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Homework Help: Lagrangian rolling cylinders + small oscillations

  1. Aug 30, 2017 #1
    1. The problem statement, all variables and given/known data
    A point mass m is fixed inside a hollow cylinder of radius R, mass M and moment of inertia I = MR^2. The cylinder rolls without slipping

    i) express the position (x2, y2) of the point mass in terms of the cylinders centre x. Choose x = 0 to be when the point mass is at the bottom.
    Show the velocity of the point mass is:

    x2' = x'(1-cos(x/R))
    y2' = x'(sin(x/R))

    ii) find the lagrangian for the generalised co-ordinate x
    and write down the Euler lagrange equation to obtain the equation of motion for x (DO NOT SOLVE)

    iii) find the frequency of small oscillations about the stable equilibrium state

    2. Relevant equations
    L = T-V

    3. The attempt at a solution
    i) can find this just by using the geometry of the situation and then differentiating
    ii) I *think* the lagrangian is:
    L = (x'^2)(M+m)(1-cos(x/R)) + mgRcos(x/R)

    using the Euler lagrange equation I think the equation of motion is:

    2x''(M+m)[1-cos(x/R)] - (x')^2((M+m)/R)(sin(x/R) + mgsin(x/R) = 0

    iii) this is where I am stuck....
    stable equilibrium is at x = 0
    if you use sin x = x then the e.o.m reduces to:

    2x''(M+m)(1-cos(x/R)) + mgx/R = 0
    I have neglected the middle term since x/R^2 is negligible...

    I was expecting to get an equation of the form x'' + w^2 x = 0 were w is the frequency ... however when expanding the cos you don't get this...

    Have I done something really wrong?

    Attached Files:

  2. jcsd
  3. Aug 30, 2017 #2


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    Please show your work. It does not look correct to me.

    This is a correct assumption for linearising the equation near the stable point. However, if ##\sin(x) = \mathcal O(x)##, then ##\cos(x) = 1 + \mathcal O(x^2)##.
  4. Aug 30, 2017 #3
    Thanks ... for the lagrangian
    i thought the KE would be rotational KE of the cylinder which is 0.5Iw^2 plus the KE of the mass = 0.5m(x2'^2+y2'^2)
    w^2 = v^2/R^2 since no slip
    I have attached my working ... am I doing this really wrong?

    Attached Files:

    • 1.jpg
      File size:
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    • 2.jpg
      File size:
      46.8 KB
  5. Aug 30, 2017 #4
    You say the cylinder rolls with out slipping, but what does it roll upon? Is it on a horizontal surface, a set of rollers, across a hill, or what? The kinematics makes all kinds of difference here.
  6. Aug 30, 2017 #5
    It is rolling on a flat horizontal surface
  7. Aug 30, 2017 #6


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    A couple of things that I noticed in your derivation of L:

    You used the same symbol V for the speed of the point mass and also for the speed of the center of the cylinder. But these speeds are different.

    You took the kinetic energy of the cylinder to be (1/2)Icω2 where Ic is the moment of inertia of the cylinder about its central axis. But the cylinder is not just rotating about this axis. It is also in translational motion. (Or, you can think of the cylinder as instantaneously rotating about the point of contact of the cylinder with the ground.)
  8. Aug 30, 2017 #7
    Ah yes!!! knew i'd made a stupid mistake somewhere - thanks for pointing out!
    Just to check... the translational velocity of the C of M is v'
    and therefore the rotational velocity w is x'/R?
    thanks so much
  9. Aug 30, 2017 #8


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    I'm a little confused with the notation. In most of your work, a prime appears to denote a time derivative. So, v' would be an acceleration.

    If x denotes the position of the CoM, then x'/R would be the rotational velocity (or, angular velocity) of the cylinder.
  10. Aug 30, 2017 #9
    Ah yes sorry .. I meant x'
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