The Geometry of Isochronal Pivot Points for a Physical Pendulum

Note that if you’re reading this on a mobile device, some of the equations don’t look right. In particular, one thing I noticed is that the square root radical over a fraction appears to only be over the numerator. All such square roots are over the denominator too. The equations look right (to me) when viewed on a non-mobile device (e.g., my laptop).

This post is about a topic in physics, which I suspect may not be of interest to many regular TIE readers. But, it could interest irregular (?) readers.

Why physics? It was my undergrad major, and I’m now helping my daughter through AP Physics. I’m finding myself going down some rabbit holes, exploring things I didn’t learn in college. Some aren’t in any textbook I have at my fingertips or online (so far as I can tell). This post is about one such thing. I used AI to check the math and logic. If you see a problem, please let me know.

What This Post Shows
Our starting point is a well-known property of a physical pendulum (also called a compound pendulum), the discovery of which is credited to Christian Huygens: you can pivot the pendulum from two different points and get the same period (they’re isochronal). Given one pivot point, Huygens and any other basic physics text that covers this topic, shows how to find “the” other isochronal pivot point.

What this post shows is that, in fact, there are an infinite number of isochronal points (hence the quotes around “the” in the previous sentence). Moreover, they all lie on two circles about the center of mass (CM).

The First Isochronal Circle
The derivation for the period of a physical pendulum can be found online or in many textbooks. The period T about pivot point P is given by

$$T = 2\pi \sqrt{\frac{I}{mgd}}$$

where I is the moment of inertia about the pivot point, m is the pendulum’s mass, g is the acceleration due to gravity, and d is the distance between P and the CM. From the parallel axis theorem

$$I = I_{\text{CM}} + m d^2$$

where the first term is the moment of inertia about the center of mass. Therefore,

$$T = 2\pi \sqrt{\frac{I_{\text{CM}} + m d^2}{m g d}}$$

which is constant for any distance d from the CM in any direction. Thus, there are infinite isochronal pivot points that lie on a circle of radius d centered on the CM. This is the first isochronal circle.​​

The Second Isochronal Circle
As discussed in many places but not shown in a simple way (that I’ve seen), there is a pivot point isochronal with P and not on the circle described above. It’s called the center of oscillation and we’ll label it point Q. The end of this post has a straightforward proof that the isochronal point Q is a distance L from P through the CM where

$$L = \frac{I}{md}$$

Define d’ = L – d as the distance between Q and the CM. The physical pendulum has a moment of inertia I’ about point Q. We can plug d’ and I’ into the equations provided in the section above (The First Isochronal Circle). The conclusion follows by the same argument as above that the period is a constant T for any point a distance d’ from the CM. We know it is T (the same period as in the previous section) because P and Q are isochronal pivot points. Thus there are an infinite isochronal pivot points that lie on a circle of radius d’ centered on the CM. This is the second isochronal circle.

Or, to sum up, Huygens and countless others open our eyes to the fact that there are two isochronal points, P and Q. What is also true is that there are two isochronal circles about the CM, one contains P, the other Q.

This is consistent with the obvious fact that the period of oscillation of a uniform density rod pivoting at one end is the same as that pivoting on the other end (an isochronal circle goes through both ends). Or, the period of oscillation is the same for any point of a circular disk of uniform density equidistant from the center (isochronal circles are concentric about the CM). These obvious facts are clear to us from symmetry.

What’s given above shows that the circular isochronal symmetry is there even for asymmetric (arbitrarily shaped, with non-uniform density) physical pendulums. This is not intuitive (not to me anyway). Yet, the only thing that isn’t fixed for all points relating to a given pendulum in the expression for T is the distance of the pivot point from the CM. With that, isochronal circularity is unavoidable.

Conclusion
Notice that d can be any positive distance, giving rise to any positive period T. So, by the above arguments, all isochronal pivot points lie on pairs of concentric circles centered about the CM. (There is a small technicality that for some (OK, an infinite number of) values of d, some isochronal points may not be within the body of the pendulum. That doesn’t mean they’re not isochronal with the other points on the circle(s) associated with d (or d’). It just means that actually getting the pendulum to pivot around such a point is challenging in practice. It’d take a massless, unbending support from that point to the pendulum’s CM.)

Proof That the Center of Oscillation is Isochronal (Or Pivot Q Has the Same Period as Pivot P)
Above, I promised this proof. Skip it if you’re willing to trust Huygens and countless other physics texts, some of which state this without proof.

With all the terms as defined in the post, start with some preliminary stuff to get an expression of d’ in terms of d that will be useful later. Recall that

$$L = \frac{I}{m d},$$

By the parallel axis theorem, and as noted in the post

$$I=I_{\text{CM}}+m{d}^2$$

Therefore,

$$L = \frac{I_{\text{CM}} + m d^2}{m d}.$$

And, simplifying,

$$L = \frac{I_{\text{CM}}}{m d} + d.$$

Using this and the definition of d’ (= L – d) from the post, we have

$$d’ = \left(\frac{I_{\text{CM}}}{m d} + d\right) – d.$$

So that

$$d’ = \frac{I_{\text{CM}}}{m d}.$$

The period of the pendulum pivoting about point Q is given by

$$T’ = 2\pi \sqrt{\frac{I’}{m g d’}},$$

The job is to show T’=T. We will do so by plugging in for I’ and d’ and doing some algebra. Here we go. The only thing we can do with I’ is apply the parallel axis theorem.

$$I’ = I_{\text{CM}} + m d’^2.$$

Plug in the expression for d’ from above. (This is why we did that preliminary work.)

$$I’ = I_{\text{CM}} + m \left(\frac{I_{\text{CM}}}{m d}\right)^2.$$

Cancel m and plug this into the numerator of T’ and use the expression for d’ in the denominator too.

$$T’ = 2\pi \sqrt{\frac{I_{\text{CM}} + \frac{I_{\text{CM}}^2}{m d^2}}{m g \cdot \frac{I_{\text{CM}}}{m d}}}.$$

There’s lots to cancel to simplify. Go ahead and do that on a scrap of paper or in your head. You’ll get

$$T’ = 2\pi \sqrt{\frac{1 + \frac{I_{\text{CM}}}{m d^2}}{\frac{g}{d}}}.$$

Multiply numerator and denominator in the square root by md² to get

$$T = 2\pi \sqrt{\frac{I}{m g d}} = 2\pi \sqrt{\frac{I_{\text{CM}} + m d^2}{m g d}}.$$

Recognize, from the parallel axis theorem, that the numerator is just I and that this is the expression for T. So, T’ = T and we are done.