# Solve this following

Question:

$\mathrm{ABC}$ is a plane lamina of the shape of an equilateral triagnle. $\mathrm{D}, \mathrm{E}$ are mid points of $\mathrm{AB}$, $\mathrm{AC}$ and $\mathrm{G}$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $\mathrm{G}$ and perpendicular to the plane $\mathrm{ABC}$ is $\mathrm{I}_{0}$. If part ADE is removed, the moment of inertia of the remaining part about the same axis

is $\frac{\mathrm{NI}_{0}}{16}$ where $\mathrm{N}$ is an integer. Value of $\mathrm{N}$ is

Solution:

Let side of triangle is a and mass is $\mathrm{m}$

MOI of plate $A B C$ about centroid

$\mathrm{I}_{0}=\frac{\mathrm{m}}{3}\left(\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2} \times 3\right)=\frac{\mathrm{ma}^{2}}{12}$

triangle ADE is also an equilateral triangle of side a/2.

Let moment of inertia of triangular plate ADE about it's centroid $\left(\mathrm{G}^{\prime}\right)$ is $\mathrm{I}_{1}$ and mass is $\mathrm{m}_{1}$

$\mathrm{m}_{1}=\frac{\mathrm{m}}{\frac{\sqrt{3} \mathrm{a}^{2}}{4}} \times \frac{\sqrt{3}}{4}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4}$

$\mathrm{I}_{1}=\frac{\mathrm{m}_{1}}{12}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4 \times 12} \frac{\mathrm{a}^{2}}{4}=\frac{\mathrm{ma}^{2}}{192}$

distance $\mathrm{GG}^{\prime}=\frac{\mathrm{a}}{\sqrt{3}}-\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{\mathrm{a}}{2 \sqrt{3}}$

$\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{m}_{1}\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2}=\frac{\mathrm{ma}^{2}}{192}+\frac{\mathrm{m}}{4} \cdot \frac{\mathrm{a}^{2}}{12}$

$=\frac{5 m a^{2}}{192}$

now MOI of remaining part

$=\frac{\mathrm{ma}^{2}}{12}-\frac{5 \mathrm{ma}^{2}}{192}=\frac{11 \mathrm{ma}^{2}}{12 \times 16}=\frac{11 \mathrm{I}_{0}}{16}$

$\Rightarrow \quad \mathrm{N}=11$