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 $I_0$ It the smaller triangle $DEF$ is removed from $ABC$, the moment of inertia of the remaining figure about the same axis is $I$. Then:

• A. $I=\frac{9}{16}{I}_0$
• B. $I=\frac{3}{4}{I}_0$
• C. $I=\frac{1}{4}{I}_0$
• D. $I=\frac{15}{16}{I}_0$

Answer 1

The correct option is D

$I=\frac{15}{16}{I}_0$

Suppose the mass of the original triangle$△ABC$ is $M$ and the side of the original triangle $△ABC$ is $a$, the mass of the removed triangle $△DEF$ is $\frac{M}{4}$ and the side of the removed triangle $△DEF$ is $\frac{a}{4}$.

The ratio of the moment of inertia of the removed triangle, $I_{removed}$ to the moment of inertia of the original triangle, $I_0$

$\frac{I_{removed}}{I_{original}}=\frac{{\displaystyle \frac{M}{4}}{\left({\displaystyle \frac{a}{2}}\right)}^2}{M{\left(a\right)}^2}$

$I_{removed}=\frac{I_0}{16}$

Moment of inertia of the remaining part, $I$

$I=I_0-I_{removed}$

$I=I_0-\frac{I_0}{16}$

$I=\frac{15I_0}{16}$

Hence, the correct option is $\left(D\right)$.