Since 01 1 2 1 therefore r r 1 2

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Unformatted text preview: nge Equation once for each generalized coordinate. For N degrees of freedom, N generalized coordinates will yield N equations of motion. • If necessary, linearize the equations of motion by neglecting nonlinear terms in the equations of motion. Note that the linear equations of motion may not adequately describe the original equations of motion. NOTE: For the single degree-of-freedom (only one coordinate), conservative (no damping), un-forced problem (no external forces), there is a simplified method called the Energy Method. d (T + U ) = 0 dt Lecture Notes -29- 06/16/06 12:25 PM Mechanical Vibrations I Lagrange Example A In order to solve the Lagrange Equation, d ∂T ∂T ∂U ∂D + + = Fi , − dt ∂qi ∂qi ∂qi ∂qi m x(t) f(t) for the given single degree-of-freedom system, it is first necessary to formulate the energy equations: kinetic, potential and dissipative. k c First, formulating the kinetic energy yields, T = 1 mx 2 2 Second, formulating the potential energy yields, U = 1 kx 2 + mgx 2 Finally, formulating the dissipative energy yields, D = 1 cx 2 2 Next, it is necessary to evaluate each of the Lagrange terms, d ∂T dt ∂x d ∂ 1 2 d = ( 2 mx ) = ( mx ) = mx dt ∂x dt ∂T ∂ 1 2 = ( 2 mx ) = 0 ∂x ∂x ∂U ∂ 1 2 = ( 2 kx + mgx ) = kx + mg ∂x ∂x ∂D ∂ 1 2 = ( 2 cx ) = cx ∂x ∂x Which are combined to form the actual equation of motion. mx − 0 + kx + mg + cx = f mx + cx + kx = f − mg Since the mg term ends up on the right hand side with the external force, the coordinate x can be redefined to be from the static equilibrium position. As a result, the mg term is eliminated as shown. mx + cx + kx = f Lecture Notes -30- 06/16/06 12:25 PM Mechanical Vibrations I Lagrange Example B Again, in order to solve the Lagrange Equation, d ∂T ∂T ∂U ∂D + + = Fi , − dt ∂qi ∂qi ∂qi ∂qi L m, J cg for the given single degree-of-freedom system, it is necessary to formulate the energy equations: kinetic, potential and dissipative. θ First, formulating the kinetic energy yields, T = 1 J cgθ 2 + 1 m Lθ 2 2 ( ) 2 Second, formulating the potential energy yields, U = mgL (1 − cosθ ) Finally, formulating the dissipative energy yields, D = 0 Next, it is necessary to evaluate each of the Lagrange terms, d ∂T dt ∂θ d ∂ = dt ∂θ ( 1 2 J cgθ 2 + 1 m Lθ 2 2 ∂T ∂ 1 2 1 = =0 2 J cgθ + 2 m Lθ ∂θ ∂θ ∂U ∂ = ( mgL (1 − cosθ ) ) = mgL sin θ ∂θ ∂θ ∂D ∂ = ( 0) = 0 ∂θ ∂θ ( d ( ) ) = dt ( J 2 cg θ + mL2θ ) = J cgθ + mL2θ ( )) Which are combined to form the actual equation of motion. J cgθ + mL2θ − 0 + mgL sin θ + 0 = 0 This time, since the mg term ends up on the left hand side (multiplied by the coordinate), it is not possible to eliminate the mg term from the solution. So the exact equation of motion remains, (J cg + mL2 )θ + mgL sin θ = 0 Lecture Notes -31- 06/16/06 12:25 PM Mechanical Vibrations I However, because the mgL sin θ term is non-linear in the desired coordinat...
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