Question

Figure shows a thin metallic triangular sheet $ABC$. The sides $AB$ and $BC$ are of equal lengths $l$. The mass of the sheet is $M$. What is its moment of inertia about $AC$?

• A. $\frac{M{l}^2}{18}$
• B. $\frac{M{l}^2}{12}$
• C. $\frac{M{l}^2}{6}$
• D. $\frac{M{l}^2}{4}$

Answer 1

Answer: (B)

$\frac{M{l}^2}{12}$

Moment of inertia about the center of a rod of length l, $I=\frac{1}{12}M{l}^2$

$I_x=I_y=\frac{1}{12}M{l}^2$

Using perpendicular axis theorem, $I_z=I_x+I_y=\frac{1}{6}M{l}^2$

Similarly, $I_1+I_2=I_z=\frac{1}{6}M{l}^2$

But, $I_1=I_2⟹I_2=\frac{1}{12}M{l}^2$

Hence, $I_{AC}=\frac{1}{12}M{l}^2$