Three particles, each having a mass of 200 g , are placed at the...

Let the three particles be placed at the vertices A, B, and C of the equilateral triangle.

Mass of each particle, $m = 200 \text{ g} = 0.2 \text{ kg}$

Side length of the triangle, $a = 10 \text{ cm} = 0.1 \text{ m}$

The axis of rotation passes through one of the particles (say, at vertex A) and is perpendicular to the plane of the triangle.

The perpendicular distance of the particle at A from the axis is $r_A = 0$.

The perpendicular distances of the particles at B and C from the axis are $r_B = a$ and $r_C = a$.

The moment of inertia of the system about the given axis is:

$I = m r_A^2 + m r_B^2 + m r_C^2$

$I = m(0)^2 + m(a)^2 + m(a)^2$

$I = 2ma^2$

Substituting the given values:

$I = 2 \times 0.2 \times (0.1)^2$

$I = 0.4 \times 0.01 = 0.004 \text{ kg-m}^2$

$I = 4.0 \times 10^{-3} \text{ kg-m}^2$