Question

$\text{ABC}$ is a plane lamina of the shape of an equilateral triangle. $\text{D, E}$ are mid-points of $\text{AB, AC}$ and $\text{G}$ is the centroid of the lamina. The moment of inertia of the lamina about an axis passing through $\text{G}$ and perpendicular to the plane $\text{ABC}$ is $I_0$. If the $\text{ADE}$ is removed. The moment of inertia of the remaining part about the same axis is $\frac{N{I}_0}{16}$ where $N$ is an integer. The value of $N$ is

Answer 1

Let, mass of triangular lamina $=m$,
length of side $=l$,
Then the moment of inertia of the lamina about an axis passing through centroid $\text{G}$ perpendicular to the plane is,

$I_0\propto m{l}^2$
$I_0=km{l}^2[k=\text{shape factor}]$
Let us divide the given lamina $\text{ABC}$ into four equal lamina as shown,


Let the moment of inertia of $\text{(DEF)}=I_1$ about $\text{G}$
$I_1\propto \left(\frac{m}{4}\right){\left(\frac{l}{2}\right)}^2\propto \frac{m{l}^2}{16}\Rightarrow I_1=\frac{I_0}{16}$
Let $I_{ADE}=I_{BDF}=I_{EFC}=I_2$
$\therefore 3I_2+I_1=I_0\Rightarrow 3I_2+\frac{I_0}{16}=I_0\Rightarrow I_2=\frac{5I_0}{16}$
Hence, moment of inertia of $\text{DECB}$ i.e., after removal part $\text{ADE}$
$I^{\prime }=2I_2+I_1=2\left(\frac{5I_0}{16}\right)+\left(\frac{I_0}{16}\right)=\frac{11I_0}{16}=\frac{N{I}_0}{16}$
Therefore, value of $N=11$