A flywheel of moment of inertia 0.32 kg-m² is rotated steadily at 120 rad /sec by a 50 W electric motor. The kinetic energy of the flywheel is

The correct option is 2304 J.

Step-by-step Explanation:

To find the rotational kinetic energy of the flywheel, we can use the formula for the kinetic energy of a rotating body:

$E_k=\frac{1}{2}I{\omega }^2$

Where:
- $I$ is the moment of inertia of the flywheel.
- $\omega$ is the angular velocity of the flywheel.

From the given problem, we have the following values:
- Moment of inertia, $I=0.32{\text{kg·m}}^2$
- Angular velocity, $\omega =120\text{rad/s}$

Note: The power of the motor (50 W) is the power required to maintain this steady speed against frictional losses, but it is not needed to calculate the instantaneous kinetic energy of the flywheel.

Now, substitute the given values into the kinetic energy formula:

$E_k=\frac{1}{2}\times 0.32\times {\left(120\right)}^2$

First, calculate the square of the angular velocity:

${\left(120\right)}^2=14400$

Now, substitute this value back into the equation:

$E_k=0.16\times 14400$

Multiply the numbers to get the final kinetic energy:

$E_k=2304\text{J}$

Thus, the kinetic energy of the flywheel is 2304 J.