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Falling (toppling) rigid tower (uniform rod)

  1. Jan 14, 2013 #1
    I am working on an animation, which involves a rigid, vertical tower falling (toppling) to the ground, and I am stuck at its core physics.

    Actually this is the same as the thin uniform rod initially positioned in the vertical direction, with its lower end attached to a frictionless axis.

    I would need the angle (compared to the ground) of the rod in a given time.

    The tower is 50meters long. (It is a simple animation, the effect of gravity only is enough: no friction, no radial acceleration, no stress forces etc. is needed)

    Thanks in advance
  2. jcsd
  3. Jan 14, 2013 #2


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    Gold Member

    So what have you tried to do to solve your problem?
    Here is something to grit your teeth on with regards to moment of inertia.
    http://www.uta.edu/physics/courses/wkim/lctr_notes/phys1443-fall06-1116-20(F).pdf [Broken]
    Last edited by a moderator: May 6, 2017
  4. Jan 14, 2013 #3
    Your tower is a pendulum, whose initial angle from "pointing down" is 180 degree. Starting from the pendulum's differential equation,


    x'' + a \sin x = 0


    x''x' + a (\sin x)x' = 0


    \frac {x'^2 - {x'}_0^2} {2} - a(\cos x - \cos x_0) = 0

    ## x_0 = \pi ## (the pendulum is upward from "pointing down"), so


    x' = \sqrt {{x'}_0^2 + 2a(\cos x + 1)}


    \int_{\pi}^x ({x'}_0^2 + 2a(\cos x + 1))^{-1/2} dx = t

    The latter integral, as far as I can tell, does not exist in the closed form, but it can be tabulated between ## \pi ## (upward) and ## \pi/2 ## (toppled), which will give you the trajectory you want. You will need ## {x'}_0 ## which is the initial angular velocity, and you will need ## a ##, which is ## \frac {3g} {2L} ## for a uniform rod, ## L ## being the length.
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