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A uniform rod AB of mass 2 kg is hinged ...

A uniform rod `AB` of mass `2 kg` is hinged at one end `A`. The rod is kept in the horizontal position by a massless string tied to point `B`. Find the reaction of the hinge (in `N`) on end `A` of the rod at the instant when string is cut. `(g = 10 m//s^(2))`

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The correct Answer is:
5
When the string is cut, the weight of the rod constitutes torque about the hinge, so
`tau_(A)=mgl/2`……..i
According to Newton's second law
`tau_(A)=Ialpha`……….ii
Where `alpha` is the angular acceleration of the rod about the end `A`. From ean i and ii we get
`Ialpha=mgl/2`
or `alpha=(mgl/2)/I`
Here `I=ml^(2)//3`, therefore
`:. alpha=(mgl//2)/(ml^(2)//3)=3/2 g/l`
Acceleration of the `CM` of the rod is
`a_(CM)=alphar = 3/2 g/l xxl/2=(3g)/4`
Again by Newtons' second law
`mg-R_(A)=ma_(CM)`
or `mg-R_(A)=mxx(3g)/4`
`:. R_(A)=(Mg)/4=(2xx10)/4=5N`
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