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530_343lecture14

# 530_343lecture14 - ME 530.343 Design and Analysis of...

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ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 33 - Lagrange’s Equation with Damping & Example Friday, April 3, 2009 Today’s Objectives Lagrange’s equations for systems with damping Double Pendulum Example Reminder: Modal analysis only applies to undamped systems. You can figure out modes with no damping, then add damping to see the actual response. Lagrange with Damping Note: this is a trick to get the damping correct, but technically this should just be thought of as an external force. We introduce a function R : Rayleigh’s dissipation function R = 1 2 ˙ x T B ˙ x B is the damping matrix Re-write Lagrange’s equation: d dt ( ∂L ˙ x i ) - ∂L ∂x i + ∂R ˙ x i = Q i The third term is new. Substitute V = 1 2 x T K x T = 1 2 ˙ x T M ˙ x R = 1 2 ˙ x T B ˙ x L = T - V And you will get M ¨ x + B ˙ x + K x = Q 1

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Double Pendulum Example Velocity of m 1 : v 1 = l 1 ˙ θ 1 Velocity of m 2 : v 2 = ( v 2 2 x + v 2 2 y ) 1 2 v 2 x = l 1 ˙ θ 1 cos( θ 1 ) + l 2 ˙ θ 2 cos( θ 2 ) v 2 y = l 1 ˙ θ 1 sin( θ 1 ) + l 2
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530_343lecture14 - ME 530.343 Design and Analysis of...

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