Philosophy 167: Class 10 - Part 9 - Huygens' Theory of Centrifugal Force: Solving for the Tension in a String.

Smith, George E. (George Edwin), 1938-


  • Synopsis: Reviews Huygen's paper De Vi Centrifuga, which were put as an appendix to Horologium Oscillatorium.

    Opening line: "All right, next topic is De Vi Centrifuga, this paper which, again, appeared in 1703."

    Duration: 9:28 minutes.

    Segment: Class 10, Part 9.
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All right, next topic is De Vi Centrifuga, this paper which, again, appeared in 1703. The results all date from 1659, and almost all the theorems in De Vi Centrifuga are put at the end as an appendix to Horologium Oscillatorium. And I added them as additional reading, what I stuck you with on Saturday, five more pages, but done without proofs.
So the first occasion anybody had to publish a proof of the solution for uniform circular motion is actually Newton, in the Principia. There was no question in Newton's mind that Huygens had it. But he just didn't publish it at the time, and that meant others couldn't figure out how Huygens got his results.
So Wren and Hooke and then Halley couldn't really get this result. Newton could. Newton had, in fact, gotten it independently of Huygens, but let's start looking at it. So the first thing Huygens worries about is he wants to be using circular motion to measure surface gravity. Surface gravity, the tension in a string, goes as time squared, so to speak.
That is, the distance you actually travel, from one unit of time to the next, grows as time squared. We know that from Galileo. To compare it, the same thing ought to be true in circular motion. So he looks at where is the object going to be if it goes off in a tangent?
And it's gonna be at K, L and N. But then he says, if the angle is very small, the difference between S and N, L and D, and K and C gets very small. And if you do that, so that these differ very little from one another, then you find out that the distance EC, the distance between where the object would be if it were not confined to a circle versus where it is on the circle, sure enough goes as 1 over the cosine of the angle minus 1.
Which, in the limit, becomes the angle squared, which is time squared. And therefore, just as if you take an object hanging from a string, break the string and it starts accelerating, it will move a distance covered proportional to time squared. Exactly the same thing is happening here. It's proportional to time squared.
Therefore, we can compare the two. I'll come right back. Here's another example. This is a Cartesian problem. What's the tension in the string in circular motion? The one I emphasized so strongly last week. How's he approaching it? He's reducing it to a Galilean problem, a uniform acceleration problem.
The actual solution I'll summarize, it's quite beautiful. So what we want is the tension in the string that retains a body in uniform circular motion, and here's the idea. It's gonna be measured by the limit as EG over time squared. Okay, cuz the tension is going to be, what's the tension in the string hanging?
It's the distance divided by the time squared, incremental distance. Having shown it goes as time squared, we're going to do the same thing here. Now what's beautiful, and I'll show you this in just a moment, here's a theorem of Euclid. Look at EG, look at GC, look at AG.
Here's the theorem. GC is the mean proportional between GE and GA, regardless of how you draw the chord and the tangent. So I can substitute, here, EG for GC squared over AG. What is GC squared? It's the velocity squared in a given time. What is AG? It's essentially the radius.
And that's what it becomes as I go to the limit. So I end up that the tension in the string, this is our modern result, is proportional to V square over R, which is the same thing as the radius over a period square. And you have to put in the weight of the body, because the heavier the body, the greater the tension in the string, whether you're hanging it or spinning it.
So that's the solution for uniform circular motion, V square over R is the force that's generated in the string that's holding it. Remember, that was Descartes's problem, except he was gonna measure the tendency to recede by the tension in the string. What Huygens has done is take the tendency to recede and solve for the tension in the string.
So he simply flipped the Descartes problem and solved the problem of uniform circular motion. It's our solution. There's nothing, other than mass replacing weight. It's entirely our structure. This is the Euclid proof. You see the first case has the chord going as a diameter through the center of the circle, and then the second case does it for a totally arbitrary one.
You will see Newton using this same proposition of Euclid's. It's the culmination of book three. It's proposition 36, book three of Euclid. Newton, on his own, gets the same solution for uniform circular motion using the same theorem from Euclid. So it's all nice and clean. So I've given you the Euclid you could look at yourself.
There's a striking feature here, though, when he's doing the circular motion case in order to get to a conical pendulum. So how's he think about a conical pendulum? Well, he's working from static, specifically from his countryman Stevin. So you look at the top. Tension in a vertically suspended string is produced by a body hanging from it varies as the density and volume of the body and the strength of the tendency bodies at the location in question have to descent.
Put them on an inclined plane, and the amount of weight you have to have here to balance a weight on the inclined plane is proportional to the cosine of that angle theta. That's again a result from Stevin. It's called Stevin's triangle. So it was known, and what's the tension in the string then, connecting these two?
It's this weight hanging down. I said this last week. I'll drive the point home again. Mersenne in his Harmonie Universelle, the theory of musical instruments. How do you control the tension in a stringed instrument? You hang a weight on the end. What do you then call the magnitude of the tension?
The weight that you hang, okay? So this is all straight from statics he's doing. This is the diagram that's stunning, and the editors of his posthumous papers chose not to include it in the paper you read. I've made a big deal of it, so has Joella Yoder. Joella Yoder I should tell you about.
Pass her dissertation around. Joella Yoder is the one who's made the study of Huygens possible for all the rest of us. She wrote her dissertation in the 1980s at University of Wisconsin and then chose not to pursue an academic career. Her husband was head of programming at Boeing Aircraft.
She lived in Seattle with him till he died. She's still in Seattle. I have begged her so many times to pursue an academic career, but she's just chosen not to. In Holland she's referred to as Mrs. Huygens. She is the world authority on him. She's the one person to have gone through all the notebooks really carefully in our generation.
So she makes a big deal of this. I make a big deal of it. Look what he's done. This is a problem in statics. I'm gonna hang. I've got a thread here, I've got a pulley here. I'm gonna hang a weight here, and I'm gonna hang another weight on a pulley pulling the thing out.
Okay, so I've got two weights on this body, B. One is in this direction, one is in this direction. And there are two questions I can answer. What's the relationship between this angle and the ratio of the two weights? That's what I want for a conical pendulum, right?
But even neater, what's the amount of weight here that's needed to compensate for the two? And of course it's going to be what we would call the vector sum of the two.