In the figure given in the text if mass of the rod is `m` then find hinge force.
(a). Just after the rod is released from the horizontal position.
(b). When rod becomes vertical

Explanation

(a).
Just after the release `omega=0`
`thereforea_(r)=romega^(2)=0`
`alpha=(tau_(mg))/(I)`
`=((mg)(r_(bot)))/(I_(0))=((mg)((l)/(2)))/((ml^(2)//3))=(3)/(2)(g)/(l)`
`a_(t)=ralpha=((l)/(2))((3)/(2)(g)/(l))=(3)/(4)g`
From eq (iii) the hinge force is
`F=m(a_(c)-g)`
`=m[a_(t)(hatj)-(-ghatj)]`
`=m[-(3)/(4)ghatj+ghatj]`
`=(mg)/(4)hatj`
Therefore, hinge force is `(mg)/(4)` in vertically upward direction
(b). When the rod becomes vertical height fallen by centre of mass is `h=(l)/(2)`
therefore from Eq. (i)
`omega^(2)=(2mgh)/(I)` (`I=I_(0)=(ml^(2))/(3)`)
`=(2mg((l)/(2)))/((ml^(2)//3))=((3g)/(l))`
`a_(r)=romega^(2)=((l)/(2))((3g)/(l))=(3)/(2)g` (towards O)
At this moment `mg` also passes through `O`. Therefore its torque about O is also zero, so from Eq, (ii)
`alpha=0impliesa_(t)=ralpha=0`
Now, substituting proper values in Eq. (iii). The hinge force is
`F=m(a_(c)-g)`
`=m[a_(r)hatj-(-ghatj)]`
`=m[((3)/(2)g)hatj+(g)hatj]`
`=((5)/(2)mg)hatj`
Therefore hinge force is `(5)/(2)` mg in vertically upward direction.