Question

Moment of inertia of triangular lamina along its center of mass

Answer 1

Inertia is a property of a body to resist the change in linear state of motion. It is measured by the mass of the body. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion.

Consider a triangular lamina of base (b), altitude (h) and mass (M). Surface mass density is mass per unit area of the lamina.

σ=M(12bh)=2Mbh.

Consider a rectangular differential strip parallel to the base of width (dy), at a distance (y) from the axis of rotation:

Moment of Inertia of a Triangular Lamina about its Base

From similar triangles ΔADE and ΔABC, we have:

DEBC=h−yh.

DE=(h−y)bh.

Area of strip DE, dA=(h−y)bhdy.

Moment of Inertia of the strip DE about the axis of rotation BC is given by:

dIBC = (Mass of strip DE) (Distance from axis)².

dIBC=(2Mbh(h−y)bhdy)y2.

dIBC=2Mh2y2(h−y)dy.

Integrate the above expression between the limits from y = 0 to y = h.

IBC=∫0h2Mh2(hy2−y3)dy=2Mh2[(hy33)−(y44)]h0=Mh26.

IBC=Mh26.

Therefore, the Moment of Inertia of a Triangular Lamina about its base (IBC)=Mh26.

Mh²/6

Answer 2

$\textbf{Answer is in Attachment !}$