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

  1. #1
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    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

    [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 \dot{\theta}^2 where i is the rotational innertia about the centre of mass of the ladder... back soon
    Attached Thumbnails Attached Thumbnails Lagrangian function of a ladder on a smooth wall-ladder.bmp  
    Last edited by punkstart; February 20th 2010 at 09:36 AM. Reason: clarity
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  2. #2
    Junior Member
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    Ok,here's my solution,please correct me if i'm wrong!

    Quote Originally Posted by punkstart View Post
    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

    [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 \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
    Last edited by punkstart; February 20th 2010 at 09:34 AM.
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