Chpter Constrints, Lgrnge s equtions
Section Constrints The position of the prticle or system follows certin rules due to constrints: Holonomic constrint: f (r. r,... r n, t) = 0 Constrints tht re not expressible s the bove re clled nonholonomic. Exmples: Rigid body: r,b = constnt Rolling without slipping: V CM = ωr CM prticle moving on circle. Generlized coordintes: q i with i = 3N C, where C is the number of constrints. r = r (q i ) Note tht q i, q i re independent vribles.. Principle of Virtul work System under equilibrium: The totl force on ech prticle f = 0 Virtul displcement δr : Arbitrry infinitesiml chnge in the position of the -th prticle keeping the constrints. This is clled virtul displcement. Therefore, the sum of virtul work is zero: f δr = 0 Note tht f = f,ext + f,int. We choose δr such tht f,int δr = 0, 5
then f,ext δr = 0. The internl forces between the molecules of the plnk does not do ny work under displcement δθ. The norml forces do no work. This is the principle of virtul work. Exmple: A plnk resting gins t wll. The bottom surfce is frictionl with the friction force = f. Work done by the frictionl force: W = fd x = f lδ(sin θ) = f l cos θδθ Work done by mg: y N W = mgδy = mg(l /)δ(cos θ) = mg(l /)sin θδθ Using principle of virtul work: W + W = 0. Therefore tn θ = f mg. θ. D Alembert s Principle N y mg For dynmics x f x Hence f = p Virtul displcement: δθ. (f p ) δr = 0. 6
Agin choose δr such tht the virtul work done by the internl forces is zero. Hence (f,ext p ) δr = 0 [.] m v d T,i,i δr,i = j dt [ q j ] T where δq j Now some lgebr: [.],i f,ext,i δr,i =,i, j U r,ext,i r,i δq j = j,i m v,i δr,i =,i, j m v,i r,i, j δq j U δq j T = mv,i is the kinetic energy of the system. The displcement δqj is rbitrry. Therefore, using Eqs. [.,.] we obtin d T dt [ ] T = U = j m d dt,i v,i r,i d r,i m v,i dt [ ] δq j Typiclly, U/ q j = 0. Then d dt [ ] = 0 Note: r,i = dr,i dt = j r,i qj + r,i t where L=T-U is the Lgrngin of the system. Hence, r,i = r,i q j Advntges of the Lgrngin formlism No need to worry bout constrint forces, simpler d r,i dt [ ] = k r,i q k qk + r,i t = r,i Anlyticl, For exmple, Mécnique nlytique by Lgrnge does not hve single figure. substitution of which in the bove yields 7
Exmples: () free prticle () Verify D Alembert s principle for block of mss M sliding down wedge with n ngle of θ. () prticle in D () Consider the plnk discussed before. Let us ssume the ground surfce to be frictionless. Generlized coordinte = θ The KE = T = m( x + y ) + ml θ = 6 ml θ The potentil energy U = mgy = mgl sin θ The Lgrngin L = T U The eqution of motion yields θ = 3 sin θ (3) Construct Lgrngin for cylinder rolling down n incline. Exercises: () A prticle is sliding on uniformly rotting wire. Write down the Lgrngin of the prticle. Derive its eqution of motion. 8
Chpter 3 Principle of Lest Action
Section Vritionl Clculus Function of functions L = L(q, q, t) L is function of q(t), which itself is function of t. Objective: Extremize ction S = L(q, q, t)dt We will derive n eqution for the required function q(t) tht extremizes the ction. We will compute ction for nother function q(t, α) = q(t,0) + αη where αη is the devition from the required function. Here α is number nd η(q, q, t). The chnge in ction due to the bove is δs = [δl(q, q, t)]dt = [ q αη + q α η + HOT ] dt with the ends fixed t (, q ) nd (, q ). q q where HOT stnds for the higher order terms. For extremiztion, we tke the limit α 0 (ignore HOT). An integrtion by prts yields [ q α η dt = ] [ q αη ] [ d dt ( q )] αηdt t t The vrition of q t the ends must vnish, tht is η=0 t the ends. Hence, the boundry term vnishes. Therefore, 0
δs = [ Since η is rbitrry, d dt ( q ) = q q d dt ( q )] αηdt Note: The following Lgrngin L (q, q, t) = L(q, q, t) + d dt f (q, t) yields the sme eqution of motion. Proof (): S = L (q, q, t)dt = Hence, δs = δs. QED Proof (b): = S + f (q, ) f (q, ) L(q, q, t)dt + df dt dt Therefore, nd d dt ( q df dt ) = f q f + q q t df q ( dt ) = f q f + q q t Hence the dditionl terms cncel ech other. Q.E.D. NOTE: On mny occsions, the dependent vrible is x rther thn time. On those cses, we replce q by q. For Multi Vribles Here the generlized vribles re q i s. Hence L = L(q i, q i, t) For this cse, q i (t, α) = q i (t,0) + αη i. Hence Eq. () becomes d dt f = f q Hence q + f t df q ( dt ) = f q δs = [δl(q i, q i, t)]dt = αη t i { q i + i q α η i i } + HOT dt
δs = i d q i dt ( q i ) αη i dt Exmples: () Minimize the distnce between two points in 3D: Since it is vlid for rbitrry η i, we obtin d dt ( q i ) = q i Beltrmi Identity If L is not n explicit function of time t, then D = x: independent vrible y,z: Dependent vribles L = dy dz + + d x ( d x ) ( d x ) dy dz + + ( d x ) ( d x ) L i qi q i = const Since y = z = 0, Proof: LHS = dl dt i = t + i = 0 d dt ( q i q i ) qi + q i q q i q i i q q d i i dt q i Therefore, y = C nd z = C y + y + z = C nd z + y + z = C Hence, y + z = constnt. Therefore, y nd z re constnts. Hence, the prticle moves on stright line. Here we hve used the eqution of motion. () Minimize the time of descent between two points in grvittionl field:
y = Cx Cx. A y A substitution of Cx = sin θ yields y = tn θ. Therefore, d x dθ = sin θ C dy nd dθ cos θ =, C B whose prmetric solution with initil condition (x=0, y=0) is x = A( cos ϕ) nd y = A(ϕ sin ϕ) x where ϕ = θ. The bove is n eqution of cycloid. T = B A ds v = B A + y gx d x y Hence, the Lgrngin is x L = + y x Since / y = 0, / y = Hence y x( + y ) = C. C, constnt, which yields Note: We chose the verticl xis s x, so tht L is independent of y. It helps simplify the solution. If we interchnged the xes, the time will be B ds T = v = A B A + y gy d x. 3
Hence the Lgrngin will be + y L = y Lgrnge proposed n lterntive. He suggests tht the vribles x, y nd new vrible λ be mde independent. Ide is to look for contour of f (x, y) tht is tngent to the g(x, y) = c curve. See figure below. Using Beltrmi identity, we obtin L q q = y( + y ) = C Therefore, y = C y, y whose solution is sme s before except the chnge of xis. From Wikipedi The intersection point is the desired extremum point. Here Extremiztion under constrints Detour to Lgrnge multiplier We illustrte using n exmple. Suppose we wnt to Extremize f (x, y) under the constrint tht g(x, y) = c. The constrint would mke f (x, y) function of single vrible (sy x) tht cn be mximized using the stndrd method. However solving constrint eqution could be tricky. Also, this method is not convenient when we hve more constrints nd vribles. f (x, y) = λ g(x, y) nd g(x, y) = c. The bove equtions cn be derived by extremizing F(x, y, λ) = f (x, y) + λ[g(x, y) c] wrt x, y, λ tht yields f x = λg x f y = λg y 4
g(x, y) = c Exmple: () Find minimum of the function x + y under the constrint tht y x = 0. Solution: We minimize the function F = (x + y ) + λ(y x ) By tking derivtives wrt x, y, λ we obtin x λ = 0; y + λ = 0; y = x + whose solution is y = /,x = /,λ = /. Appliction to vrition clculus S = (x y y x)dt under the constrint tht I = ( x + y )dt Hence we extremize L = (x y y x) + λ x + y which yields y d dt ( y + λ x x + y ) = 0 Extremize functions under constrints vritionlly. We illustrte using n exmple. x d dt ( x + λ y x + y ) = 0 () Prmetric curve x = x(t), y = y(t) We return to the originl point, but with constnt perimeter. Tht is, x( ) = x( ) = x 0 nd y( ) = y( ) = y 0. We wnt to know function tht yields mximum re. tht yields y which yields λ x x + y = C ; x + (x C ) + (y C ) = λ λ y x + y = C 5
which is n eqution of circle. The prmeter λ is determined by the perimeter of the circle. b S[y] = (Py + Qy )d x Find the extrem of the bove subject to the condition tht Exercises: () A rope of liner density γ nd length L is hnging by two supports tht re locted horizontlly prt. Assuming equilibrium position for the rope, compute its eqution. () On sphere, the gret rc is defined s the curves tht minimizes the distnce trvelled between the given two points. Compute the eqution of gret rc. b y d x = The resulting eqution is clled Sturm-Liouville problem. Relte this eqution to the Schrodinger s eqution. (3) Anlyze the vritionl problems corresponding to the following functionls. In ech cse tke y(0) = 0 nd y() =. () y d x 0 (b) yy d x 0 (c) xyy d x 0 (4) Consider the functionl 6