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Falling (toppling) rigid tower (uniform rod) 
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#1
Jan1413, 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 


#2
Jan1413, 10:10 AM

P: 1,474

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...111620(F).pdf 


#3
Jan1413, 10:35 AM

Thanks
P: 5,791

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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