Question

Exercise 9.34 5 of 14 An airplane propter is 1.87 min length from tip to tipl with mass 900 kg and is roting at 2300 rpm rev/min) about an axis through its center. You can model the propeller as a slender fod Part A What is its rotational kinetic energy VOALA R- J Submit Pero Answers Theuns Answer * IncorrectTry Again, a attempts remaining Part 8 Suppose that due to weight constraints, you had to reduce the propeller's mass to 70% 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? VO AXO ? rpm Submit Rest A

Answer 1

Part A Rotational kinetic Energy, k= 12 I w? = $(M12) w? = ( 90-91.87 2300 X 21 60 12 = 1760724.47 J Part B Kinetic Energy . Kinetic E. Ewages of 2 1 1/2 Mile w Wi MEL & We (2 & M; L;" wi²= Me he web -) 90x187*(2300) = (0.75X90) (187) (Wp)? - Wf 2655 5.81 rpm