Classical Mechanics Examples (Lagrange Multipliers) Dipan Kumar Ghosh Physics Department, Inian Institute of Technology Bombay Powai, Mumbai 400076 September 3, 015 1 Introuction We have seen that the effect of holonomic constraints is to reuce the number of egrees of freeom. If there are N particles, we expect 3N egrees of freeom if the motion is unconstraine. If there are k holonomic constraints, the number of egrees of freeom becomes 3N k an we nee this many inepenent generalise coorinates to escribe the system. For non-holonomic constraints, however, the number of generalise coorinates require is more than the number of egrees of freeom. If there are m non-holonomic constraints, in aition to the k holonomic constraints, the number of generalise coronates is still 3N k, i.e. m more than the number of egrees of freeom. These aitional quantities are eliminate from the equations of motion by metho of Lagrange Multipliers. 1.1 A mathematical Example: Fin extremal value of f(x, y) = xy subject to the constraint h(x, y) = x 8 + y 1 = 0 This is an equation to al ellipse with major axis an minor axis. The level curves for f(x, y) = xy are hyperbola. Suppose the level curve for f(x, y) intersects the constraint curve h(x, y) at P. If we move along the constraint curve to its right, the value of xy increases whereas if we move to the left its value ecreases. Thus the point P cannot be a point of minimum or a maximum. Clearly, when the point is an extremum, the level curve of f an the constraint curve touch each other an they have a common tangent. Hence at the common point their normals are parallel (or antiparallel) to each other. Thus the set of points S which satisfies h(x, y) = 0 an f + λ h = 0 for some λ contains the extremal value of f, subject to the given constraint.
c D. K. Ghosh, IIT Bombay We efine Setting graient of F equal to zero implies f x + λ h ˆx + x which gives us F (x, y, λ) = f(x, y) + λh(x, y) ( f y + λ h y f x + λ h x = 0 f y + λ h y = 0 ) ŷ + hλ = 0 For the present problem, we have, x F (x, y, h) = xy λ 8 + y 1 In this case from (1) we get h = 0 (1) y λ 4 x = 0 x λy = 0 x 8 + y 1 = 0
c D. K. Ghosh, IIT Bombay 3 which gives λ = ±. Substituting the possible values of λ into the above equations, we fin two maximum values of f = xy = at (1, ), ( 1, ) an two minimum values xy = at (1, ), ( 1, ). Euler Lagrange Equation with constraint The metho of Lagrange multiplier is use in classical mechanics in hanling situations where the number of ynamical variables happens to be more than the number of egrees of freeom. We ha seen that the principle of least action is expresse in the form δ L = 0 from which we ha obtaine the Euler Lagrange equation as j [ L ] δq j = 0 Using the fact that the generalise coorinates are inepenent, we erive the Euler Lagrange equation for each pair pf generalise coorinate an velocity L = 0 () In the presence of non-holonomic constraints, the generalise coorinates are not inepenent as their number is greater than the number of egrees of freeom an consequently () is not vali. In such situations (an also in case care is not taken to reuce the number of generalise coorinates using the holonomic constraints), the metho of unetermine multiplier is useful. (The metho has limitations, for instance, it cannot be use in cases where the constraints are state as inequalities) Suppose there are m number of non-holonomic constraints involving the generalise coorinates in ifferential form n a r,j q j + b r,j = 0 (3) j=1 a r,j an b r,j may epen on j an t. Here r is an inex which runs from 1 to m an (3) is actually m equations, one for each value of r. We can get correct equations motion if the varie paths are virtual isplacements from actual motion in which case the constraint is n j=1 a r,jq = 0 because vital isplacements take place over constant time. We can then rewrite the principle of least action as n j j=1 [ L + ] λ r a r,j δq j = 0 Note that the aitional term n j=1 m λ ra r,j δq j is actually zero an hence we can put it insie the integral. δq j s are not inepenent an satisfy (3). Since we have m
c D. K. Ghosh, IIT Bombay 4 unetermine multipliers λ r, we can choose them such that the first m terms, i.e. j = 1 to m is each zero. Suppose we choose λ j such that for j = 1,,... m, the equation to be satisfie is L + λ r a r,j = 0 (4) Note that the last term in the above equation is no longer zero as the sum over j is missing an we have simply reistribute the term which was zero in various ways. With the λ r etermine by (4), we are left with n j=m+1 [ L + ] λ r a r,j δq j = 0 However, now our q j are inepenent an we have, as a consequence, for j = m + 1,..., n L + (4) an (5) allows us to write a single equation for j = 1,,... n, L + λ r a r,j = 0 (5) λ r a r,j = 0 (6) (4) etermines the m values of λ an (5) gives us n equations of motion. Define Q j = λ r a r,j (7) Equation (5) can now be written as for j = 1,,... n. L = Q j (8) The right han sie of the above is seen to be the generalise force corresponing to the constraint conitions..1 Example : A sliing mass on a paraboloi of revolution Consier a particle of mass m moving frictionlessly on a paraboloi of revolution given by x + y = az.
c D. K. Ghosh, IIT Bombay 5 z φ ^z ^φ m ρ^ x The Lagrangian of the system is L = T V = 1 m[ ρ + ρ ϕ + ż ] + mgz (9) where ρ = x + y is the particle from the z axis. The system is holonomic with two egrees of freeom since the constraint ρ = az can be use to eliminate one of the coorinates. Let us write the constraint as ρδρ aδz = 0 (10) Comparing (10) with (3) we ientify a ρ = ρ, a ϕ = 0 an a z = a. As there is only one constraint equation, there is only one Lagrange multiplier λ an the Euler Lagrange equation is L = Q j = λa j The Euler Lagrange equations for the three coorinates are L ρ ρ = λ ρ L ϕ ϕ = 0 L ż z = λ a These equations simplify to
c D. K. Ghosh, IIT Bombay 6 m( ρ ρ ϕ ) = λρ m (ρ ϕ) = 0 = mρ ω 0 = L m z = mg λa We can solve these equations to etermine ρ(t), z(t), ϕ(t) an λ. Consier the case where the mass goes aroun in a horizontal circe at a height h from the bottom. z = h. This also implies ρ = constant = ha. We then have (from the last equation) λ = mg a. ϕ = constant = ω 0 = g (from the first equation, using ρ = 0) a λ = mg a = m ω 0 The force which is responsible for the circular motion is the component N sin ψ of the normal force. tan ψ = z ρ = ρ The component of the normal force towars the centre of a the circle is ρ N sin ψ = N 4ρ + a mρω 0 The vertical component of N must be equal to mg a N cos ψ = N 4ρ + a = mg Thus which is what we expecte. λ = mg a = N 4ρ + a = 1 mω 0. A hoop rolling own the top of a cyliner without slipping Consier the case of a hoop rolling own the top of a cyliner without slipping. The constraint is holonomic till the mass slies off. Let r be the istance from O to C. θ is the polar coorinate of O an ϕ is the angle of rotation of the hoop about its own axis. R is the raius of the cyliner. O θ R C V=0
c D. K. Ghosh, IIT Bombay 7 The Lagrangian is L = 1 m(ṙ + r θ ) + 1 ma ϕ mgr cos θ Constraints are : (i) r = R + a (ii) (R + a) θ = a ϕ The first is a holonomic constraint : f 1 = r (R + a) an the secon is a non-holonomic (though integrable) constraint f = (R + a) θ + a ϕ = 0. With the former we associate λ (associate with normal reaction) while with the latter we associate µ (associate with the tangential force require for rolling motion). The Euler Lagrange equations (for r, θ an ϕ) are L The equations give = λ f 1 + µ f m r mr θ + mg cos θ = λ mr θ + mrṙ θ mgr sin θ = µ(r + a) ma ϕ = µa Hence µ = ma ϕ = m(r + a) θ Substitute this in the secon equation an use ṙ = 0. θ = g sin θ (R + a) We multiply the above equation by θ an integrate θ = Use this in the raial equation (put ṙ = 0) g (1 cos θ) R + a λ = mr θ + mg cos θ g = m(r + a) (1 cos θ) + mg cos θ R + a = mg( cos θ 1) For θ = 0, λ = mg, the normal force. For θ > 60, λ becomes negative an contact with the cyliner is lost..3 Example : A Rolling hoop on an incline Consier a hoop rolling own without slipping on an incline. The problem is one of holonomic constraint an has been solve earlier. However, we will use this problem to illustrate the metho of Lagrange multiplier. In this case there is just one egree of freeom an there is just one generalise coorinate, i.e. the angle θ by which the hoop rolls. The istance move along the incline is x = Rθ.
c D. K. Ghosh, IIT Bombay 8 x R θ α Let us, however, use both x an θ are our generalise coorinates. We then have one holonomic constraint h(x, θ) = x Rθ (11) An observation may be mae here that (11) is not in the form (3). A holonomic constraint is expresse in the form f(q 1, q,..., q n, t) = 0 We can write it in a ifferential form n f δq k + f δt = 0 (1) q k t Comparing this with (3) we have k=1 a r,k = f q k b r,t = f t Thus in the present case, where we have only one constraint (11) (we rop the inex r) The Lagrangian is then given by b = 0 a x = (x Rθ) = 1 x a θ = (x Rθ) = R θ L = 1 mẋ + 1 mr θ mg(l x) sin α where l is the length of the incline. The Euler-Lagrange equations for the coorinate x an θ are given by L ẋ x λ 1 = 0 i.e. mẍ mg sin α λ = 0 (14) an L θ θ + λ R = 0 (13)
c D. K. Ghosh, IIT Bombay 9 i.e. Equations (14), (15) along with the equation to the constraint mr θ + λr = 0 (15) ẋ R θ = 0 (16) provie the necessary equations for solution of the problem. From (15) an (16) we get λ = mẍ. Substituting these in (14) we get ẍ = g sin α mg sin α λ = The force of constraint corresponing to the motion along x irection is given by λ h x = mg sin α mg sin α 1 =.4 Rolling with Spinning introucing a non-holonomic constraint Let us now allow the hoop to spin about the vertical axis, in aition to rolling. Let us take the irection own the incline as the x-irection an the irection perpenicular to it along the plane as the y- irection. (17) (18) R θ ϕ The kinetic energy of the hoop is given by α T = 1 mv + 1 mr θ + 1 mr = mr θ + 1 4 mr ϕ Here we have use the fact that the hoop spins about its iameter an the moment of inertia about the iameter is mr /. The Lagrangian is L = mr θ + 1 4 mr ϕ mg(l x) sin α ϕ
c D. K. Ghosh, IIT Bombay 10 Here y is cyclic an ẏ oes not enter into the Hamiltonian. Hence we ignore the equation of motion for y. Since the hoop rolls without sliing, the velocity of the rim is v = R θ. When the hoop has spanne by an angle ϕ, we have ẋ = R θ cos ϕ (19) ẏ = R θ sin ϕ (0) These are non-integrable constraints because these cannot be use to connect x with θ an ϕ. The problem has two egrees of freeom. Let us freeze the motion an make virtual isplacement δθ of the angle θ(a isplacement in tilt angle is of secon orer). We then have δx = R cos ϕδθ As before we can get a x = 1 an a θ = R cos ϕ. We can now write the Lagrange equations of motion an get the following equations: mg sin α λ = 0 (1) mr θ + λr cos ϕ = 0 () mr ϕ = 0 (3) These equations etermine the motion completely in terms of θ an ϕ. Note that x has been alienate from the equations of motion.