Show that if a rod held at angle $\mathrm{\theta }$ to the horizontal and released, its lower end will not slip if the friction coefficient between rod and ground is greater than $\frac{3\mathrm{sin\theta cos\theta }}{1+3{\sin }^2\mathrm{\theta }}$.

Point A is momentarily at rest.

$$\begin{gathered} \mathrm{\alpha }=\frac{\mathrm{mg}\frac{\mathrm{l}}{2}\mathrm{cos\theta }}{\frac{{\mathrm{ml}}^2}{3}}=\frac{3}{2}\frac{\mathrm{gcos\theta }}{\mathrm{l}} \\ \therefore {\mathrm{a}}_{\mathrm{C}}=\frac{\mathrm{l}}{2}\mathrm{\alpha }=\frac{3}{4}\mathrm{gcos\theta } \\ \text{ Now } \mathrm{\mu N}={\mathrm{ma}}_{\mathrm{x}}\text{ or }\mathrm{\mu N}=\mathrm{mqsin\theta } \\ \text{ or } \mathrm{\mu N}=\frac{3}{4}\mathrm{mgsin\theta cos\theta } \\ \text{ Further, } \mathrm{mg}-\mathrm{N}={\mathrm{ma}}_{\mathrm{y}} \\ \text{ or } \mathrm{N}=\mathrm{mg}-{\mathrm{ma}}_{\mathrm{c}}\mathrm{cos\theta } \\ \text{ or } \mathrm{N}=\mathrm{mg}-\frac{3}{4}{\mathrm{mgcos}}^2\mathrm{\theta } \end{gathered}$$

Dividing Eq. (i) by Eqs. (ii), we have

$$\begin{gathered} \mathrm{\mu }\frac{\frac{3}{4}\mathrm{sin\theta cos\theta }}{1-\frac{3}{4}{\cos }^2\mathrm{\theta }}=\frac{3\mathrm{sin\theta cos\theta }}{4-3{\cos }^2\mathrm{\theta }} \\ =\frac{3\mathrm{sin\theta cos\theta }}{1+3{\sin }^2\mathrm{\theta }} \end{gathered}$$