Q. 93.5( 2 Votes )

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 :

Given data-



Mass of the particle = 200 g = 0.20 kg.


Length of the equilateral triangle r = 10 cm or 0.10 m.


Perpendicular distance on axis


AD = √3/2 × 10 = 5√3 m


(a)Let’s take the side BC-


Now, moment of inertia along BC



Where,


m=mass of the particle


r=distance or radius


Substituting the values-






(b) Let’s take AD as perpendicular side.


Now, moment of inertia along BC



Substituting the values-






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