Sect. 1.6: Simple Applications of the Lagrangian Formulation Lagrangian formulation: 2 scalar functions, T & V Newtonian formulation: MANY vector forces & accelerations. (Advantage of Lagrangian over Newtonian!) “Recipe” for application of the Lagrangian method: Choose appropriate generalized coordinates Write T & V in terms of these coordinates Form the Lagrangian L = T - V Apply: Lagrange’s Eqtns: (d/dt)[(L/qj)] - (L/qj) = 0 (j = 1,2,3, … n) Equivalently D’Alembert’s Principle: (d/dt)[T/qj] - (T/qj) = Qj (j = 1,2,3, … n)

  vi  (dri/dt) = ∑j(ri/qj)(dqj/dt) + (ri/t)  T  (½)∑imi(vi)2 Sometimes, T & V are easily obtained in generalized coordinates qj & velocities qj & sometimes not. If not, write in Cartesian coordinates & transform to generalized coordinates. Use transformation eqtns: ri = ri (q1,q2,q3,.,t) (i = 1,2,3,…n)   vi  (dri/dt) = ∑j(ri/qj)(dqj/dt) + (ri/t)  T  (½)∑imi(vi)2 = (½)∑imi[∑j(ri/qj)(dqj/dt) + (ri/t)]2 Squaring gives: T = M0 + ∑jMjqj + ∑jMjkqjqk M0  (½)∑imi(ri/t)2, Mj  ∑imi(ri/t)(ri/qj) Mjk  ∑imi(ri/qj)(ri/qk)

Always: T = M0 + ∑jMjqj + ∑jMjkqjqk Or: T0 + T1 + T2 T0  M0 independent of generalized velocities T1  ∑jMjqj linear in generalized velocities T2  ∑jMjkqjqk quadratic in generalized velocities NOTE: From previous eqtns, if (ri/t) = 0 (if transformation eqtns do not contain time explicitly), then T0 = T1 = 0  T = T2  If the transformation eqtns from Cartesian to generalized coords do not contain the time explicitly, the kinetic energy is a homogeneous, quadratic function of the generalized velocities.

Examples Simple examples (for some, the Lagrangian method is “overkill”): 1. A single particle in space (subject to force F): a. Cartesian coords b. Plane polar coords. 2. The Atwood’s machine 3. Time dependent constraint: A bead sliding on rotating wire

Particle in Space (Cartesian Coords) The Lagrangian method is “overkill” for this problem! Mass m, force F: Generalized coordinates qj are Cartesian coordinates x, y, z! q1 = x, etc. Generalized forces Qj are Cartesian components of force Q1 = Fx, etc. Kinetic energy: T = (½)m[(x)2 + (y)2 + (z)2] Lagrange eqtns which contain generalized forces (D’Alembert’s Principle): (d[T/qj]/dt) - (T/qj) = Qj (j = 1,2,3 or x,y,z)

(d[T/qj]/dt) - (T/qj) = Qj (j = 1,2,3 or x,y,z) T = (½)m[(x)2 + (y)2 + (z)2] (d[T/qj]/dt) - (T/qj) = Qj (j = 1,2,3 or x,y,z) (T/x) = (T/y) = (T/z) = 0 (T/x) = mx, (T/y) = my, (T/z) = mz  d(mx)/dt = mx = Fx ; d(my)/dt = my = Fy d(mz)/dt = mz = Fz Identical results (of course!) to Newton’s 2nd Law.

Particle in Plane (Plane Polar Coords) Plane Polar Coordinates: q1 = r, q2 = θ Transformation eqtns: x = r cosθ, y = r sinθ  x = r cosθ – rθ sinθ y = r sinθ + rθ cosθ  Kinetic energy: T = (½)m[(x)2 + (y)2] = (½)m[(r)2 + (rθ)2] Lagrange: (d[T/qj]/dt) - (T/qj) = Qj (j = 1,2 or r, θ) Generalized forces: Qj  ∑iFi(ri/qj)  Q1 = Qr = F(r/r) = Fr = Fr Q2 = Qθ = F(r/θ) = Frθ = rFθ

 mr - mr(θ)2 = Fr (1)  mr2θ + 2mrrθ = rFθ (2) T = (½)m[(r)2 + (rθ)2] Forces: Qr = Fr , Qθ = rFθ Lagrange: (d[T/qj]/dt) - (T/qj) = Qj (j = r, θ) Physical interpretation: Qr = Fr = radial force component. Qr = Fr = radial component of force. Qθ = rFθ = torque about axis  plane through origin r: (T/r) = mr(θ)2; (T/r) = mr; (d[T/r]/dt) = mr  mr - mr(θ)2 = Fr (1) Physical interpretation: - mr(θ)2 = centripetal force θ: (T/θ) = 0; (T/θ) = mr2θ; (Note: L = mr2θ) (d[T/θ]/dt) = mr2θ + 2mrrθ = (dL/dt) = N  mr2θ + 2mrrθ = rFθ (2) Physical interpretation: mr2θ = L = angular momentum about axis through origin  (2)  (dL/dt) = N = rFθ

Atwood’s Machine PE: V = -M1gx - M2g( - x) KE: T = (½)(M1 + M2)(x)2 M1 & M2 connected over a massless, frictionless pulley by a massless, extensionless string, length . Gravity acts, of course!  Conservative system, holonomic, scleronomous constraints 1 indep. coord. (1 deg. of freedom). Position x of M1. Constraint keeps const. length . PE: V = -M1gx - M2g( - x) KE: T = (½)(M1 + M2)(x)2 Lagrangian: L =T-V = (½)(M1+M2)(x)2-M1gx- M2g( - x)

L = (½)(M1+M2)(x)2-M1gx- M2g( - x) Lagrange: (d/dt)[(L/x)] - (L/x) = 0 (L/x) = (M2 - M1)g ; (L/x) = (M1+M2)x  (M1+M2)x = (M2 - M1)g Or: x = [(M2 - M1)/(M1+M2)] g Same as obtained in freshman physics! Force of constraint = tension. Compute using Lagrange multiplier method (later!).