Falling (toppling) rigid tower (uniform rod)


by zaphat
Tags: falling, rigid, toppling, tower, uniform
zaphat
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#1
Jan14-13, 07:42 AM
P: 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
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#2
Jan14-13, 10:10 AM
P: 1,260
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/w...1116-20(F).pdf
voko
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#3
Jan14-13, 10:35 AM
Thanks
P: 5,484
Your tower is a pendulum, whose initial angle from "pointing down" is 180 degree. Starting from the pendulum's differential equation,

[tex]

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
[/tex]

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

[tex]
\\

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

\\

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

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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