Question

A uniform rod of length $2a$ is placed horizontally on the edge of a table. Initially, the centre of mass of the rod is at a distance $a/3$ from the edge. The rod is released from rest. If the rod slips after it has turned through an angle $\theta$, the coefficient of friction between the rod and the table is $\beta \tan \theta$. The value of $\beta$ is ___.

Answer 1

Answer: 2

From the law of conservation of energy,
$K{E}_i+GP{E}_i=K{E}_f+GP{E}_f$

$\therefore 0+0=\frac{1}{2}I{\omega }^2+\left(-mg\frac{a}{3}\sin \theta \right)...(i)$
From parallel axes theorem, $I=\frac{m(2a{)}^2}{12}+m{\left(\frac{a}{3}\right)}^2$
Using this in Eq. (i) will get ${\omega }^2=\frac{3g\sin \theta }{2a}$
Differentiating w.r.t. $\theta$, we get $2\omega \frac{d\omega }{d\theta }=\frac{3g}{2a}\cos \theta$
As, $\alpha =\omega \frac{d\omega }{d\theta },\alpha =\frac{3g}{4a}\cos \theta$
Along $t$-axis, $\sum F_t=mg\cos \theta -N=m\frac{a}{3}\alpha$
Or $mg\cos \theta -N=m\cdot \frac{a}{3}\cdot \frac{3g}{4a}\cos \theta$
$\Rightarrow N=\frac{3mg}{4}\cos \theta$
Along $n$-axis, $\sum F_n=\mu N-mg\sin \theta =m\left(\frac{a}{3}\right){\omega }^2$
(this is centripetal force)
$\therefore \mu \left(\frac{3mg}{4}\cos \theta \right)-mg\sin \theta =m\left(\frac{a}{3}\right)\left(\frac{3g\sin \theta }{2a}\right)$
$\mu \frac{3mg}{4}\cos \theta =\frac{3}{2}mg\sin \theta$
$\Rightarrow \mu =2\tan \theta$
$\beta =2$