A rod of mass m and length l is hinged at 'its' end and released from rest in position shown. Find the force exerted by the hinge when the rod becomes horizontal

Options

A. `(mg)/4`
B. `(3mg)/2 cos theta`
C. `(mg)/4sqrt(9cos^(2)theta+1/4)`
D. `(mg)/2 sqrt(9 cos^(2)theta+1/4)`

Answer

D

Explanation

`wg=(mgl)/2 cos theta=1/2 xx(ml^(2))/3 omega^(2)`
`omega^(2)=(3 g cos theta)/l`
`a_(CR)=(omega^(2)l)/2=(3g)/2 cos theta`
`mg l/2 =Ialpha=((ml^(2))/3) xx alpha`
`alpha=(3g)/(2l) " " a_(ct)=l/2 alpha=(3g)/4`
`mg-Ny=ma_(ct)rArr N_(y)=(mg)/4`
`N_(x)=m omega^(2)l/2=(3mg)/2 cos theta`
`F=sqrt(N_(x)^(2)+N_(y)^(2)) =(mg)/2sqrt(9 cos^(2)theta+1/4)`