Philosophy 167: Class 10 - Part 14 - Horologium Oscillatorium: the Center of Oscillation for Cycloidal Pendulums.

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


  • Synopsis: Concludes with an overview of the Horologium Oscillatorium.

    Opening line: "hese are two of the last three paragraphs of part four."

    Duration: 4:16 minutes.

    Segment: Class 10, Part 14.
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These are two of the last three paragraphs of part four. And they're stunning. So start. It is not possible to determine the center of oscillation for pendula suspended between cycloids and how to overcome the difficulty with which this causes. First, why is it not possible? The center of oscillation depends on the length of the pendulum as well as the masses of the box.
In a cycloidal pendulum, the length is changing. So there's no one answer to the center of oscillation. Okay? That makes the cycloidal pendulum not as attractive as a small arc pendulum for measuring surface gravity, okay? And he recognizes that. Let's see what he says. If one very closely compares what we have demonstrated, above concerning pendulums suspended between cycloids.
With our discussion at the center of oscillation, he will see that these oscillations fall short of the perfect equality which we would prefer. First he will have doubts in determining the generating circle of a cycloid as to whether the length of the pendulum should be measured from the point of suspension to the center of gravity of the attached lead weight, or to the center of oscillation.
I've skipped a little now. If we say that the length should be measured to the center of oscillation, then it will not be clear how what was proven about the center of oscillation applies to a pendulum which is constantly changing its length. As is the case for a pendulum, which moves between cycloids.
For it would seem that its center of oscillation changes for each different link. So, we don't have a simple correction for center of oscillation with cycloidal pendulums. It works for circular pendulums because the length stays the same. Now listen to how he's going to fix this. However, if we wish to escape these problems completely, we can succeed if we make the sphere or lentil of the pendulum move around its own horizontal axis.
This is done by inserting both ends of that axis into the bottom of the rod of the pendulum. The rod having been split in half for the purpose. For in this way, the nature of motion is such that the sphere of the pendulum will maintain perpetually the same position in respect of the horizontal plane.
And any point in it, as well as its center, will cross the same cycloids. Hence a consideration of the centers of oscillation is no longer relevant. Such a pendulum will maintain an equality of times which is no less perfect. Then if all its weight were contained in one point.
What's he say? If I've got to take a spherical bob. As it goes up and down. The sphere's rotating. Put it on a bevel, so that its vertical axis always remains the same. It's not rotating anymore. That's the reason the center of oscilation is not the center of gravity.
Some of the motion is going into rotation, rather than translation. He's picked up the distinction between rolling versus falling. He didn't realize the whole thing til 1693. He was the first that I know of anywhere, just to state it and give the right number, but he's got exactly the right idea here.
That what is happening, and he's thinking of it in what we would call energy terms. Some of the energy's going into rotation. How do you prevent that? Have it not rotate. Then the problem's gone, okay? Then the very next paragraph explains air resistance really doesn't have any significant effect here because it cancels going down and coming back.