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Problem 1. An aircraft's propeller consists of a rigid hub of radius R and a flexible tapered blade of length 3R that has density p (kg/m³). The propeller is rotating about the center of the hub with a steady angular velocity, thereby creating an effective axial distributed force in the blade. Neglect forces due to gravity. The cross sectional area distribution of the blade is A(r), defined as A(r) = (5-±)4₂ : if R

Problem 1. An aircraft's propeller consists of a rigid hub of radius R and a flexible tapered blade of length 3R that has density p (kg/m³). The propeller is rotating about the center of the hub with a steady angular velocity, thereby creating an effective axial distributed force in the blade. Neglect forces due to gravity. The cross sectional area distribution of the blade is A(r), defined as A(r) = (5-±)4₂ : if R

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Problem 1. An aircraft's propeller consists of a rigid hub of radius R and a flexible tapered blade of length 3R that has density p (kg/m³). The propeller is rotating about the center of the hub with a steady angular velocity, thereby creating an effective axial distributed force in the blade. Neglect forces due to gravity. The cross sectional area distribution of the blade is A(r), defined as A(r) = (5-±)4₂ : if R <r<4R R (a) Write down the expression for body force due to rotational acceleration. (b) In the 1D equilibrium equation derived in class, the area of cross section was constant. Show that the 1D equilibrium equation for the case of a changing cross sectional area is d(σA) dr · + fA = 0.
Transcribed Image Text:Problem 1. An aircraft's propeller consists of a rigid hub of radius R and a flexible tapered blade of length 3R that has density p (kg/m³). The propeller is rotating about the center of the hub with a steady angular velocity, thereby creating an effective axial distributed force in the blade. Neglect forces due to gravity. The cross sectional area distribution of the blade is A(r), defined as A(r) = (5-±)4₂ : if R <r<4R R (a) Write down the expression for body force due to rotational acceleration. (b) In the 1D equilibrium equation derived in class, the area of cross section was constant. Show that the 1D equilibrium equation for the case of a changing cross sectional area is d(σA) dr · + fA = 0.
hub blade R 3R
Transcribed Image Text:hub blade R 3R
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