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Thread: 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 $\displaystyle \alpha$ 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 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= $\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
    Attached Thumbnails Attached Thumbnails Lagrangian function of a ladder on a smooth wall-ladder.bmp  
    Last edited by punkstart; Feb 20th 2010 at 08:36 AM. Reason: clarity
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  2. #2
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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 $\displaystyle \alpha$ 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
    So here's how i will be sending it in to my lecturer;

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

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

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

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

    Hope this (if correct) can help others.
    Prof Lesame, my student number is 36025909
    Last edited by punkstart; Feb 20th 2010 at 08:34 AM.
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