# Lecture L20 - Energy Methods: Lagrange s Equations

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1 S. Widnall 6.07 Dynamics Fall 009 Version 3.0 Lecture L0 - Energy Methods: Lagrange s Equations The motion of particles and rigid bodies is governed by ewton s law. In this section, we will derive an alternate approach, placing ewton s law into a form particularly convenient for multiple degree of freedom systems or systems in complex coordinate systems. This approach results in a set of equations called Lagrange s equations. They are the beginning of a complex, more mathematical approach to mechanics called analytical dynamics. In this course we will only deal with this method at an elementary level. Even at this simplified level, it is clear that considerable simplification occurs in deriving the equations of motion for complex systems. These two approaches ewton s Law and Lagrange s Equations are totally compatible. o new physical laws result for one approach vs. the other. Many have argued that Lagrange s Equations, based upon conservation of energy, are a more fundamental statement of the laws governing the motion of particles and rigid bodies. We shall not enter into this debate. Derivation of Lagrange s Equations in Cartesian Coordinates We begin by considering the conservation equations for a large number () of particles in a conservative force field using cartesian coordinates of position x i. For this system, we write the total kinetic energy as M T = m i ẋ i. () n= where M is the number of degrees of freedom of the system. For particles traveling only in one direction, only one x i is required to define the position of each particle, so that the number of degrees of freedom M =. For particles traveling in three dimensions, each particle requires 3 x i coordinates, so that M = 3. The momentum of a given particle in a given direction can be obtained by differentiating this expression with respect to the appropriate x i coordinate. This gives the momentum p i for this particular particle in this coordinate direction. The time derivative of the momentum is T = pi () ẋ i d T = m i ẍ i (3) dt ẋ i

2 For a conservative force field, the force on a particle is given by the derivative of the potential at the particle position in the desired direction. From ewton s law we have V F i = xi (4) F i = dp i dt Equating like terms from our manipulations on kinetic energy and the potential of a conservative force field, we write d T V dt ẋ i = xi (6) ow we make use of the fact that and Using these results, we can rewrite Equation (6) as d (T V ) (T V ) = 0 (9) dt x i x (5) T = 0 (7) x i V = 0 (8) ẋ i We now define L = T V : L is called the Lagrangian. Equation (9) takes the final form: Lagrange s equations in cartesian coordinates. dt ẋ i xi = 0 (0) where i is taken over all of the degrees of freedom of the system. Before moving on to more general coordinate systems, we will look at the application of Equation(0) to some simple systems. i Mass-spring System We first consider a simple mass spring system. This is a one degree of freedom system, with one x i. Its kinetic energy is T = /mẋ ; its potential is V = /kx ; its Lagrangian is L = /mẋ /kx. Applying Equation (0) to the Lagrangian of this simple system, we obtain the familiar differential equation for the mass-spring oscillator. d x m + kx = 0 () dt Clearly, we would not go through a process of such complexity to derive this simple equation. However, let s consider a more complex system, governed by the same laws.

3 This is a degree of freedom system, governed by differential equations. The number of springs for this configuration is 3. These governing equations could be obtained by applying ewton s Law to the force balance that exists at each mass due to the deflection of the springs as was done in Lecture 9. The deflection of springs and 3 are influenced by the boundary condition at either end of the slot; in this case the deflection is zero. The governing equations can also be obtained by direct application of Lagrange s Equation. This approach is quite straightforward. The expression for kinetic energy is the expression for the potential is T = m i ẋ i ; () n= Applying Lagrange s equation to T V = L d L dt ẋ d L dt ẋ we obtain the governing equations as V = /k x + /k (x x ) + /k 3 x (3) L x = 0 (4) L x = 0 (5) d x m = k x + k (x x ) (6) dt m d x = k (x x ) k 3 x (7) dt Clearly, for multi degree of freedom systems, this approach has advantages over the force balancing approach using ewton s law. Extension to General Coordinate Systems A significant advantage of the Lagrangian approach to developing equations of motion for complex systems comes as we leave the cartesian x i coordinate system and move into a general coordinate system. An 3

4 example would be polar coordinates where for a two-dimensional position of a mass particle, x and x could be given by r and θ. A two-degree of freedom system remains two-degree so that the number of coordinate variables required remains two. r and θ and their counterparts in other coordinate systems will be referred to as generalized coordinates. We introduce quite general notation for the relationship between the n cartesian variables of position x i and their description in generalized coordinates. (For some systems, the number of generalized coordinates is larger than the number of degrees of freedom and this is accounted for by introducing constraints on the system. This is an important part of the discussion of the Lagrange formulation. We shall not however develop these relations but will work directly with the number of variables equal to the number of degrees of freedom of the system.) We express the cartesian variable x i using generalized coordinates q j. (Polar coordinates r, θ would be an example.) In the general case, each x i could be dependent upon every q j. x i = x i (q,..q j,...q n ). (8) What is remarkable about the Lagrange formulation, is that (0) holds in a general coordinate system with x i replaced by q i. = 0 (9) dt q i qi Before showing how this result can be derived from ewton s Law, we show two applications in polar coordinates to demonstrate the power of the approach. Simple Pendulum by Lagrange s Equations We first apply Lagrange s equation to derive the equations of motion of a simple pendulum in polar coordinates. This is a one degree of freedom system. However, it is convenient for later analysis of the double pendulum, to begin by describing the position of the mass point m with cartesian coordinates x and y and then express the Lagrangian in the polar angle θ. Referring to a) in the figure below we have x = h sin θ (0) y = h cos θ () so that the kinetic energy is The potential energy is The Lagrangian is T = m (ẋ + ẏ ) = m h θ () V = m gy = m gh cos θ (3) L = T V = m h θ + m gh cos θ (4) 4

5 Applying (0) with q = θ, we obtain the differential equation governing the motion. m h θ + m gh sin θ = 0 (5) Again, for such a simple system, we would typically not go through this formalism to obtain this result. However, this framework will enable us to derive the equations of motion for the more complex systems such as the double pendulum shown in b). Double Pendulum by Lagrange s Equations Consider the double pendulum shown in b) consisting of two rods of length h and h with mass points m and m hung from a pivot. This systems has two degrees of freedom: θ and θ. To apply Lagrange s equations, we determine expressions for the kinetic energy and the potential as the system moves in angular displacement through the independent angles θ and θ. From the geometry we have x = h sin θ y = h cos θ x = h sin θ + h sin θ y = h cos θ h cos θ (6) (7) (8) (9) The kinetic energy is T = m ẋ + ẏ + m ẋ + ẏ. Expressed in variables θ and θ, the kinetic energy of the system is T = m h θ + m h θ + h θ + h h θ θ cos(θ θ ) (30) 5

6 The potential energy of the system is V = m gy + m gy = (m + m )gh cos θ m gh cos θ. (3) The Lagrangian is then L = T V = (m +m )h θ + m h θ +m h h θ θ cos(θ θ )+(m +m )gh cos θ +m gh cos θ (3) Since the generalized coordinates are now θ and θ, Lagrange s equation becomes = 0 (33) dt θ i θi for both i = and i =. = 0 (34) dt θ θ = 0 (35) dt θ θ Working out the details, we have d L = (m + m )h dt θ θ + m h h θ cos(θ θ ) m h h θ sin(θ θ )(θ θ ) (36) L = h g(m + m ) sin(θ ) m h h θ θ sin(θ θ ) (37) θ d L = m h dt θ θ + m h h θ cos(θ θ ) m h h θ sin(θ θ )(θ θ ) (38) L = h gm sin(θ ) + m h h θ θ sin(θ θ ) (39) θ From Equation(35-39), we obtain the final form of the governing equations for the double pendulum (m + m )h θ + m h θ cos(θ θ ) + m h θ sin(θ θ ) + g(m + m ) sin θ = 0 (40) m h θ + m h θ cos(θ θ ) m h θ sin(θ θ ) + m g sin θ = 0 (4) The double pendulum is a system of great interest, displaying conventional linear multi degree of freedom system behavior for small θ and θ, but displaying chaotic behavior for large θ. A chaotic system is a deterministic system that exhibits great sensitivity to the initial conditions: the butterfly effect. A simulation of the motion of a double pendulum is available on For a particular choice of initial conditions, the position of m with time is shown in the figure. 6

7 Derivation of Lagrange s Equation for General Coordinate Systems We now follow the earlier procedure we used to derive Lagrange s equation from ewton s law but using generalized coordinates instead of cartesian coordinates. (See additional reading: Slater and Frank and/or Marion and Thorton.) The cartesian variable are expressed in terms of the generalized coordinates as x i = x i (q,..q j,...q n ). (4) To obtain the velocity ẋ i, we take the time derivative of (4) applying the chain rule. n xi ẋ i = q j. (43) q j j= Taking the partial derivative of (43) with respect to q j we obtain a relation between these two derivatives, ẋ i q j x i =, (44) q j a result which we shall need shortly. In deriving equation (44), we take advantage of the fact that since the generalized coordinates q i are independent, qi = 0 for i j. q j Following the approach leading to Equations (-0), we define a generalized momentum as T p i = (45) q i M with T = m j ẋ j as given by Equation (). n= We are now ready to express ewton s law in the generalized coordinates q i that we have introduced. 7

8 For the generalized momentum we have where we have made use of (44). T n p i = = q i j= m j ẋ j ẋ j q i = For a conservative force field, the work done during a displacement dx i is given by n j= m j ẋ j x j q i, (46) or expressed in the generalized coordinates q j dw = F i dx i (47) i= x i dw = F i dx i = F i dq j (48) q i= j= i= j We identify x i F i dq j (49) q i= j as the work done by a displacement through dq j and define a generalized force Q j as so that the work is expressed as. For a conservative system, the work done by a small displacement dq j is x i Q j = F i (50) q i= j dw = Q j dq j (5) j= dw = dv = V dq j, (5) q j where V is the potential function expressed in the coordinate system of the generalized coordinates and V q j is the change in potential due to a change in the generalized coordinate q j. So that the generalized force is V Q j = qj (53) in analogy with (4). We now examine the time derivative of the generalized momentum, p i. dp i = d ( T ) = (m j ẍ x j d x j j + m j ẋ j ). (54) dt dt q i q i dt q j= i Since x j / q i is a function of the q s which are functions of time, we have by the chain rule From ewton s law, m j ẍ j = F j, so that the first term in (54) is given by d x j x j = q k (55) dt q i q i q k k= x j m j ẍ j = Q i (56) q j= i 8

9 ow let us consider the second term of (54). Consider the expression for T/ q i and use (43, 44). T = mj ẋ j ẋ j = m j ẋ j x i q k (57) q i q i q i q k k= This is precisely the second term in equation (54). We have now identified simpler forms for the two expressions in the equation for the time derivative of the generalized momentum. We may now write dp i d T T V T dt = = Q i + = + (58) dt q i qi qi qi Since for a conservative system, the potential is independent of the velocities, we can place this equation into final form by defining the Lagrangian as L = T V to obtain the final form of Lagrange s equation as = 0 (59) dt q i qi in agreement with (9). Equation (59) is Lagrange s Equations in generalized coordinates. In our previous example, we applied this equation to simple and double pendulums in polar coordinates using q = θ and q = θ. What is significant about this equation, adding to its power, is that each i equation contains only derivatives with respect to that q i and q i. ADDITIOAL READIG Slater and Frank, Mechanics, Chapter IV Marion and Thorton, Chapter 7 9

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