# Pendulum Lagrangian

#### The Simple Pendulum

You are given a pendulum that is made from a rod of length b having at its end two masses m and M. Mass M can move on a horizontal surface without friction. Please refer to the diagram. We use $theta$ (the angle with vertical) $X$ (the distance of $M$ from origin). Initial conditions are: at t=0, M is at X=0 and speed is $v0$. $theta(0)=theta0$ and $theta'(0)=0$.

a) Please write the pendulum Lagrangian of these two masses.

b) Please write the Euler-Lagrange equations.

c) What quantity is conserved for this pendulum lagrangian?. You can find simple examples of Lagrangian here.

a)

X and $theta$ are the generalized coordinates of this pendulum lagrangian

$x = X +b*sin(theta)$

$y =-b*cos(theta)$

The Lagrangian is thus

$L = T- U$   (=kinetic energy – potential energy)

We make the following notations

$dX/dt = dot{X}$

$d(theta)/dt = dot{theta}$

$Ek(M) = M*(dot{X})^2 / 2$

$Ek(m) = Ek (translation) + Ek(rotation) = m(dot{X})^2 / 2 + m*(dot{theta})^2*b^2 / 2$

because:

-for translation $m$ has the same translational speed as $M$

-for rotation $m$ has speed $v = omega*b = (dot{theta})*b$

Potential energy is only for $m$

$U = -m*g*b*cos(theta)$

$L = (M+m)*(dot{X})^2 /2 + m*b^2*(dot{theta})^2 /2 – m*g*b*cos(theta)$

$dL/dX =0$   ($X$ does not appear explicitly, but $dot{X}$)

$dL/d(dot{X}) =(M+m)*(dot{X})$

$d/dt (dL/d(dot{X})) = (M+m)*d/dt (dot{X}) = (M+m)*(ddot{X})$

The first Euler Lagrange equation is

$dL/dX = d/dt (dL/d(dot{X}))$

$0 = (m+M)*(ddot{X})$

$dL/d(theta) = m*g*b*sin(theta)$

$dL/d(dot{theta}) = m*b^2*(dot{theta})$

$d/dt (dL/d(dot{theta})) =m*b^2*d/dt (dot{theta}) = m*b^2*(ddot{theta})$

The second Euler Lagrange equation is

$dL/d(theta) = d/dt (dL/d(dot{theta}))$

$m*g*b*sin(theta) = m*b^2*(ddot{theta})$

b) Hamiltonian $H=T+U$ and total energy $Etot$ are the same if Lagrangian does not depend EXPLICITLY on time t.

In our case this is true and $L$ does not depend on t.

Hence, YES $H$ and $Etot$ are the same.

YES, $H$ is conserved since it does not depend EXPLICITLY on time $t$

YES, $Etot$ is conserved since it is equal to $H$