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An airplane propeller is 1.98 m in length (from tip to tip) with mass 122 kg and is rotating at 3000 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy?

Answer :

The rotational kinetic energy of the propeller, modeled as a slender rod, is approximately 1,936,904 Joules, based on its moment of inertia and angular velocity.

To find the rotational kinetic energy, we use the formula:

Rotational Kinetic Energy (K) = (1/2) * I * ω²

First, we need to determine the moment of inertia (I) of the slender rod about its center:

I = (1/12) * m * L²

Where:

m = 122 kg

L = 1.98 m

I = (1/12) * 122 kg * (1.98 m)² = 39.19 kg•m²

Next, convert the angular velocity (ω) from rpm to rad/s:

ω = (3000 rev/min) * (2π rad/rev) * (1 min/60 s) = 314 rad/s

Now, we can calculate the rotational kinetic energy:

K = (1/2) * 39.19 kg•m² * (314 rad/s)² = 1,936,904 J

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