Lagrangian formulation provides an alternative but equivalent way to derive equations of motion compared to Newtonian mechanics. The document provides examples of deriving equations of motion for simple harmonic oscillators, Atwood's machine, and a spring pendulum using the Lagrangian formulation. It also shows the equivalence between Lagrange's equations and Newton's second law. Specifically, it demonstrates that for a conservative system using generalized coordinates, Lagrange's equations reduce to F=ma, where the generalized forces are equal to the negative gradient of the potential energy.

1. Lagrangian Formulation: Examples and
equivalence of Lagrange’s and Newton’s
equations
Dillip K. Satapathy
Textbook:
H. Goldstein, C. Poole and J. Safko, Classical Mechanics, Third Edition
L. D. Landau and E. M. Lifshitz, Mechanics, Volume 1, Third Edition;
S. T. Thornton and J. B. Marion, Classical dynamics of particles and systems
1/15

2. Simple harmonic
oscillator
Example-1: Use Lagrange’s formalism to write the equations
of motion for a simple harmonic oscillator.
The kinetic energy is given by, The potential energy is given by
,
1
T = mx˙2
2
So the Lagrangian can be written
as,
2
1
V = kx2
1
2
2 1
2
L = T −V = mx˙− kx 2
Thus,
∂L = −kx;
∂x
∂L =mx˙
∂x˙
So the Lagrange’s equation turns out to
be: d ∂L
dt ∂x˙
. Σ
=
∂L
∂x
mx¨= −kx
This is the familiar form of Newton’s second law for a simple
harmonic oscillator. 2/15

3. Atwood’s
machine
Example-2: Two weights with mass m1 and m2 are connected by a
light, inextensible string of length l, that passes over a pulley of
radius r and moment of inertia I supported on a frictionless axis as
illustrated in the figure. What is the acceleration of the mass
m1?
• A single coordinate x specifies the
position of both masses and its time
derivative x˙describes the speed of
both masses and the rim of the
pulley.
• We assume that the string does
not slip around the pulley.
• The angular speed of the pulley
ω= x˙/r.
3/15

4. Atwood’s
machine
The kinetic energy
is,
1 1
2 2 1 ẋ
. Σ2
T = m1x˙ + m2x˙ + I
2 2 2 r
and the potential energy is,
V= +m1gx + m2g(X20 −x)
where, X20 depends upon the arbitrary choice of level 0 from
which we measure the potential energy. Thus, the Lagrangian
is,
2
1 2
1 I
r2
. Σ
2
L = m + m + x˙+ g(m −m )x + m gX
2 1 2 20
The last term is a constant and vanishes from derivatives.
4/15

5. Atwood’s
machine
∂L
∂x
2 1
= g(m − m );
∂L
∂x˙
1 2
I
r2
. Σ
= m + m + ẋ
So the Lagrange’s equation
becomes;
d . ∂L Σ ∂L
dt ∂ẋ ∂x
= ;
. I Σ
m1 + m2 +
r2 x
¨= g(m2 −m1)
The acceleration
is,
x
¨=
g(m2 −m1)
I
m1 + m2 +
r2
For m2 > m1, the acceleration is positive and x increases (m1 moves
up) when released from rest. For m1 > m2, the acceleration is
negative.
5/15

6. Atwood’s machine -
Constraints
Usually, one thinks that 3 coordinates is required to specify the
locations of both masses and the orientation of the pulley.
• The inextensible string of
length l that connects the
masses provide one
constraint
• The lack of slipping
between the string and
pulley provides another
constraint
These constraints can be written in terms of velocities as
x˙= rθ (1) x˙2= −x˙1= −x˙
Note that, we now choose x to measure the distance down. This is
convenient, and the potential energy of mass m1 must carry a
negative
(2)
sign
,
V1 = −m1gx 6/15

7. Atwood’s machine -
Constraints
The Eqs. 1 and 2 can readily be integrated to
x = rθ + c
and
x2 = C −x
The term C introduces an arbitrary constant into the potential energy
V2 corresponding to mass m2. Note here, the constraints are given in
terms of velocities (or derivatives) but they are integrated to obtain
equations relating the generalized coordinates.
Such constraints are also calledholonomic; each constraint reduces
the number of degrees of freedom by one.
The the Atwood’s machine completely specified by the single coordinate
x.
7/15

8. Compound Atwood’s
machine
Consider a compound Atwood’s machine shown schematically in
the figure.
The kinetic energy is given by,
1
T =
2
m1x˙ +
2
I1
1 ẋ1
r1
. Σ2 1
1 2
2
+ m2(−x˙ + x˙) + 2
1 2
.
1 ẋ2
M x˙1+ I2
2 2 r2
Σ2 1
+ m3(−x˙1 − x˙2) 2 2
The potential energy is given
by
V= −m1gx1 − m2g(c1 − x1 +x2)
−m3g(c1 −x1 + c2 −x2) −Mg(c1 −x −1)
∂L ∂L ∂L
1 2 1
Express L = T − V Compute
∂x
;
∂x
;
∂x˙
and
∂L
∂x˙2
8/15

9. Spring
Pendulum
Example-3 (From David Morin): Consider the pendulum made of
a spring with a mass m on the end (see figure). The spring is
arranged to lie in a straight line. The equilibrium length of the
spring is l. Let the spring have length l +
x(t), and let its angle with the verticalbe θ(t). Assuming that the
motion is confined to a vertical plane, find the equation of motion for x
andθ.
The kinetic energy has two parts, i.e. the
radial and tangential parts. Therefore,
1
2
˙
T = m x˙+ (l + x) θ
2 2 2
. Σ
Similarly, the potential energy coming from
two parts, i.e. from gravity and thespring,
1
2
V(x, θ) = −mg(l + x)cosθ + kx 2
9/15

10. Spring
Pendulum
The Lagrangian is
therefore 1 ˙
2 2 2
. Σ 1
2 2
L = T −V = m x˙ + (l +x) θ + mg(l + x) cos θ− kx 2
The two Euler-Lagrange equations for the two variables, x and
θ d ∂L
dt ∂x˙
. Σ
=
∂L
∂x
=⇒ mx¨= m(l + x)θ˙2 + mg cosθ −kx, (3)
an
d
dt
d ∂L
. Σ
∂θ˙
=
∂L
∂θ
=⇒
d
dt
2 ˙
. Σ
m(l + x) θ= −mg(l + x) sin θ. (4)
• The Eq. 3 is simple radial F = ma equation complete
with centripetal acceleration −(l +x)θ˙2.
• The Eq. 4 tells us that the torque is equal to the rate of change
of angular momentum.
• Equation 4 can be simplified to m(x + l)θ¨+ 2mx˙θ˙= −mg sinθ,
the tangential F = ma equation, complete with the Coriolis force
−2mx˙θ˙.[arises, since we work in rotating frame of reference]
10/15

11. Newton’s equations from Lagrange’s
equations
For simplicity lets choose the generalized coordinates to be
the rectangular coordinates.
−
∂L d ∂L
∂xi dt ∂x˙i
. Σ
= 0, i = 1,2,3
∂(T − V) d ∂(T − V)
∂xi dt ∂x˙i
. Σ
− = 0, i = 1,2,3
For a conservative system in rectangular coordinates the T = T (x˙i)
and
V= V(xi). Therefore,
∂T
∂xi
= 0; and
∂V
∂x˙i
=0
Therefore, the Lagrange’s equation
becomes,
∂V d ∂T
−
∂x
=
dt ∂x˙
i i
11/15

12. Newton’s equations from Lagrange’s
equations
Also for the conservative
system,
∂V
−
∂x i
= Fi (5)
an
d
=
d ∂T d ∂
dt ∂x˙i dt ∂x˙i
. 3
Σ
j =1
1
2 j
Σ d
dt
mx˙2 = (mx˙i) =p˙i (6)
Combining Eq. 5 and Eq. 6, we
get
Fi = p˙i
The Lagrangian and the Newtonian equations are identical, if
generalized coordinates are rectangular coordinates.
Now let’s derive Lagrange’s equations from Newtonian concepts:
12/15

13. Equivalence of Lagrange’s and Newton’s
equations
Again, for simplicity, consider only a single particle. We need to
transform the rectangular coordinates (xi) to generalized
coordinates (qj ). As we know,
xi = xi(qj, t)
i
x˙=
Σ
q˙+
∂x ∂x
i i
∂qj
j
∂t
an
d
∂x˙i
=
∂xi
∂q˙j ∂qj
The generalized momentum pj associated with qj is determined
by,
(7)
∂T
pj =
∂q˙ j
(8)
[Consistent with the Newtonian concepts. For example, the
kinetic
energy of a particle in plane polar coordinate, T = 1 m(r˙2 + r2θ˙2). The
2
linear momentum, pr = ∂T/∂r˙ = mr˙ and the angular momentum
pθ = ∂T/∂θ˙ =mr2θ˙]
13/15

14. Equivalence of Lagrange’s and Newton’s
equations
The generalized force can be defined by, considering the virtual work
δW
done by the varied path δxi,
Σ Σ ∂xi
i
∂qj
i i j
Σ
i i,j j
δW = F δx = F δq = Q δq
j j
The generalized force associated with Qj
is
j
Σ
Q = F
i
∂xi
i
∂qj
For a conservative system, Qj can be derived from the potential
energy, ∂V
Qj = −
∂q j
(9)
Lets calculate pj,
j
p = =
∂T ∂
∂q˙j ∂q˙j
. Σ
i
1
2
2
i i
Σ Σ
m ẋ = mẋ
i
∂x˙i
i
∂q˙j
Σ
= mẋ
i
∂xi
i
∂qj
(10)
(using Eq. 7) 14/15

15. Equivalence of Lagrange’s and Newton’s
equations
Taking time derivative Eq. 10 and simplifying
further,
j
Σ
p˙ = mx¨ +mx˙
i
∂x d ∂x
i i
i
dt∂qj
. Σ
i
∂qj
∂T
= Qj +
∂q j
Using Eq. 8 and
9,
dt ∂q˙j
d
.
∂T
Σ
−
∂T
∂qj
j
= Q = −
∂V
∂qj
Since Vdoes not depend on generalized velocities (q˙j
),
dt ∂q˙j
Σ Σ
d ∂(T − V) ∂(T −V)
∂qj
− =0
Using L = T −V,
d ∂L
. Σ
∂L
dt ∂q˙j ∂qj
− = 0
These are the Lagrange’s equations of
motion.
15/15