A uniform slender bar `B` of mass `m` and length `L` supported by a firctionless `m` and length `L` supported by a firctionless pivot at `A` is released from rest at its vertical position as shown the figure. Calculate the reaction at pivot when the bar just acquires the horizontal position shown dotted.

Explanation

As there is no friction the mechanical energy will remain constant. Let us applying the conservation of mechanical energy between the vertuical position `A` and the horizontal position `B`.
Using `/_\K+/_\U=0`
`(1/2I_(A)omega^(2)-0)+(-mgL/2)=0`
which gives
`omega=sqrt((3g)/L)`
Considering the free body diagram of the bar at this instant, the velocity of `COM` of bar at this instant is give below.
Let `alpha` be the angular acceleration of bar about `A` at the instant when it becomes horizontal.
`alpha=(tau_(A))/(I_(A))=(mgL/2)/((mL^(2))/3)=3/2 g/L`

Let `N_(x)` and `N_(y)` are the components of reaction at `A`. Then in horizontal direction `N_(x)=ma_(r)=momega^(2)(L/2)` in vertical direction
`and mg-N_(y)=ma_(t)` .........i
`a_(t)=` acceleration of `COM` in `y` direction `=(L/2)a`

or `N_(y)=mg-m(L/2alpha)=mg-(3mg)/4=(mg)/4`..........ii
Therefore, resultant reaction at support `A` is
`N=sqrt(N_(x)^(2)+N_(y)^(2))`
`N=sqrt(((3mg)/2)^(2)+((mg)/4)^(2))=(sqrt(37))/4mg`
`tanalpha = (N_(y))/(N_(x))=(mg/4)/(3mg/2)=1/6impliesalpha=tan^(-1)(1/6)`
Therefore, reaction at pivot is `(sqrt(37))/(4mg)` at an angle `alpha=tan^(-1)(1/6)` with horizontal.