Question

A uniform rod of length $2 a$ 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

From the law of conservation of energy,
$KE _{i}+ GPE _{i}= KE _{f}+ GPE _{f}$

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