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Homework Help: Lagrange - Cylinder rolling on a Fixed Cylinder

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform solid cylinder of mass `m' and radius `a' rolls on the rough outer surface of a fixed horizontal cylinder of radius `b'. Let `theta' be the angle between the plane containing the cylinder axes and the upward vertical (generalized coord.)

    Deduce that the cylinder will leave the fixed cylinder at [tex]\theta=\cos^{-1}(4/7)[/tex]

    3. The attempt at a solution

    I got,

    [tex]\ddot{\theta} + \frac{2g}{3(a+b)} \sin(\theta) = 0[/tex]

    I just don't know the best way to approach answering the last bit about when the cylinder leaves the fixed on it's rolling on.
    Or how do I solve this ODE?
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2
    Answer the following questions, so we can help you:

    1. What happens when the cylinder 'leaves' the surface of the fixed cylinder?

    2. Unrelated - The DE you have is nonlinear, and does not have a transcendental solution. Hint for the solution: look up any standard book on classical mechanics for the 'exact' solution of an unforced simple pendulum.
  4. May 26, 2009 #3
    Normal force will be zero i.e.,

    [tex]mg\cos \theta - N = m \frac{v^2}{a+b}[/tex]

    => [tex] v^2 = g(a+b) \cos \theta[/tex]

    So if I knew velocity I could solve for theta but I can't get [itex]\theta(t)[/itex] because I can't solve the ODE. Saying [itex]\sin \theta \approx \theta[/itex] doesn't make sense here.
  5. May 26, 2009 #4
    Ok I got this part. Was my original ODE for [tex]\ddot{\theta}[/tex] correct?

  6. May 26, 2009 #5
    Show us your working here, leading up to the DE. I think its okay, but I haven't checked (I remember the 2/3 factor from memory, having solved something like this a long time ago).
  7. May 26, 2009 #6
    [tex]V = -mg(a+b)\cos \theta[/tex]

    [tex]T = 1/2mv^2 + 1/2I\omega^2 = 1/2mv^2 + 1/2(1/2ma^2)(v/a)^2[/tex]

    [tex]= 3/4mv^2 = 3/4m(a+b)^2\dot{\theta}^2[/tex]

    [tex]L = T-V[/tex] So,

    [tex]\frac{d}{dt}\frac{dL}{d\dot{\theta}} = 3/2m(a+b)^2\ddot{\theta}[/tex] (*)

    [tex]\frac{dL}{d\theta} = -mg(a+b)\sin \theta[/tex] (**)

    Then subtracting (**) from (*) and equating to zero gives you my result.
  8. May 26, 2009 #7
    I'm curious - have you been studying the use of Lagrange multipliers for solving problems with constraints? This problem is a classic example from that subject, where you would use two generalized coordinates (e.g x and y) that are constrained in some way - in this case the path of the center of the rolling cylinder follows a circular path while it remains in contact with the lower cylinder. One of your eqs. of motion is then the eq. of constraint, and you can solve for when the force of constraint goes to zero.

    I believe it's equivalent to what you're doing, so it's not necessarily a better approach, but I was just wondering if perhaps that's what you were meant to do here.
  9. May 27, 2009 #8
    Seems alright to me. As an added exercise, try energy conservation to get the same result.
  10. May 27, 2009 #9
    No we haven't done Langrange multipliers yet.
  11. May 27, 2009 #10


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    This is a little roundabout, but technically, while you cannot solve for [tex]{\theta}[/tex], you can solve your differential equation for [tex]\dot{\theta}[/tex] from your equation of motion.

    Multiply your equation of motion throughout by [tex]\dot{\theta}[/tex], and integrate. You will get the Hamiltonian, which in this case, could have been gotten from T+V.

    I suppose this would qualify as 'Using the Lagrangian' as you did not immediately jump to E = T + V

    From that, you can solve for [tex]\dot{\theta}[/tex], and find v, and then apply normal force = 0.

    For a more technical approach, in mechanics, for a variable [tex]{x}[/tex], the momenta (conjugate variable of x) is
    [tex]{p}=\frac{\partial L}{\partial \dot x}[/tex]

    The Hamiltonian is then defined as [tex]{\dot x p} - L[/tex], which in this case, gives T + V.

    So you would technically be working from the Lagrangian.
    Last edited: May 27, 2009
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