2. 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) = (j = 1,2,3, … n)
Equivalently D’Alembert’s Principle:
(d/dt)[T/qj] - (T/qj) = Qj (j = 1,2,3, … n)

3.   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)

4. 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.

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 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)

7. (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.

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 – 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

9.  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

10. 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)

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!).