# Math Help - Lagrangian function of a ladder on a smooth wall

1. ## Lagrangian function of a ladder on a smooth wall

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 $\alpha$ 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 $\theta$ as my generalised coordinate,since i believe the ladder's 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= $(1/2)m\dot{s}^2$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 $\theta$ and $\dot{s} = a\dot{\theta}$] and so T= $1/2ma^2\dot{\theta}^2$
and U= $mgasin\theta$ so that
L $(\theta,\dot{\theta})$= $1/2ma^2\dot{\theta}^2-mgasin\theta$

 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 $\dot{\theta}^2$ where i is the rotational innertia about the centre of mass of the ladder... back soon

2. ## Ok,here's my solution,please correct me if i'm wrong!

Originally Posted by 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 $\alpha$ 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 $\theta$ 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= $(1/2)m\dot{s}^2$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 $\theta$ and $\dot{s} = a\dot{\theta}$] and so T= $1/2ma^2\dot{\theta}^2$
and U= $mgasin\theta$ so that
L $(\theta,\dot{\theta})$= $1/2ma^2\dot{\theta}^2-mgasin\theta$

 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 $\dot{\theta}^2$ where i is the rotational innertia about the centre of mass of the ladder... back soon
So here's how i will be sending it in to my lecturer;

I have one generalised coordinate, $\theta$ since i can find the x and y intercepts of the ladder from $\theta$ alone.

We have two kinetic energies for this motion, namely
$T_{T}$=Translational kinetic energy of the centre of mass along the curve s, and $T_{R}$= Rotational kinetic energy of the ladder's rotation about it's centre of mass, with the final kinetic energy given by $T=T_{T}+T_{R}.$

Here $T_{T} = (1/2)m\dot{s}^2=(1/2)ma^2\dot{\theta}^2\; (since\; s=a\theta)$
And $T_{R}= (1/2)I\dot{\theta}^2\; where\;I=(m/3)a^2\;so\;that\; T_{R}=(m/6)a^2\dot{\theta}^2$
Thus $T=(1/2)ma^2\dot{\theta}^2+(m/6)a^2\dot{\theta}^2=(2/3)ma^2\dot{\theta}^2$

Now the potential is $U= mg(Y_{G})\;where\;Y_{G}$ is the y component of the centre of mass of the ladder,and $Y_{G} = asin\theta$
Thus $U= mgasin\theta$ and the lagrangian is given by
$L(\theta,\dot{\theta}) = (2/3)ma^2\dot{\theta}^2 - mgasin\theta$

Hope this (if correct) can help others.
Prof Lesame, my student number is 36025909