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Classical Mechanics (Lagrangian)

  1. Oct 12, 2005 #1

    I posted this yesterday, but I figured I'd better post it again since the other thread can't be replied to at this time since its in the archives.

    I'm looking for some advice on whether or not I'm doing a problem correctly.

    The problem is:
    A particle of mass m rests on a smooth plane. (the particle starts at r) The plane is raised to an inclination [tex]\theta[/tex], at a constant rate [tex]\alpha[/tex], with [tex]\theta = 0[/tex] at t=0, causing the particle to move down the plane.

    So, I'm taking the x to be the distance the particle travels down the slope.

    I come up with the following as the Lagrangian:

    [tex]L = \frac{1}{2} m\dot{x}^2 - mg(r-x)sin\theta[/tex]

    I'm not sure if this is correct.

    I would then get the equations of motion to be [tex]mgsin\theta - m\ddot{x}=0[/tex] and [tex]-mgsin(r-x)cos\theta=0[/tex].
  2. jcsd
  3. Oct 12, 2005 #2


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    You don't seem to have explicitly factored in that the slope is changing in time. (There is no alpha dependence in your equations.) What this also means is that the velocity of the particle is not as simple as it seems, and is not along the inclined plane in the lab reference frame anyway. For example, the particle on the slope has velocity component perpendicular to the incline, that matches the incline's velocity at that point perpendicular to its surface.

    Be careful about which reference frame you are using when you write down the equations.

    One way to suspect that you don't quite have the right answer is that the solution you give would be the same for a static incline - do you really think the solution should be the same?
  4. Oct 21, 2005 #3
    I'm also just learning about Lagrangians, and I probably would have made the same mistake without the above hint, but would the Lagrangian instead be:
    [tex] L= \frac{1}{2}m \dot{r}^2 + \frac{1}{2}mr^{2}\dot{\theta}^{2} -mgrsin\theta[/tex]
    Is that correct?
  5. Oct 21, 2005 #4
    I mean a much more elementary way to look at it is simple conservation of energy to find out what energy contributions there are.

    KE: T trans + T rot
    PE: V(y)

    Ttrans: .5mr^2
    T rot: .5Iw^2
    V: mg(sin[theta])

    where you would figure out I as a particle on a massless string.
  6. Oct 21, 2005 #5
    I haven't had any formal training in solving Lagrange's equations, but here's what I would do:

    [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{r}} -\frac{\partial L}{\partial r}=0[/tex] and [tex]\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} -\frac{\partial L}{\partial \theta}=0 [/tex]

    Which I'd then solve to get:

    [tex]m \ddot{r}+mgsin\theta=0[/tex] and [tex]mr^2 \ddot{\theta}+mgrcos\theta=0[/tex]
    Is that the right way to do it?
    Last edited: Oct 21, 2005
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