Question

An equilateral triangle $ABC$ is cut from a thin solid sheet of wood. (See figure) $D, \, E$ and $F$ are the mid-points of its sides as shown and $G$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $G$ and perpendicular to the plane of the triangle is $I_{0}$ . If the smaller triangle $DEF$ is removed from $ABC,$ the moment of inertia of the remaining figure about the same axis is $I.$ Then:

• A. $I=\frac{3}{4}I_{0}$
• B. $I=\frac{15}{16}I_{0}$
• C. $I=\frac{9}{16}I_{0}$
• D. $I=\frac{I_{0}}{4}$

Answer 1

Answer: (B)

$I=\frac{15}{16}I_{0}$

Answer 2

Dimension analysis $I_{o}=KMa^{2}$ Now for small lamina $I^{'}=K\frac{M}{4}\left(\frac{a}{2}\right)^{2}=\frac{k m a^{2}}{16}$ $I^{'}=\frac{I_{o}}{16}$ So moment of Inertia of remaining part $I_{L}=I_{o}-I^{'}$ $=I_{o}-\frac{I_{o}}{16}$ $I_{L}=\frac{15 I_{o}}{16}$