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 $\frac{NI_0}{16}$ where N is an integer. Value of N is

Answer 1

7

Answer 2

The moment of inertia of a uniform equilateral triangular lamina of mass $M$ and side $a$ about an axis through its centroid and perpendicular to its plane is $I_c = \frac{Ma^2}{24}$. Given $I_0$ is the moment of inertia of the full triangle ABC about its centroid G, so $I_0 = \frac{Ma^2}{24}$. The removed part ADE is an equilateral triangle with side $a/2$ and mass $M/4$. The distance between the centroid G of ABC and the centroid $G_{ADE}$ of ADE is $d = \frac{\sqrt{3}}{6}a$. Using the parallel axis theorem, the moment of inertia of ADE about G is $I_{ADE, G} = \frac{(M/4)(a/2)^2}{24} + (M/4)d^2 = \frac{Ma^2}{384} + \frac{M}{4}\frac{a^2}{12} = \frac{9Ma^2}{384}$. Substituting $Ma^2 = 24I_0$, we get $I_{ADE, G} = \frac{9}{384}(24I_0) = \frac{9}{16}I_0$. The moment of inertia of the remaining part is $I_{remaining} = I_0 - I_{ADE, G} = I_0 - \frac{9}{16}I_0 = \frac{7}{16}I_0$. Comparing this with $\frac{NI_0}{16}$, we find $N=7$.