A thin rod of mass `m` and length `2l` is placed horizontally and perpendicular to a horizontal rough nail, as shown in figure and set free. The point of contanct of the rod with the nails is `l//3` distance away form the centre of rod. If the rod starts slipping when it forms an angle `theta` with the horizontal and the coefficient of friction of rod with nails is `mu`, then find `(mu)/(tan theta)`.

Answer

2

Explanation

Till the rod does'tn slip the centre of mass of the rod in circular motion around the peg. Thus from `COE`
`mg 1/2 sin theta=1/2 l omega^(2)`
Where `I=(4ml^(2))/12+(ml^(2))/9=(4ml^(2))/9`
thus `omega=sqrt((3g)/(2l) sin theta)`
Now `m1/3 omega^(2)=F_(r)-mg sin theta.......(1)`
or `F_(r)=3/2 mg sin theta`
angular acceleration of rod at an angle3 `theta`
`(mg(1/3) cos theta)/((4//9)ml^(2))rArralpha=3/4 g/l cos theta`
and linear acceleration of centre of mass perpendicular to rod`rArrmg cos theta-N=ma` from pure rolling `a=(1/3)alpha`
thus`a=g/4cos theta`
`rArrN=(3mg)/4 cos theta.....(ii)`
`mu=F_(r)//N=2 tan theta`