Talk:Pendulum (derivations)

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I think it'd be good to add a way to derive the Pendulum using Lagrangian formulation.

Using Lagrangian mechanics the pendulum can be derived through the use of the Euler-Lagrange equation of motion. Defining Lagrangian as the difference between kinetic and potential energies,

L=T-V\,

where T is kinetic energy and V is potential.

T=\frac {m}{2} \left ( \dot x^2 + \dot y^2 \right )
V=mgh=mgl( 1- \cos\theta )\,
L=\frac {m}{2} \left ( \dot x^2 + \dot y^2 \right ) - mgl(1-\cos\theta)
x = lsinθ

where x is the displacement in the horizontal direction

y = lcosθ

y is the displacement in the vertical direction

\dot x^2 = l^2 \dot \theta^2 \cos^2 \theta
\dot y^2 = l^2 \dot \theta^2 \sin^2 \theta
L=\frac {m}{2} \left ( l^2 \dot \theta^2 \cos^2 \theta + l^2 \dot \theta^2 \sin^2 \theta \right ) - mgl(1-\cos\theta)

Using Lagrange's Equation

\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = \frac{\partial L}{\partial \theta}

we get

\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = m \left ( l^2 \ddot \theta^2 \cos^2 \theta + l^2 \ddot \theta^2 \sin^2 \theta \right ) = ml^2 \ddot \theta^2
\frac{\partial L}{\partial {\theta}} = -mgl\sin\theta
ml^2 \ddot \theta = -mgl\sin\theta
\ddot \theta = - \frac{g}{l}\sin\theta

--Nefreat

[edit] Wrong negative signal in some formulas of the derivations

The explanation for the negative signal in equations:

F = - mg\sin\theta\,
a = -g\sin\theta\,

namely, that g\, is negative because it is pointing downward, is simply not true.
If the formulas refer to absolute values, then there should be no minus signal.

The reason behind these negative signs is the authors wish to obtain the correct signal in equation:

\ell{d^2\theta\over dt^2} = -g\sin\theta\,

My argument for this minus signal is as follows. The starting equation is

a_t = g\sin\theta\,

where a_t\, is the tangencial acceleration.
Now, in Figure 2, let us suppose that the bob is going UP. In this case we should write:

a_t = - {dv\over dt}

because the tangencial acceleration and the velocity are in opposite directions. And,

v = {ds\over dt},

because velocity is in the direction of the growing of the arc s\,. This gives:

a_t = - {dv\over dt} = - {d^2s\over dt^2}

and since s = \ell\theta\,, we finally have

a_t = - \ell{d^2s\over dt^2}

The same argument goes when the bob is going DOWN. In this case we have:

a_t = {dv\over dt}, and v = - {ds\over dt}

But the final result is, as before,

a_t = - \ell{d^2s\over dt^2}

Rui Ferreira 18:05, 12 April 2007 (UTC)