Question

Three identical rods each of mass 'm' and length 'l' form an equilateral triangle. Find the moment of inertia.

Answer 1

The moment of inertia is $\bf{\frac{1}{2}ML^2}$

Explanation:

The moment of inertia of a single rod about an axis passing through its center and perpendicular to it is $\frac{1}{12}ML^2$

Mass per unit length of rod is denoted by m/L.

Each side of  an equilateral triangle about an axis passing through the center of a triangle and perpendicular to its plane is $\frac{1}{12}ML^2 + M(\frac {L}{2\sqrt{3}})^2$

$=\frac{1}{6}ML^2$

The moment of inertia of the triangle about this axis (according to parallel axis theory) = $3 \times \frac{1}{6}ML^2 = \frac{1}{2}ML^2$

Thus,  the moment of inertia is $\bf{\frac{1}{2}ML^2}$