A unifrom rod `AB` of length `L` and mass `M` is held horizontally with the help of one light string and a hinge as shown. If string at `B` is burnt suddenly, find out the initial angular acceleration of the rod and the intial reaction force at the hinge.

Explanation

Since `A` is a stationary point, we can apply Newton's second law in rotational form about it.
Torque about `A` is
`tau=Mg.(L)/(2)`……..`(i)`
Also `tau=l alpha`……`(ii)`
`impliesl alpha=Mg.(L)/(2)`
`implies((1)/(3)ML^(2))alpha=Mg.(L)/(2)`
`impliesalpha=(3g)/(2L)`
Let `a` be acceleration of `C.M`.
`impliesa=alpha.R=(3g)/(2L)xx(L)/(2)=(3g)/(4)`
Now, from rods `F.D.D`
`Mg-N=Ma` (Newton's `2^(nd)` Law)
`=ltMg-N=M.(3g)/(4)`
`impliesN=Mg-(3Mg)/(4)=(Mg)/(4)`