G j dq i + G j. q i. = a jt. and

Transcription

1 Lagrange Multipliers Wenesay, 8 September 011 Sometimes it is convenient to use reunant coorinates, an to effect the variation of the action consistent with the constraints via the metho of Lagrange unetermine multipliers. As a bonus, we obtain the generalize forces of constraint. Consier a mechanical system with N P egrees of freeom, which we escribe with N generalize coorinates an P equations of constraint. We will assume that issipation may be neglecte so that the system may be escribe by a Lagrangian. Let us suppose that the equations of constraint may be written in the form G j (q i, t) = 0 G j = i for j = 1,,..., P. For notational convenience 1, efine Physics 111 q i + t = 0 t = a ji an t = a jt Then the jth constraint equation may be written a ji q i + a jt t = 0 or a ji q i + a jt = 0 (P equations) Note that the coefficients a ji an a jt may be functions of the generalize coorinates q i an the time t, but not the generalize velocities q i. Hamilton s principle says that of all possible paths, the one the system follows is that which minimizes the action, which is the time integral of the Lagrangian: t b δs = δ L(q i, q i, t) t = 0 (1) t a Expaning the variation in the Lagrangian an integrating by parts, we obtain t a t b t ( L )] δq i t which must vanish on the minimum path. Remember that we are using the summation convention; we sum over i. When the coorinates q i forme a minimal complete set, we argue that the virtual isplacements δq i were arbitrary an that the only way to ensure that the action be minimum is for each term in the square brackets to vanish. 1 Translation: just to confuse you. Physics Peter N. Saeta

2 Now, however, an arbitrary variation in the coorinates will sen us off the constraint surface, leaing to an impossible solution. One way to trick Prof. Hamilton into fining the right solution (the solution consistent with the constraint equations) woul be to sneak something insie the brackets that woul make the variation in the action zero for isplacements that take us off the surface of constraint. That s harly cheating; it just takes away the incentive to cheat! If we manage to fin such terms, then it wouln t matter which way we varie the path; we get zero change in the action, to first orer in the variation. We coul then treat all the variations δq i as inepenent. Before running that little operation, consier what it might mean for those terms to represent the generalize forces of constraint. Since the allowe virtual isplacements of the system are all orthogonal to the constraint forces, those forces o no work an they on t change the value of the action integral. In fact, they are just the necessary forces to ensure that the motion follows the constraine path. So, if we can figure out what terms we nee to a to the Lagrangian to make the illegal variations vanish, we will have also foun the forces of constraint. Each of the constraint equations is of the form G j (q i, t) = 0, so if we were to a a multiple of each constraint equation to the Lagrangian, it woul leave the action unchange. So, we form the augmente Lagrangian: L = L + λ j G j where I m using the summation convention an the λ j are Lagrange s unetermine multiplier, one per equation of constraint; all may be functions of the time. Then the (augmente) action is t S = L t (but since we ae zero, it s the same as the un-augmente action). By Hamilton s principle, the action is minimize along the true path followe by the system. We may now effect the variation of the action an force it to vanish: t δq i + L δ q i + λ j δq i ] t = 0 Integrate the mile term by parts (an remember we re using the summation convention): t t ( L ) + λ j a ji ] δq i t = 0 We now have N variations δq i, only N P of which are inepenent (since the system has only N P egrees of freeom). Using the P inepenent Lagrange multipliers, we may ensure that all N terms insie the square bracket vanish, so that no matter what the variations δq i the value of the integral oesn t vary. Thus we have t ( L ) L = λ j a ji (N equations) a ji q i + a jt t = 0 or a ji q i + a jt = 0 (P equations) Peter N. Saeta Physics 111

3 Since there are N generalize coorinates an P Lagrange multipliers, we now have a close algebraic system. When we erive Lagrangian mechanics starting from Newton s laws, we showe that t ( T ) T = Fi tot = F tot where F is the total force on the particle an Fi tot is the generalize force corresponing to the ith generalize coorinate. If we separate the forces into those expressible in terms of a scalar potential epening only on positions (not velocities), the forces of constraint, an anything left over, then this becomes t ( L ) L = Fi constraint + Fi noncons () Comparing with the N Lagrange equations above, we see that when all the forces are conservative, Fi constraint = F constraint r = λ j a ji = λ j j where I have written the sum explicitly in the last expression, just to remin you. In other wors, the sum λ j a ji is the generalize constraint force. r Example 1 A hoop of mass m an raius R rolls without slipping own a plane incline at angle α with respect to the horizontal. Solve for the motion, as well as the generalize constraint forces. Using the inicate coorinate system, we have no motion in y, but coorinate motion between x an θ, which are linke by the constraint conition y θ x α R θ = x or R θ x = 0 Therefore, a 1θ = R, a 1x = 1, a 1t = 0 The kinetic energy is T = m ẋ + mr θ an the potential energy is V = mgx sin α, so the Lagrangian is L = T V = m ẋ + mr θ + mgx sin α Physics Peter N. Saeta

4 We will first solve by using the constraint equation to eliminate θ: R θ = ẋ, so L = m ẋ + m ẋ + mgx sin α = mẋ + mgx sin α This Lagrangian has a single generalize coorinate, x, an thus we obtain the equation of motion t ( L ẋ ) L x = 0 mẍ mg sin α = 0 ẍ = g sin α which is half as fast as it woul accelerate if it sli without friction. If we elay the gratification of inserting the constraint an instea use the Lagrangian with two generalize coorinates, we get t ( L ẋ ) L x = λ 1a 1x mẍ mg sin α = λ 1 ( 1) t ( L L ) θ θ = λ 1a 1θ mr θ 0 = λ1 R From the secon equation, we obtain λ 1 = mr θ = mẍ, where I have use the constraint equation R θ = ẍ in the last step. Substituting into the first equation, we again obtain ẍ = g/ sin α. What about the constraint forces? The generalize constraint force in x is F x = λ 1 a 1x = mẍ = mg sin α. This is the force heaing up the slope prouce by friction; it is responsible for the slowe motion of the center of mass. The generalize constraint force in θ is F θ = λ 1 R = mrẍ = mgr sin α. This is the torque about the center of mass of the hoop cause by the frictional force. Summary When you wish to use reunant coorinates, or when you wish to etermine forces of constraint using the Lagrangian approach, here s the recipe: 1. Write the equations of constraint, G j (q i, t) = 0, in the form a ji q i + a jt t = 0 where a ji =.. Write own the N Lagrange equations, t ( L ) L = λ j a ji (summation convention) where the λ j (t) are the Lagrange unetermine multipliers an F i generalize force of constraint in the q i irection. = λ j a ji is the Peter N. Saeta 4 Physics 111

5 3. Solve, using the N Lagrange equations an the P constraint equations. 4. Compute the generalize constraint forces, F i, if esire. Problem 1 Use the metho of Lagrange unetermine multipliers to calculate the generalize constraint forces on our venerable bea, which is force to move without friction on a hoop of raius R whose normal is horizontal an force to rotate at angular velocity ω about a vertical axis through its center. Interpret these generalize forces. What o they correspon to physically? Physics Peter N. Saeta