Built on Facts

An exploration of physics, and the search to understand our universe

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Falling chimneys

May 31st, 2008 · 1 Comment

A while back the county government in my home town was considering a request from a phone company to construct a new cell tower to bring cellular service to a rural area. The government balked, largely for aesthetic reasons, though they couched their opinion in safety terms. They said they were concerned about the tower falling in inclement weather. Judging by the location, the maximum possible damage might be to one of the poor cows in the field surrounding the site. But that’s politics for you. Modern tower engineering is very well developed and your average cell tower will not be brought down by much short of a direct tornado hit. But what happens when a tower does fall?

Make a nice tall tower out of something. Lego bricks, building blocks, Oreos, anything. Then let it topple over. If you pay particular attention, you may notice that the tower has probably failed to fall in one piece. Long before it hit the ground, it likely snapped. The taller the tower, the more likely this is to happen.

Why? Presumably some kind of bending force is involved, since that’s usually what snaps things in half. Let’s see if we can track down its source. Maybe we’ll get somewhere by calculating how fast the tower should fall under the influence of gravity.

There’s a couple ways to attack this problem, but as is often the case it’s easiest to look at things from an energy perspective. We’d like to know the gravitational potential energy of the tower as it’s standing before it falls. We might guess it’s an average between 0 and mgL where L is the full height, and that guess would be right. To show this, we can take each bit (mass per length λ) of the tower with potential energy gλx dx and integrate between the bottom and top of the tower, sure enough:

and when it hits the ground all that will have been converted into kinetic energy of rotation about the tower base. In that case, K = (1/2)Iω2, where I is the moment of inertia and ω is the angular velocity - revolutions per second times 2π. Don’t worry about the 2π if you don’t know why it’s there. Just be assured that in a lot of problems it makes the math much easier to include it. The moment of inertia is basically just a measure of how much inertia a rotating object has. We can look it up in a table or calculate it in the following way (again, don’t worry about the details if you’re not familiar with the procedure)

Now use that in the rotational kinetic energy equation and set it equal to the gravitational potential energy.

And solve for ω, the angular velocity.

Ah ha! Disaster! If the tower wants to naturally stay in one piece, the whole thing needs to fall at the same angular speed. But it doesn’t! The taller the tower is (and thus the bigger L), the smaller ω is. This means taller towers will take longer to fall down - just what we expect. But that means the upper part of the tower has to hold back the lower parts of the tower from reaching the ground as quickly as they would naturally. This requires a force, and if the tower is long enough that force will break the tower.

With a little more effort a clever person can calculate where the tower will break. Here’s a site where some physics students from LMU did that calculation and experiment, which you may well find pretty interesting. Think about it the next time you see a smokestack!

Tags: College Physics 101 · Physical Concepts · Worked Problems

1 response so far ↓

  • 1 Nick // Jun 1, 2008 at 2:54 pm

    This is really fascinating - these ‘worked problems’ posts are one of the reasons that Built on Facts is one of my new favorite blogs! What inspired you to start writing it?


    Matt replies: I was sitting in a graduate Quantum II class taught by a brilliant older Russian professor (one of Landau’s grad students, actually). He spent all his time on derivations and pure theory. I thought to myself that though the derivations and mathematical formalism are vital, there was never enough time left in lecture for the professor to actually use the method to do some examples. This is true in most physics classes I’ve taken. Though the hard work necessary to translate the formalism into homework solutions is what makes a person good at physics, some help from meaningful examples also can do a lot of good. So I thought that maybe I could present some that other people may find useful.

    Mostly it’s been undergrad-level stuff so far. I figure once class starts again this fall I’ll write up some of the solutions to whatever problems I get assigned, so there will be some harder problems represented here as well.

    I’m very glad you like the blog, by the way! Any time you have suggestions, feel free to let me know since I’m still new at all this online writing stuff.

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