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. `0.6 kg- m^(2)`
B. `0.15 kg- m^(2)`
C. `0.8 kg- m^(2)`
D. `0.7.5 kg- m^(2)`

Answer

To find the moment of inertia of the flywheel, we will use the formula for rotational kinetic energy, which is given by:

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

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

### Step 1: Write down the given values
We are given:
- Kinetic Energy, \( KE = 360 \, J \)
- Angular Speed, \( \omega = 30 \, rad/s \)

### Step 2: Substitute the values into the kinetic energy formula
Substituting the given values into the kinetic energy formula:

\[ 360 = \frac{1}{2} I (30)^2 \]

### Step 3: Simplify the equation
First, calculate \( (30)^2 \):

\[ (30)^2 = 900 \]

Now, substitute this back into the equation:

\[ 360 = \frac{1}{2} I (900) \]

### Step 4: Solve for \( I \)
Rearranging the equation to solve for \( I \):

\[ 360 = 450 I \]

Now, divide both sides by 450:

\[ I = \frac{360}{450} \]

### Step 5: Simplify the fraction
Now simplify \( \frac{360}{450} \):

\[ I = \frac{360 \div 90}{450 \div 90} = \frac{4}{5} = 0.8 \, kg \cdot m^2 \]

### Final Answer
The moment of inertia of the flywheel about the axis of rotation is:

\[ I = 0.8 \, kg \cdot m^2 \]

---