Falling (toppling) rigid tower (uniform rod)

1. Jan 14, 2013

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)

2. Jan 14, 2013

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 [Broken]

Last edited by a moderator: May 6, 2017
3. Jan 14, 2013

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.