A physical pendulum consists of two meter-long sticks joined together as shown in above figure. What is the pendulums period of oscillation about a pin inserted through point $\mathrm{A}$ at the center of the horizontal stick?

To use Eq. $T=2\pi \sqrt{\frac{I}{mgh}}$ we need to locate the center of mass and we need to compute the rotational inertia about $A$. The center of mass of the stick shown horizontal in the figure is at $A$, and the center of mass of the other stick is $0.50\mathrm{m}$ below $A$. The two sticks are of equal mass, so the center of mass of the system is $h=\frac{1}{2}(0.50m)=0.25m$ below $A$, as shown in the figure. Now, the rotational inertia of the system is the sum of the rotational inertia $I_1$ of the stick shown horizontal in the figure and the rotational inertia $I_2$ of the stick shown vertical. Thus, we have
$I=I_1+I_2=\frac{1}{12}M{L}^2+\frac{1}{3}M{L}^2=\frac{5}{12}M{L}^2$
where, $L=1.00m$ and $M$ is the mass of a meter stick (which cancels in the next step). Now, with $m=2M$ (the total mass), yields
$T=2\pi \sqrt{\frac{\frac{5}{12}M{L}^2}{3Mgh}}=2\pi \sqrt{\frac{5L}{6g}}$
where, $h=L/4$ was used. Thus, $T=1.83\mathrm{s}$.