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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 degreeoffreedom (only one coordinate), conservative (no damping), unforced 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 degreeoffreedom 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 degreeoffreedom 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 nonlinear in the desired coordinat...
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 Fall '11
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