The moment of inertia of the system about one rod as axis `I=(mL^(2))/3+mL^(2), I=4/3 mL^(2)`
Potential energy decreases for `B` and `C`
`(mgL)/2+mgL=3/2mgL`
By conservation of mechanical energy, the loss in `PE` of body is equal to the gain in rotational `KE`
`:. 3/2 mgL=1/2(4/3 mL^(2))omega^(2)` on solving `omega=3/2sqrt(g/L)`