Generalized Coordinates, Lagrange’s Equations, and Constraints CEE 541. Structural Dynamics Department of Civil and Environmental Engineering Duke University Henri P. Gavin Fall, 2016

1 Cartesian Coordinates and Generalized Coordinates

The set of coordinates used to describe the motion of a dynamic system is not unique. For example, consider an elastic pendulum (a mass on the end of a spring). The position of the mass at any point in time may be expressed in Cartesian coordinates (x(t), y(t)) or in terms of the angle of the pendulum and the stretch of the spring (θ(t), u(t)). Of course, these two coordinate systems are related. For Cartesian coordinates centered at the pivot point,

y x(t) = (l + u(t)) sin θ(t) (1) y(t) = −(l + u(t)) cos θ(t) (2) x

θ l θ l where is the un-stretched length of the spring. Let’s define " # " # " # r1(t) x(θ(t), u(t)) (l + u(t)) sin θ(t) u r(t) = = = (3) K r2(t) y(θ(t), u(t)) −(l + u(t)) cos θ(t)

y = −(l+u)cos M g and " # " # q1(t) θ(t) x = (l+u)sin θ q(t) = = (4) q2(t) u(t) so that r(t) is a function of q(t). The potential energy, V , may be expressed in terms of r or in terms of q. In Cartesian coordinates, the velocities are " # " # r˙1(t) u˙(t) sin θ(t) + (l + u(t)) cos θ(t)θ˙(t) r˙(t) = = (5) r˙2(t) −u˙(t) cos θ(t) + (l + u(t)) sin θ(t)θ˙(t)

So, in general, Cartesian velocities r˙(t) can be a function of both the velocity and position of some other coordinates (q˙(t) and q(t)). Such coordinates q are called generalized coordinates. The kinetic energy, T , may be expressed in terms of either r˙ or, more generally, in terms of q˙ and q. Small changes (or variations) in the rectangular coordinates, (δx, δy) consistent with all displacement constraints, can be found from variations in the generalized coordinates (δθ, δu). ∂x ∂x δx = δθ + δu = (l + u(t)) cos θ(t)δθ + sin θ(t)δu (6) ∂θ ∂u 2 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

∂y ∂y δy = δθ + δu = (l + u(t)) sin θ(t)δθ − cos θ(t)δu (7) ∂θ ∂u

2 Principle of Virtual Displacements

Virtual displacements δri are any displacements consistent with the constraints of the system. The principle of virtual displacements1 says that the work of real external forces through virtual external displacements equals the work of the real internal forces arising from the real external forces through virtual internal displacements consistent with the real external displacements. In system described by n coordinates ri, with n internal inertial forces mir¨i(t), potential energy V (r), and n external forces pi collocated with coordinates ri, the principle of virtual dispalcements says,

n ! n X ∂V X mir¨i + δri = (pi(t)) δri (8) i=1 ∂ri i=1

3 Derivation of d’Alembert’s principle from Minimum Total Potential Energy

The total potential energy, Π, is defined as the internal potential energy, V , minus the potential energy of external forces2. In terms of the r (Cartesian) coordinate system and n forces fi collocated with the n displacement coordinates, ri, the total potential energy is given by equation (??). n X Π(r) = V (r) − firi . (9) i=1 The principle of minimum total potential energy states that the total potential energy is minimized in a condition of equilibrium. Minimizing the total potential energy is the same as setting the variation of the total potential energy to zero. For any arbitrary set of “vari- ational” displacements δri(t), consistent with displacement constraints,

n n n X ∂Π X ∂V X min Π(r) ⇒ δΠ(r) = 0 ⇔ δri = 0 ⇔ δri − fi δri = 0 , (10) r i=1 ∂ri i=1 ∂ri i=1

In a dynamic situation, the forces fi can include inertial forces, damping forces, and external loads. For example, fi(t) = −mir¨i(t) + pi(t) , (11) where pi is an external force collocated with the displacement coordinate ri. Substituting equation (??) into equation (??), and rearranging slightly, results in the d’Alembert equa- tions, the same as equation (??) found from the principle of virtual displacements,

n ! X ∂V mir¨i(t) + − pi(t) δri = 0 . (12) i=1 ∂ri

1d’Alembert, J-B le R, 1743 2In solid mechanics, internal strain energy is conventionally assigned the variable U, whereas the potential energy of external forces is conventionally assigned the variable V . In Lagrange’s equations potential energy is assigned the variable V and kinetic energy is denoted by T .

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 3

In equations (??) and (??) the virtual displacements (i.e., the variations) δri must be ar- bitrary and independent of one another; these equations must hold for each coordinate ri individually. ∂V mir¨i(t) + − pi(t) = 0 . (13) ∂ri

4 Derivation of Lagrange’s equations from d’Alembert’s principle

For many problems equation (??) is enough to determine equations of motion. However, in coordinate systems where the kinetic energy depends on the position and velocity of some generalized coordinates, q(t) and q˙(t), expressions for inertial forces become more complicated. P The first goal, then, is to relate the work of inertial forces ( i mir¨i δri) to the kinetic energy in terms of a set of generalized coordinates. To do this requires a change of coordinates from variations in n Cartesian coordinates δr to variations in m generalized coordinates δq,

m m m X ∂ri X ∂r˙i d X ∂r˙i δri = δqj = δqj and δri = δr˙i = δqj . (14) j=1 ∂qj j=1 ∂q˙j dt j=1 ∂qj

Now, consider the kinetic energy of a constant mass m. The kinetic energy in terms of 1 2 2 2 velocities in Cartesian coordinates is given by T (r˙) = 2 m(r ˙1 +r ˙2 +r ˙3), and ∂T δri = mir˙i δri . (15) ∂r˙i

Applying the product and chain rules of calculus to equation (??),

d ∂T ! d δri = mir¨i δri + mir˙i δri (16) dt ∂r˙i dt  m  d ∂T X ∂r˙i ∂T  δqj = mir¨i δri + δr˙i (17) dt ∂r˙i j=1 ∂q˙j ∂r˙i  m  m d ∂T X ∂r˙i ∂T X ∂r˙i mir¨i δri =  δqj − δqj . (18) dt ∂r˙i j=1 ∂q˙j ∂r˙i j=1 ∂qj

P Equation (??) is a step to getting the work of inertial forces ( i mir¨i δri) in terms of a set of generalized coordinates qj. Next, changing the derivatives of the potential energy from ri to qj, m ∂V X ∂V ∂qj ∂ri δri = δqj . (19) ∂ri j=1 ∂qj ∂ri ∂qj

Finally, expressing the work of non-conservative forces pi in terms of the new coordinates qj,

n n m m n m X X X ∂ri X X ∂ri X pi(t) δri = pi(t) δqj = pi(t) δqj = Qj(t) δqj , (20) i=1 i=1 j=1 ∂qj j=1 i=1 ∂qj j=1

cbnd H.P. Gavin September 1, 2020 4 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

where Qj(t) are generalized forces, collocated with the generalized coordinates, qj(t). Substi- tuting equations (??), (??) and (??) into d’Alembert’s equation (??), rearranging the order of the summations, factoring out the common δqj, canceling the ∂ri and ∂r˙i terms, and eliminating the sum over i, leaves the equation in terms of qj andq ˙j,

n   m  m m m  X d ∂T X ∂r˙i ∂T X ∂r˙i X ∂V ∂qj ∂ri X ∂ri   δqj − δqj + δqj − pi(t) δqj = 0 i=1 dt ∂r˙i j=1 ∂q˙j ∂r˙i j=1 ∂qj j=1 ∂qj ∂ri ∂qj j=1 ∂qj m n " ! # X X d ∂T ∂r˙i ∂T ∂r˙i ∂V ∂qj ∂ri ∂ri − + − pi δqj = 0 j=1 i=1 dt ∂r˙i ∂q˙j ∂r˙i ∂qj ∂qj ∂ri ∂qj ∂qj m " ! # X d ∂T ∂T ∂V − + − Qj δqj = 0 j=1 dt ∂q˙j ∂qj ∂qj

The variations δqj must be arbitrary and independent of one another; this equation must hold for each generalized coordinate qj individually. Removing the summation over j and 3 canceling out the common δqj factor results in Lagrange’s equations,

d ∂T ! ∂T ∂V − + − Qj = 0 (21) dt ∂q˙j ∂qj ∂qj are which are applicable in any coordinate system, Cartesian or not.

5 The Lagrangian

Lagrange’s equations may be expressed more compactly in terms of the Lagrangian of the energies, L(q, q˙, t) ≡ T (q, q˙, t) − V (q, t) (22) Since the potential energy V depends only on the positions, q, and not on the velocities, q˙, Lagrange’s equations may be written,

d ∂L ! ∂L − − Qj = 0 (23) dt ∂q˙j ∂qj

3Lagrange, J.L., Mecanique analytique, Mm Ve Courcier, 1811.

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6 Derivation of Hamilton’s principle from d’Alembert’s principle

The variation of the potentential energy V (r) may be expressed in terms of variations of the coordinates ri n n X ∂V X δV = δri = fi δri . (24) i=1 ∂ri i=1

where fi are potential forces collocated with coordiantes ri. In Cartesian coordinates, the variation of the kinetic energy T (r˙)

n X 1 2 T = mir˙i (25) i=1 2 may be expressed in terms of variations of the coordinate velocities,r ˙i

n n X ∂T X δT = δr˙i = mr˙i δr˙i . (26) i=1 ∂r˙i i=1

For a system of n particle masses mi acted on by n internal forces fi of the potential V , d’Alembert’s principle (?? or ??), is

n n X X mir¨i δri + fi δri = 0 (27) i=1 i=1

Integrating d’Alembert’s equation over a finite time period,

n n Z t2 X Z t2 X mir¨i δri dt + fi δri dt = 0 t1 i=1 t1 i=1 n X Z t2 d Z t2 mi r˙i δri dt + δV dt = 0 i=1 t1 dt t1 n " # Z t2 d Z t2 X m r δr |t2 − m r δr dt δV dt i ˙i i t1 i ˙i i + = 0 i=1 t1 dt t1 n Z t2 X Z t2 − mir˙i δr˙i dt + δV dt = 0 t1 i=1 t1 Z t2 Z t2 − δT dt + δV dt = 0 t1 t1 Z t2 δ (T − V ) dt = 0 (28) t1

In this derivation we consider a variation of the coordinate motions δr(t) from t1 to t2. That is, δr(t1) = 0 and δr(t2) = 0, which eliminates the first term in the third line. The fourth line involves a transposition of the variation and the derivitive (d(δr)/dt = δr˙). The last line is a statement of Hamilton’s principle, which is presented formally in the next section. Note that kinetic energy and potential energy are scalar valued quantities, invariant to changes in coordiante systems. So, while Hamilton’s principle is derived here in the context of Cartesian coordinates, it applies to generalized coordinates, as well.

cbnd H.P. Gavin September 1, 2020 6 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

7 Derivation of Lagrange’s equations from Hamilton’s principle

Define a Lagrangian of kinetic and potential energies

L(q, q˙, t) ≡ T (q, q˙, t) − V (q, t) (29)

and define an action potential functional

Z t2 S[q(t)] = L(q, q˙, t) dt (30) t1 with end points q1 = q(t1) and q2 = q(t2). Consider the true path of q(t) from t1 to t2 and a variation of the path, δq(t) such that δq(t1) = 0 and δq(t2) = 0. Hamilton’s principle:4 The solution q(t) is an extremum of the action potential S[q(t)] ⇐⇒ δS[q(t)] = 0 Z t2 ⇐⇒ δL(q, q˙, t) dt = 0 t1 Substituting the Lagrangian into Hamilton’s principle, ( " #) Z t2 X ∂T ∂T ∂V δq˙i + δqi − δqi dt = 0 t1 i ∂q˙i ∂qi ∂qi

We wish to factor out the independent variations δqi, however the first term contains the variation of the derivative, δq˙i. If the conditions for admissible variations in position δq fully specify the conditions for admissible variations in velocity δq˙, the variation and the differentiation can be transposed, d δqi = δq˙i, (31) dt and we can integrate the first term by parts,

" #t2 ! Z t2 ∂T ∂T Z t2 d ∂T δq˙i = δqi − δqi dt t1 ∂q˙i ∂q˙i t1 dt ∂q˙i t1

Since δq(t1) = 0 and δq(t2) = 0, ( " ! #) Z t2 X d ∂T ∂T ∂V − δqi + δqi − δqi dt = 0 t1 i dt ∂q˙i ∂qi ∂qi

The variations δqi must be arbitrary, so the term within the square brackets must be zero for all i. d ∂T ! ∂T ∂V − + = 0. dt ∂q˙i ∂qi ∂qi

4Hamilton, W.R., “On a General Method in Dynamics,” Phil. Trans. of the Royal Society Part II (1834) pp. 247-308; Part I (1835) pp. 95-144

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8 A Little History

Sir Jean-Baptiste Carl William Isaac le Rond Joseph-Louis Fredrich Rowan Newton d’Alembert Lagrange Gauss Hamilton

1642-1727 1717-1783 1736-1813 1777-1855 1805-1865

Second d’Alembert’s M´echanique Gauss’s Hamilton’s Law Principle Analytique Principle Principle 1687 1743 1788 1829 1834-1835 fi dt = d(mivi)

P (fi − miai)δri = 0

d ∂L ∂L − = Qi dt ∂q˙i ∂qi

1 T G = 2 (q¨ − a) M(q¨ − a) δG = 0

S R t2 L q, q, t dt = t1 ( ˙ ) δS = 0

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9 Example: an elastic pendulum

For an elastic pendulum (a mass swinging on the end of a spring), it is much easier to express the kinetic energy and the potential energy in terms of θ and u than x and y. 1 1 T (θ, u, θ,˙ u˙) = M((l + u)θ˙)2 + Mu˙ 2 (32) 2 2 1 V (θ, u) = Mg(l − (l + u) cos θ) + Ku2 (33) 2

Lagrange’s equations require the following derivatives (where q1 = θ and q2 = u): d ∂T d ∂T d = = (M(l + u)2θ˙) = M(2(l + u)u ˙θ˙ + (l + u)2θ¨) (34) dt ∂q˙1 dt ∂θ˙ dt ∂T ∂T = = 0 (35) ∂q1 ∂θ ∂V ∂V = = Mg(l + u) sin θ (36) ∂q1 ∂θ d ∂T d ∂T d = = Mu˙ = Mu¨ (37) dt ∂q˙2 dt ∂u˙ dt ∂T ∂T = = M(l + u)θ˙2 (38) ∂q2 ∂u ∂V ∂V = = Ku − Mg cos θ (39) ∂q2 ∂u

Substituting these derivatives into Lagrange’s equations for q1 gives M(2(l + u)u ˙θ˙ + (l + u)2θ¨) + Mg(l + u) sin θ = 0 . (40)

Substituting these derivatives into Lagrange’s equations for q2 gives Mu¨ − M(l + u)θ˙2 + Ku − Mg cos θ = 0 . (41) These last two equations represent a pair of coupled nonlinear ordinary differential equa- tions describing the unconstrained motion of an elastic pendulum. Equation (??) represents moments (N.m) collocated with θ. Equation (??) represents forces (N) collocated with u. They may be written in matrix form as " #" # " # M(l + u)2 0 θ¨ −Mg(l + u) sin θ − 2M(l + u)u ˙θ˙ = (42) 0 M u¨ M(l + u)θ˙2 − Ku + Mg cos θ or M(q(t), t) q¨(t) = Q(q(t), q˙(t), t) (43)

As can be seen, the inertial terms (involving mass) are more complicated than just Mr¨, and can involve position, velocity, and acceleration of the generalized coordinates. By carefully relating forces in Cartisian coordinates to those in generalized coordinates through free-body diagrams the same equations of motion may be derived, but doing so with La- grange’s equations is often more straight-forward once the kinetic and potential energies are derived.

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10 Maple and Mathematica software to compute derivatives for Lagrange’s equations 10.1 A forced spring-mass-damper oscillator

Consider a forced mass-spring-damper oscillator . . . basically the same as the elastic pendulum with just the u(t) coordinate, (without the θ coordinate), but with some linear viscous damping in parallel with the spring, and external forcing f(t), collocated with the displacement u(t). 1 T (u, u˙) = Mu˙(t)2 2 1 V (u) = Ku(t)2 2 p(u, u˙) δu = −cu˙(t) δu + f(t) δu where (p δu) is the work of real non-conservative forces through a virtual displacement δu, in which the damping force is cu˙ and the external driving force is f(t).

10.1.1 Using Maple

The Maple software may be used to apply Lagrange’s equations to these expressions for kinetic energy, T , potential energy V , and the work of non-conservative forces and external forcing, W . Here are the Maple commands:

> with(Physics): > Setup(mathematicalnotation=true)

> v := diff(u(t),t); d v := -- u(t) dt > T := (1/2) * M * vˆ2; /d \2 T := 1/2 M |-- u(t)| \dt /

> V := (1/2) * K * u(t)ˆ2; 2 V := 1/2 K u(t)

> p := -C * diff(u(t),t) + f(t) ; /d \ p := -C |-- u(t)| + f(t) \dt /

The lines above setup Maple to invoke the desired functional notation, define the ve- locities v as the derivatives of time-dependent displacements u(t), and represent the kinetic

cbnd H.P. Gavin September 1, 2020 10 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin energy T, potential energy V, and the external and non-conservative forces p. The subsequent lines evaluate the derivatives and combine the derivatives into Lagrange’s equations to give us the equations of motion.

> dTdv := diff(T, v); /d \ dTdv := M |-- u(t)| \dt / > ddt_dTdv := diff(dTdv,t); / 2 \ |d | ddt_dTdv := M |--- u(t)| | 2 | \dt / > dTdu := diff(T,u(t)); dTdu := 0

> dVdu := diff(V,u(t)); dVdu := K u(t) > Q := p*diff(u(t),u(t)); /d \ Q := -C |-- u(t)| + f(t) \dt /

> EOM := ddt_dTdv - dTdu + dVdu = Q;

/ 2 \ |d | /d \ EOM := M |--- u(t)| + K u(t) = -C |-- u(t)| + f(t) | 2 | \dt / \dt / > quit

. . . giving us the expected equation of motion (EOM)...

Mu¨(t) + Cu˙(t) + Ku(t) = f(t)

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10.1.2 Using Mathematica

The same problem using Mathematica software is solved as follows:

v[t] = D[u[t], t] u ’[t]

T[t] = 1 / 2 * M * v[t]ˆ2 (1/2) M u ’ [t] 2

V[t] = 1 / 2 * K * u[t]ˆ2 (1/2) K u[t] 2

Q[t] = - C D[u[t], t] + f[t] f[t] - C u ’ [t] dTdv[t] = D[T[t], v[t]] M u ’ [t] ddTdvdt[t] = D[dTdv[t], t] M u" [t] dVdu[t] = D[V[t], u[t]] K u[t] dTdu[t] = D[U[t], u[t]] 0

EOM = ddTdvdt[t] - dTdu[t] + dVdu[t] == Q[t] K u[t] + M u"[t] == f[t] - C u’[t]

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10.2 An elastic pendulum

For the elastic pendulum problem considered earlier, the first coordinate is θ(t), and is called q(t) in the Maple commands and q[t] in the Mathematica commands.. The second coordinate is u(t) . . . called u(t) in Maple and u[t] in Mathematica.

10.2.1 Using Maple

> with(Physics): > Setup(mathematicalnotation=true)

> v1 := diff(q(t),t); d v1 := -- q(t) dt

> v2 := diff(u(t),t); d v2 := -- u(t) dt > T := (1/2) * M * (( l+u(t))*v1)ˆ2 + (1/2) * M * v2ˆ2;

2 /d \2 /d \2 T := 1/2 M (l + u(t)) |-- q(t)| + 1/2 M |-- u(t)| \dt / \dt /

> V := M * g * ( l - (l+u(t)) * cos(q(t))) + (1/2) * K * u(t)ˆ2;

2 V := M g (l - (l + u(t)) cos(q(t))) + 1/2 K u(t)

> EOMq := diff( diff(T,v1) , t) - diff(T,q(t)) + diff(V,q(t));

/ 2 \ /d \ /d \ 2 |d | EOMq := 2 M (l + u(t)) |-- q(t)| |-- u(t)| + M (l + u(t)) |--- q(t)| \dt / \dt / | 2 | \dt / + M g (l + u(t)) sin(q(t))

> EOMu := diff( diff(T,v2) , t) - diff(T,u(t)) + diff(V,u(t));

/ 2 \ |d | /d \2 EOMu := M |--- u(t)| - M (l + u(t)) |-- q(t)| - M g cos(q(t)) + K u(t) | 2 | \dt / \dt /

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10.2.2 Using Mathematica

KE = 1/2 * M * ((l + u[t]) * D[q[t], t])ˆ2 + 1/2 * M * D[u[t], t]ˆ2 (1/2) M (l + u[t])ˆ2 q’[t]ˆ2 + (1/2) M u’[t]ˆ2

PE = M * g * (l - (l + u[t]) * Cos[q[t]]) + 1/2 * K * u[t]ˆ2 (1/2) K u[t]ˆ2 + g M (l - Cos[q[t]] (l + u[t]))

Collect[EOMq = D[D[KE, D[q[t], t]], t] + D[PE - KE, q[t]], {q’’[t], q’[t], q[t]}] g M Sin[q[t]] (l + u[t]) + 2 M (l + u[t]) q’[t] u’[t] + M (l + u[t])ˆ2 q"[t]

Collect[EOMu = D[D[KE, D[u[t], t]], t] + D[PE - KE, u[t]], {u ’’[t], u ’[t], u[t]}] -g M Cos[q[t]] - l M q’[t]ˆ2 + u[t] (K - M q’[t]ˆ2) + M u"[t]

The Maple and Mathematica expressions EOMq and EOMu are the same equations of motion as the previously-derived equations (??) and (??).

cbnd H.P. Gavin September 1, 2020 14 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

11 Matlab simulation of an elastic pendulum

To simulate the dynamic response of a system described by a set of ordinary differential equations, the system equations may be written in a state variable form, in which the highest- order derivatives of each equation (θ¨ andu ¨ in this example) are written in terms of the lower-order derivatives (θ˙,u ˙, θ, and u in this example). The set of lower-order derivatives is called the state vector. In this example, the equations of motion are re-expressed as θ¨ = −(2u ˙θ˙ + g sin θ)/(l + u) (44) u¨ = (l + u)θ˙2 − (K/M)u + g cos θ (45) In general, the state derivativex ˙ = [θ,˙ u,˙ θ,¨ u¨]T is written as a function of the state, x = [θ, u, θ,˙ u˙]T,x ˙(t) = f(t, x). The equations of motion are written in this way in the following Matlab simulation.

1 % elastic p e n d u l u m s i m 2 % simulate the free response of an elastic pendulum 3 4 animate = 1; % 1: animate the response 5 MakeMovie = 1; % 1: makea movie, 0: don’t. 6 7 8 % system constants are global variables... 9 10 global l g K M 11 12 l = 1.0; % un−s t r e t c h e d length of the pendulum,m 13 g = 9.8; % gravitational acceleration,m/sˆ2 14 K = 25.6; % elastic spring constant,N/m 15 M = 1.0; % pendulum mass, kg 16 17 fprintf (’spring-mass frequency = %f Hz\n’, sqrt (K/M )/(2* pi )) 18 fprintf (’pendulum frequency = %f Hz\n’, sqrt (g/l )/(2* pi )) 19 20 t_final = 36; % simulation duration,s 21 delta_t = 0.02; % time step increment,s 22 points = floor (t_final/delta_t); % number of data points 23 t = [1:points]*delta_t; % time data,s 24 25 % initial conditions 26 27 theta_o = 0.0; % initial rotation angle, rad 28 u_o = l; % initial spring stretch,m 29 theta_dot_o = 0.5; % initial rotation rate, rad/s 30 u_dot_o = 0.0; % initial spring stretch rate,m/s 31 32 x_o = [ theta_o ; u_o ; theta_dot_o ; u_dot_o ]; % initial state 33 34 % solve the equations of motion 35 [t,x,dxdt,TV] = ode4u ( ’elastic_pendulum_sys’ , t , x_o ); 36 37 theta = x(1,:); 38 u = x (2 ,:); 39 theta_dot = x(3,:); 40 u_dot = x(4,:); 41 42 T = TV (1 ,:); % kinetic energy 43 V = TV (2 ,:); % potential energy 44 45 % convert from”q” coordinates to”r” coordinates... 46 47 x = (l+u ).* sin ( theta ); % eq’n (1) 48 y = -(l+u ).* cos( theta ); % eq’n (2)

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 15

1 function [dxdt,TV] = elastic_pendulum_sys(t,x) 2 %[dxdt,TV]= elastic p e n d u l u m s y s(t,x) 3 % system equations for an elastic pendulum 4 % compute the state derivative, dxdt, the kinetic energy,T, and the 5 % potential energy,V, of an elastic pendulum. 6 7 % system constants are pre−d e f i n e d global variables... 8 9 global l g K M 10 11 % define the state vector 12 13 theta = x(1); % rotation angle, rad 14 u = x (2); % spring stretch,m 15 theta_dot = x(3); % rotation rate, rad/s 16 u_dot = x(4); % spring stretch rate,m/s 17 18 % compute the acceleration of theta andu 19 20 theta_ddot = -(2*u_dot*theta_dot + g* sin (theta)) / (l+u); % eq’n (35) 21 22 u_ddot = (l+u)*theta_dotˆ2 - (K/M)*u + g*cos( theta ); % eq’n (36) 23 24 % assemble the state derivative 25 26 dxdt = [ theta_dot ; u_dot ; theta_ddot ; u_ddot ]; 27 28 TV = [ (1/2) * M * ((l+u).*theta_dot).ˆ2 + (1/2) * M * u_dot.ˆ2 ; %K.E. (25) 29 M*g*(l-(l+u).* cos(theta)) + (1/2) * K * u.ˆ2 ]; %P.E. (26) 30 31 32 % −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− e l a s t i c p e n d u l u m s y s

cbnd H.P. Gavin September 1, 2020 16 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

1

0.5

0 (t) and u(t) θ -0.5 θ swing, q1(t) = (t), rad stretch, q (t) = u(t), m -1 2 0 5 10 15 20 time, s 6 K.E. 5 P.E. 4 K.E.+P.E. 3 2 1 0 energies, Joules -1 -2 0 5 10 15 20 time, s

0

t = 36.000 s

-0.5

-1 y position, m

-1.5

-2 (x0, y0) -1 -0.5 0 0.5 1 x position, m

Figure 1. Free response of an elastic pendulum from an initial rotational velocity, θ˙(0) of 2 p 0.5 rad/s for l = 1.0 m, g = 9.8 m/s , K = 25.6 N/m, and M = 1 kg. Tk = 2π M/K = p √ 1.242 s. Tl = 2π l/g = 2.007 s. Even though Tk/Tl ≈ ( 5 − 1)/2, the least rational number, the record repeats every 24 seconds. Note that in the absence of internal damping and external forcing, the sum of kinetic energy and potential energy is constant. Why is it that the internal potential energy becomes negative as compared to it’s initial value? What is the initial configuration of the system (θ(0), u(0)) in this example? About what values do θ(t) and u(t) oscillate for t > 0? (spyrograph!) An animation of this motion is here: http://www.duke.edu/∼hpgavin/cee541/elastic pendulum sim.mp4

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 17

12 Constraints

Suppose that in a dynamical system described by m generalized dynamic coordinates, h i q(t) = q1(t), q2(t), ··· , qm(t)

the coordinate trajectories must satisfy additional (constraining) relationships g(q) = 0. For example, suppose the elastic pendulum must move along a particular line, g(θ(t), u(t)) = 0, or that the pendulum can not move past a particular line, g(θ(t), u(t)) ≤ 0. If these constraints to the motion of the system can be written pureley in terms of the positions of the coordinates, then the system (including the differential equations and the constraints) is called a holonomic system. There are a number of intriguing terms connected to constrained dynamical systems.

• A system with constraints that depend only on the position of the coordinates,

g(q(t), t) = 0

is holonomic. • A system with constraints that depend on the position and velocity of the coordinates,

g(q(t), q˙(t), t) = 0

(in which g(q(t), q˙(t), t) can not be integrated into a holonomic form) is nonholonomic. • A system with constraints that are independent of time,

g(q) = 0 or g(q, q˙) = 0

is scleronomic. • A system with constraints that are explicitely dependent on time, as in the constraints listed under the first two definitions is rheonomic. • A system with constraints that are linear in the velocities,

g(q(t), q˙(t)) = f(q(t)) · q˙(t)

is pfafian.

Any unconstrained system must be forced to adhere to a prescribed constraint. The required constraint forces QC are collocated with the generalized coordinates q, and the virtual work of the constraint forces acting through virtual displacements is zero5

QC · δq = 0.

Geometrically, the constraint forces are normal to the displacement variations. 5Lanczos, C., The Variational Principles of Mechanics, 4th ed., Dover 1986

cbnd H.P. Gavin September 1, 2020 18 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

12.1 Holonomic systems

Consider a constraint for the elastic pendulum in which the pendulum must move along a curve g(θ(t), u(t)) = 0. For example, consider the constraint

u(t) = l − lθ2(t) ⇐⇒ g(θ(t), u(t)) = 1 − θ2(t) − u(t)/l .

Clearly the elastic pendulum will not follow the path g = 0 all by itself; it needs to be forced, somehow, to follow the prescribed trajectory. In a holonomic system (in which the constraints are on the coordinate positions) it can be helpful to think of a frictionless guide that enforces the dynamics to evolve along a partiular line, g(q) = 0. The guide exerts constraint forces QC in a direction transverse to the guide, but not along the guide. For relatively simple systems such as this, incorporating a constraint can be as simple as solving the constraint equation for one of the coordinates, for example, u(t) = l − lθ2(t) and substituting that expression into the Lagrangian or the unconstrained equations of motion. With the substitution of u(t) = l−lθ2(t) into the expressions for kinetic energy and potential energy, Lagranges equations can be written in terms of the remaining coordinate, θ. This is a perfectly acceptable means of incorporating holonomic constraints into an analysis. However, in general, a set of c constraint equations g(q) = 0 need not be re-arranged to express the position of c coordinates in terms of the remaining (m − c) coordinates. Furthermore, reducing the dimension of the system by using the constraint equation to eliminate variables does not give us the forces required to enforce the constraints, and therefore misses an important aspect of the behavior of the system. So a more general approach can be followed in order to derive the equations of motion for constrained systems. Recall that an admissible variation, δq must adhere to all constraints. For example, the solution to the constrained elastic pendulum, perterbued by [δθ, δu], must lie along the curve 1 − θ2 − u/l = 0. In order for the variation δq to be admissible, the perturbed solution must also lie on the line of the constraint. A constraint evaluated at a perturbed solution q + δq, in general, can be approximated via a truncated Taylor series

"∂g # g(q + δq, t) = g(q, t) + δq + h.o.t.. ∂q

The constraints at the perturbed solution are satisfied for infinitessimal pertubations as long as the variation δq is orthogonal to the gradient of g with respect to q,

"∂g # δq = 0 . (46) ∂q

Because admissible variations δq are normal to [∂g/∂q] and the constraint force QC is normal to δq, the constraint force must lie within the subspace spanned by [∂g/∂q],

"∂g # QC = λT . ∂q

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 19

q . 2 q2 dg dg dq dq.

δq δq q q. 1 ..1 g(q1 , q 2 ) = 0 g(q1 , q 2 , q 1 , q 2) = 0

Figure 2. Admissibile variations δq for holonomic (left) and nonholonomic (right) systems. In holonomic systems δq adheres to the constraint g(q) = 0 since δq is normal to the gradient of g with respect to q. In nonholonomic systems δq need only be normal to ∂g/∂q˙.

C The constraint forces Qj may now be added into Lagrange’s equations as the forces required to adhere the motion to the constraints gi(q) = 0.

! d ∂T ∂T ∂V X ∂gi − + − λi − Qj = 0 (47) dt ∂q˙j ∂qj ∂qj i ∂qj

in which X ∂gi λi i ∂qj

is the generalized force on coordinate qj applied through the system of constraints,

g(q, t) = 0 . (48)

These forces are precisely the actions that enforce the constraints. Equations (??) and (??) uniquely describe the dynamics of the system. In numerical simulations these two systems of equations are solved simultaneously for the accelerations, q¨(t), and the Lagrange multipliers, λ(t), from which the constraint forces, QC(t), can be found. A system with m coordinates and c holonomic constraints can be reduced to a system of m − c equations of motion by substituting constraint equations into the equations of motion and eliminating coordinates. Alternatively, the Lagrange multipliers can be treated as dynamic variables, and the constraint equations can be enforced in the acceleration form, by differentiating them twice with respect to time. In either case, the initial conditions must be admissible, and numerical integration errors that would lead to dynamic responses that violate g(q, t) = 0 must be numerically corrected.

cbnd H.P. Gavin September 1, 2020 20 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

Let’s apply this to the elastic pendulum, constrained to move along a path u(t) = l − lθ2(t) ⇐⇒ g(θ(t), u(t)) = 1 − θ2(t) − u(t)/l . (49) The derivitives of T and V are as they were. The new derivitives required to model the actions of the constraints are ∂g ∂g = = −2θ (50) ∂q1 ∂θ ∂g ∂g = = −1/l (51) ∂q2 ∂u Note that these derivites are related to, but are not the same as, the gradient of g(q). Equation (??) now includes a new term for the constraint moment in the θ direction, +λ(2θ), and the new term for the constraint force in the u direction in equation (??) is +λ(1/l). (This problem has one constraint, and therefore one Lagrange multiplier, but two coordinates, and therefore two generalized constraint forces.) The Lagrange multiplier in this problem has units of moment (N.m). There are no other non-conservative forces Qj applied to this system. The problem now involves three equations (two equations of motion and the constraint) and three unknowns θ, u (or their derivitives) and λ. In principle, a solution can be found. In solving the equations of motion numerically, as a system of first-order o.d.e’s, we solve for the highest order derivitives in each equation in terms of the lower-order derivitives. In this case the highest order derivitives are θ¨ andu ¨. In the case of constrained dynamics, we also need to solve for the Lagrange multiplier(s), λ. The constraint equation(s) give(s) the additional equation(s) to do so. But in this problem the constraint equation is in terms of positions θ and u. However, by differentiating the constraint we can put this in terms of accelerations θ¨ andu ¨. So doing, with some rearrangement, 2lθθ¨ +u ¨ = −2lθ˙2 . (52) Now we have three equations and three unknowns for θ¨,u ¨, and λ. M(l + u)2θ¨ + 2θλ = −Mg(l + u) sin θ − 2M(l + u)u ˙θ˙ (53) 1 Mu¨ + λ = M(l + u)θ˙2 − Ku + Mg cos θ (54) l 2lθθ¨ +u ¨ = −2lθ˙2 (55) The constraint moment in the θ direction is λ(2θ) and the constraint force in the u direction is λ/l. The constraint forces acting along g(q) = 0 are λ/l in the u direction and 2θλ/(l + u) perpendicular to the u direction. The three equations may be written in matrix form . . .  M(l + u)2 0 2θ   θ¨   −Mg(l + u) sin θ − 2M(l + u)u ˙θ˙       ˙2   0 M 1/l   u¨  =  M(l + u)θ − Ku + Mg cos θ  (56) 2θ 1/l 0 λ −2θ˙2 Note that the initial condition of the system must also adhere to the constraints,

2 lθ0 + u0 = l (57)

2lθ0θ˙0 +u ˙ 0 = 0 (58)

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 21

These equations can be integrated numerically as was shown in the previous Matlab exam- ple, except for the fact that in the presence of constraints, the accelerations and Lagrange multiplier need to be evaluated as a solution of three equations with three unknowns.

cbnd H.P. Gavin September 1, 2020 22 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

1

0.5

0

-0.5 (t) and u(t) θ -1 θ swing, q1(t) = (t), rad stretch, q (t) = u(t), m -1.5 2 0 5 10 15 20 time, s 5 K.E. P.E. 4 K.E.+P.E. 3

2

1 energies, Joules 0 0 5 10 15 20 time, s

Figure 3. Free response of an undamped elastic pendulum from an initial condition uo = l and 2 θ˙o = 1 constrained to move along the arc u = l − lθ . The motion is periodic and the total energy is conserved exactly.

0

t = 20.000 s

-0.5

-1 y position, m

-1.5

-2 (x0, y0) -1 -0.5 0 0.5 1 x position, m

Figure 4. The constrained motion of the pendulum is seen to satisfy the equation u = l − lθ2 for l = 1 m. An animation of this motion and the dynamic constraint forces is here: http: //www.duke.edu/∼hpgavin/cee541/elastic pendulum H sim.mp4

15

10

5

0

-5

θ constraint force, Qθ u constraint force, Q -10 u 0 5 10 15 20

Figure 5. The constraint moment in the θ direction (blue) and force in the u direction (green) required to enforce the constraint u(t) = l − lθ2(t).

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 23

12.2 Nonholonomic systems

C A nonholonomic system has internal constraint forces Qj which adhere the responses to a non-integrable relationship involving the positions and velocities of the coordinates,

g(q, q˙, t) = 0 . (59)

C The constraint forces Qj are normal to the constraint surface g(q, q,t˙ ) = 0. As always, any admissible variations must satisfy the constraints

"∂g # "∂g # g(q + δq, q˙ + δq˙, t) = g(q, q˙, t) + δq + δq˙ + h.o.t. = 0 ∂q ∂q˙

So, for infinitessimal variations,

"∂g ∂g # " δq # , · = 0 . ∂q ∂q˙ δq˙

It may be shown6 that this condition is equivalent to

"∂g # · δq = 0 (60) ∂q˙

which is the constraint on the displacement variations for nonholonomic systems. Since admissible variations δq are normal to [∂g/∂q˙], the constraint forces must there- fore lie within [∂g/∂q˙], "∂g(q, q˙)# QC = λT . ∂q˙ Including these constraint forces into Lagrange’s equations gives the nonholonomic form of Lagrange’s equations,78

! d ∂T ∂T ∂V X ∂gi − + − λi − Qj = 0 , (61) dt ∂q˙j ∂qj ∂qj i ∂q˙j in which X ∂gi λi i ∂q˙j

is the generalized force on coordinate qj applied through the system of constraints g = 0. Equations (??) and (??) uniquely prescribe the constraint forces (Lagrange multipliers λ) and the dynamics of a system constrained by a function of velocity and displacement.

6P.S. Harvey, Jr., Rolling Isolation Systems: Modeling, Analysis, and Assessment, Ph.D. dissertation, Duke Univ, 2013 7M.R. Flannery, “The enigma of nonholonomic constraints,” Am. J. Physics 73(3) 265-272 (2005) 8M.R. Flannery, “d’Alembert-Lagrange analytical dynamics for nonholonomic systems,” J. Mathematical Physics 52 032705 (2011)

cbnd H.P. Gavin September 1, 2020 24 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

As an example, suppose the dynamics of the elastic pendulum are controlled by some internal forces so that the direction of motion arctan(dy/dx) is actively steered to an angle φ that is proportional to the stretch in the spring, φ = −bu/l. The velocity transvese to the steered angle must be zero, giving the constraint, u˙ = tan(θ + bu/l) ⇐⇒ g(θ, u, θ,˙ u˙) =u ˙ cos(θ + bu/l) − lθ˙ sin(θ + bu/l) (62) lθ˙ The additional derivitives required for the nonholonomic form of Lagranges equations are ∂g ∂g = = −l sin(θ + bu/l) (63) ∂q˙1 ∂θ˙ ∂g ∂g = = cos(θ + bu/l) (64) ∂q˙2 ∂u˙ Equation (??) now includes a new term for the constraint moment in the θ direction, −λ(ls), and the new term for the constraint force in the u direction in equation (??) is +λ(c), where s = sin(θ + bu/l) and c = cos(θ + bu/l). Note that in this case, the constraint forces are dependent upon the position of the pendulum (θ and u), but not on the velocities. The constraint equation along with the equations of motion uniquely define the solution θ(t), u(t), and λ(t). Since, for numerical simulation purposes, we are interested in solving for the accelerations, θ¨(t) andu ¨(t), we can differentiate the constraint equation to transform it into a form that is linear in the accelerations, g˙ = −bu˙ 2s/l +uc ¨ − lθ˙2c − lθs¨ − θ˙u˙(s + bc) (65) Our three equations are now, M(l + u)2θ¨ − (ls)λ = −Mg(l + u) sin θ − 2M(l + u)u ˙θ˙ (66) Mu¨ + (c)λ = M(l + u)θ˙2 − Ku + Mg cos θ (67) −(ls)θ¨ + (c)¨u = lθ˙2c + θ˙u˙(s + bc) + bu˙ 2s/l (68)

with an initial condition that also satisfies the constraint,u ˙ 0 = lθ˙0 tan(θ + bu/l). The three equations may be written in matrix form . . .  M(l + u)2 0 −ls   θ¨   −Mg(l + u) sin θ − 2M(l + u)u ˙θ˙       ˙2   0 M c   u¨  =  M(l + u)θ − Ku + Mg cos θ  (69) −ls c 0 λ lθ˙2c + θ˙u˙(s + bc) + bu˙ 2s/l Note that the upper-left 2 × 2 blocks in the matrices of equations (??) and (??) are the same as the corresponding matrix M in the unconstrained system (??). The same is true for the first two elements of the right-hand-side vectors of equations (??) and (??) and the right-hand-side vector Q of the unconstrained system (??). Furthermore, if the differentiated constraint equations (??) and (??) are written as gˆ = A(q(t), q˙(t)) q¨(t) − b(q(t), q˙(t), t) then both equations (??) and (??) may be written " #" # " # MAT q¨ Q = . (70) A 0 λ b This expression can be obtained directly from Gauss’s principle of least constraint, which provides an appealing interpretation of constrained dynamical systems.

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 25

1

0.5

0

-0.5 swing, q1(t) = θ(t), rad θ (t) and u(t) -1 stretch, q2(t) = u(t), m

0 5 10 15 20 25 30 35 40 time, s 5 4 3 2 K.E. 1 0 P.E. -1 K.E.+P.E. energies, Joules -2 0 5 10 15 20 25 30 35 40 time, s

Figure 6. Free response of an undamped elastic pendulum from an initial condition uo = l/2, θo = 0.2 rad, θ˙o = 1 rad/s, with trajectory constrained by lθ/˙ u˙ = tan(θ + bu/l), with b = 5. The motion is not periodic but total energy is conserved exactly and the constraint is satisfied at all times. 2

1.5

1

0.5

0 t = 40.000 s

y position, m -0.5

-1

-1.5 (x0, y0)

-2 -2 -1 0 1 2 x position, m

Figure 7. The constrained motion of the pendulum does not follow a fixed relation between u and θ — the constraint is nonholonomic. An animation of this motion and the dynamic constraint forces is here: http://www.duke.edu/∼hpgavin/cee541/elastic pendulum NH sim.mp4 (ying & yang)

40

20

0

-20

-40

θ constraint moment, Qθ

u constraint force, Qu -60 0 5 10 15 20 25 30 35 40

Figure 8. The constraint moment in the θ direction (blue) and the force u direction (green) ˙ directions required to enforce the constraint lθ/u˙ = tan(θ + bu/l). cbnd H.P. Gavin September 1, 2020 26 CEE 541. Structural Dynamics – Duke University – Fall 2020 – H.P. Gavin

13 Gauss’s Principle of Least Constraint

Consider the equations of motion of the unconstrained system written as equation (??), M(q, t) a = Q(q, q˙, t) . The matrix M is assumed to be symmetric and positive definite. Define the accelerations of the unconstrained system as a ≡ M −1Q and write the differentiated constraints in terms of the constrained accelerations q¨ as A(q, q˙, t) q¨ = b(q, q˙, t). The constrained system requires additional actions QC to enforce the constraint, so the equations of motion of the constrained system are M(q, t) q¨ = Q(q, q˙, t) + QC(q, q˙, t) . Lastly, define a quadratic form of the accelerations, 1 G = (q¨ − a)T M (q¨ − a) 2 Gauss’s Principle of Least Constraint910 states that the accelerations of the constrained sys- tem q¨minimize G subject to the constraints Aq¨ = b. The naturally-constrained accelerations q¨ are as close as possible to the unconstrained accelerations a in a least-squares sense and at each instant in time while satisfying the constraint Aq¨ = b. To minimize a quadratic objective subject to a linear constraint the objective can be augmented via Lagrange multipliers λ,

1 T T GA = (q¨ − a) M (q¨ − a) + λ (A q¨ − b) (71) 2 1 1 = q¨TMq¨ + q¨TMa + aTMa + λTAq¨ − λTb 2 2 minimized with respect to q¨, !T ∂GA = 0 ⇐⇒ Mq¨ = Q − ATλ (72) ∂q¨ and maximized with respect to λ, !T ∂GA = 0 ⇐⇒ Aq¨ = b (73) ∂λ These conditions are met in equation (??), as derived earlier for both holonomic and non- holonomic systems, " #" # " # MAT q¨ Q = . A 0 λ b

9Hofrath and Gauss, C.F., “Uber ein Neues Allgemeines Grundgesatz der Mechanik,”, J. Reine Ange- wandte Mathematik, 4:232-235. (1829) 10Udwadia, F.E. and Kalaba, R.E., “A New Perspective on Constrained Motion,” Proc. Mathematical and Physical Sciences, 439(1906):407-410. (1992).

cbnd H.P. Gavin September 1, 2020 Generalized Coordinates and Lagrange’s Equations 27 (t) ) 2 (t), q" 1 G( q" Aq"=b

a2(t) q"2(t) fC(t)=-AT y a1(t) q"1(t)

The force of the constraints is now apparent, QC = −ATλ. While equation (??) is sufficient for analyzing the behavior and simulating the response of constrained dynamical systems, it is also useful in building an understanding of their nature. Solving the first of the equations for q¨, q¨ = M −1Q − M −1ATλ and inserting this expression into the constraint equation, AM −1Q − AM −1ATλ = b the Lagrange multipliers may be found, λ = −(AM −1AT)−1(b − AM −1Q) , and substituting a = M −1Q, the constraint force, QC = −ATλ, can be found as well, QC = −AT(AM −1AT)−1(Aa − b) . or QC = K(Aa − b) . The difference (Aa − b) is the amount of the constraint violation associated with the uncon- strained accelerations a. If the unconstrained accelerations satisfy the constraint, then the constraint force is zero. The matrix K = −AT(AM −1AT)−1 is a kind of natural nonlinear feedback gain matrix that relates the constraint violation of the unconstrained accelerations (Aa − b) to the constriant force required to bring the constraint to zero, QC. The constrained minimum of the quadratic objective G gives the correct equations of motion.

cbnd H.P. Gavin September 1, 2020