A flywheel is making `(3000)/(pi)` revolutions per minute about its axis. If the moment of inertia of the flywheel about that axis is 400 `kgm^(2)`, its rotational kinetic energy is :

Options

A. `2xx10^(6)J`
B. `3xx10^(3)J`
C. `5xx10^(3)J`
D. `12xx10^(3)J`

Answer

To solve the problem, we need to calculate the rotational kinetic energy of the flywheel using the formula for rotational kinetic energy:

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

where:
- \( K_R \) is the rotational kinetic energy,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular velocity in radians per second.

### Step 1: Convert the angular velocity from revolutions per minute (rpm) to radians per second.

Given:
\[ \text{Angular velocity} \, \omega = \frac{3000}{\pi} \text{ rpm} \]

To convert rpm to radians per second, we use the conversion factor:
\[ 1 \text{ revolution} = 2\pi \text{ radians} \]
\[ 1 \text{ minute} = 60 \text{ seconds} \]

Thus, the conversion is:
\[ \omega = \frac{3000}{\pi} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \]

Calculating this gives:
\[ \omega = \frac{3000 \times 2\pi}{\pi \times 60} = \frac{3000 \times 2}{60} = \frac{6000}{60} = 100 \text{ radians/second} \]

### Step 2: Substitute the values into the rotational kinetic energy formula.

We have:
- Moment of inertia \( I = 400 \, \text{kgm}^2 \)
- Angular velocity \( \omega = 100 \, \text{rad/s} \)

Now substituting these values into the formula:
\[ K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 400 \times (100)^2 \]

### Step 3: Calculate \( K_R \).

Calculating \( (100)^2 \):
\[ (100)^2 = 10000 \]

Now substituting this back into the equation:
\[ K_R = \frac{1}{2} \times 400 \times 10000 = 200 \times 10000 = 2000000 \, \text{Joules} \]

### Final Answer:
The rotational kinetic energy of the flywheel is:
\[ K_R = 2 \times 10^6 \, \text{Joules} \]