Three particles each of mass100g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendiculr to its plane.

Explanation

`m=100g=100xx10^(-3)kg`
Side `a=10cm`

Moment of mertia, `I=3mr^(2)`
`=3xx100xx10^(-3)xx[(10)/(sqrt(3))xx10^(-2)]^(2)`
`=(3xx10^(-1)xx10^(2)xx10^(-4))/(3)=10^(-3)kgm^(2)`.