Chapter 2 Lagrangian Modeling

Chapter 2 Lagrangian Moeling The basic laws of physics are use to moel every system whether it is electrical, mechanical, hyraulic, or any other energy omain. In mechanics, Newton s laws of motion provie key concepts to moel-relate physical phenomenon. The Lagrangian formulation of moeling erives from the basic work energy principle an Newton s laws of motion. The basic law states that the force acting on a boy is irectly proportional to its acceleration equate with constant mass. F ¼ ma ¼ m v ¼ m 2 x 2 ð2:1þ In this equation, force is irectly proportional to the erivative of velocity v or ouble erivative of isplacement x with respect to time. If a boy moves from point A to point B then the work one by a boy is a ot prouct of force an isplace path integrate from point A to B. W ¼ ð B A W ¼ ð B A F x m 2 x x ¼ 2 ð B A m x x ð2:2þ ð2:3þ We now erive the equation as xx ¼ _x x x ¼ _x ¼ _x_x ð2:4þ Springer International Publishing Switzerlan 2016 A.M. Mughal, Real Time Moeling, Simulation an Control of Dynamical Systems, DOI 10.1007/978-3-319-33906-1_2 19

20 2 Lagrangian Moeling So Eq. (2.3) formulates as ð B W ¼ m _x_x ¼ m _x 2 B ¼ m A 2 A 2 _x 2 B _x 2 A W ¼ K B K A ¼ ΔK ð2:5þ ð2:6þ It is important to know that here we are iscussing motion in a conservative fiel where work one is inepenent of path an epens upon the ifference between the initial an final value of kinetic energy K. We note that work one is represente as change in K between two points. By the law of conversation of energy an work energy principle we know that a change in kinetic energy shoul also change the potential energy U, as the total energy of the system remains constant. So a change in kinetic an change in potential energies shoul have a net effect of zero as ΔK þ ΔU ¼ 0 ð2:7þ So ΔU ¼W ¼ ð B A F x ð2:8þ Using the funamental theorem of calculus for antierivatives, we express Eq. (2.8) as F ¼ U x ð2:9þ This shows forces require to change the potential energy, whereas the force require to change the kinetic energy is etermine from the efinition of K. K ¼ 1 2 mv2 ¼ 1 2 m _x 2 ð2:10þ So K _x ¼ m _x ð2:11þ Differentiating Eq. (2.11) gives a force of kinetic energy F ¼ K _x ¼ ð m _x Þ ¼ ma ð2:12þ

2.1 Moeling in Three Axes 21 As sum of forces equal to zero so K _x U x ¼ 0 ð2:13þ Now we efine a Lagrangian function L as the ifference of kinetic an potential energies L ¼ K U ð2:14þ We observe that kinetic energy K is a function of velocity _x, an potential energy U is a function of isplacement x. Accoringly the Lagrangian function L has two terms as a function of velocity x _ : an isplacement x inepenently. As we know that ¼ K an _x _x x ¼ U ð2:15þ x Lagrangian function L now represents Eq. (2.13) as _x x ¼ 0 ð2:16þ This is Lagrangian formulation for a system with motion only in single axis. 2.1 Moeling in Three Axes Now consier a motion in three axes an every equation now to be represente in three imensional vector notations. F * ¼ m 2 x * 2 ð2:17þ The force F * has three components, F x, F y, an F z in x, y an z axes respectively. The potential energy U is a scalar fiel in three imensions an the negative graient of this fiel provies a force vector in three axes. Now Eq. (2.9) represents as F * ¼ Ux; ð y; zþ ¼ U T U U ð2:18þ x y z

22 2 Lagrangian Moeling The kinetic energy K in three axes is given as K ¼ 1 2 m _x* 2 ¼ 1 T 2 m _x* * _x ¼ 1 2 m _x 2 þ _y 2 þ _z 2 ð2:19þ This provies the three components of a force through kinetic energy Eq. (2.19) as F * ¼ T K K K ð2:20þ _x _y _z By equating forces we can write Eq. (2.13) for three axes as T K K K U T U U ¼ 0 ð2:21þ _x _y _z x y z This gives us Lagrangian formulation of rigi boy (particle) for each axis in Cartesian coorinates as _x _y _z x ¼ 0 y ¼ 0 z ¼ 0 ð2:22þ 2.2 Moeling in Cartesian an Spherical/Polar Coorinates In the above section, we evelope a moeling scheme in only a Cartesian coorinate system but often the moel is represente in spherical coorinates in 3D space or with polar coorinates in 2D space. The translation an rotational motion also become part of a moel in ynamical equations. Now, we iscuss the metho in 3D space with spherical coorinates, which can also be use in polar coorinates by eliminating an axis. We know that x ¼ fðr; θ; φ; tþ y ¼ fðr; θ; φ; tþ z ¼ fðr; θ; φ; tþ ð2:23þ We efine our system in Cartesian coorinates by using the transformation in Eq. (2.23). Now let us consier that we want to generalize the concept of spherical,

2.2 Moeling in Cartesian an Spherical/Polar Coorinates 23 polar, or cylinrical coorinates transforme into Cartesian coorinates by using a generalize set of variables instea of r, θ, φ. We now efine a generalize vector * q ¼ ½ q1 q 2 q 3 T an the position in Cartesian coorinates with a vector * p ¼ ½x y z T ¼ ½p 1 p 2 p 3 T. 2 3 p 1 ¼ fðq 1 ; q 2 ; q 3 ; tþ 6 7 4 p 2 ¼ fðq 1 ; q 2 ; q 3 ; tþ5 ¼ * p ¼ f * q ; t ð2:24þ p 3 ¼ fðq 1 ; q 2 ; q 3 ; tþ The functions in Eq. (2.24) follow the chain rule for respective erivatives because x, y, z are ifferentiable with respect to time as well as with respect to each generalize coorinates. Now we follow the convention that _x ¼ x ; a ot on top of the variables represents its ifferentiation with respect to time for both Cartesian an generalize coorinates. The change with respect to time in Cartesian coorinates is efine as _p 1 ¼ p 1 q 1 _q 1 þ p 1 q 2 _q 2 þ p 1 q 3 _q 3 _p 2 ¼ p 2 q 1 _q 1 þ p 2 q 2 _q 2 þ p 2 q 3 _q 3 ð2:25þ _p 3 ¼ p 3 q 1 _q 1 þ p 3 q 2 _q 2 þ p 3 q 3 _q 3 Now consier a generalize notation for any element p i (i.e. x, y, z) of a position vector * p with respect to time erivates an generalize coorinates _p i ¼ X3 j¼1 _q j ð2:26þ The change with respect to generalize coorinates is given as δp i ¼ X3 j¼1 δq j ð2:27þ We observe that _p i ¼ p i þ _p i ð2:28þ

24 2 Lagrangian Moeling Solving for the secon erivative of position coorinate, we get p i ¼ _p i _p i ð2:29þ Differentiating Eq. (2.26) with respect to q j where j ¼ f1; 2; 3g _p i _q j ¼ ð2:30þ As p i is a function of q j then the partial erivatives with respect to time are given as ¼ X3 k¼1 q k _q q k j ð2:31þ We know that in complex or real fiels, secon-orer cross partial erivatives are equal for any position variable with respect to other generalize coorinates q k ¼ q k ð2:32þ Now taking partial erivative of Eq. (2.26) with respect to q j we get _p i ¼ X3 k¼1 _q q k ¼ X3 k By comparing Eqs. (2.31) an (2.33) we get k¼1 ¼ _p i q k Now substituting Eqs. (2.30) an (2.34) in Eq. (2.29) p i ¼ _p i _p i _q j _p i _p i _q q k j ð2:33þ ð2:34þ ð2:35þ Now we write Eq. (2.35) as p i ¼ 2 _p i 2 _p i _q j 2 2 ð2:36þ

2.3 Work an Energy Formulation 25 2.3 Work an Energy Formulation We know that change in work one is the ot prouct of force applie an corresponing change in position given as δw ¼ F * δp * ð2:37þ Equivalently, force vector F * has three components F P1, F P2 ; an F P3, such as δw ¼ X3 F Pi δ Pi ð2:38þ Each component of force is expresse with Newton s law as Equating δw in Eq. (2.38) we have F Pi ¼ m p i ð2:39þ δw ¼ X3 m p i δ Pi ¼ X3 F Pi δ Pi ð2:40þ Substituting Eq. (2.27) in Eq. (2.40) we get δw ¼ X3 j¼1 X 3 m p i Pi δ qj ¼ X3 j¼1 X 3 F Pi Pi qj δ qj ð2:41þ Let F qj ¼ X3 F Pi Pi qj ð2:42þ Equation (2.41) becomes δw ¼ X3 j¼1 X 3 m p i Pi δ qj ¼ X3 j¼1 F qj δ qj ð2:43þ

26 2 Lagrangian Moeling Now by substituting Eq. (2.36) in Eq. (2.43) we get δw ¼ X3 j¼1 m X3 2 _p i 2 _p i δ qj ¼ X3 F qj δ qj _q j 2 2 j¼1 ð2:44þ By efinition of kinetic energy of a particle of mass m is given as K ¼ m 2 X 3 _p 2 i ð2:45þ There are three Lagrange equations which generate from Eq. (2.44) for each axis ðj ¼ 1, 2, 3Þ as m X3 2 _p i 2 _p i ¼ F qj ð2:46þ _q j 2 2 Each value of j represents a Lagrange equation in its axis. The purpose of keeping δw in Eq. (2.44) is that we nee to equate change in energy through work one by other physical interpretations. The Lagrange equation eals with the kinetic energy of an object in motion in Eq. (2.44). Work one is also represente by change in potential energy as given in Eq. (2.8). There may be non-conservative forces that are acting on the boy or internal energy of the system, but ue to the law of conservation of energy these forces must balance each other. The negative graient of potential energy provies forces acting upon the boy which may be causing it to move or stop. So if we efine potential energy U in Cartesian coorinates, forces in Cartesian coorinates are represente as F Pi ¼ U ð2:47þ Equation (2.42) represent as F qj ¼ X3 U Pi qj ¼ U Now we write Eq. (2.46) as a Lagrange equation in q j coorinates m X3 2 _p i 2 _p i ¼ U _q j 2 2 ð2:48þ ð2:49þ Again using the efinition of kinetic energy from Eq. (2.45) to represent Eq. (2.49) in energy variables

2.3 Work an Energy Formulation 27 K K ¼ U _q j ð2:50þ Kinetic energy is a function of both position an velocity of a particle in generalize coorinate q j an the potential energy is only a function of generalize position coorinates, ¼ _q j K _q j an ¼ U ð2:51þ Now using a Lagrange variable L from Eq. (2.14) in three axes system an using Eq. (2.51) in each coorinates we get the following equation for a conservative system in each q j coorinate _q j ¼ 0 ð2:52þ In a conservative fiel Eq. (2.7) hols for change in both potential an kinetic energy, but in a non-conservative fiel the change is also ue to non-conservative, external, or internal forces in each coorinate system. E nc is a non-conservative energy ue to all of these forces acting upon the boy, so the energy equation becomes ΔK þ ΔU ¼ E nc ð2:53þ In this case the Lagrange equation (2.52) oes not satisfy the ue non-conservative force acting upon a boy. Let F ncqj is the sum of all non-conservative forces acting in q j irection. The generalize Lagrange equation in generalize coorinates is _q j ¼ F ncqj ð2:54þ Depening upon the space in which coorinates are efine, the value of j can be change. For a motion with two egrees of freeom there will be only two Lagrange equations but in a motion with three egrees of freeom there will be three Lagrange equations to escribe the system. In certain cases, there is motion in both Cartesian coorinates an generalize coorinates; then accoringly a irection of motion in Cartesian is also treate as a variable q j.

28 2 Lagrangian Moeling 2.4 State Space Representation The Lagrange equation itself is a nonlinear or linear state space representation. All non-conservative forces are treate as input to the system whether these are controllable or exogenous. The generalize coorinates an their first-orer erivatives constitute the state vector. Example 2.1: A Simple Penulum The simple penulum is a boy with mass m attache at length l of massless string from the origin (or groun) to move between two en points as shown in Fig. 2.1. If no external force is applie then the movement will epen upon the starting point an eventually comes to en at mass exactly below the origin. In Cartesian coorinates x an y mass move in a semicircle, which can also be relate through polar coorinates l an θ, which are fixe length of a string an angle from the origin respectively. There is only one egree of freeom, so there is only one generalize coorinate q 1 ¼ θ. Conventionally these simple examples are represente with θ instea of q 1. Now we efine the position an velocity of Cartesian coorinate as a function of a generalize coorinate as xðθþ ¼ l sin θ yðθþ ¼ l cos θ _x θ; θ _ ¼ l θ _ cos θ _y θ; _θ ¼ l _θ sin θ ð2:55þ ð2:56þ The kinetic energy of the mass m in Cartesian coorinate is given as K ¼ 1 2 mv2 ¼ 1 2 m _x 2 þ _y 2 1 ¼ 2 ml2 _θ 2 ð2:57þ The potential energy U is ue to gravitational pull an it is lowest when θ ¼ 0 where kinetic energy is highest an vice versa when θ ¼ π 2 or π 2.So or U ¼ mglð1 cos θþ ð2:58þ Fig. 2.1 A simple penulum

2.4 State Space Representation 29 The Lagrangian L ¼ K U is given as L ¼ 1 2 ml2 θ _ 2 mglð1 cos θþ ð2:59þ The Lagrange equation of the system erives as follows _θ θ ¼ 0 ð2:60þ ml 2 _θ mgl sin θ ¼ 0 ð2:61aþ ml 2 θ þ mgl sin θ ¼ 0 ð2:61bþ The final equation with one egree of freeom now appears as θ þ g l sin θ ¼ 0 ð2:62þ State space formulation of the system represent Eq. (2.62) with efinition of state variables given as x 1 ¼ θ an x 2 ¼ _θ. We have state vector * x ¼ θ _θ an the nonlinear state space representation as _x 1 ¼ x 2 _x 2 ¼ g l sin x 1 ð2:63þ The system can be linearize at relaxe equilibrium point " # θ e T _θ e ¼ ½0 0 with only one matrix A ¼ 0 1 g 0 l Example 2.2: Inverte Penulum on a Moving Cart A cart of mass M is moving by an applie force F acting along x-axis. An inverte penulum of mass m is attache on the center of cart with a ro of length l (Fig. 2.2). The motion of the cart is efine from origin an the angle is measure from vertical y-axis. The system can either be represente with horizontal an vertical variables of cart an bob or by using horizontal isplacement of cart an angular position of bob. There are only two egrees of freeom; an angle can be measure from the horizontal but in orer to be conversant with remaining literature we chose the angle θ measure from y-axis. So for 2-DoF system, there are two generalize coorinates x 1 ¼ x, the cart s isplacement an θ the angle of a bob from vertical. The position of bob (x 2, y 2 ) can be given in terms of x 1 an θ as

30 2 Lagrangian Moeling Fig. 2.2 Inverte penulum on a moving cart 2 m 2 1 ( ) M x 2 ¼ x þ l sin θ y 2 ¼ l cos θ ð2:64þ The velocity components of bob are given as _x 2 ¼ _x þ l _θ cos θ _y 2 ¼l _θ sin θ ð2:65þ The kinetic energies of both masses are given as K 1 ¼ 1 2 M _x 2 K 2 ¼ 1 2 m _x 2 2 2 þ _y 2 K 2 ¼ 1 h 2 m _x þ l 2 2 i _θ cos θ þ l _θ sin θ K 2 ¼ 1 2 m _x 2 þ 2l _x θ _ cos θ þ l 2 θ _ 2 ð2:66þ ð2:67aþ ð2:67bþ ð2:67cþ So the total kinetic energy of the system from Eqs. (2.66) an (2.67c) is given as K ¼ K 1 þ K 2 ¼ 1 2 ðm þ mþ_x 2 þ ml _x _θ cos θ þ 1 2 ml2 _θ 2 ð2:68þ The potential energy in a bob is given as U ¼ mgy 2 ¼ mgl cos θ ð2:69þ

2.4 State Space Representation 31 The Lagrangian L is given as L ¼ K U ¼ 1 2 ðm þ mþ_x 2 þ ml _x θ _ cos θ þ 1 2 ml2 θ _ 2 mgl cos θ ð2:70þ we represent two equations of motion sing the variable L an two generalize coorinates (x, θ) as _x x ¼ F ð2:71þ an _θ θ ¼ 0 ð2:72þ From Eqs. (2.70), (2.71), an (2.72) are solve as So equations of motion are _x ¼ þ mþ_x θ _ cos θ ð2:73aþ _θ ¼ ml _x cos θ þ ml2 _θ ð2:73bþ x ¼ 0 ¼ mgl sin θ θ ð2:74þ ð2:73cþ ðm þ mþ x þ ml θ cos θ ml _θ 2 sin θ ¼ F ð2:75þ ml x cos θ ml _x _θ sin θ þ ml 2 θ mgl sin θ ¼ 0 ð2:76þ These are nonlinear equations for the system of a moving cart with an inverte penulum. In orer to make a state space system, first we nee to solve variables with ouble erivatives simultaneously. Each generalize coorinate of the system is secon-orer so a total orer of the system is 4. We have both x an θ, so we nee to solve these equations simultaneously. Eqs. (2.75) an (2.76) are represente as ml 2 l θ ¼ F ðm þ mþ x þ ml θ cos θ _ 2 sin θ ð2:77þ x ¼ _x _θ tan θ l θ cos θ þ g tan θ ð2:78þ

32 2 Lagrangian Moeling Inserting value of x from Eq. (2.78) ineq.(2.75) an then inserting the value of ml 2 θ from Eq. (2.77) in Eq. (2.76) simultaneously solve the equations for state space T representation of a state vector x θ _x _θ as x ¼ m cos θ M þ m 1 m _x _θ sin θ þ mg sin θ þ ml _θ 2 tan θ þ F cos θ cos θ θ ¼ 1 l M cos θ Problems m cos θ þ m cos θ ð2:79þ 1 ðm þ mþ _x θ _ tan θ þ g tan θ þ ml θ _ 2 sin θ þ F P2.1 Consier spherical coorinates of mass m in motion x ¼ r cos θ sin φ y ¼ r sin θ sin φ z ¼ r cos φ ð2:80þ Formulate Lagrangian equation of motion in three axis for three generalize coorinates an represent as state space moel. P2.2 A penulum given in Example 2.1 is rotating with angular velocity ω in a circular path of raius r. The motion is uner the influence of gravity with no external forces acting upon it. The Cartesian coorinates are given as xt ðþ¼rcos ðωtþþl cos θ yt ðþ¼rsin ðωtþþ l sin θ Fin the state space representation of a system through Lagrangian formulation. P2.3 Define a generalize coorinate system for the problem given in Example 2.2 an represent its nonlinear state space formulation. Use the symbolic math toolbox to obtain Jacobian matrices A an B given in Eq. (1.19) an Eq. (1.20) by linearizing at relaxe position. P2.4 Nonlinear systems are also linearize by approximating nonlinear terms within a specifie range on equilibrium point. The inverte penulum on a cart can be linearize by approximating a small angle an negligible velocities at equilibrium point for trigonometric functions i.e., sin θ θ, cos θ 1. Fin a linear state space equation by these assumptions an obtain a state space representation of the inverte penulum on a cart system.

References 33 P2.5 The position of bob of the inverte penulum on a moving cart can be controlle by applying external torque τ at the hinge changing the Eq. (2.78) as ml x cos θ ml _x _θ sin θ þ ml 2 θ mgl sin θ ¼ τ Linearize the complete system by the assumption given in P2.4 an obtain state space representation to monitor all state variables at output inepenently. References 1. Frielan, Bernar. 2005. Control System Design An Introuction to State Space Methos. Mineola: Dover Publications. 2. Zak, Stanislaw H. 2002. Systems an Control. Oxfor: Oxfor University Press. 3. Brizar, Alain J. 2008. An Introuction to Lagrangian Mechanics. Singapore: Worl Scientific Publishing. 4. Wellstea, P.E. 1979. Introuction to Physical System Moelling. Lonon: Acaemic Press. 5. Amirouche, Fari. 2006. Funementals of Multiboy Dynamics Theory an Applications. Basel: Birkhäuser.

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