Answer
$$ \tau=0.970 \mathrm{~s}
$$
Work Step by Step
Datum at initial level of center of gravity of disk.
$$
\begin{aligned}
\Delta & =\bar{r}(1-\cos \theta) \\
E & =T+V \\
& =\frac{1}{2} I_{I C}(\dot{\theta})^2+W \bar{r}(1-\cos \theta) \\
\dot{E} & =\dot{\theta}\left(I_{I C} \ddot{\theta}+W \bar{r} \sin \theta\right)=0
\end{aligned}
$$
For small $\theta$,
$$
\begin{aligned}
& \sin \theta=\theta \\
& \ddot{\theta}+\frac{W \bar{r}}{I_{I C}} \theta=0 \\
& \bar{r}=\frac{4(0.5)}{3 \pi}=0.212 \mathrm{ft} \\
& I_A=I_G+m \bar{r}^2 \\
& \frac{1}{2}\left(\frac{10}{32.2}\right)(0.5)^2=I_G+\frac{10}{32.2}(0.212)^2 \\
& I_G=0.02483 \mathrm{slug} \cdot \mathrm{ft}^2 \\
& I_{I C}=I_G+m(r-\bar{r})^2 \\
&=0.02483+\frac{10}{32.2}(0.5-0.212)^2 \\
&=0.05056 \mathrm{slug} \cdot \mathrm{ft}^2 \\
& \tau=\frac{2 \pi}{\omega_n}=2 \pi \sqrt{\frac{I_{I C}}{W \bar{r}}}=2 \pi \sqrt{\frac{0.05056}{10(0.212)}} \\
& \tau=0.970 \mathrm{~s}
\end{aligned}
$$
