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Unformatted text preview: AE 353: HW 3 Solutions 3.50 Hint Provided Say y (0) = 0 and ˙ y (0) = 0. Will the conductor move at all? Note that when ˙ y (0) = then α = 0, so L ( α ) = 0. Hence ¨ y = 0. Conductor stays at rest. Only reason why conductor moves is, if there is an initial velocity or non-zero initial position. Assume for simplicity y (0) = 0. and ˙ y (0) = ǫ (very small since oscillations are of low frequency). Since ǫ is small, we can do linearization about ˙ y = 0. Do linearization term by term and then do laplace transform. Though ǫ is small if poles of Y ( s ) are in RHP, then y ( t ) increases with t . We want poles of Y ( s ) to be in LHP i.e. roots of denominator of Y ( s ) should be in LHP and Routh criterion gives the su ffi cient conditions for this. solution If ˙ y is small compared to ν , then α = − tan − 1 ( ˙ y ν ) ≈ − ˙ y ν . Now, L ( α ) ≈ L (0) + ∂ L ∂α vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle α = α ≈ L (0) − ∂ L ∂α...
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This note was uploaded on 04/12/2010 for the course AE 353 taught by Professor Voulgaris,p during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08