# The Toppling of a transmitter mast

1. Sep 13, 2004

### elivian

About a week ago a big transmitter mast has fallen. The transmitter mast was about 100 meters high, is it possible to calculate how long it took before it touched the ground without knowing it's weight? The time starts as soon as the mast starts falling. I need the answer for my school exam and since I couldn't find the answer anywhere on the net I hope you can help solve this...

(the air friction is negligible as well as the thickness of the mast)

2. Sep 13, 2004

### HallsofIvy

Without knowing exactly HOW it fell it's impossible to say. Certainly one can calculate that the time required for an object to fall directly down from 100 m (ignoring air resistance) is $\sqrt{200/9.8}$ which is about 4.5 seconds.

3. Sep 13, 2004

### elivian

Let's assume it was just nearly standing sraight upright when it fell. The bottom of the mast is the center of rotation. So the top will move in a quarter of a circle to the ground.

I did some calculations, in the end I found this formula but I don't know if it's correct/fully correct.

9.81 * cos(h) = a

the h is the angle between the ground and the transmitting mast
the a is the acceleration at a specific point (h)

I guess this is not all, isn't it?

4. Sep 13, 2004

### Tide

The problem is that the time it takes to fall depends critically on the initial deviation of the tower from its vertical position. You'll find that as the initial deviation is made arbitrarily small the time it takes to fall becomes arbitrarily large.

5. Sep 13, 2004

### elivian

And what if the mast starts at 45% with no speed? Is there any formula about this all?

Since this will just be a start for a formula about a chain of toppling domino's I need to a formula for the falling of a domino or comparable a transmitter mast...

6. Sep 13, 2004

### Tide

Yes, it can be done but it's not really trivial!

If the base is fixed at a point and R is the distance from the base to the center of gravity of the transmitter then the torque is $\tau = -R mg \cos \theta$. But this is equal to the moment of inertia (I) times the angular acceleration so that
$$I \frac {d^2 \theta}{dt^2} = -R m g \cos \theta$$
which you will need to solve to find the time it takes to fall. Basically, it's like a pendulum with very large amplitude.

7. Sep 15, 2004

### elivian

The latex graphics won't load, i tried to deciver the formula but it's too difficult, what's theta? whats mg mass * 9.81? Can you please explain it in a easy way?