Chapter 10 Notes: Lagrangian Mechanics

Thus far we have solved problems by using Newton’s Laws (a vector approach) or energyconservation (a scalar approach.) In this chapter we approach the subject in a very differentfashion, and one that initially seems far from evident. We will still use kinetic and potentialenergies, but will define the Lagrangian, L = T − V from which we can generate theequations of motion by doing simple derivatives

The power of this approach will become evident in examples. The development of thisapproach spanned the work of Wilhelm von Leibniz(1646-1716), Johann Bernoulli (1667-1748), Jean LeRond D’Alembert (1717-1783), and Joseph Louis de Lagrange(1736-1813).We begin with William Rowan Hamilton’s (1805-1865) Variational Principle espoused sub-sequent to the work of those previously mentioned.

When the Lagrangian method works it is very slick. So why use Newton’s methods atall? The Lagrangian method is particularly powerful when dealing with a conservativesystem (and indeed that is the only application we will do.) If what we want is an equationof motion, Lagrange works well—when we also want to find a constraint force such as anormal force, Lagrangian life may be tougher than Newtonian life.

When non-conservative forces or velocity dependent forces are present Newtonian methodsare generally preferred.

1 Hamilton’s Variational Principle

Consider a conservative system. Define the Lagrangian of a system as

L = T − V (1)

the difference between kinetic and potential energies. To do this we must already knowexpressions for the energies, and this has developed from the Newtonian perspective.

Next suppose that the system evolves from time t1 to t2 along a particular trajectory, thatis a particular y(t), y(t) where for simplicity I am considering 1D motion. The starting and

1

ending points are fixed, but we can imagine that there are many paths that can be takenbetween these fixed points as shown in Figure 10.1.1.

Hamilton’s Variational Principle states that the integral of the Lagrangian along the actualpath of motion is an extremum (maximum or minimum) along the actual path taken. Insymbols we evaluate the integral

J =∫ t2

t1

Ldt (2)

and say that

δJ = δ

∫ t2

t1

Ldt = 0 (3)

The notation may seem a little odd. Recall that if we have a function of a variable, F (x),its extrema are found by dF/dx = 0.

We are now expressing something slightly different in Equation 3: the integral is a functionof two parameters, y, y (position and velocity) represented in 1D by paths on the y−t, y−tgraphs. A change in the parameter, i.e. a change in the path taken between fixed startand end points, will change the result of the integral, and this integral is an extremum(maximum or minimum) for the actual path traveled.

E.g. Consider the case of an object falling near the earth in 1D motion. We will show firstthat Hamilton’s Principle leads to the usual differential equation. We have kinetic energyT = my2/2 and potential energy V = mgy so

L =12my2 −mgy (4)

To evaluate the variation we consider δy, δy to be slight variations in position and timefrom the extremum value. Thus1

δJ = δ

∫ t2

t1

[12my2 −mgy

]dt =

∫ t2

t1

[my δy −mg δy] dt (5)

where this equation illustrates how to bring the variation inside the integral.

Nowδy =

d(δy)dt

(6)

We evaluate the first term using integration by parts as∫ t2

t1

my δy dt = my δy∣∣∣t2t1−∫ t2

t1

my δy dt (7)

1This is sort of like a total differential. dF = ∂F∂y

dy + ∂F∂y

dy.

2

Since the endpoints are fixed, with no variation, the first term on the right must be zero.Hence

δJ =∫ t2

t1

[−my −mg] δy dt = 0 (8)

Since δy is arbitrary, this variation can be zero only if

−my −mg = 0 (9)

This is our usual differential equation of motion.

E.g Come again? How does the δy stuff work?

If we have initial conditions of t = 0, y = 0, y = 0 the solution to the differential equationis y(t) = −gt2/2, y = −gt. Now we write the variation from this solution as

y(α, t) = y(0, t) + αη(t) (10)

Here η(t) represents the variation, and is any function of time that (a) has a continuousfirst derivative (velocity) on the time interval [t1, t2], and (b) has values η(t1) = 0 = η(t2)so that the starting and ending points are fixed.

The parameter α is the strength of the variation. When α = 0 we have our true solu-tion.

From Equation 10 we havey(α, t) = y(0, t) + αη(t) (11)

and η(t1) = 0 = η(t2)

Using these expressions we can get the kinetic and potential energies and hence the La-grangian.

T =12m [y(0, t) + αη(t)]2 =

12m [−gt+ αη(t)]2 =

12mg2t2−mg tα η(t) +

12mα2η2(t) (12)

V = mg [y(0, t) + αη(t)] = −12mg2t2 +mg αη(t) (13)

L = m

{g2t2 − αg [tη(t) + η(t)] +

12α2η2(t)

}(14)

Now evaluate the integral of the Lagrangian. First evaluate the term linear in α, integratingby parts. ∫ t2

t1

[tη(t) + η(t)] dt = tη(t)∣∣∣t2t1−∫ t2

t1

η(t)dt+∫ t2

t1

η(t)dt = 0 (15)

This means that

J(α) =13g2(t32 − t31) +

12α2

∫ t2

t1

η2(t)dt (16)

3

Now for any function η(t), η2(t) is positive, and the latter integral must be positive, soJ(α) must be parabolic in α. The minimum in this function occurs for α = 0 as expectedfrom Hamilton’s Principle.

E.g. Motion of a particle in force-free region. We have Newton’s First Law that saysthe motion should be a straight line. We will use Hamilton’s Principle to show that if weassume that the motion is sinusoidal, the sinusoid must have zero amplitude, and hence bestraight line motion between the start and end.

We suppose the particle to move from t = 0, x = 0, y = 0 to t = x1/vx, x = x1, y = 0. Thecomponents of the position during motion are

x = vxt (17)

y = ±η sinπvxt

x1(18)

where η, the amplitude of the sinusoid, can be varied. (Like α in the previous exam-ple.)

The potential energy V is constant, so

L =12m

[v2x +

(ηπvxx1

)2

cos2 πvxt

x1

]− V (19)

Evaluating J(η),

J(η) =mvxx1

2− V x1

vx+ η2mvxπ

2

4x1(20)

and this is a minimum for η = 0.

2 Generalized Coordinates, Degrees of Freedom, Holonomicand Non-Holonomic Constraints

Next in the development is a discussion of generalized coordinates.

Recall Chapter 1 when we defined various 3D orthogonal coordinate systems, cartesian(x, y, z), cylindrical polar (r, θ, z) and spherical polar (r, θ, φ). These are all orthogonalmeaning that ea · eb = 0 where a and b are two of the unit vectors for a particular coordinatesystem.

Generalized coordinates by contrast do not need to be orthogonal, but are chosen to bestrepresent a complicated system.

4

E.g. A pendulum constrained to move in the x − y plane. Initially we might choosecartesian (x, y, z) coordinates, however they do not encapsulate two constraints: the planeof oscillation is x−y and the length of the pendulum, r, is fixed: i.e. z = 0 and r2−x2−y2 =0. It is more convenient to choose cylindrical polar coordinates with z = 0, i.e. plane polarcoordinates. The second constraint is r = constant meaning that there is only one variable,θ, and we can describe the motion as “one dimensional” meaning a single variable describesthe motion. Similarly motion of a roller coaster car is one dimensional.

Generalized coordinates qi are a set of coordinates that are independent, and just sufficientto uniquely specify the configuration of a system.

Degrees of freedom are the number of variables that are needed to describe a system. Thusa single particle in 3D space has 3 degrees of freedom (x, y, z) while two particles in 3Dspace have 6 degrees of freedom, (x1, y1, z1, x2, y2, z2). However a rigid dumbbell composedof two particles separated by a rigid rod has only 5 degrees of freedom. A good choiceof generalized coordinates for the dumbbell would be (x, y, z) for the center of mass plus(θ, φ) for the orientation of the rod joining the particles (see Figure 10.2.2).

Conditions of constraint are statements about limitations on motion of a system. Theyare categorized into holonomic and non-holonomic types.

Holonomic constraints can be represented by an equality involving coordinates such asd2 −

[(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

]= 0.

Being on a surface, or having a fixed separation between two particles are examples.

Non-holonomic constraints include those that involve an inequality (being outside theearth) or involving velocity constraints. We will focus on holonomic constraints and leavenon-holonomic constraints for an advanced mechanics course.

If we have N particles and m holonomic constraints, there are 3N −m degrees of freedomand the same number of generalized coordinates.

3 Many Paths to One Answer: Getting T and V in terms ofgeneralized coordinates

Here we will look at a more complicated system (one we would NOT want to tackle withjust Newton’s Laws) and get the same result three different ways.

The system consists of two particles. One has mass M and is constrained to move in astraight line on a frictionless surface. Attached to it is a simple pendulum of length r withmass m constrained to swing in a plane. We want to find kinetic and potential energies sothat we can construct the Lagrangian.

5

The natural choice for generalized coordinates are the position of the first mass, X, andthe angle that the pendulum makes with the vertical, θ. Refer to Figure 10.3.1.

Path 1 Start with the cartesian coordinates, (X,Y, Z) and (x, y, z). Constraints can bewritten Y = Z = z = 0 and (x − X)2 + y2 − r2 = 0. This means we have twogeneralized coordinates (duh, we just said that!)

In terms of cartesian coordinates,

T =12MX2 +

12m(x2 + y2)

V = mgy (21)

Now we write transformation equations from the cartesian coordinates to the gener-alized coordinates.

X = X (22)x = X + r sin θ (23)y = −r cos θ (24)

(25)

Hence

X = X (26)x = X + rθ cos θ (27)y = rθ sin θ (28)

(29)

and using these in Equations 21

T =12MX2 +

12m[X2 + (rθ)2 + 2Xrθ cos θ

](30)

V = −mgr cos θ (31)

Using the coordinate transformations is usually necessary to get the potential energy.There are alternate ways to get the kinetic energy.

Notice that

• The kinetic energy term has a cross term in it. The generalized coordinates arenot orthogonal.

• The potential energy depends on a single generalized coordinate (this is not truein general.)

6

• The variable X does not appear.

Path 2 Using generalized coordinates at the start to get velocities.

The velocity of M is ~VM = iX and the velocity of m is the sum of the velocity of Mand the velocity of m relative to M .

~vm = ~VM + ~vm,rel = iX + eθrθ (32)

Then since eθ = cos θ i+ sin θ j, x = X + rθ cos θ and y = rθ sin θ as in Path 1.

Path 3 Similar to Path 2

If we writeT =

12M~VM · ~VM +

12m~vm · ~vm (33)

then using i · eθ = cos θ we find the expression for kinetic energy in Equation 30.

Usually either Path 1 or Path 3 will be helpful.

4 Lagrange’s Equations of Motion

Now we apply Hamilton’s Variational Principle to the situation of a Lagrangian as a func-tion of generalized coordinates (qi, qi, i = 1 · · ·N) and generate the Lagrangian Equationsof motion.

δJ = δ

∫ t2

t1

Ldt =∫ t2

t1

δL dt =∫ t2

t1

∑i

(∂L

∂qiδqi +

∂L

∂qiδqi

)dt = 0 (34)

The qi are parameters that are functions of time, and we can write

δqi =d(δqi)dt

(35)

We can use this to evaluate the second term of Equation 34 using integration by parts.∫ t2

t1

∑i

(∂L

∂qiδqi

)dt =

∑i

∂L

∂qiδqi

∣∣∣t2t1−∫ t2

t1

∑i

d

dt

(∂L

∂qi

)δqi dt (36)

The first term vanishes at the endpoints (δq1 = δq2 = 0) and so

δ

∫ t2

t1

Ldt =∫ t2

t1

∑i

[∂L

∂qi− d

dt

(∂L

∂qi

)]δqi dt = 0 (37)

7

The generalized coordinates are independent, and so each δqi is independent of the othervariations. Thus to ensure that the integral is identically zero for all variations the item insquare brackets must be zero for each generalized coordinate. The Lagrange equations ofmotion are thus

∂L

∂qi− d

dt

(∂L

∂qi

)= 0 i = 1, 2, · · · , N (38)

This can also be written as

∂L

∂qi=

d

dt

(∂L

∂qi

)i = 1, 2, · · · , N (39)

In Section 6 we will define a conjugate generalized momentum for each generalized coordi-nate, pi = ∂L

∂qiand the equations of motion can be written

pi ≡dpidt

=∂L

∂qi(40)

Equations 38, 39, and 40 are all equivalent. Now we need to see how to use them.

The equations are sometimes called the Euler-Lagrange

5 Some Applications of the Lagrange Formulation

The general strategy consists of

1. Select a suitable set of generalized coordinates

2. Find transformation equations between the cartesian and generalized coordinates

3. Write the kinetic energy as a function of generalized coordinates using the ideas fromSection 3, Path 1 or Path 3

4. Write the potential energy as a function of generalized coordinates

5. Construct the Lagrangian and do derivatives to get the equations of motion (use oneof Equations 38, 39, and 40.)

5.1 Using Lagragian to get things we already know

E.g. 1 Free particle Yup this is boring, but let’s make sure it works for a particle movingin 1D with no forces, therefore no potential energy. There is only one generalizedcoordinate, x. The Lagrangian is simply L = T = 1

2mx2.

8

Then the partial derivatives are ∂L∂xi

= 0 and ∂L∂xi

= mx and the equation of motion is

0 =d

dt(mx) (41)

i.e. linear momentum is constant, just a restatement of Newton’s First Law.

E.g. 2 Projectile Motion in Uniform g Imagine motion in the x− z plane. The gen-eralized coordinates are just (x, z) and we can quickly write the Lagrangian

L =12m(x2 + z2)−mgz (42)

The partials are

∂L

∂x= 0

∂L

∂z= −mg ∂L

∂xi= mx

∂L

∂zi= mz (43)

so the equations of motion are

0 =d

dt(mx) i.e. horizontal momentum is constant (44)

−mg =d

dt(mz) = mz (45)

and these are just what we get from Newton’s Laws

E.g. 3 Harmonic Oscillator in 1D We have a single generalized coordinate x repre-senting the displacement of the mass from the unstretched location of the spring.

The Lagrangian is L = 12mx

2 − 12kx

2 and the partials are

∂L

∂x= −kx ∂L

∂x= mx (46)

and our equation of motion is simply

−kx = mx (47)

which is just our usual differential equation.

E.g. 4 Central Force in 2D Here we have a constraint, z = 0 so we need (3-1) = 2 gen-eralized coordinates. We choose the polar coordinates (r, θ) and have transformationequations (Path 1)

x = r cos θ y = r sin θ x = r cos θ − rθ sin θ y = r sin θ + rθ cos θ (48)

9

Since the potential energy comes from a central force, it only depends on r. Thekinetic energy can be calculated by Path 1 by (1/2)(x2 + y2) using the above trans-formations (cross terms cancel) resulting in the Lagrangian

L =12m(r2 + r2θ2

)− V (r) (49)

The text illustrates getting the Lagrangian using Path 3. The partials are

∂L

∂r= mr

∂L

∂r= mrθ2 − ∂V

∂r

∂L

∂θ= mr2θ

∂L

∂θ= 0 (50)

Recognizing that f(r) = −∂V/∂r we write the equations of motion as

mr = mrθ2 + f(r)d

dt(mr2θ) = 0 (51)

These are familiar from Chapter 6.

Notice that anytime a generalized coordinate (like θ here) does not appear in theLagrangian, then the conjugate momentum, ∂L/∂qi (yet to be discussed) is constant.In this example the angular momentum, mr2θ is constant.

5.2 More difficult problems

E.g. 5 The Complete Atwood’s Machine Consider a pulley mounted on a fixed, fric-tionless bearing having moment of inertia I and radius a. A string of length ` connectstwo blocks of masses m1,m2 that can only move vertically. We neglect any air resis-tance and assume an ideal string (massless and non-stretchy) that does not slip onthe pulley. This is shown in figure 10.5.1.

There are 5 holonomic constraints, y1 = z1 = y2 = z2 = 0 produce 1D motion, and` = constant for the string length. So we need (3×2−5 = 1) generalized coordinates.(Or if you include the pulley, with three holonomic constraints on position of pulley,(3× 3− 8 = 1) generalized coordinates.)

We use x for the position of m1 and then the position of m2 is (`− πa− x). We alsoknow that since the string does not slip, ω = x/a.

The Lagrangian is then

L =12

(m1 +m2 +

I

a2

)x2 +m1gx+m2g(`− πa− x) (52)

The partials are

∂L

∂x= (m1 −m2)g

∂L

∂x=(m1 +m2 +

I

a2

)x (53)

10

Leading to the equation of motion(m1 +m2 +

I

a2

)x = (m1 −m2)g (54)

meaning that the acceleration is constant with value

(m1 −m2)gm1 +m2 + I

a2

(55)

This should be familiar from Physics 312.

E. g. 6 The Double Atwood Machine Replace the second object by a second Atwood’smachine with masses m2,m3. Simplify the problem by making the two moments ofinertia zero, and the pulley radii negligible. The string lengths will be ` and `′. SeeFigure 10.5.2 in text.

The generalized coordinates will be x, the position of m1 relative to the first pulley,and x′, the position of m2 relative to the moving pulley.

The holonomic constraints are for the three masses and the moving pulley, with nomotion in the y, z directions (8 constraints) plus constraints on the string lengths (2constraints.) Verifying, we need (3(4) - 8 - 2 = 2) generalized coordinates.

Path 1 is easy.

x1 = x xp = `−x x2 = xp+x′ = `−x+x′ x3 = xp+(`′−x′) = `+`′−x−x′(56)

The text writes the constraints as

x1 + xp − ` = 0 (57)x2 + x3 − 2xp − `′ = 0

(2x1 + x2 + x3)− (2`+ `′) = 0 (58)

Here we have defined down as positive x, so the potential energy is V = −m1gx1 −m2gx2 −m3gx3.

Hence we get the Lagrangian

L =12m1x

2+12m2(−x+x′)2+

12m3(x+x′)2+(m1−m2−m3)gx+(m2−m3)gx′+Constants

(59)The partials are

∂L

∂x= (m1 −m2 −m3)g

∂L

∂x= m1x−m2(−x+ x′) +m3(x+ x′)

11

∂L

∂x′= (m2 −m3)g

∂L

∂x′= m2(−x+ x′) +m3(x+ x′) (60)

and the equations of motion are

(m1 +m2 +m3)x+ (−m2 +m3)x′ = (m1 −m2 −m3)g (61)

(−m2 +m3)x+ (m2 +m3)x′ = (m2 −m3)g (62)

These simultaneous equations can be easily solved, although the reult is messy withsymbols. Try it with numbers. Let m1 = 5 kg, m2 = 2 kg, and m3 = 3 kg. Find theaccelerations x and x′and then find the acceleration of each mass. Next try reversingthe values for m2,m3.

E. g. 7 Block Sliding on a Frictionless Wedge that is Free to Move on Frictionless TableA wedge of mass M and angle θ can slide freely in 1D on a horizontal frictionlesstable. A box of mass m slides along the frictionless incline of the wedge. Find theequations of motion from the Lagrangian.

As suggested in Figure 10.5.3 we will have two generalized coordinates, x for thehorizontal location of the wedge and x′ for the position of the box relative to the topof the wedge. We will use Path 3 to get the kinetic energy.

The velocity of the wedge is ~V = xi and the velocity of the box is ~v = xi+ x′i′.

The kinetic energies are

TM =12M~V · ~V =

12Mx2 Tm =

12m~v · ~v =

12mx2 +

12mx′ 2 +mxx′ cos θ (63)

The potential energy is V = −mg x′ sin θ and the Lagrangian is

L =12Mx2 +

12mx2 +

12mx′ 2 +mxx′ cos θ +mg x′ sin θ (64)

with partials∂L

∂x= 0

∂L

∂x= (M +m)x+mx′ cos θ (65)

∂L

∂x′= mg sin θ

∂L

∂x′= mx′ +mx cos θ (66)

With the first partial being zero, this means that (M +m)x+mx′ cos θ = Constant.In a moment we will refer to this a a generalized momentum (conjugate to x) which inthis case is conserved. For this problem this means that the horizontal net momentumis constant.

The two equations of motion are

(m+M)x+ (m cos θ)x′ = 0 x′ + x cos θ = g sin θ (67)

12

and we can solve these simultaneous equations to yield

x =−g sin θ cos θ

(1 +M/m)− cos2 θx′ =

g sin θ1−m cos2 θ/(m+M)

(68)

Try to think of all limiting cases to check the validity of these equations.

6 Generalized Momenta and Ignorable Coordinates

We now have several examples of Lagrangian formulation. In several of the examples theLagrangian did not include one of the generalized coordinates (look at example 7, x doesnot appear in L), and in these cases the text refers to the coordinate as “ignorable”.

Whenever we have an ignorable coordinate q, ∂L/∂q = 0 and therefore ∂L/∂q is constant.We define the generalized momentum conjugate to the generalized variable q (conjugatemomentum for short) as

pq =∂L

∂q(69)

and thus can write the equation of motion as

pq ≡dpqdt

=∂L

∂q(70)

If q has units of length, its conjugate momentum has units of momentum, kg·m/s. If qis an angle, the conjugate momentum has units of angular momentum, kg·m2/s. In somecases they clearly represent a momentum or momentum component that we might writedown directly (example 1-4, in the latter case pr is the radial component of momentum, pθis the angular momentum) while in other cases the conjugate momentum is not as obvious(Examples 5-7).

Generalized coordinates and their conjugate momenta become central in our definition ofthe Hamiltonian at the end of this chapter.

E.g. 8 Pendulum Attached to Movable Support In Section 3 we began discussing amass M constrained to move in a straight line on a frictionless surface with a simplependulum of length r, mass m attached to it. The generalized coordinates were Xand θ and the Lagrangian was

L =12MX2 +

12m[X2 + (rθ)2 + 2Xrθ cos θ

]+mgr cos θ (71)

X is ignorable so

pX = (M +m)X +mrθ cos θ = Constant (72)

13

and the equations of motion are

(M +m)X +mrθ cos θ −mrθ2 sin θ = 0 (73)

d

dt

(r2θ + Xr cos θ

)= −(Xrθ + gr) sin θ

or

θ +X

rcos θ +

g

rsin θ = 0 (74)

The last two equations can be combined to eliminate X and yield a rather uglylooking second order differential equation in θ.

The text makes some simplifying assumptions to see if these equations make sense.Read and understand these

E.g. 9 Spherical Pendulum Including Approximations Return to the simple pen-dulum, a mass m attached to a string of length ` that s fixed at the other end, butnow let the pendulum be free to move in 3D.

This problem is identical to that of a spherical bowl of radius r = ` with a frictionlessbar of soap free to move in it.

There are two degrees of freedom, (θ, φ), the usual spherical polar variables. Thevelocity (Equation 1.12.12) is

~v = eθ`θ + eφ` sin θ φ (75)

If we choose the lowest position for V = 0, the height is y = `(1 − cos θ) and theLagrangian is

L =12m`2(θ2 + φ2 sin2 θ)−mg`(1− cos θ) (76)

Since φ is ignorable we have conservation of the conjugate momentum,

pφ = m`2φ sin2 θ = constant ≡ m`2S (77)φ sin2 θ = S (78)

where we have defined a new constant S = φ sin2 θ.

The remaining equation of motion is then

d

dt

(m`2θ

)= m`2θ = m`2φ2 sin θ cos θ −mg` sin θ

orθ +

g

`sin θ − S2 cos θ

sin3 θ= 0 (79)

We shall NOT try to solve this in general, but will look at three special cases: (a) sim-ple plane pendulum, φ = 0, (b) Conical pendulum, θ = 0, and (c) slight perturbationfrom a conical pendulum.

14

1. For the simple plane pendulum, φ = 0, we have S = 0 and the equation ofmotion becomes

θ +g

`sin θ = 0 (80)

which is just what we have seen before for the plane pendulum. So we gainconfidence that our derivation is correct.

2. For a conical pendulum, θ = 0, the bob moves in a circle with the string havinga constant angle θ0 and the equation of motion becomes

g

`sin θ − S2 cos θ

sin3 θ= 0

or, putting in the expression for S

φ2 =g

` cos θ=g

`sec θ (81)

and this gives the required angular velocity φ for the conical motion.

3. Now suppose we disturb the conical pendulum slightly from the above value.To indicate the perfect conical pendulum value we will add a subscript

φ20 =

g

` cos θ0=g

`sec θ0 (82)

Also we will express the ideal conical value of

S20 =

g

`sin4 θ0 sec θ0 (83)

Also we will indicate the period of the circular motion of the conical pendulumas

T0 =2πφ

= 2π

√` cos θ0g

(84)

We assume that the perturbation will not change the value of S significantly(nor will it change the period T0) and write the equation of motion as

θ +g

`

(sin θ − sin4 θ0

cos θ0cos θsin3 θ

)= 0 (85)

Now consider the expression in parentheses

f(θ) = sin θ − sin4 θ0cos θ0

cos θsin3 θ

(86)

Do a Taylor series expansion about θ = θ0 to first order to get

f(θ) ≈ (3 cos θ0 + sec θ0)(θ − θ0) (87)

15

Introduce the variable (pronounced ksi) ξ = θ − θ0, with ξ = θ so that theequation of motion is

ξ +g

`(3 cos θ0 + sec θ0)ξ (88)

This is our familiar friend the harmonic oscillator with period

T1 = 2π

√`

g(3 cos θ0 + sec θ0)(89)

So as the pendulum makes its primarily circular motion with period T0, it bobsup and down with period T1.

E.g. Suppose that ` = 60 cm and θ0 = 35◦. Then T0 = 1.407 s and T1 = 0.811s.

The maxima in θ are a good visual reference to the motion. In the time T1

between successive maxima, the bob rotates through an angle

φ ≈ φ0T1 =2π√

3 cos2 θ0 + 1> π (90)

and so the point of maximum θ precesses in the direction of increasing φ asshown in Figure 10.6.2. The text integrates the equation of motion once toget an effective potential that it plots versus angle to show the angular turningpoints.

7 Forces of Constraint: Lagrange Multipliers

The previous sections have shown how the Lagrangian lets us determine constants of motionand equations of motion. We may also want to determine the size of the constraint forcessuch as normal forces or tensions. Rather than return to Newton’s Methods, we introduceanother new powerful method, Lagrange multipliers.

In the Lagrangian method used previously we used the constraints to reduce the num-ber of variables needed to describe the system, and to make all the remaining variablesindependent.

7.1 Text Development

Consider a system with two generalized coordinates, (q1, q2) connected by a single equationof constraint, f(q1, q2, t) = 0 where a time dependent constraint has been allowed. In theLagrangian approach already used we would use the constraint to reduce the system to a

16

single generalized coordinate. Here we will NOT do this step so that we can determine aforce of constraint.

Start with Hamilton’s Variational Principle from Section 4,

δ

∫ t2

t1

Ldt =∫ t2

t1

∑i

[∂L

∂qi− d

dt

(∂L

∂qi

)]δqi dt = 0 (91)

In Section 4 the constraints had already been used, so the δq’s were all independent.In this section we are starting earlier and have not used the constraint, so the δq areNOT independent. We write the variation in the constraint, which at any time must bezero.

δf =(∂f

∂q1δq1 +

∂f

∂q2δq2

)= 0 (92)

Hence

δq2 = −(∂f/∂q1∂f/∂q2

)δq1 (93)

Putting this into Equation 91,∫ t2

t1

[(∂L

∂q1− d

dt

∂L

∂q1

)−(∂L

∂q2− d

dt

∂L

∂q2

)(∂f/∂q1∂f/∂q2

)]δq1 dt = 0 (94)

Now we are varying a single coordinate, and the variation can take any value. To makethe integral zero, therefore, the item in square brackets must be zero, or

(∂L/∂q1)− (d/dt)(∂L/∂q1)(∂f/∂q1)

=(∂L/∂q2)− (d/dt)(∂L/∂q2)

(∂f/∂q2)(95)

The text states that the left hand side is an expression that is a function of (q1, q1, t) withtime being either implicitly (via q1(t), q1(t)) or explicitly involved. Likewise the right handside is a function of (q2, q2, t). For these to be equal for arbitrary choices of the generalizedvariables, they must both equal a function of time alone, that we call −λ(t).

Then we can write

∂L

∂q1− d

dt

∂L

∂q1+ λ(t)

∂f

∂q1= 0 (96)

∂L

∂q2− d

dt

∂L

∂q2+ λ(t)

∂f

∂q2= 0 (97)

These two equations plus the constraint condition provide three equations in the threeunknowns (q1(t), q2(t), λ(t)) that can be solved.

The quantities

Qi = λ(t)∂f

∂qi(98)

17

are called the generalized forces and are either a force for a spacelike q or a torque for anangular q.

The text generalizes to more than 2 generalized coordinates and more than one con-straint.

7.2 Alternate Method from Analytical Mechanics By Louis N. Hand,Janet D. Finch, available as a Google Book

We will start with an example that is more straightforward calculus: Consider the functionF (x, y) = x2+y2. What values of (x, y) will minimize this function subject to the constrainty = 2x+ 1.

Geometrically we can see that the constraint describes a straight line on the x − y plane,and F is the square of the distance from the origin to the line.

Standard calculus would use the constraint equation to eliminate y in F , then dF/dx = 0could be solved for the desired value, x = −0.4. Then y = 0.2.

Instead we introduce a Lagrange multiplier λ and write F ′ = x2 + y2 + λ(y − 2x− 1) andget three equations, ∂F ′/∂x = 0, ∂F ′/∂y = 0 ∂F ′/∂λ = 0 or in this case

2x− 2λ = 0 (99)2y + λ = 0 (100)

y − 2x− 1 = 0 (101)

Solving the simultaneous equations we get x = λ = −0.4, y = 0.2.

Try this one: Let F (x) = 3x2 + 2y2 and the constraint be x = cos(2y). Use Lagrangemultipliers to minimize F and show that the answers are x = 0.2293, y = 0.6697, λ =1.3761.

Returning to the mechanics realm, and for the case of a single constraint equation f(q1, q2, t) =0, we introduce a modified Lagrangian,

L′ = L+ λf (102)

Then the equation of motion becomes

∂L′

∂qi− d

dt

∂L′

∂qi= 0 (103)

18

7.3 Examples

E.g. 1 Ball rolling down a ramp A ball of radius R and moment of inertia I = mk2

rolls without slipping down a slope inclined at α to the horizontal. Find the equationsof motion and the generalized forces.

Choose up the incline to be positive x and positive θ to be consistent with this. Theconstraint between these variables is x−Rθ = 0.

The modified Lagrangian is then

L′ =12mx2 +

12Iθ2 −mgx sinα+ λ(x−Rθ) (104)

We have two equations of motion that come from this modified Lagrangian, plus theconstraint equation,

−mg sin θ + λ−mx = 0 (105)−λR− Iθ = 0 (106)x−Rθ = 0 (107)

These are solved in a trivial fashion to get (using the radius of gyration k)

x = −g sinα(

R2

R2 + k2

)(108)

θ = −g sinα(

R

R2 + k2

)(109)

λ = mg sinα(

k2

R2 + k2

)(110)

The generalized forces are a friction force

Fx = λ∂f

∂x= mg sinα

(k2

R2 + k2

)(111)

and a torque about the center of mass due to friction

Fθ ≡ N = λ∂f

∂θ= −mgR sinα

(k2

R2 + k2

)(112)

E.g. The yo-yo A solid disk, mass m radius a, is used in a yo-yo. Use Lagrangians andLagrange multipliers to get the acceleration and the tension in the string.

19

Using y (down positive) and φ (clockwise positive) as variables, we can write theLagrangian

L =12my2 +

12Icmφ

2 +mgy (113)

There is a constraint relating y and φ, y − aφ = 0 and for a disk Icm = 12ma

2. If weuse these two facts the Lagrangian becomes

L =34my2 +mgy (114)

with the resulting equation of motion quickly reducing to

y =23g (115)

Now we solve the problem without using the constraint to eliminate a dependentvariable. The constraint is f = y − aφ = 0. Start with the modified Lagrangian

L′ =12my2 +

12Icmφ

2 +mgy + λ(y − aφ) (116)

Then we get three Lagrange equations using variables y, φ and λ

∂L′

∂y=

d

dt

∂L′

∂y

mg + λ = mφ (117)

∂L′

∂φ=

d

dt

∂L′

∂φ

−λa = Icmφ (118)

∂L′

∂λ=

d

dt

∂L′

∂λy − aφ = 0 (119)

Take the second derivative of the last equation, y − aφ = 0, insert the moment ofinertia, and solve the equations to get

λ = −13mg (120)

φ =23g

a(121)

y =23g (122)

20

The generalized force associated with y is

Qy = λ∂f

∂y= λ = −1

3mg (123)

This is the tension in the string. The generalized force associated with φ is

Qφ = λ∂f

∂φ= −λa =

13mga (124)

and this is the torque on the disk.

8 D’Alembert’s Principle: Generalized Forces

This section extends the Lagrangian formulation to systems that might include non-conservative forces. We will not go into the details, but here is what I understand about themeaning of the approach. Newton focuses on vector forces. The Europeans contemporariesof Newton focused on scalar quantities like energy. Bernoulli and D’Alembert made theconnection between the two approaches by introducing the ideas of virtual work.

If the objects in a system are instantaneously displaced from equilibrium by infinitesimaldisplacements, subject to satisfying constraint equations, then the virtual work done iszero. Since we want to deal with non-equilibrium situations, we move the rate of changeof momentum to the force side of the equation and write D’Alembert’s Principle,∑

(~Fi − ~p ) · δ~ri = 0 (125)

9 The Hamiltonian Function: Hamilton’s Equations

Again we will not go into any detail (and really, neither does the text).

Define a new function, the Hamiltonian which is

H =∑i

qipi − L (126)

It can be shown that this function is just the total energy of a system, H = T + V .

In the Lagrangian formulation we had L = T (qi, q1) − V (qi) as a function of generalizedcoordinates and their derivatives. We also defined momenta conjugate to the coordinates,pi = ∂L/∂qi.

21

This equation can be solved (at least in principle) for qi = qi(q1, p1) and then by usingHamilton’s Variational Principle we find Hamilton’s canonical equations of motion

∂H

∂pi= qi (127)

∂H

∂qi= −pi (128)

While Lagrangians for a system of n degrees of freedom yields n second order differentialequations, Hamilton’s equations yield 2n first order differential equations.

E.g. 1D Simple Harmonic Oscillator The energies are

T =12mx2 V =

12kx2 (129)

hence L = mx2/2 − kx2/2. The conjugate momentum is p = ∂L/∂x = mx and weinvert to get x = p/m. Rewrite the energies in terms of x and p, and add to get H.

H = T + V =p2

2m+kx2

2(130)

Hamilton’s canonical equations are

∂H

∂p= x

∂H

∂x= −p (131)

and doing the derivatives,p

m= x kx = −p (132)

The first equation is a restatement of the definition of momentum, and the second isthe same as our Newton’s Law result.

Particle on Cylindrical Surface Subject to a Central Force Imagine an origin sur-rounded symmetrically by a cylinder of radius R extending in the z-direction. Aparticle of mass m is constrained to move on the surface of the cylinder. A force~F = −k~r acts on the particle.

The constraint is x2 + y2 = R2. The force has an associated potential energy

V =12kr2 =

12k(x2 + y2 + z2) =

12k(R2 + z2) (133)

Symmetry suggests the use of cylindrical coordinates, in which v2 = r2 + r2θ2 + z2,and since r = R = Constant,

T =12m(R2θ2 + z2) (134)

22

The Lagrangian is

L =12m(R2θ2 + z2)− 1

2k(R2 + z2) (135)

whence we can get expressions for the generalized momenta

pθ = mR2θ pz = mz (136)

Now we write the Hamiltonian H in terms of variables θ, z, pθ and pz,

H =p2θ

2mR2+

p2z

2m+

12kz2 +

12kR2 (137)

Hamilton’s canonical equations are then

pθ = −∂H∂θ

= 0 pz = −∂H∂z

= −kz (138)

θ =∂H

∂pθ=

pθmR2

z =∂H

∂pz=pzm

(139)

In this case, the last two equations just duplicate the definitions of the generalizedmomenta, angular momentum and linear momentum respectively. The first equationtells us the angular momentum is constant, while the second equation can be writtenas

mz = −kz (140)

and this is the equation for simple harmonic motion. Thus the particle moves up anddown across the z = 0 plane in SHM while rotating around the z-axis at a constantrate.

Although Hamilton’s canonical equations are of some use in classical mechanics, they playvery important roles in both statistical mechanics and quantum mechanics.

10 Another view of Hamilton’s Variational Principle

(From Classical Dynamics of Particles and Systems, Jerry B. Marion) Here is a bit morerigorous discussion of Euler’s equations.

We imagine two fixed end points, x1, x2 and a function y(x) that describes the path takenbetween these points. In the Brachistochrone problem this describes the shape of a wireconnecting starting and ending points.

We define a functional that is evaluated on this path. For the Brachistochrone problemthis is the time taken to travel the wire. A useful notation for the functional is

f [y(x), y′(x);x] (141)

23

This means the functional depends on the function y as well as its derivative dy/dx, andthe parametric variable is x.

We integrate the functional over the range,

J =∫ x2

x1

f [y, y′;x]dx (142)

Our goal in the calculus of variations is to find a function y that will produce a stationarystate, that is an extremum for the integral. This desired solution will be called y(0, x).Introduce variations to this desired solution as

y(α, x) = y(0, x) + αη(x) (143)y′(α, x) = y′(0, x) + αη′(x) (144)

With fixed endpoints η(x1) = η(x2) = 0.

A necessary condition for the selection of the proper y(x) is the following. For anyvariation function η(x), the integral will be a minimum with respect to variations in α, inthe limit of α→ 0.

∂J

∂α

∣∣∣α=0

= 0 (145)

Now we can do some calculus. From above we can write (notice the total derivative of ηin the second equation)

∂y

∂α= η(x)

∂y′

∂α= η′ =

dη

dx(146)

∂J

∂α=

∂

∂α

∫ x2

x1

f [y, y′;x]dx (147)

=∫ x2

x1

(∂f

∂y

∂y

∂α+∂f

∂y′∂y′

∂α

)dx (148)

=∫ x2

x1

(∂f

∂yη +

∂f

∂y′dη

dx

)dx (149)

We can evaluate the second term using integration by parts with u = ∂f/∂y′ and dv =dηdxdx. Then

∂J

∂α=

∫ x2

x1

∂f

∂yη dx+

∂f

∂y′η∣∣∣x2

x1

−∫ x2

x1

d

dx

(∂f

∂yη dx

)(150)

At the limits the middle term disappears and

∂J

∂α=∫ x2

x1

η

(∂f

∂y− d

dx

∂f

∂y

)dx (151)

24

The variable α is hidden implicitly in y and y′. For this to be true for all variationsη(x),

∂f

∂y− d

dx

(∂f

∂y

)= 0 (152)

This is Euler’s Equation. When applied in mechanics to the Lagrangian it is called La-grange’s Equation. Or collectively refer to it as the Euler-Lagrange equation.

E.g. The Brachistochrone problem. Imagine two points with coordinates (x1, y1) and(x2, y2) connected by a wire of arbitrary shape. A bead can slide along the wire with nofriction. Find the shape of the wire (i.e. y(x)) that minimizes the time taken by the beadto move between the points. Choose down as positive-x and to the right as positive-y andassume the bead is at rest at the starting point.

We will determine what the functional is that we wish to evaluate, then use Euler’s rela-tion.

Since the bead is released from rest, x = 0, when it is at x we can use simple kinematicsto write the speed, v2 = 2gx. In a short time the bead will travel a distance

ds =√dx2 + dy2 =

√1 +

(dy

dx

)2

dx =√

1 + y′2dx (153)

The time taken to travel this distance is

dt =ds

v=

√1 + y′2dx√

2gx(154)

Define the functional f [y, y′;x] =√

1+y′2

x

Now get the Euler derivatives

∂f

∂y= 0

∂f

∂y′=

y′√x(1 + y′2

(155)

This means thaty′√

x(1 + y′2)= Constant ≡ 1√

2a(156)

where we define the constant a for convenience. Squaring,

y′2

x(1 + y′2)=

12a

(157)

y′ =dy

dx=√

x

2a− x(158)

25

This relation defines the relation between y and x that is required to minimize the time.Now we will show that the relation defines a cycloid. Rearrange and integrate to get y,using initial condition that x = y = 0 at release.

y =∫

x√2ax− x2

dx (159)

We talked about cycloids last quarter, and had the parameterization x = a(1 − cos θ).Inserting this and doing some algebra we get

y =∫a(1− cos θ)dθ = aθ − a sin θ (160)

which is the parameterized form of y for a cycloid.

10.1 Notation

Fowles and Cassiday use δJ for the variation if the integral. Marion indicates that this isshort hand for the following. With y(α, x) = y(0, x) + αη(x),

δJ =∂J

∂αdα =

∫ x2

x1

(∂f

∂y− d

dx

∂f

∂y

)∂y

∂αdα dx (161)

In Fowles and Cassiday notation this is

δJ =∫ x2

x1

(∂f

∂y− d

dx

∂f

∂y

)δy dx (162)

meaning that∂J

∂αdα ≡ δJ ∂y

∂αdα ≡ δy (163)

26