Question

One-fourth length of a uniform rod of mass m and length $l$ is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread as shown in the figure. The thread is then burnt and the rod starts rotating about the edge. The coefficient of friction between the rod and the surface is $\mu$. If the angle between the rod and the horizontal when it is about to slide on the edge $\theta ={\tan }^{-1}\left(\frac{x\mu }{13}\right)$, find the value of $x$.

Answer 1

Figure (a) and (b)
$\omega :$ Decrease in gravitational potential energy $=$ increase in rotational kinetic energy
$\therefore mg\frac{l}{4}\sin \theta =\frac{1}{2}{I}_0{\omega }^2$
$=\frac{1}{2}[\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2]{\omega }^2$
$\therefore \omega =\sqrt{\left[\frac{24g\sin \theta }{7l}\right]}......\left(i\right)$
$\alpha :\alpha =\frac{\tau }{I}$
$=\frac{mg\frac{l}{4}\cos \theta }{[\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2]}$
$=\frac{12g\cos \theta }{7l}$....(ii)
$\sum F_y=m{a}_y$ or $mg\cos \theta -N=m{a}_t$
or $N=mg\cos \theta -m{a}_t$ $=mg\cos \theta -m\frac{l}{4}\alpha$
Substituting value of $\alpha$ from Eq. (il), we get
$N=\frac{4}{7}mg\cos \theta ....\left(iii\right)$
Rod begins to slip when
$\mu N-mg\sin \theta =m{a}_n$ or $\frac{4}{7}\mu mg\cos \theta -mg\sin \theta =m\frac{l}{4}{\omega }^2$
Substitution value ofro from Eq. (i), we get
$\tan \theta =\frac{4\mu }{13}$
$\theta ={\tan }^{-1}\left(\frac{4\mu }{13}\right)$