4 thin rods of length L and mass M are joined together to make a square. Find the moment of inertia of the system about any side.
My approach:-
I am finding the MI about the left side. The MI of the top and bottom rods(each) will be (ML2/12)+ 1/4(ML2) about left side rod. MI of the right side rod will be 2(ML2/12)+ ML2. MI of left side rod will be 2(ML2/12).
Now I found the total.
There is something wrong in my approach. Please explain me the correct approach.
The answer is (5/3)ML2 .
21yes sudipto ur answer is wrng.look,the axis is passing through one of the of sides so m.i due to that side=0 since u have mentioned them thin rods .and the m.i of the side opposite to it = (ml^2) and the m.i due to other two sides would be equal to (ml^2)/3.now since m.i is a scalar quantity just add them.(ml^2)+(2(ml^2)/3)=5ml^2/3. :)
