lagranian mechanics

Quote:

Originally Posted by edeffect

Hey

Going to be honest I am struggling to get this question done. Havent been keeping up with my study and I practically have no idea how to do this question. But Im looking through some notes and trying to work it out. But if you are willing to help I would really appreciate it!

Cheers (Nod)

Ed

I presume the problem is in setting up the Lagrangian? I will use two generalized coordinates: x and $\theta$ . Take a look at the mass M first. I am setting the 0 point for the GPE to be at the equilibrium position of the spring. So
$V = -Mgx$
and
$T = \frac{1}{2}M \dot{x}^2$

The swinging mass m is a bit more of a problem. The potential energy is easy enough:
$V = -mg(x + y~cos( \theta ))$
where y is the length of the string.

For the kinetic energy we have the downward motion of the mass m to contend with as well as the radial motion of the swing. So we have two terms:
$T = \frac{1}{2}m \dot{x}^2 + \frac{1}{2}my^2 \dot{ \theta } ^2$

The Lagrangian will be total T minus total V:
$L = \frac{1}{2}M \dot{x}^2 + \frac{1}{2}m \dot{x}^2 + \frac{1}{2}my^2 \dot{ \theta } ^2 + Mgx + mg(x + y~cos( \theta ))$

-Dan