Sect. 1.6: Simple Applications of the Lagrangian Formulation

2. Sect. 1.6: Simple Applications of the Lagrangian Formulation • Lagrangian formulation: 2scalar functions,T & V • Newtonian formulation:MANYvector 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 LagrangianL = 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)

3. 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)

4. Always:T = M0 + ∑jMjqj + ∑jMjkqjqk Or: T0 +T1 + T2 T0  M0independent of generalized velocities T1  ∑jMjqjlinear 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.

5. 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

6. 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 Qjare 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)

7. 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.

8. Particle in Plane (Plane Polar Coords) • Plane Polar Coordinates: q1 = r, q2 = θ • Transformation eqtns: x = r cosθ, y = r sinθ  x = r cosθ – rθsinθ y = r sinθ+ rθcosθ 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

9. 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

10. Atwood’s Machine • 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)

11. 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!).