Two men support a uniform rod of mass `M` and length `L` at its two ends. If one of them suddenly withdraws, find the force excerted by the rod on the other man
(`a`) immediately after withdrawl and
(`b`) when the rod makes angle `theta` with vertical.

Explanation

(`a`)
Taking torque about `A`
`tau_(A)=MgL/2=I_(A)alpha=(ML^(2))/(3)alpha implies alpha=(3g)/(2L)`
`a_(c)=(L)/(2)alpha=(3g)/(4)`
`Mg-N=Ma=(3Mg)/(4) implies N=(Mg)/(4)`
where `N` is the force exerted on rod at `A`.
(`b`) When the rod makes `theta` with vertical

`tau_(A)=Mgsintheta(L)/(2)=I_(A)alpha=(ML^(2))/(3)alpha implies alpha= (3gsintheta)/(2L)`
`a_(c)=(L)/(2)alpha=(3gsintheta)/(4)`
`Mgsintheta-N=Ma_(c)=(3Mg)/(4)sintheta`
`N=(1)/(4)Mgsintheta`