A uniform thin bar of mass 6kg and length 2 . 4 meter is bent to...

$61 = 2 . 4$ $ℓ = 0 . 4m$

$\sin 60 ° = \dfrac{r}{ℓ}$

$r = 1\sin 60 ° = \dfrac{ℓ\sqrt{3}}{2}$

MOI, $I = \left[\dfrac{m{ℓ}^{2}}{12} + m{r}^{2}\right]6$

$= \left[\dfrac{m{ℓ}^{2}}{12} + m{\left(\dfrac{ℓ\sqrt{3}}{2}\right)}^{2}\right]6$

$= 5m{l}^{2}$

$= 5 \times 1 \times 0 . 16$

$= 0 . 8$

$I = 8 \times {10}^{- 1}kg{m}^{2}$