Lecture 13: Forces in the Lagrangian Approach • In regular Cartesian coordinates, the Lagrangian for a single particle is: 3 = − = 1 2 − ( ) L T U mƒ x"l U xi 2 l=1 • Given this, we can readily interpret the physical significance of the quantity ∂L : ∂ x"i ∂L = mx" = p ∂ i i It’s the momentum! x"i • Since there’s nothing special about rectangular Cartesian coordinates in Lagrangian mechanics, we can define a generalized momentum associated with any generalized coordinate: ∂L p ≡ i ∂ q"i • Keep in mind that, just as a generalized coordinate doesn’t have to have dimensions of length, a generalized momentum doesn’t have to have the usual units for momentum – If the generalized coordinate corresponds to an angle, for example, the generalized momentum associated with it will be an angular momentum • With this definition of generalized momentum, Lagrange’s Equation of Motion can be written as:
∂L d − p = 0 ∂ j q j dt Just like Newton’s ∂ Laws, if we call ∂L = L p" j ∂ ∂ a “generalized q j q j force” Lagrangian Mechanics with Constraints
• In claiming that we could write x and x" as functions of q and q " (and time) we were making an assumption that the constraints on the system could be expressed as relations among the coordinates – In other words, the constraint equations are of the form ( )= g xα ,i ,t 0 No dependence on velocity!
• Such constraints are called holonomic, and the Lagrange Equations of Motion we’ve written are only true for this type of constraint – Non-holonomic constraints can be dealt with, but that’s beyond the scope of this course Using Undetermined Multipliers • One great advantage of the Lagrangian method, as we’ve seen, is that it allows us to solve for the motion of particles under constraints, even if we don’t know the force causing the constraint • In some cases, though, we’d like to determine the forces of constraint – i.e., an engineer designing a mechanical device to provide a constraint needs to know how strong it must be • We can do this using the form of Lagrange’s equations with the constraints taken into account explicitly (i.e., with undetermined multipliers λ(t) ): ∂L d ∂L m ∂g − + λ (t) k = 0 ∂ ∂ ƒ k ∂ q j dt q" j k=1 q j • Note that there are 3n such equations for an n-particle system – The set of Lagrange equations in which the variables are already related to each other consists of 3n-m equations • In total, we have 3n+m equations – since the m equations of constraint are still available to us • This means we can solve for all the q’s and all the λ’s • But what do the λ’s represent, physically? – Recall that they’re called “undetermined multipliers” since we don’t know the value beforehand – But they have the values needed to make sure the constraint is obeyed – a hint that they are closely related to the forces of constraint! Generalized Force of Constraint • To see what the λ’s mean, imagine that whatever part of the system that causes the constraints is removed
• Now apply a set of external forces Qj to the system, such that the system moves as though it were constrained • Then the equation of motion would be: ∂L p" = + Q j ∂ j q j ∂L d ∂L − + Q = 0 ∂ ∂ j q j dt q" j • But we already have an equation of motion for the constrained system ∂L d ∂L m ∂g − + λ (t) k = 0 ∂ ∂ ƒ k ∂ q j dt q" j k=1 q j • For this to be be consistent ,we see that Qj must be given by: m ∂g Q = λ k j ƒ k ∂ k=1 q j
• So finding the λ’s is a key to determining the forces of constraint • Of course, when I say “force”, I really mean “generalized force” • As with all the “generalized” quantities, some care needs to be taken in the interpretation – i.e., the units won’t always be that for a force – for example, if the generalized coordinate represents an angle, the generalized force will be a torque Example: Tension in a pendulum’s string • Given the following simple pendulum, find the tension in the string:
O l I use the following generalized θ coordinates: θ : angle as shown in diagram m r: distance from O to m
• In this case, we have a simple equation of constraint: g = r − l = 0 • The kinetic and potential energies are: 1 2 1 T = m(rθ") + mr"2 ; U = −mgr cosθ 2 2 • With this, the equations of motion are: ∂L d ∂L ∂g − + λ (t) = 0 ∂r dt ∂r" ∂r ∂L d ∂L ∂g − + λ (t) = 0 ∂θ dt ∂θ" ∂θ • Using what we know about L and g these become: mrθ"2 + mg cosθ − m"r"+ λ (t)(1) = 0 −mgr sinθ − mr2θ""− 2mrr"θ" + λ (t)(0) = 0
• But we know from the equation of constraint that r = l and r" = "r" = 0so the first equation tells us that: λ (t) = −mg cosθ − mlθ"2 • And the tension is just the generalized force for r: ∂g T = Q = λ = −mg cosθ − mlθ"2 r ∂r