitute the values of \( KE \) and \( \omega \) into the equation:

\[ I = \frac{2 \cdot 225 \, \text{J}}{(30 \, \text{rad/s})^2} \]

**Step 3: Calculate \( \omega^2 \)**

Calculating \( \omega^2 \):

\[ \omega^2 = 30^2 = 900 \, \text{(rad/s)}^2 \]

**Step 4: Substitute \( \omega^2 \) back into the equation**

Now substituting \( \omega^2 \) back into the equation for \( I \):

\[ I = \frac{450}{900} \]

**Step 5: Simplify the fraction**

Now simplify the fraction:

\[ I = 0.5 \, \text{kg m}^2 \]

Thus, the moment of inertia of the flywheel about its axis of rotation is:

\[ \boxed{0.5 \, \text{kg m}^2} \]

---