Question

An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

Answer 1

a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.

b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.

How to determine the angular speed of a airplane propeller

Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:

K = 0.5 · I · ω²  (1)

Where:

• I - Moment of inertia of the airplane propeller, in kilogram-square meters.
• ω - Angular speed, in radians per second

In addition, the moment of inertia of a slender rod rotating around its center is:

I = 0.0833 · M · L²   (2)

Where:

• M - Mass of the propeller, in kilograms
• L - Length of the propeller, in meters

a) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:

K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²

K = 1792152.287 J

The rotational kinetic energy of the airplane propeller is 1792152.287 joules. $\blacksquare$

b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:

$M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}$

$\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}$   (3)

If we know that $\omega_{o} = 2900\,\frac{rev}{min}$, $M_{o} = 100\,kg$ and $M_{f} = 75\,kg$, then the angular speed of the airplane propeller is:

$\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }$

$\omega_{f} \approx 3348.631\, \frac{rev}{min}$

The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. $\blacksquare$

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