# Lagrange's Foramlism for Classical Mechanics

## Hamilton's principle

Hamilton's principle falls out of the calculus of variations in that seeking the shortest time interval is the focus of the variations.

Of all possible paths along which a dynamical system may move from on point to another within a specified time interval, the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies.

Casting this in the language of the calculus of variations

if you want the above "action" integral to be stationary then according to the calculus of variations you want the Euler-Lagrange equation to be satisfied where

or

here

the generalized force if I have conservative forces

Newton's second law in an Inertial reference frame = time derivative of the generalized momentum

thus

## Lagrange's Equations in generalized coordinates

Generalized coordinates are a set of parameters that uniquely specify the instantaneous state of a dynamical system.

The number of independent generalized coordinates is given by subtracting the number of constraints from the number of degrees of freedom .

### Pedulum example

Consider the 2-D pendulum where an object of mass is constrained by a rod of length . The object is at one end of the rod and the rod is fixed to rotate about the other end.

There are 32 degrees of freedom for the 2-D problem
The particle is constrained to a rod
The motion of the particle may be described using one component

The constraint may be expressed in cartesian coordinates as

you can express the position of the object on the end of a rod as a function of just one generalized coordinate

is the generalized coordinate

You could also express the position as a function of the deflection angle in cartesian coordinates

is the generalized coordinate

Note
If you were to start using Polar coordinates right away such that

then the dependence of the function on would not be obvious as this dependence is implicit to changes in the direction

The number of generalized coordinates becomes more obvious when you begin expressing the Potential and Kinetic Energy

### Motion on Sphere example

Consider a particle constrained to move on a sphere of radius .

There are 3 degrees of freedom for the 3-D problem
The particle is constrained to the surface of the sphere
The motion of the particle may be described using two components

The constraint expressed in terms of cartesian coordinates is

The above constraint equation can be used to reduce the degrees of freedom by using the constraint to eliminate one of the above components

for Example

or if you chose the angles and from spherical coordinates

### Generalized Force and Momentum

Generalized Force
Generalized Momentum

As shown above, Hamilton's principle leads to a re-expression of Newton's second law through the Euler-Lagrange Equation in a differential form known as Lagrange's equations.

Below is the Euler-Lagrange equations expressed in terms of generalized coordinates.

### Holonomic

If each coordinate used to describe a system can vary independently of the others, then the system is said to be holonomic.

A Non-holonomic system, has a lest one coordinate that depends on one of the others thereby reducing the number of degrees of freedom.

Example: Consider a sphere constrained to roll on a plane (ball on the floor). A spheres position on the plane can be specified using 2 coordinates and the orientation ( rotation) of a point on the sphere can be described by 3 coordinates. Thus, this system has 5 degrees of freedom.

The coordinates are not independent since as the sphere rolls without slipping then at least two coordinates must change making the system nonholonomic.

other examples: car, bicycle

## Example: Lagrangian for object in 2-D moving in a conservative field

### In cartesian coordinates

In cartesian coordinates the generalized coordinates are represented as the parameters x and y

for the generalized coordinate labeled
for the generalized coordinate labeled

### In polar coordinates

In cartesian coordinates the generalized coordinates are represented as the parameters and

recall

#### The parameter

In polar coordinates

## Examples using Lagrange's Equation

### Pendulum

Returning to the problem of an object of mass attached to a weightless rod of length that is constrained to rotate at the end of the rod where there is no mass.

The number of generalized coordinates = 2-1 = 1.

Selecting polar coordinates to describe this system allows a description using generalized coordinates of only one parameter, .

### Atwoods Machine

Let represent the distance of from the pulley and the distance of from the pulley.

The problem has 2 masses hat can move up and down giving us 2 degrees of freedom.

There is a constraint due to the masses being joined by a string of length l.

The number of generalized coordinate = 2-1 = 1.

One may define the potential arbitrarily such that the above constant is zero.

### Block Sliding on a Wedge

A block slides down an incline of elevation that is free to move on the table.

How long does it take the block to reach the bottom if it is released a distance from the top.

Let

distance of block from the top of the incline
distance incline moves from its starting point.

Kinetic energy of the block of mass m
kinetic energy of the incline of mass M

The kinetic energy of the block is a combination of the velocity of the block down the incline and the velocity of the incline. As the block moves down the inline, the incline moves.

for q_1

for q_2

substituting q_2 into q_1

Note
If is 90 degrees then

The acceleration is constant so one may use constant acceleration equations to find the fall time

Note
If M is infinite then

## Lagrange Multipliers

The method of Lagrange multipliers is a means to incorporate the constraints of a system into the Euler-Lagrange equation.

consider the problem of a disk rolling down an incline plane without slipping.

If we assume the incline does not move then we have 2 degrees of freedom for the disk. Let represent the distance from the top of the incline and an angle of disk rotation.

The "no-slip" constraint means that there is a force of friction; which does no work by the way.

To roll without slipping requires that

let

then

The above constraint is holonomic since the constraint on the velocities can be integrated to give a relationship between the coordinates. Non-holonomic constraints do not have this property.

### Constraints in Differential form

Using the above example we can generalize the constraint equation to take the form of a differential equation.

let

where

then

since the total derivative with respect to time is zero , the function is a constant in time

A holonomic constraint (doesn't depend on other variables)

The advantage of having the constraint in differential form is that instead of incorporating their integrated form into the problem we add them to the Euler-Lagrange equation.

### Incorporate the constraints into Euler-Lagrange Equations

Consider a system having constraints with generalized coordinate parameters.

For the case of time independent constraints, the -th constraint equation equation for the constraint function may be represented as

let be an undetermined coefficient such that

The Euler-Lagrange equation is

we could add (or subtract) the Lagrange multiplier to the above equation without changing it

but this is more than just equating terms that are equal to zero.

consider a system in cartesian coordinates such that

the above coordinates may be written in term of generalized coordinates or in other words they can be expressed as functions of q_i such that

the same is try for y and z

the total derivative with respect to time is given by the chain rule as

since

then

only the jth term survives the derivative of the sum

take derivative of both sides with respect to time

since

here the total and partial derivatives are interchangeable

and

Then

Lagrange's equations of motion

where

Generalized forces of constraint

### disk rolling down incline place using multipliers

Returning to the problem of a disk rolling without slipping down an incline place

substituting

and

consider a ladder of length L leaning against a frictionless wall and sitting on a frictionless floor.

Since everything is frictionless the ladder will begin to fall.

Find the equation of motion for the ladder.

There are 3 degrees of freedom; The the and position of the ladder and the angle, , the ladder makes with respect to the floor.

There are 2 constraint equations

for :

But from the constraint equation f_1

thus

For :

But from the constraint equation f_2

thus

for :

substituting in for and

= torque resulting from normal force of wall on ladder ( choose point of ladder's contact with floor as the point of rotation)

## More Examples

### 2-D Central Force

Consider a particle of mass that is constrained to move in 2-D due to a potential U(2).

Since this is a 2-D problem, lets use polar coordinates.

For the generalized coordinate parameter

constant

for the generalized coordinate parameter

let

constant

then

### Particle confined on a cylinder

Consider a particle of mass that is experiencing a force and is constrained to move on a frictionless cylinder of radius .

Find the equation(s) to describe its motion.

Since the particle is constrained by a cylinder it seems worthwhile to use cylindrical coordinates to describe its motion.

to determine if the force is conservative.

Test if

The potential energy for this force may be defined according to the work integral

simple harmonic motion in the z coordinate
angular momentum is conserved

### ball in a bowl

consider a ball of mass and radius is constrained to roll without slipping spherical bowl of radius that is in a gravitational field.

The generalized coordinate functions for this problem are and where is the rotation angle of the ball and is the angle of inclination of the ball with respect to the cylinders central axis.

constraint equation

for the generalized coordinate function:

for the generalized coordinate function:

The constraint equation

oscillation frequency

### Double Pendulum

The double pendulum problem in 2-D has two objects (4 degrees of freedom) and two constraints.

The number of generalized coordinates = 4-2 = 2

Choosing the two angles and

In cartesian coordinates one may write the position and velocity of the two objects as

Object 1
Object 2 is a little more complicated as its position depends on object 1

for the generalize coordinate paramter

if I assume small oscillations

where

for the generalize coordinate paramter

assuming small oscillations

#### Solving coupled ODEs

The double pendulum's motion for small angles is described by a set of coupled , non-linear, second order ordinary differential equations.

where

create a linear combination of the ODEs by adding them together using an arbitrary constant

Now require that the raio of the coefficients for the non-differential terms be equal to the ratio of the coefficients for the double differential terms

The differential equation for this linear combination may be written as

A normal coordinate is defined such that

Then the ODE looks like an oscillator

We have two possible solutions

"in phase"

"out of phase"

if &

Then

the "in phase" oscillation

the "out of phase" oscillation

Note
The frequency for the case when the two pendulums are oscillating in phase is lower than when they are out of phase.

### Bead on a spinning hoop

A bead of mass is threaded on a frictionless circular wire hoop or radius that is rotating at a constant velocity

The bead makes and angle with respect to the vertical.

Find equilibrium positions for the bead where it remains at a constant in terms of the centripetal acceleration ( )

The bead has two velocity components.

If the bead is moving up or down the circular hoop then

The bead is moving with the circular hoop as the hoop is rotated at a constant angular velocity

The Lagrangian for the bead is

For the bead to be in equilibrium the acceleration should be zero

if degrees then the bead is at the top or bottom of the hoop

another way is if

or

if

Then only and are equilibrium positions, is too small , the circular ring is rotating too slowly.

if

then

#### stability

There are two cases depending on

When and

then

oscillating motion due to gravity pulling it down any movement away from the bottom of the hoop will result in a net force back towards the bottom of the hoop to
oscillation frequency if bead is displaced a small amount from the bottom of the hoop

If

Then

If the bead moves away from then you will have oscillation again but you won't return to as gravity is pulling the bead down and the normal force only cancels this at . This is an unstable critical point.

When and

then

consider a small displacement from the above angle by and amount \epsilon

:: is close to zero

similarly

:: is close to zero

but

given equilibrium condition

since

given equilibrium condition

#### Bifurcation

Bifurcation: The splitting of a main body into two parts.

In the example above, as the hoop begins to rotate in the presence of a gravitational field we have one stable equilibrium at the bottom of the hoop. When the hoops rotating frequency reaches a critical value then the bead will be unstable at the bottom and seek to move up the hoop. There are two possible directions equilibrium points on either side of the hoop. The system is said to have bifricated at this point.