Question

An aircraft is coming in for a landing at a height of 300 m when the propeller falls off. The aircraft is flying at 43.0 m/s horizontally. The propeller can be modeled by three uniform rods, each 100 kg and 90 cm long, rotating about their ends. It has a rotation rate of 20.0 rev/s.

a. How much rotational kinetic energy does the propeller have when it hits the ground?

b. What about translational kinetic energy (when it hits the ground)?

Answer 1

a)

net mass of 3 rods

m = 3 * 100 = 300 kg

net rotational kE

kEr = 0.5 I w^2 = 0.5* (1/3) m L^2 * w^2

kEr = 0.5* (1/3)* 300* 0.9* 0.9* (20* 2*3.14)^2

kEr = ≈ 638902 J

kEr = 640 kJ

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b)

net mass of 3 rods

m = 3 * 100 = 300 kg

net translation kE

kEt = 0.5 m v^2 + mgh

kEt = 0.5* 300* 43^2 + 300* 9.8*300

kEt = 1159350

kEt = 1160 kJ

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