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 then 3sinthetacostheta/1+3sin^2theta.

Answer 1

$\alpha= \frac{mg\frac{1}{2}cos\theta}{ \frac{ml^2}{3}}= \frac {3 g\cos\theta}{2\times3}$

$a_C= \frac {1}{2}\alpha= \frac {3}{4}g \cos\theta$

Now $\mu N=ma_C \sin\theta or \mu N= \frac {3}{4} mg \sin\theta \cos\theta..........(i)$

Further, $mg-N=ma_y$ or $N=mg- ma_y \cos \theta$ or $N= mg- \frac {3}{4} mg \cos^2 \theta..........(ii)$

Dividing equation (i) and (ii) we have

$\mu= \frac {\frac{3}{4}\cos\theta\sin\theta} {1-\frac{3}{4}\cos^2\theta}$ = $\frac {3\sin\theta \cos \theta}{4-3 \cos^2\theta}$

= $\frac {3\sin\theta \cos\theta}{1+3\sin^2\theta}$ Hence Proved.