Question

$ABC$ is a plane lamina of the shape of an equilateral triagnle. $D,E$ are mid points of $AB$ $AC$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and perpendicular to the plane $ABC$ is $I_0$. If part 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. Value of $N$ is _______.

Answer 1

Answer: 11

Let side of triangle is a and mass is $m$

MOI of plate $ABC$ about centroid
$I_0=\frac{m}{3}\left({\left(\frac{a}{2\sqrt{3}}\right)}^2\times 3\right)=\frac{m{a}^2}{12}$
triangle ADE is also an equilateral triangle of side $a/2$
Let moment of inertia of triangular plate ADE about it's centroid (G') is $I_1$ and mass is $m_1$
$m_1=\frac{m}{\frac{\sqrt{3}{a}^2}{4}}\times \frac{\sqrt{3}}{4}{\left(\frac{a}{2}\right)}^2=\frac{m}{4}$
$I_1=\frac{m_1}{12}{\left(\frac{a}{2}\right)}^2=\frac{m}{4\times 12}\frac{a^2}{4}=\frac{m{a}^2}{192}$

distance $G{G}^{\prime }=\frac{a}{\sqrt{3}}-\frac{a}{2\sqrt{3}}=\frac{a}{2\sqrt{3}}$
so MOI of part ADE about centroid G is
$I_2=I_1+m_1{\left(\frac{a}{2\sqrt{3}}\right)}^2$
$=\frac{m{a}^2}{192}+\frac{m}{4}\cdot \frac{a^2}{12}$
$=\frac{5m{a}^2}{192}$
now MOI of remaining part
$=\frac{m{a}^2}{12}-\frac{5m{a}^2}{192}$
$=\frac{11m{a}^2}{12\times 16}=\frac{11I_0}{16}$
$\Rightarrow N=11$