Question

Three particles each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertia of the system about an axis?
(a) joining two of the particles and (b) passing through one of the particles and perpendicular to the plane of the particles.

Answer 1

Let us draw the diagram first ,Refer to the attachment

Given in the question 

Mass of the particle(each) m = 200 g or  0.20 kg.
Side of the equilateral triangle r = 10 cm or 0.10 m.
Perpendicular distance on axis 
AD = √3/2 × 10
AD = 5√3 m.

(a) Joining of the particle :-

Now, moment of inertia along BC.
$l = mr^2$
l = 200 K (5√3)²
l = 200 × 25 × 3
l = 15000 gm.cm²
l = 1.5 × 10⁻³ kg.m²

(b) Passing through the particle and perpendicular to the plane .

Here assume that the line passes from A and perpendicular to the Δ .
Hence masses B and C produces Torque in it.

∴ Moment of Inertia,

$I = mr^2 + mr^2$
$I = 2mr^2$ 
l = 2 × 200 × 10²
I = 40000 gm.cm²
l = 4 × 10 ⁻³ kg.m²


Hope it Helps :-)