Question

3 point particles of masses 2kg each placed on 3 corners pf an equilateral triangle of perimeter 12m . This triangular structure suppose to have niglible mass. Find moment of inertia of this system about an axis passing through one of its side.


plz help me guys....​

Answer 1

perimeter = 12

side = 12/3 = 4

let axis pass through BD.its perpendicular bisector and its equilateral triangle so it will pasd through A

As particle is at A also so its moment of inertia along axis is 0

So Moment is due to B and C

BC= CD = 4/2 = 2

SO Moment of inertia = 2 M( BC)^2

= 2 (2)( 2)^2

= 16