# Three Particles, Each Of Mass 200G Are Kept At The Corners Of An Equilateral Triangle Of Side 10Cm. 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.

Sol:

Given:

Mass = 200g = 0.2 Kg

Sides = 10cm = 0.1 m

The perpendicular distance, AD = $$\frac{\sqrt{3}}{2} * 10$$ $$\Rightarrow 5\sqrt{3 cm}$$

(A) Joining Two Of The Particles

Moment of inertia along BC

I = mr2

= $$0.2 (5\sqrt{3})^{2} * 10^{-4}$$

I = 1.5 – 10-3 Kg m2

(B) Passing Through One Of The Particles And Perpendicular To The Plane Of The Particles.

I = mr2 + mr2

I = 4 * 10-3 Kg m2

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