Question

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

Answer 1

Point $A$ is momentarily at rest
$\alpha =\frac{mg\frac{l}{2}\cos \theta }{\frac{m{l}^2}{3}}=\frac{3}{2}\frac{g\cos \theta }{l}$
$\therefore a_C=\frac{l}{2}\alpha =\frac{3}{4}g\cos \theta$
Now $\mu N=m{a}_x$
or $\mu N=m{a}_C\sin \theta$ or $\mu N=\frac{3}{4}mg\sin \theta \cos \theta ...\left(i\right)$
Further, $mg-N=m{a}_y$
or $N=mg-m{a}_C\cos \theta$
or $N=mg-\frac{3}{4}mg{\cos }^2\theta ...\left(ii\right)$
Dividing Eq. (i ) by Eqs. (ii). we have
$\mu =\frac{\frac{3}{4}\sin \theta \cos \theta }{1-\frac{3}{4}{\cos }^2\theta }=\frac{3\sin \theta \cos \theta }{4-3{\cos }^2\theta }$
$=\frac{3\sin \theta \cos \theta }{1+3{\sin }^2\theta }$
Hence proved.