A thin uniform rod mass `M` and length `L` is hinged at its upper end. And released from rest from a horizontal position. The tenstion at a point located at a distance `L//3` from the hinge point, when the rod become vertical will

Options

A. `22Mg//27`
B. `11Mg//13`
C. `6Mg//11`
D. `2Mg`

Explanation

Decrease in gravitational potential energy
`=` increase in rotatinal kinetic energy
`=` increase in rotational energy
`thereforeMg(L)/(2)=(1)/(2)((ML^(2))/(3)).omega^(2)thereforeomega^(2)=(3g)/(L)`
`r=(L)/(3)+(2L//3)/(2)=(2L)/(3)`
`(T-2Mg)/(3)=((2M)/(3))a_(C)=((2M)/(3))(romega^(2))`
`thereforeT=(2Mg)/(3)+((2Mg)/(3))((2L)/(3))((3g)/(L))`
`=2Mg`