Derivation Of Lagrange's Equation Of Motion For Nonholonomic Constraints Using Lagrange s Multiplier

Transcription

1 Derivation Of Lagrange's Equation Of Motion For Nonholonomic Constraints Using Lagrange s Multiplier We consier a path ABC in x, y, x., y.,t space which a system traverse from time to time. Now we consier another ajacent path that iffers from the ABC path by a arbitrary function δx. Now we write the ifference of the action integral ( L ) of the two path as δs.we then try to make δs zero (so that s is extremum) to get the correct path the system will traverse consiering only x an y coorinates t2 δs = δ L = 0 L = L(x, y, x., y., t) t2 δs = δ L = Now the secon term δx +. δx. + y δy + y. δy.. δx. =. δx =. δx. δx =. δx this term is zero as at the two en points t = an t =, δx = 0 same as that

2 y. δy. =. δy y So δs =. δx + y. δy = 0 (i) y Say x an y are inepenent. Now δx is arbitrary. So even when we change the function δx, the value of the integral remains unchange, it still remains zero. So, we can say. δx = 0 Then from eq (i) we can say y y. δy = 0 But if x an y are epenent on each other by some constraint, even if we change δx (as δx is arbitrary), δy will also change with it (an the integral will remain zero). So we can only say for the arbitrariness of δx Now our constraint equation is. δx + y A 1 x + A 2 y + A = 0 Here A 1, A 2, A can be a function of x, y, x., y.,t For instantaneous constraints = 0, so. δy = 0.. (ii) y [which is the same as saying A 1 x. + A 2 y. + A = 0 ] A 1 δx + A 2 δy = 0 Or, λ 1 A 1 δx + λ 1 A 2 δy = 0... (iii) Where λ 1 is a constant, its calle lagrange's multiplier From eq (ii) an (iii) we can write. λ 1A 1 δx + y. y λ 1A 2 δy = 0... (iv) We can choose the value of λ 1 such that

3 . λ 1A 1 = 0 Then from eq (iv) y. y λ 1A 2 = 0 So the equations of motion are. = + λ 1A 1 (v) along with the constraint equation y. = y + λ 1A 2 (vi) A 1 x + A 2 y + A = 0 Now if there were no constraints the terms λ 1 A 1 an λ 1 A 2 in equation (v) an (vi) an woul not be present. So the terms an must represent the constraint forces (in x an y irection respectively) Further if a constraint is f x, y = c (say a particle moves on the surface f x, y = c ) then we can write f x + f y y = 0 A 1 A 2 Then the equations of motion woul be [from eq (v) an (vi)]. = + λ f 1 y. = y + λ f 1 y Then the constraint force is λ 1 f x^ + λ 1 f y y^ = λ 1 f x, y an f x, y is perpenicular to f x, y = c. So, the constraint force term is perpenicular to the surface which again proves λ 1 A 1 an λ 1 A 2 are the constraint forces.

4 Example 1. Atwoo's machine Here the string (l=length of string) is inextensible, so y 1 + ΠR + y 2 = l The lagrangian is Constraint equation y. 1 + y. 2 = 0 L = 1 2 m 1y m 2y m1 gy 1 + m 2 gy 2 1 y y. 2 = 0 [this is a holonomic constraint] A 1 A 2 y. 1 = y 1 + λa 1 m 1 y 1 = m 1 g + λ (vii) this is the constraint force (or the tension) on mass m 1

5 y. 2 = y 2 + λa 2 m 2 y 2 = m 2 g + λ (viii) this is the constraint force (or the tension) on mass m 2 As y. 1 + y. 2 = 0 y 1 = y 2 Putting this in eq (viii) m 2 y 1 = m 2g + λ ix From (vii) an (ix) m 1 + m 2 y 1 = m 1 m 2 g y 1 = m 1 m 2 g m 1 + m 2 Which is the acceleration of the mass m 1 Again from eq (vii) an (ix) g + λ m 1 = g λ m 2 2g = λ 1 m m 2 λ = 2gm 1m 2 m 1 + m 2 This λ is the tension of the string! For the irection of the constraint force put this in eq (vii) m 2 y 2 = m 2 g + λ m 2 y 2 = m 2 g 2gm 1m 2 m 1 + m 2 So,the tension works in opposite irection of weight mg

6 Example 2 A coin or isc is rolling freely on a planar surface. For simplicity we restrict the isc to a vertical position. So, it oes not tilt. ABCD is the path the coin traverses when it moves freely. At D point its velocity points in DE irection. If the coin rolls without slipping then the velocity is Rθ.. If the velocity makes φ irection with x axis the x component of velocity is Rθ. cos φ an y componen of velocity is Rθ. sin φ So the constraint equations are x. = Rθ. cos φ xa y. = Rθ. sin φ xb [these two constraint are non holonomic as they are not integrable as φ can change] 1x. + 0y. + Rcosφ θ. + 0φ. = 0 (for this,multiplier is λ A ) A 1 A 2 A 3 A 4 0x. + 1y. + Rsinφ θ. + 0φ. = 0 (for this,multiplier is λ B ) B 1 B 2 B 3 B 4

7 Lagrangian is L = 1 2 m x. 2 + y Iθ Jφ. 2 moment of inertia with respect to AB axis is J moment of inertia with respect to CD axis is I. = + λ AA 1 + λ B B 1 mx = λ A (xia) y. = y + λ AA 2 + λ B B 2 my = λ B (xib) θ. = θ + λ AA 3 + λ B B 3 Iθ = λ A Rcosφ + λ B Rsinφ (xii) φ. = φ + λ AA 4 + λ B B 4 Jφ = 0 (xiii) From eq (xiii) φ = 0, so φ. = ω,an φ = ωt where ω is constant From eq (xii) Iθ = mx Rcosφ + my Rsinφ as mx = λ A an my = λ B

8 = m Rθ. cos φ Rcos φ m Rθ. sin φ Rsinφ = mr 2 θ sin 2 φ + cos 2 φ = mr 2 θ So, I + mr 2 θ = 0,or θ = 0,or θ. = Ω where Ω is constant So, from constraint equations (xa) an (xb) x = Rθ. cos φ t = RΩcosωt = RΩ ω 0 sin ωt t y = Rθ. sin φ = RΩsinωt = RΩ ω cos ωt + RΩ ω x 2 + y RΩ ω = RΩ ω So, the isc will move in a circle.