Question

An aircraft is coming in for a landing at 300 meters height when the propeller falls off. The aircraft is flying at 40.0 m/s horizontally. The propeller has a rotation rate of 20 rev/s, a moment of inertia of 70.0kg-m2 , and a mass of 200 kg. Neglect air resistance. (a) With what translational velocity does the propeller hit the ground? (b) What is the rotation rate of the propeller at impact?

Answer 1

Answer

given,

height = 300 m

speed of aircraft = 40 m/s

propeller rotation = 20 rev/s

moment of inertia = I = 70 Kg/m²

mass = 200 Kg

since no non conservative force  act on system, the mechanical energy is conserved after propeller fall off.

by conservation of energy

       $E_i = E_f$

       $mgh + \dfrac{1}{2}mv_i^2 + \dfrac{1}{2}I\omega^2_i=\dfrac{1}{2}mv_f^2 + \dfrac{1}{2}I\omega^2_f$

as there is no air resistance ωi = ωf

       $mgh + \dfrac{1}{2}mv_i^2=\dfrac{1}{2}mv_f^2$

       $v_f=\sqrt{v_i^2 + 2gh}$

       $v_f=\sqrt{40^2+ 2\times 9.8 \times 300}$

       $v_f=86.48\ m/s$

b) the rotation rate of the propeller remains same as there is no air resistance.