Forest UCM LEq

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

[math]S \equiv \int_{t_1}^{t_2} \left ( T(\dot x) - U(x) \right ) dt = \int_{t_1}^{t_2} f(x,\dot x ; t) dt = \int_{t_1}^{t_2} \mathcal {L} ( x, \dot x;t) dt [/math]

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

[math] \left [ \left ( \frac{\partial f}{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial f}{\partial \dot x} \right ) \right ] =0 [/math]

or

[math] \left [ \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot x} \right ) \right ] =0 [/math]

here [math]x^{\prime} = \frac{d x}{d t} \equiv \dot x[/math]

[math] \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) = \frac{\partial }{\partial x} \left ( T(\dot x) - U(x) \right ) = -\frac{\partial U(x)}{\partial x} = F_x[/math] the generalized force if I have conservative forces

[math] \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot x} \right ) = \frac{d}{dt} \left ( \frac{\partial }{\partial \dot x} \right ) \left ( T(\dot x) - U(x) \right ) [/math]

[math] = \frac{d}{dt} \left ( \frac{\partial }{\partial \dot x} \right ) T(\dot x) = \frac{d}{dt} m \dot x = \frac{d}{dt} p_x = \dot{p}_x = F_x [/math] Newton's second law in an Inertial reference frame = time derivative of the generalized momentum

thus

[math] \left [ \left ( \frac{\partial \mathcal {L} }{\partial x} \right ) - \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot x} \right ) \right ] = F_x - F_x = 0 [/math]

Lagrange's Equations in generalized coordinates

Generalized coordinates [math](q_i)[/math] are a set of parameters that uniquely specify the instantaneous state of a dynamical system.

The number of independent generalized coordinates [math](N_q)[/math] is given by subtracting the number of constraints [math](N_C)[/math] from the number of degrees of freedom [math]N_{DF}[/math].

[math]N_q = N_{DF} - N_C[/math]

Pedulum example

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

[math]N_{DF} =2 [/math] There are 32 degrees of freedom for the 2-D problem

[math] N_C =1 [/math] The particle is constrained to a rod

[math]N_q = 2-1 = 1 [/math] The motion of the particle may be described using one component

The constraint may be expressed in cartesian coordinates as

[math] x^2 + y^2 = l^2[/math]

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

[math]\vec r = x \hat i + y \hat j = x \hat i + \sqrt{l^2-x^2} \hat j = \vec r(x)[/math] [math]x[/math] is the generalized coordinate

You could also express the position as a function of the deflection angle [math]\phi[/math] in cartesian coordinates

[math]\vec r = l \cos \phi \hat i + l \sin \phi \hat j = \vec r (\phi)[/math] [math]\phi[/math] is the generalized coordinate

Note

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

[math]\vec r = l \hat r[/math]

then the dependence of the function on [math]\phi[/math] would not be obvious as this dependence is implicit to changes in the [math]\hat r[/math] direction

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

[math]U = mgh = l( 1-\cos \phi)[/math]

[math] T = \frac{1}{2} m v^2 = \frac{1}{2} l^2 \dot \phi^2[/math]

Motion on Sphere example

Consider a particle constrained to move on a sphere of radius [math]R[/math].

[math]N_{DF} =3 [/math] There are 3 degrees of freedom for the 3-D problem

[math] N_C =1 [/math] The particle is constrained to the surface of the sphere

[math]N_q = 3-1 = 2 [/math] The motion of the particle may be described using two components

The constraint expressed in terms of cartesian coordinates is

[math] x^2 + y^2 + z^2 = R^2[/math]

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

[math] z^2 = R^2-x^2 - y^2[/math]

[math]\vec r = x \hat i + y \hat j \pm \sqrt{R^2-x^2 - y^2} \hat k = \vec r ( x,y)[/math]

or if you chose the angles [math]\theta[/math] and [math]\phi[/math] from spherical coordinates

[math] \vec r = R \cos \phi \sin \theta \hat i + R \sin \phi \sin \theta \hat j + R \cos \theta \hat k = \vec r (\phi, \theta) [/math]

Generalized Force and Momentum

Generalized Force [math]\equiv \frac{\partial \mathcal {L} }{\partial q}[/math]

Generalized Momentum [math]\equiv \frac{\partial \mathcal {L} }{\partial \dot q}[/math]

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.

[math] \left [ \left ( \frac{\partial \mathcal {L} }{\partial q} \right ) = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot q} \right ) \right ] [/math]

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

[math]\mathcal L(x,\dot x, y, \dot y;t) = T - U = \left ( \frac{1}{2} m(\dot x^2 + \dot y^2) \right ) - \left ( U(x,y) \right ) [/math]

[math] \frac{\partial \mathcal {L} }{\partial q} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot q} \right ) [/math]

for the generalized coordinate labeled [math]x[/math]

[math] \frac{\partial \mathcal {L} }{\partial x} = - \frac{\partial U }{\partial x} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot x} \right ) = \frac{d}{dt}m \dot x = m\ddot x \Rightarrow F_x = ma_x[/math]

for the generalized coordinate labeled [math]y[/math]

[math] \frac{\partial \mathcal {L} }{\partial y} = - \frac{\partial U }{\partial y} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot y} \right ) = \frac{d}{dt}m \dot y = m\ddot y \Rightarrow F_y = ma_y[/math]

In polar coordinates

In cartesian coordinates the generalized coordinates are represented as the parameters [math]r[/math] and [math]\phi[/math]

[math]\mathcal L(r, \dot r, \phi, \dot \phi;t) = T - U = \left ( \frac{1}{2} m(\dot r^2 + r^2 \dot \phi^2) \right ) - \left ( U(r,\phi) \right ) [/math]

[math] \left ( \frac{\partial \mathcal {L} }{\partial q} \right ) = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot q} \right ) [/math]

The [math]r[/math] parameter

[math] \frac{\partial \mathcal {L} }{\partial r} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot r} \right ) [/math]

[math] mr\dot \phi^2-\frac{\partial U }{\partial r} = m \ddot r [/math]

[math] mr\dot \phi^2+F_r = m \ddot r [/math]

[math] \Rightarrow F_r = m\left ( \ddot r - r\dot \phi^2\right ) [/math]

recall

[math]\vec{a} = \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi} [/math]

The [math]\phi[/math] parameter

[math] \frac{\partial \mathcal {L} }{\partial \phi} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot \phi} \right ) [/math]

[math] -\frac{\partial U }{\partial \phi} = \frac{d}{dt} \left ( mr^2 \dot \phi \right ) [/math]

In polar coordinates

[math]\vec \nabla = \frac{ \partial}{\partial r } \hat r + \frac{1}{r} \frac{ \partial}{\partial \phi } \hat \phi[/math]

[math]\Rightarrow F_{\phi} = \left . -\nabla U \right |_{\hat \phi} = - \frac{1}{r} \frac{ \partial U }{\partial \phi } \hat \phi[/math]

[math] -\frac{\partial U }{\partial \phi} = \frac{d}{dt} \left ( mr^2 \dot \phi \right ) [/math]

[math] rF_{\phi} = \frac{d}{dt} \left ( L \phi \right ) [/math]

[math] \mathcal \Tau = \frac{d L }{dt} [/math]

Examples using Lagrange's Equation

Pendulum

Returning to the problem of an object of mass [math]m[/math] attached to a weightless rod of length [math]l[/math] 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, [math]\phi[/math].

[math]U = mgh = mgl( 1-\cos \phi)[/math]

[math] T = \frac{1}{2} m v^2 = \frac{1}{2} l^2 \dot \phi^2[/math]

[math]\mathcal L(\phi, \dot \phi;t) = T - U = \left ( \frac{1}{2} mr^2 \dot \phi^2 \right ) - mgl( 1-\cos \phi) [/math]

[math] \left ( \frac{\partial \mathcal {L} }{\partial q} \right ) = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot q} \right ) [/math]

[math] \frac{\partial \mathcal {L} }{\partial \phi} = \frac{d}{dt} \left ( \frac{\partial \mathcal {L} }{\partial \dot \phi} \right ) [/math]

[math] -mgl \sin \phi = \frac{d}{dt} \left (ml^2 \dot \phi \right ) [/math]

[math] = ml^2 \ddot \phi [/math]

[math] \ddot \phi + \frac{g}{l} \sin \phi = 0[/math]

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 [math]\phi_1[/math] and [math]\phi_2[/math]

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

Object 1

[math]x_1 = l_1 \sin \phi_1 \;\;\;\;\; y_1 = -l_1 \cos\phi_1[/math]

[math] \dot x_1 = l_1 \cos \phi_1 \dot \phi_1 \;\;\;\;\; \dot y_1 = l_1 \sin \phi_1 \dot \phi_1[/math]

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

[math]x_2 = l_1 \sin \phi_1 + l_2 \sin \phi_2 \;\;\;\;\; y_1 = -l_1 \cos \phi_1-l_2 \cos \phi_2[/math]

[math] \dot x_1 = l_1 \sin \phi_1 \dot \phi + l_2 \sin \phi_2 \dot \phi_2\;\;\;\;\; \dot y_1 = l_1 \sin \phi_1 \dot \phi_1 + l_2 \sin \phi_2 \dot \phi_2[/math]

[math]T = \frac{1}{2}( m_1 \dot x_1^2 + m_2 \dot x_2^2)[/math]

[math]= \frac{1}{2} (m_1 l_1^2 \dot \phi_1^2 + m_2 ( l_1^2 \dot \phi_1^2 + 2 l_1 l_2 \cos(\phi_1-\phi_1) \dot \phi_1 \dot \phi_2 + l_2^2 \dot \phi_2^2))[/math]

Bead on a hoop

http://www-physics.ucsd.edu/students/courses/fall2010/physics110a/LECTURES/CH06.pdf

Forest_Ugrad_ClassicalMechanics