$L=1 \mathrm{~m}$ long rods are used to construct a regular $n$-sided polygon, and we swing it as a physical pendulum in its plane about an axis passing through one of its vertices. Calculate the period of oscillation and examine some special cases!
The period of a physical pendulum for small oscillations is given by the equation

$$
T=2 \pi \sqrt{\frac{\Theta_{0}}{m g s}}
$$

where $\Theta_{0}$ is the moment of inertia of the oscillating body about the suspension point, $m$ is the mass of the oscillating body, and $s$ is the distance between the suspension point and the center of mass ($g$ is the acceleration due to gravity).

Our task is to determine $\Theta_{0}$ (the moment of inertia about one vertex of the polygon, which is the suspension point). For this, we need to use the Steiner theorem, which is as follows:

$$
\Theta_{0}=\Theta_{s}+m d^{2}
$$

where $\Theta_{s}$ is the moment of inertia about an axis passing through the center of mass, $\Theta_{0}$ is the moment of inertia about another axis parallel to the first and at a distance $d$ from it, and $m$ is the mass of the body.

The moment of inertia of a thin rod of mass $m$ and length $L$ about an axis perpendicular to the rod and passing through its center of mass is

$$
(1 / 12) m L^{2}
$$

According to the Steiner theorem, the moment of inertia of a rod of mass $m$ forming one side of an $n$-sided polygon about the center of the polygon is

$$
(1 / 12) m L^{2}+m((L / 2) \operatorname{ctg} \gamma)^{2}=(1 / 4) m L^{2}\left(1 / 3+\operatorname{ctg}^{2} \gamma\right)
$$

where $\gamma=180^{\circ} / n$, and $(L / 2) \operatorname{ctg} \gamma$ is the distance between the center and the midpoint of the side. The moment of inertia of the polygon about the center is thus

$$
\Theta_{s}=(1 / 4) m n L^{2}\left(1 / 3+\operatorname{ctg}^{2} \gamma\right)
$$

Since the center is also the center of mass, we can apply the Steiner theorem again. The distance between a vertex and the center is $L /(2 \sin \gamma)$, so

$$
\Theta_{0}=\Theta_{s}+n m\left(\frac{L}{2 \sin \gamma}\right)^{2}=\frac{1}{2} m n L^{2}\left(\frac{1}{\sin ^{2} \gamma}-\frac{1}{3}\right)
$$

Thus, the period of the physical pendulum for small oscillations is

$$
T=2 \pi \sqrt{\frac{\Theta_{0}}{m g L /(2 \sin \gamma)}}=2 \pi \sqrt{n \frac{L}{g}\left(\frac{1}{\sin \gamma}-\frac{\sin \gamma}{3}\right)}
$$

$\left(\gamma=180^{\circ} / n\right)$

For some cases $(L=1 \mathrm{~m})$, the period is:

$$
\begin{array}{ll}
n=2 & T \approx 2.32 \mathrm{~s}, \\
n=3 & T \approx 3.23 \mathrm{~s}, \\
n=4 & T \approx 4.35 \mathrm{~s}, \\
n=5 & T \approx 5.50 \mathrm{~s}, \\
n=6 & T \approx 6.65 \mathrm{~s}, \\
n=8 & T \approx 8.94 \mathrm{~s} .
\end{array}
$$

As the number of sides of the polygon increases, $T$ becomes larger, and as $n \rightarrow \infty$, $T$ approaches infinity. This is because the side length of the polygon is always a fixed $L$. However, if we do not fix the value of $L$ but the radius $R$ of the circumscribed circle, then

$$
L=2 R \sin \gamma
$$

Substituting this value, in the limit $n \rightarrow \infty(\gamma \rightarrow 0)$, we obtain

$$
T=2 \pi \sqrt{2 R / g}
$$

which is the same as the period of a thin circular ring of radius $R$ when suspended from its circumference.

Based on the work of Albert Vass (Csongrád, Batsányi J. Gymnasium, 3rd grade).