A rod of mass `m` and length `l` is released from rest from vertical position as shown in the figure. The normal force as a function of `theta`, which is exerted on the rod by the ground as it falls downwards, assuming that it does not slip is `mg((3 cos theta-1)/n)^(2)` then `n=`

Answer

2

Explanation

At angle `theta`
`1/2Iomega^(2)=mgl/2(1-cos theta)`or, `omega^(2)=(3g)/l(1-cos theta)...(i)`
Differentiating w.r.t q.,
`alpha=(mgl/2 sin theta)/((ml^(2))/3)=3/2(g sin theta)/l.....(ii)`
`a_(n)=l/2 omega^(2)=(3g)/2(1-cos theta)` and `a_(t)=l/2 alpha=3/4 g sin theta`
`f=ma_(x)=m(a_(t)cos theta-a_(n) sin theta)`
`=m(3/4 g sin thetacos theta-(3 g sin theta)/2(1-cos theta))`
`=3/2 mg sin theta[3/2cos theta-1]`

Further, `mg-N=ma_(y)`or,
`N=m(g-a_(t))`
`N=m[g-3/4 g sin^(2)-(3g cos theta)/2(1-cos theta)]`
`=(mg)/4[4-3 sin^(2)theta-6 cos theta+6 cos^(2) theta]`
`=(mg)/2(1-3 cos theta)^(2)`
. The rod does not slip until `N=0`
i.e.,
`theta=cos^(-1)(1/3)`