Question

An airplane propeller is 2.00 m in length (from tip to tip) with mass 99.0 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

m = mass of propeller = 99 kg

L = length = 2 m

I = moment of inertia of propeller = m L²/12 = 99 (2)²/12 = 33 kgm²

w = angular speed = 2900 rpm = 2900 (2 x 3.14 /60) = 303.53 rad/s

Rotational KE is given as

KE = (0.5) I w² = (0.5) (33) (303.53)² = 1.52 x 10⁶ J

m = mass of propeller = 0.75 x 99 = 74.25 kg

L = length = 2 m

I = moment of inertia of propeller = m L²/12 = (74.25) (2)²/12 = 24.75 kgm²

w = angular speed = ?

Rotational KE is given as

KE = (0.5) I w²

1.52 x 10⁶ = (0.5) (24.75) w²

w = 350.5 rad/s = 350.5 (60/6.28) = 3348.73 rpm