posted by hriday .
two small objects of masses 'm' & '2m' each are attaached to the ends of a light rigid rod of length 'L' . the rod is pivoted at a point where centre of mass of the two objects lie. the moment of inertia of the arrangement about an axis through the centre of mass , perpendicularly to the rod is ..........
The center of mass is L/3 from mass 2m and 2L/3 from mass m. Add the moment of inertia contributions of each mass to get
m*(2L/3)^2 + (2m)(L/3)^2
= mL^2*(4/9 + 2/9)
= (2/3) mL^2