INERTIAL MECHANICSLagrange’s equation for mechanisms Aside:Power:Power continuity:Aside:Key point:Generalized force is not the derivative of generalized momentumA scalar example:SUMMARIZING:The key ideas behind the Lagrangian formulation:Advantages:Disadvantages:INERTIAL MECHANICS Neville Hogan The inertial behavior of a mechanism is substantially more complicated than that of a translating rigid body. Strictly speaking, the dynamics are simple; the underlying mechanical physics is still described by Newton’s laws. The complexity arises from the kinematic constraints between the motions of its members. One powerful method to describe inertial mechanics is Lagrange’s equation, which is traditionally introduced using the variational calculus with Hamilton’s principle of stationary action. Here’s a more direct approach that may provide more insight. Inertial Mechanics page 1 Neville HoganLAGRANGE’S EQUATION FOR MECHANISMS Begin with the uncoupled members of the mechanism. x uncoupled coordinates (orientations, locations of mass centers) with respect to a non-accelerating (inertial) reference frame v velocities p momenta f forces These four fundamental quantities are related as follows. dx/dt = v dp/dt = f Inertial Mechanics page 2 Neville HoganThe constitutive equation for kinetic energy storage (inertia) is: p = Mv M diagonal matrix of inertial parameters (masses, moments of inertia, e.g. about mass centers) Kinetic co-energy is the dual of kinetic energy: Ek* = ⌡⌠ ptdv = 12 vt Mv = Ek*(v) Thus, by definition: p = ∂Ek*/∂v Inertial Mechanics page 3 Neville HoganAside: The underlying mechanical physics is fundamentally independent of choice of coordinates. Therefore, these may be regarded as tensor equations. By the usual conventions: v is a contravariant rank 1 tensor (vector) M is a twice co-variant rank 2 tensor p is covariant rank 1 tensor (vector) These observations become more useful when we consider transformations of variables. Inertial Mechanics page 4 Neville HoganNext consider the kinematically coupled mechanism. θ generalized coordinates (or configuration variables) — a (non-unique) set of independent variables that uniquely and completely define the (mechanism) configuration ω generalized velocities — the time derivatives of generalized coordinates. dθ/dt = ω τ generalized forces (moments or torques) η generalized momenta Inertial Mechanics page 5 Neville HoganThe relation between generalized forces and momenta requires care. If the kinematic constraints are holonomic, the relation between coordinates is a set of algebraic equations1. x = L(θ) Relation between velocities: dx/dt = dθ/dt ()θθL ∂∂ /)(v = J(θ)ω where J(θ) = ()θθL ∂∂ /)( 1 Non-holonomic constraints are commonplace. A typical example is a constraint between velocities that cannot be integrated to a constraint between coordinates. Inertial Mechanics page 6 Neville HoganThe relation between generalized forces may be derived from power continuity2 (a differential statement of energy conservation). Power: P = τt ω Power continuity: P = τt ω = ft v = ft J(θ)ω This must be true for all values of ω, therefore τ = J(θ)t f Aside: A common error is to mis-identify generalized forces. The relation between power, generalized force and generalized velocity is a rigorous and reliable definition of generalized forces. 2 This avoids the sometimes-confusing principle of virtual work but is completely equivalent. Inertial Mechanics page 7 Neville HoganThe relation between kinetic co-energies may be obtained by substitution using the relation between velocities. Ek* = 12 ωt J(θ)t MJ(θ)ω Kinetic energy in generalized coordinates is a quadratic form in velocity. The kernel of the quadratic form is the inertia tensor. I(θ) = J(θ)t MJ(θ) Ek* = 12 ωt I(θ)ω Note that kinetic co-energy, which previously was a function of velocity alone, is now a function of velocity and position. Ek* = Ek*(θ,ω) This is the main reason why (to paraphrase Prof. Stephen Crandall) “mechanics is hard for humans”. Inertial Mechanics page 8 Neville HoganThe relation between momenta follows directly. Generalized momenta are defined as before. η = ∂Ek*/∂ω η = I(θ)ω = J(θ)t MJ(θ)ω η = J(θ)t Mv = J(θ)t p η = J(θ)t p Inertial Mechanics page 9 Neville HoganKEY POINT: Generalized force is not the derivative of generalized momentum dη/dt ≠ τ Differentiate the relation between momenta dη/dt = J(θ)t dp/dt + ωt [∂J(θ)t /∂θ]p dη/dt = J(θ)t f + ωt [∂J(θ)t /∂θ]MJ(θ)ω The second term appears to be related to the kinetic co-energy. It is: ∂Ek*/∂θ = 12 ωt [∂J(θ)t /∂θ]MJ(θ)ω + 12 ωt J(θ)t M[∂J(θ)/∂θ]ω ∂Ek*/∂θ = ωt [∂J(θ)t /∂θ]MJ(θ)ω dη/dt = τ + ∂Ek*/∂θ This is Lagrange's equation dη/dt – ∂Ek*/∂θ = τ Inertial Mechanics page 10 Neville HoganIt may be more familiar in expanded form. Identify kinetic co-energy with the Lagrangian, L(θ,ω) L(θ,ω) = Ek*(θ,ω) ⎥⎦⎤⎢⎣⎡∂∂ω*kEdtd– θ∂∂*kE= τ ⎥⎦⎤⎢⎣⎡∂∂ωLdtd– θ∂∂L= τ Inertial Mechanics page 11 Neville HoganA SCALAR EXAMPLE: x = L(θ) v = J(θ)ω τ = J(θ)f Ek* = 12 mv2 = 12 mJ(θ)2ω2 η = ∂Ek*/∂ω = mJ(θ)2ω = J(θ)mv = J(θ)p dη/dt = J(θ)dp/dt + []θ/)θ(J∂∂ωp dη/dt = τ + ωmJ(θ)ω []θ/)θ(J ∂∂∂Ek*/∂θ = mJ(θ)ω2∂J(θ)/∂θ dη/dt = τ + ∂Ek*/∂θ Inertial Mechanics page 12 Neville HoganSUMMARIZING: inertial coordinates relation generalized coordinates displacement x x = L(θ) θ flow v v = J(θ)ω ω effort f τ = J(θ)t f τ momentum p η = J(θ)t p η inertia tensor M I(θ) = J(θ)t MJ(θ)I(θ) constitutive equation p = Mv η = I(θ)ω kinetic co-energy 12 vt Mv 12 ωt I(θ)ω dx/dt = v dθ/dt = ω dp/dt = f dη/dt = τ + ∂Ek*/∂θ Inertial Mechanics page 13 Neville HoganThe key ideas behind the Lagrangian formulation: 1. Incorporate holonomic kinematic constraints directly. 2. Write the momentum balance equation in terms of a state function, the kinetic co-energy. Advantages: 1. Velocities are easily identified and kinetic co-energy is easily computed. 2. There is no need to write explicit expressions for the forces of constraint. Disadvantages: 1. The kinetic co-energy is a quadratic form whose kernel typically contains trigonometric functions of sums of coordinates. Differentiating trigonometric functions of sums of