Final answer:
The rotational kinetic energy of the motorcycle wheel is approximately 6,144 J.
Step-by-step explanation:
The formula to calculate the rotational kinetic energy of an object is:
K = 0.5 * I * ω^2
where K is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
For the motorcycle wheel, the moment of inertia can be calculated using the formula for the moment of inertia of a ring:
I = 0.5 * m * (R2^2 - R1^2)
where m is the mass of the wheel, R2 is the outer radius, and R1 is the inner radius.
Plugging in the values given in the question, we get:
I = 0.5 * 12 kg * (0.330 m^2 - 0.280 m^2) = 0.5 * 12 kg * 0.1735 m^2 = 1.0425 kg m^2
Finally, substituting the values into the formula for rotational kinetic energy:
K = 0.5 * 1.0425 kg m^2 * (120 rad/s)^2 = 6,144 J
Therefore, the rotational kinetic energy of the motorcycle wheel is approximately 6,144 J, which makes option a) the correct answer.