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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:

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Solution

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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 IXY.

Due to symmetry

IAB=ICD AND IPQ=IRS

Let Isquare 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

Isquare=IAB+ICD=IPQ+IRS

∴Isquare=2IAB=IPQ

But, IAB=2I

∴I=12IAB=12∗(12ML26) ....(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=M2∗L2/12=112MtriangleL2

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