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Calculate the rotational kinetic energy of a 12-kg motorcycle wheel if its angular velocity is 120 rad/s, and its inner radius is 0.280 m and outer radius is 0.330 m.

a) 6,144 J
b) 8,192 J
c) 10,240 J
d) 12,288 J

1 Answer

4 votes

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.

User StichResist
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