Question

Find the moment of inertia of a thin sheet of mass $M$ in the shape of an equilateral triangle about an axis as shown in figure. The length of each side is $L$.

Answer 1

Given: a thin sheet of mass M in the shape of an equilateral triangle. The length of each side is L.

To find the moment of inertia about the vertex parallel to the base.

Solution:

The area moment of inertial of a triangle about an axis passing through its centroid and parallel to one side is $\frac{b{h}^3}{36}$.

The mass moment of inertia is $\frac{b{h}^3}{36}\times \frac{M}{A}$

where M is the mass and A is the area of the triangle, $A=\frac{1}{2}bh$,

MI of a triangle is therefore $\frac{M{h}^2}{18}$ about an axis passing through the centroid and parallel to one side

Here the axis is shifted to one vertex,i.e, moved to through a distance = $\frac{2}{3}h$.

By parallel axis theorem, the moment of inertia of an object about a parallel axis to the axis through the centroid = I about the centroid + $M{d}^2$

where d is the distance of the axis from the centroid.

Hence in the present case

$I=\frac{M{h}^2}{18}+M\times \frac{(2h{)}^2}{3^2}=M{h}^2(\frac{1}{18}+\frac{4}{9})$

$I=M{h}^2\times \frac{1+8}{18}$

$I=\frac{1}{2}M{h}^2$

In an equilateral triangle

$h=\frac{\sqrt{3}}{2}L$

where L is the side of the triangle.

$I=\frac{1}{2}M\times \frac{(\sqrt{3}L{)}^2}{2^2}⟹I=\frac{3}{8}M{L}^2$

is the moment of inertia of the equilateral triangle about the vertex parallel to the base.