Physics Moment of Inertia Equilateral Triangle

Three balls of masses 1 kg, 2 kg, and 3 kg respectively are arranged at the corners of an equilateral triangle of side l m. What will be the moment of inertia of the system about an axis through the centroid perpendicular to the plane of the triangle?

What formula is used to determine AB in the equation $AB^2 = AM^2 + BM^2$?

What is the value of 'M' in M = √3 / 2 * m?

What is the value of AO in the equation $AO = 2/ 3 * AM$

Problem Description

Two balls with masses 1 kg, 2 kg, and 3 kg are arranged at the corners of an equilateral triangle.
The side length of the triangle is 1 meter.
Calculate the moment of inertia of the system about an axis passing through the centroid and perpendicular to the plane of the triangle.

Calculations

Using the law of cosine, calculate the distance from the centroid to each corner of the equilateral triangle.
The centroid is located at a distance of (1/√3)m from each corner of the triangle.
Calculate the perpendicular distance from the centroid to each side of the triangle.
The perpendicular distance from the centroid to each side is calculated as (1/√3) / 2 = 1/√3 meter.
Based on the geometry of the system, the distances from the centroid to the sides of the triangle to points on the sides (AO = 2/3 AM, CO = 2/3 CP, BO = 2/3 BN) were calculated.
The moment of inertia about each of the sides for the masses at the corners was calculated.
The moment of inertia was calculated to be I = (1/3) (m1)(x1)^2+(2/3)(m2)(x2)^2+(3/3)(m3)(x3)^2

This quiz explores the calculation of moment of inertia for a system of balls arranged at the corners of an equilateral triangle. You'll need to apply geometric principles and formulas to derive the moment of inertia about an axis through the centroid. Test your understanding of mechanics and inertia concepts in this engaging physics quiz.