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**punkstart** The problem statement : A uniform ladder of mass m and length 2a rests with one end on a smooth horizontal floor and the other end against a smooth vertical wall.The ladder is initially at rest and makes an angle $\displaystyle \alpha$ with the horizontal. Determine the lagrangian function for the ladder.

If you see my attached picture,the centre of mass follows a circular path of radius a. I want to use the angle $\displaystyle \theta$ as my generalised coordinate,since i believe the ladders position is completely described by this angle..is this right? Can i use this angle as my generalised coordinate ?

Now if L=T-U, can i approximate the motion of the ladder as the motion of it's centre of mass along the circular path s, i.e. T= $\displaystyle (1/2)m\dot{s}^2 $or will the kinetic energy of the ladder have some rotational components ? What might my kinetic energy equation look like if i allowed for rotation?

So that my lagrangian ( without thinking about rotation and rotational inertias) is given by ... [ s=a$\displaystyle \theta$ and $\displaystyle \dot{s} = a\dot{\theta}$] and so T=$\displaystyle 1/2ma^2\dot{\theta}^2$

and U=$\displaystyle mgasin\theta$ so that

L$\displaystyle (\theta,\dot{\theta})$=$\displaystyle 1/2ma^2\dot{\theta}^2-mgasin\theta$

[edit] well it seems i'm slowly solving this one on my own ! Will put what i figure out on here, but just went through my first year mechanics book,seems the kinetic energy is best described by the translational+rotational kinetic energies,where rotational kinetic energy is given by 1/2I$\displaystyle \dot{\theta}^2$ where i is the rotational innertia about the centre of mass of the ladder... back soon