The three rods of the previous problem are connected symmetrically as shown. Find the `M.I.` about `(1)-(1), (2)-(2)` and `(3)-(3)`

Explanation

For rod `A`, `M.I.` about its own axis `I_(a-a)=0`

To find the `M.I.` of rod `A` about `(1)-(1)`, applying the parallel axes theorem.
`I_(A)=I_(a-a)+M((L)/(2))^(2)=0+(ML^(2))/(4)=(ML^(2))/(4)`
Similar analysis for rod `C`
For rod `B, I_(B)=(ML^(2))/(12)` (For rod `B`, axis `(1)-(1)` is passing through its `c.m.` and `bot^(ar)` to the length)
`I_(1-1)=I_(A)+I_(C)+I_(B)`
`=(ML^(2))/(4)xx2+(ML^(2))/(12)`
`=(7ML^(2))/(12)`
About `(2)-(2)`

`I_(2-2)=I_(A)+I_(C)+I_(B)`
`(ML^(2))/(12)+(ML^(2))/(12)+0=(ML^(2))/(6)`
About `(3)-(3)`

`I_(1-1)=I_(A)+I_(C)+I_(B)`
`=(0+ML^(2))+((ML^(2))/(3))+0`
`(4ML^(2))/(3)`
Alternatively, `O` is the center of mass of system
`I_(3-3)=(I_(1-1))_(c.m.)+3Md^(2)=(7ML^(2))/(12)+3M((L)/(2))^(2)`
`=(4ML^(2))/(3)`