Falling (toppling) rigid tower (uniform rod)

[zaphat]

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

 

[256bits]

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

 

[voko]

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.