Question

A thin uniform metallic triangular sheet of mass M has sides $AB=BC=L$. Its moment of inertia about the axis AC lying in the plane of the sheet is:

• A. $\frac{M{L}^2}{12}$
• B. $\frac{M{L}^2}{3}$
• C. $\frac{2M{L}^2}{3}$
• D. $\frac{M{L}^2}{6}$

Answer 1

Answer: (A)

The given triangular sheet has an area equal to half of a square sheet having same side length as the length of the base of the triangle.

Let MI of the triangular sheet about AC be I.

I is half of the MI of the square plate about an axis passing through the centre O and being in the plane of the plate.

Let the MI of the square about an axis XY be $I_{XY}$.

Due to symmetry

$I_{AB}=I_{CD}$ AND $I_{PQ}=I_{RS}$

Let $I_{square}$ be the MI a bout an axis passing through the centre of the square and perpendicular to the plane of the square plate.

Then using Perpendicular Axis Theorem

$I_{square}=I_{AB}+I_{CD}=I_{PQ}+I_{RS}$

$\therefore I_{square}=2I_{AB}=I_{PQ}$

But, $I_{AB}=2I$

$\therefore I=\frac{1}{2}{I}_{AB}=\frac{1}{2}\ast \left(\frac{1}{2}\frac{M{L}^2}{6}\right)$ ....(1) where M is the mass of the square plate.

But the triangular sheet has half the mass of the square plate. Writing I in terms of the mass of the plate.

$\therefore I=\frac{M}{2}\ast L^2/12=\frac{1}{12}{M}_{triangle}{L}^2$