Moment of Inertia

Let m_AB = m = 1 kg

AB = 0.4 m =  $\mathcal{l}$

d = OM = $\sqrt[]{0.16-0.04}=\sqrt[]{0.12}$

$d=2\sqrt[]{3}*10^{-1}$  

$I_{ABaboutO}=\frac{m{\mathcal{l}}^2}{12}+m{d}^2$           

$\therefore l_{hexagon,aboutO}=6\left[\frac{m{\mathcal{l}}^2}{12}+m{d}^2\right]=\frac{m{\mathcal{l}}^2}{2}+6m{d}^2$       

$=\frac{1*{\left(0.4\right)}^2}{2}+6*1*{\left(2\sqrt[]{3}*10^{-1}\right)}^2=\frac{0.16}{2}+6*4*3*10^{-2}$   

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-1 kgm²

Using parallel axis theorem:

$l=2*\left[l_{cm}+m{d}^2\right]$  

$l=2*\left[\frac{2}{5}\left(1.5\right){\left(0.5\right)}^2+\left(1.5\right){\left(2.5\right)}^2\right]$

l = 19.05 kgm²

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}*\frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$