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Pendulum in a freely moving box

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A box of mass M can slide horizontally on a frictionless surface. A simple pendulum of string length l and mass m, is suspended inside the block. Denote the coordinate of the centre of mass of the box by x and the angle that the pendulum makes with the vertical by θ . At t = 0 the pendulum displacement is θ = θ0 which is not equal to zero.

    Find the Lagrangian and the equation of motion for the generalized coordinates x and θ . Which conservation law is obtained as a result of the cyclic coordinate?

    Find the solutions for x and θ in the small angle approximation, hence show that the pendulum and the box execute SHO about their centre of mass at a frequency ω=[(M+m/M)^0.5]*(g/l)^0.5

    2. Relevant equations
    The EL equation

    3. The attempt at a solution
    The kinetic energy T=0.5M(dx/dt)2+0.5m[(dx/dt)2+l2(dθdt)2+2l(dx/dt)(dθ/dt)cosθ].

    The potential energy measured from the support point of the pendulum U=-mglcosθ.

    The Lagrangian L=T-V

    This gives the EOM as
    (also x is a cyclic coordinate, which before taking the time derivative of (∂x/∂t), shows linear momentum is conserved), and

    In the small angle approximations, sinθ->θ, cosθ->1, then the EOM become

    Then rearranging the second for (d2x/dt2)=-(gθ+l(d2θ/dt2)) and putting it into the first gives
    i.e there is a nasty (dθ/dt)2 term which if not present would give me the correct answer... However I can't justify removing it, as small θ doesn't necessarily mean small (dθ/dt) does it?

    Any clues? Thanks.
  2. jcsd
  3. Mar 6, 2014 #2


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    What happens to the kinetic energy of ##l \dot \theta \sin \theta## ?
    [edit] Never mind, you have it in.
  4. Mar 6, 2014 #3


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    At the risk of stumbling again: ##{\partial T\over \partial \dot x} = (M+m)\dot x + m \ l \dot \theta \cos\theta\quad ## so why is the first EOM so complicated ?
    Last edited: Mar 6, 2014
  5. Mar 6, 2014 #4
    Because I then need to use d/dt on that - which involves differentiating the following term by the product rule ## m \ l \dot \theta \cos\theta\quad ##
  6. Mar 6, 2014 #5
    It does. In the small oscillations approximation, equations should be linear, anything non-linear is Taylor expanded, and only the linear terms are retained.
  7. Mar 6, 2014 #6
    I'm not quite sure what you mean. Do you mean I am allowed to ignore it just on the grounds it is non-linear?
  8. Mar 6, 2014 #7
    Not because it is not linear. Because its Taylor expansion does not have any linear terms.
  9. Mar 6, 2014 #8
    Right, I'm not too sure how to go about Taylor expanding (dθ/dt)2 unfortunately - I should understand then.
  10. Mar 6, 2014 #9
    It is already its own expansion. That is why it gets ignored.
  11. Mar 6, 2014 #10
    So I can ignore (dθ/dt)2 because it is not a linear term. Why can I ignore non-linear terms? Surely I should care about whether the quantity is small or not?
  12. Mar 6, 2014 #11
    Has the method of the small oscillations approximation been explained to you? What is the theory behind it?
  13. Mar 6, 2014 #12
    Not very much really - I just know the angles are small and so I can expand things using that.
  14. Mar 6, 2014 #13
    The Lagrangian is ##L = T(q, \dot q) - U(q)##. It is further known that in all practical cases ##T(q, \dot q) = t(q)\dot q^2##. Assuming ##q = 0## in equilibrium, we must have ##U'(0) = 0##. So in small oscillations ##U(q) \approx U(0) + U''(0)q^2/2 = c + k q^2 /2 ##. ##T(q, \dot q)## is approximated with ##t(0)\dot q^2 = m \dot q^2/2##. The corresponding equation of motion is then ## m \ddot q + k q = 0 ##.
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