Question

what is moment of inertia about the vertex of isosceles triangle?

and about some distance down from that vertex?

(only mass moment of inertia)

Answer 1

he moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression

IXX=136bh3

where b
is the base width, and specifically the triangle side parallel to the axis, and h

is the triangle height (perpendicular to the axis and the base) as shown in the figure.

Inserting given values for a right angled isosceles triangle we get

IXX=136L×(L2)3

IXX=1288L4

Moment of inertia I

about a line passing through apex angle and parallel to the base, i.e., vertex C of triangle can be found with the help of parallel axis theorem.

I=ICM+Ad2

where d
is distance from CM (centroid) to Vertex C, which is 23×L2=L3 and A

is area of shape

Therefore I=1288L4+12×L×L2(L3)2

I=1288L4+136L4

I=132L4