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Lagrangian and Euler-Lagrange Equation Problem

  1. Sep 26, 2015 #1
    First off, apologies if this is in the wrong forum, if my notation is terrible, or any other signs of noobishness. I just started university and I'm having a hard time with my first Lagrange problems. Help would be very much appreciated.

    1. The problem statement, all variables and given/known data

    A body of mass m is lying on a smooth, frictionless plane. The plane, which is originally horizontal, is lifted up at one end at a constant rate such that the angle of the plane with the horizontal at time t is θ = at.
    This problem comes with a diagram which includes an arrow pointing down labled g. I presume this is a constant force g from gravity?

    2. Relevant equations
    I need to show that the lagrangian of the body, expressed in terms of the distance q from the base of the plane where it hits the horizontal, is
    L = 1/2(mqdot^2) + 1/2(ma^2q^2) - mgqsin(at)

    Also, determine the Euler-Lagrange equations for the system


    3. The attempt at a solution
    I understand that L = T - V and T = 1/2(mqdot^2), V = 1/2(kq^2); but after that, I'm just not sure what to do to make it look like it's presented in the problem. I'm not even sure what k should be for V. ma^2 ? If so, could someone explain why?

    Thank you very much for reading.
     
  2. jcsd
  3. Sep 26, 2015 #2

    Orodruin

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    Hint: The ma^2 term does not come from the potential energy.
     
  4. Sep 26, 2015 #3
    Thank you very much! Thanks to the hint, I was able to get somewhere during my break at work today.
    So, I think I've worked out that the 1/2(mqdot^2) is the kinetic energy from the ball moving along the surface and the 1/2(ma^2q^2) is from the ball being moved upwards as the slope rises.

    Since I'm working with gravity, V is obviously just mgh! xD So yeah, I've figured that part out too.

    All I need to do now is find a nice way to express how I know what the kinetic energy is. Could you please give me an example?
     
  5. Sep 27, 2015 #4

    Ray Vickson

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    If ##\vec{v}_1## and ##\vec{v}_2## are the velocity components along two perpendicular directions ##\vec{d}_1## and ##\vec{d}_2##, then isn't the kinetic energy just equal to
    [tex] \text{K.E.} = \frac{1}{2} m \left( \vec{v}_1 \cdot \vec{v}_1 + \vec{v}_2 \cdot \vec{v}_2 \right) ? [/tex]

    You can get convenient perpendicular components by looking at the positions of the particle at two nearby times ##t## and ##t + \Delta t##, as shown in the attached figure.
     

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  6. Sep 27, 2015 #5

    Orodruin

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    I think the easiest way is to simply consider polar coordinates in the plane of rotation. The rotation then implies a constraint ##\theta = at## on the angle.
     
  7. Sep 27, 2015 #6
    Part two of the question asks me to find the Euler-Lagrange Equation of motion of the mass. I'll just put up my answer here and if one of you could please tell me if I'm right or off the mark, I'd be grateful. Once again, thanks for all the help so far.

    mqdoubledot + mgSinat - ma^2q = 0
     
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