One end of a uniform rod of mass `M` and length `L` is supproted by a frictionless hinge which can with stand a tension of `1.75 Mg`. The rod is free to rotate in a vertical plane. The maximum angle should the rod be rotated from the vertical position so that when left, the hinge does not brek is `(pi)/n`

Answer

3

Explanation

`1.75 Mg =Mg+(2M)/L((Lomega)/2)^(2)....(i)`
`MgL/2(1-cos theta)=1/2(ML^(2))/3 omega^(2)....(ii)`
Solving (i) and (ii) we get
`theta=60^(@)`
`a_(x)=0`