A flywheel having a radius of gyration of 2m and mass 10 kg rotates at an angular speed of `5 rad s^(-1)` about an axis perpendicular to it through its centre. The kinetic energy of rotation is

Options

A. 500 J
B. 2000 J
C. 1000 J
D. 250 J

Explanation

To find the kinetic energy of rotation of the flywheel, we will use the formula for rotational kinetic energy:

\[ K.E. = \frac{1}{2} I \omega^2 \]

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

### Step 1: Calculate the Moment of Inertia (I)

The moment of inertia \( I \) can be calculated using the radius of gyration \( k \) and mass \( m \):

\[ I = m \cdot k^2 \]

Given:
- Mass \( m = 10 \, \text{kg} \)
- Radius of gyration \( k = 2 \, \text{m} \)

Substituting the values:

\[ I = 10 \cdot (2)^2 = 10 \cdot 4 = 40 \, \text{kg} \cdot \text{m}^2 \]

### Step 2: Calculate the Kinetic Energy (K.E.)

Now we can substitute the moment of inertia \( I \) and the angular speed \( \omega \) into the kinetic energy formula. The angular speed is given as:

\[ \omega = 5 \, \text{rad/s} \]

Now substituting \( I \) and \( \omega \) into the kinetic energy formula:

\[ K.E. = \frac{1}{2} \cdot 40 \cdot (5)^2 \]

Calculating \( (5)^2 \):

\[ (5)^2 = 25 \]

Now substituting this back into the equation:

\[ K.E. = \frac{1}{2} \cdot 40 \cdot 25 = 20 \cdot 25 = 500 \, \text{Joules} \]

### Conclusion

The kinetic energy of rotation of the flywheel is:

\[ \boxed{500 \, \text{Joules}} \]