4 rods of equal mass are arranged in the form of square than find out moment of inertia about an axis which passes through center and perpendicular to the plane.
Answers
We have to use the theorem of parallel axes: I = ICM +Ma2 where ICM is the the moment of inertia about an axis through the centre of mass of the rod and I is the moment of inertia about a parallel axis at distance a.
The moment of inertia of a thin rod of mass M and length L about an axis through the mid point of the rod and perpendicular to its length is ML2/12. The moment of inertia of the rod about a parallel axis at a distance L/2 (which is the distance of the centre of the frame from each ro.is ML2/12 + M (L/2)2 = ML2/3
You must remember that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. Therefore Moment of inertia of one part is (4/3) ML2.
But diagonals are also mutually perpendicular.
Hence
ID + ID = (4/3) ML^2
Hence ID = (2/3) ML^2