Modeling Math

Homework 6 92.450 / 550 Math Modeling Due July 6 1. A uniform string of length L hangs straight down from a pivot which begins to rotate with frequency ω. If ω is not too large the string stays vertical, but at a critical frequency ωc the string begins to flail. Newton’s Second Law gives the equation of the rotating string as ∂ 2u ∂ t 2 = g ∂ ∂ x x ∂u ∂ x + ω 2u, u(L,t ) = 0 where u(x ,t ) is the lateral displacement from equilibrium at distance x from the bottom of the string and t is time. This is the hanging chain problem with the addition of the centripetal acceleration term. (a) Show u(x ,t ) = e r t X (x ) is a solution if X satisfies the boundary value problem (x X ′ ) ′ + λX = 0, X (L) = 0 where λ = (ω 2 − r 2 )/g . (1) (b) Use (1) and information from class to find λL and then r . For instability r > 0 — use this to find ωc . 2. Wind turbine. The equation governing the lateral displacement u(x ,t ) at height x and time t of a wind turbine tower is ∂ 2u ∂ t 2 = −k 2 ∂ 4u ∂ x 4 , 0 < x < L, t > 0 (2) together with the boundary conditions u(0,t ) = ∂u ∂ x (0,t ) = ∂ 2u ∂ x 2 (L,t ) = 0, t > 0 (3) and ∂ 2u ∂ t 2 (L,t ) = k 2µ L ∂ 3u ∂ x 3 (L,t ), t > 0. (4) Here k 2 = E I /Aρ (where E is Young’s modulus, ρ is the linear density, and A,I are the area and second moment of area of the cross-section) and µ = ρAL/m is the ratio of the mass of the tower to the blade assembly. The first two boundary conditions say the tower is fixed to the ground, the third says there is no bending moment at the top and the fourth is Newton’s Second Law applied to the blade assembly. (a) Show u(x ,t ) = X (x )cosωt satisfies (2) provided X ′′′′ = (β/L) 4X where β = L p ω/k. The general solution of this equation is X (x ) = A cos β L x + B sin β L x + C cosh β L x +D sinh β L x . (b) The boundary conditions (3) say X (0) = X ′ (0) = X ′′(L) = 0. Find the solution X which satisfies these conditions. (c) Show (4) implies β 4X (L) + µL 3X ′′′(L) = 0 and use this show β satisfies the equation β sinβ coshβ − cosβ sinhβ 1 + cosβ coshβ = µ. (5) (d) Solve (5) to find the first positive value of β for the GE 1.5s turbine with tower mass 1.7×106 kg and blade assembly mass 1.6×106 kg (you will need a computer to do this). (e) Find the corresponding value of ω in revolutions per minute — use values for the GE 1.5s L = 70 m, I /A = 12.5 m2 , E = 2.0×1011 N/m2 and ρ = 7.8×103 kg/m3 .

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