A rod of mass $m$ and length $\mathcal{l}$ 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 downward, assuming that it does not slip is  $mg{\left(\frac{3\cos \theta -1}{n}\right)}^2$  then $n=$

 

At angle $\theta$

$\frac{1}{2}I{\omega }^2=mg\frac{\mathcal{l}}{2}(1-\cos \theta ) or, {\omega }^2=\frac{3g}{t}(1-\cos \theta ) ......\left(i\right)$

Differentiating w.r.t. q,

$\alpha =\frac{mg{\displaystyle \frac{\mathcal{l}}{2}}\sin \theta }{{\displaystyle \frac{m{\mathcal{l}}^2}{3}}} = \frac{3}{2}\frac{g \sin \theta }{\mathcal{l}} ......\left(ii\right)$

$$\begin{gathered} a_n=\frac{\mathcal{l}}{2}{\omega }^2=\frac{3g}{2}(1-\cos \theta ) and a_t=\frac{\mathcal{l}}{2}\alpha =\frac{3}{4}g \sin \theta \\ f=m{a}_x =m(a_1 \cos \theta -a_n\sin \theta ) \\ =m\left(\frac{3}{4}g \sin \theta \cos \theta -\frac{3g \sin \theta }{2}(1-\cos \theta )\right) \\ =\frac{3}{2}mg \sin \theta \left[\frac{3}{2}\cos \theta -1\right] \end{gathered}$$

further,

          , $$\begin{gathered} mg-N=m{a}_y or, \\ N=,(g-a_1) \\ N=m\left[g-(a_t \sin \theta +a_n\cos \theta )\right] \\ =m\left[g-\frac{3}{4}g {\sin }^2-\frac{3g \cos \theta }{2}(1-\cos \theta )\right] \\ =\frac{mg}{4}\left[4-3{\sin }^2\theta -6\cos \theta +6{\cos }^2\theta \right] \\ =\frac{mg}{4}{(1-3\cos \theta )}^2. \end{gathered}$$

The rod does not slip until $N=0$

i.e.,          $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$