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:

Engineering

Physics

Moment of Inertia

Calculus

Forces and Types of Forces

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:

$\frac{{ML}^{2}}{12}$

$\frac{{ML}^{2}}{6}$

$\frac{{ML}^{2}}{3}$

$\frac{2 {ML}^{2}}{3}$

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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 l_xy.

Due to symmetry

l_AB = I_CD AND I_PQ = I_RS

Let l_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

l_square = l_AB + l_CD = l_PQ + l_RS

∴ l_square = 2l_AB = l_PQ

But, l_AB = 2l

∴ I = \frac{1}{2} I_{AB} = \frac{1}{2} * (\frac{1}{2} \frac{{ML}^{2}}{6})

....(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.

∴ I = \frac{M}{2} * L^{2} / 12 = \frac{1}{12} M_{triangle} L^{2}