A uniform rod of mass `M` and length `L` is hinged at its end. The rod is released from its vertical position by slightly pushing it. What is the reaction at the hinge when the rod becomes horizontal, again vertical.

Explanation

Applying energy consevation between position (`1`) and (`2`)
`mg(L/2)=(1)/(2)I_(0)omega^(2)=(1)/(2)(ML^(2))/(3)omega^(2)impliesomega=sqrt((3g)/(L))`
On a hinge, two perpendicular forces act.

`downarrow:Mg-F_(2)=Ma`
Torque about `O`
`Mg(L/2)=I_(0)alpha=(ML^(2))/(3)alphaimpliesalpha=(3g)/(2L)`
`a=(L)/(2)alpha=(3g)/(4)`
`Mg-F_(2)=Ma=(3Mg)/(4)implies F_(2)=(Mg)/(4)`
In the horizontal direction

`dF=dmomega^(2)x`
`F_(1)=(M)/(L)omega^(2)int_(0)^(L)xdx=(1)/(2)Momega^(2)L`
`(1)/(2)Mxx(3g)(L)L=(3Mg)/(2)`
Reaction at the hinge
`R=sqrt(F_(1)^(2)+F_(2)^(2))=sqrt(((3Mg)/(2))^(2)+((Mg)/(4))^(2))`
`=(sqrt37Mg)/(4)`
When the rod again becomes vertical,
`Mg(L/2)=-Mg(L/2)+(1)/(2)I_(0)omega^(2)`
`MgL=(1)/(2)xx(ML^(2))/(3)omega^(2)impliesomega'=sqrt((6g)/(L))`
`R=Mg+(1)/(2)Momega^(2)L=Mg+(1)/(2)M(6g)/(L)=4Mg`