Physical Pendulum and Center of Oscillation

Back in high school Physics II, we tied a string to the ceiling, hung an iron ball off it, and watched it swing~ swing~. But here, we’re gonna hang a rigid body off a pivot and let that swing.

(I mean, you could just skewer a sweet potato and let it dangle. But to keep things clean, let’s say we’ve hung a 2D flat plate on a skewer.)

So the rigid plate is hanging on a skewer like so. At this exact moment, what’s the magnitude of the torque about the rotation axis, from gravity?

$mgl\sin\theta$.

Right? Torque $\vec{N} = \vec{r}\times\vec{F}$ is a vector, but here we only care about the magnitude.

And that equation from before,

$$N = I\dot{\omega}$$

plug it in:

$$I\dot{\omega} = I\ddot{\theta} = -mgl\sin\theta$$$$\text{if}\quad \theta \approx 0 \\ I\ddot{\theta} = -mgl\theta \\ \ddot{\theta} + \frac{mgl\theta}{I} = 0 \\ \text{equation of motion — done.}$$

Now this isn’t the time to solve the ODE from scratch, so I’m just gonna write down the general solution for $\theta$ directly. heh

$$\theta = \theta_{0}\cos(2\pi f_{0}t - \Phi)$$

Stuff like $\theta_0$ and the $\Phi$ afterwards — those get pinned down by initial conditions, so don’t sweat them. heh

So,

$$f_{0} = \frac{1}{2\pi}\sqrt{\frac{mgl}{I}}\quad \text{which gives} \\ T_{0}(\text{period}) = \frac{1}{f_{0}} = 2\pi\sqrt{\frac{I}{mgl}} = 2\pi\sqrt{\frac{k^{2}}{gl}}$$

Ahh — the radius of gyration $k$ from earlier is hiding in there~~

So the period of a rigid body whose CM is distance $l$ from the axis, swinging as a pendulum (only when the swing angle is absurdly small) — looks like that.

But wait??? That looks suspiciously similar to the pendulum period we memorized in high school physics!!

Remember those two formulas we drilled into our heads back then?!

$$T_{0} = 2\pi\sqrt{\frac{l}{g}}\quad (\text{simple pendulum})\quad l:\text{length},\ g:\text{gravitational acceleration} \\ T_{0} = 2\pi\sqrt{\frac{m}{k}}\quad (\text{mass on a spring})\quad m:\text{mass},\ k:\text{spring constant}$$

That’s the one!!!

Oh ho~~~ There’s a clean analogy between the pendulum motion of a rigid body and the simple pendulum we saw way back in high school!!!

Namely: the period of a physical pendulum equals the period of a simple pendulum of length $k^2/l$.

OK OK, let’s verify.

Check: what’s the square of the radius of gyration of a long~ thin rod of length $a$?

$$I(\text{moment of inertia}) = \int_{0}^{a}{r^{2}\,dm} \\ = \int_{0}^{a}{r^{2}\,\frac{m}{a}\,dr} = \frac{1}{3}ma^{2}$$$$\therefore\ k^{2} = \frac{I}{m} = \frac{a^{2}}{3} \\ T_{0} = 2\pi\sqrt{\frac{k^{2}}{lg}} = 2\pi\sqrt{\frac{a^{2}/3}{g\cdot a/2}}\quad (\text{the $l$ from before is the distance to the CM, so $a/2$}) \\ \text{rod's period} = 2\pi\sqrt{\frac{2a}{3g}}$$

Yeaaaah okay. Checks out.

The two pendulums have the same period, I’m telling you~ hahahaha (wild, heh)

You might be wondering why I’m going on and on about just the period…

It’s because we’re chasing down the “center of oscillation.”

Ever taken swings at one of those coin-op batting cages where balls come firing at you?

Sometimes you swing and the bat rattles like crazy — your hands go completely numb. Other times, the ball clearly cracks off the bat, flush as anything…

(you know that feeling? the clean shot??) …and your hands feel nothing.

@@@ That’s NOT because you hit it “too cleanly”!

It’s because you hit it on the center of oscillation. Let’s crack open the secret of this “center of oscillation” business.

First, let’s express $I$ using the parallel axis theorem.

(Parallel axis theorem works for any rigid body, whereas the perpendicular axis theorem only works for flat flat flat flat flat flat rigid plates — you remember that, right??)

$$I = I_{cm} + ml^{2}\quad (l:\text{distance from the rotation axis to the axis through the CM}) \\ \text{substituting }I = mk^2: \\ mk^{2} = mk_{cm}^{2} + ml^{2} \\ \therefore\ k^{2} = k_{cm}^{2} + l^{2}$$

OK OK OK OK, so then:

$$\text{When the axis is at O:} \\ T_{0} = 2\pi\sqrt{\frac{k^{2}}{lg}} = 2\pi\sqrt{\frac{k_{cm}^{2} + l^{2}}{lg}} \\ \text{When the axis is at O':} \\ T_{0} = 2\pi\sqrt{\frac{k_{cm}^{2} + l'^{2}}{l'g}}$$

One’s the period when the axis sits a distance $l$ from the CM, the other’s when the axis sits $l'$ from the CM~

(What we always have to remember: this is only valid for tiny swing angles;;)

$$\text{When the two periods are equal????} \\ 2\pi\sqrt{\frac{k_{cm}^{2} + l^{2}}{lg}} = 2\pi\sqrt{\frac{k_{cm}^{2} + l'^{2}}{l'g}} \\ ll' = k_{cm}^{2} \\ \text{The point O' satisfying this equation} \\ \text{is called the 'center of oscillation' relative to O.}$$

When you swing a baseball bat — if the ball hits this center of oscillation, the batter feels way less shock.

So you grip point O on the bat and go pow at the ball, and if the ball happens to land right at point O’, that’s the spot you feel as a clean shot. haha

You’re gripping the bat at that point, so that grip becomes one axis. And the ball hits exactly at O’, so that becomes another axis, right? (center of percussion?)

Apparently among baseball players they call this the sweet spot (I don’t really know baseball (crying))

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Originally written in Korean on my Naver blog (2015-01). Translated to English for gdpark.blog.