An equilateral triangle $\mathrm{ABC}$ is cut from a thin solid sheet of wood. (See figure) $\mathrm{D},…
An equilateral triangle $\mathrm{ABC}$ is cut from a thin solid sheet of wood. (See figure) $\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are the mid-points of its sides as shown and $\mathrm{G}$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $\mathrm{G}$ and perpendicular to the plane of the triangle is $\mathrm{I}_{0}$. If the smaller triangle $\mathrm{DEF}$ is removed from $\mathrm{ABC}$, the moment of inertia of the remaining figure about the same axis is $I$. Then
$\mathrm{I}=\frac{15}{16} \mathrm{I}_{0}$
$\mathrm{I}=\frac{3}{4} \mathrm{I}_{0}$
$\mathrm{I}=\frac{9}{16} \mathrm{I}_{0}$
$\mathrm{I}=\frac{\mathrm{I}_{0}}{4}$
Solution
Let mass of the larger triangle $=\mathrm{M}$ Side of larger triangle $=\ell$ Moment of inertia of larger triangle $=\mathrm{ma}^{2}$
Mass of smaller triangle $=\frac{\mathrm{M}}{4}$
Length of smaller triangle $=\frac{\ell}{2}$
Moment of inertia of removed triangle $=\frac{\mathrm{M}}{4}\left(\frac{\mathrm{a}}{2}\right)^{2}$
$\therefore \frac{\mathrm{I}_{\text {removed }}}{\mathrm{I}_{\text {original }}}=\frac{\frac{\mathrm{M}}{4}}{\mathrm{M}} \cdot \frac{\left(\frac{\mathrm{a}}{2}\right)^{2}}{(\mathrm{a})^{2}}$
$I_{\text {removed }}=\frac{I_{0}}{16}$
$\mathrm{So}, \mathrm{I}=\mathrm{I}_{0}-\frac{\mathrm{I}_{0}}{16}=\frac{15 \mathrm{I}_{0}}{16}$
