Moment Of Inertia

Given: Three particles of mass m grams each are placed at the vertices of an equilateral triangle ABC with side length a cm.
Required: Find the moment of inertia of the system about the line AX, which is:
- Perpendicular to side AB
- Lies in the plane of triangle
Coordinate Geometry Setup:
- Let point $A%20%3D%20(0%2C%200)$
- Then $B%20%3D%20(a%2C%200)$
- Point $C$ lies above the midpoint of $AB$ , so:
  Midpoint of $AB%20%3D%20%5Cleft(%5Cfrac%7Ba%7D%7B2%7D%2C%200%5Cright)$
  Height of equilateral triangle = $%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Da$
  So $C%20%3D%20%5Cleft(%5Cfrac%7Ba%7D%7B2%7D%2C%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Da%5Cright)$
Moment of Inertia Calculation:
- About point A: Distance = 0 → $I_A%20%3D%200$
- Point B: Perpendicular distance from AX = $a$ → $I_B%20%3D%20m%20a%5E2$
- Point C: Perpendicular distance from AX = $%5Cfrac%7Ba%7D%7B2%7D$ → $I_C%20%3D%20m%20%5Cleft(%5Cfrac%7Ba%7D%7B2%7D%5Cright)%5E2%20%3D%20%5Cfrac%7B1%7D%7B4%7Dma%5E2$
Total Moment of Inertia:
$I%20%3D%20I_A%20%2B%20I_B%20%2B%20I_C%20%3D%200%20%2B%20ma%5E2%20%2B%20%5Cfrac%7B1%7D%7B4%7Dma%5E2%20%3D%20%5Cfrac%7B5%7D%7B4%7Dma%5E2$
Final Answer: $%5Cboxed%7B%5Cfrac%7B5%7D%7B4%7Dma%5E2%7D$

Chapter Questions
1 MCQs