Question

Fig shows a thin metallic triangular sheet $ABC$. The mass of the sheet is $M$. The moment of inertia of the sheet about side $AC$ is :

• 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

The correct option is B $\frac{M{l}^2}{12}$

Moment of inertia about the center of the rod of length = $\frac{1}{12}\times M{L}^2$

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

Using perpendicular axis theorem,

$I_z=I_x+I_y=\frac{1}{6}\times M{L}^2$

Similarly, $I_1+I_2=I_z$

But, $I_1=I_2,\Rightarrow I_1=\frac{1}{2}\times \frac{1}{6}M{L}^2$

$\Rightarrow I_1=\frac{1}{12}\times M{L}^2$

Hence, the correct option is B.