Question

A uniform bar with mass $m$ lies symmetrically across two rapidly rotating fixed rollers A and B, with distance ${}^{\prime }{l}^{\prime }$ between the bar's centre of mass and each roller. The rollers whose direction of rotation are shown in figure slip against the bar with coefficient of friction $\mu$. Suppose the bar is displaced horizontally by a small distance ${}^{\prime }{x}^{\prime }$ and then released, find the time period of oscillation.

• A. $T=\pi \sqrt{\frac{l}{\mu g}}$
• B. $T=\pi \sqrt{\frac{l}{4\mu g}}$
• C. $T=2\pi \sqrt{\frac{2l}{\mu g}}$
• D. $T=2\pi \sqrt{\frac{l}{\mu g}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{l}{\mu g}}$

Let $N_1$ and $N_2$ be the normal contact forces between the bar and the rollers A, B respectively. Suppose CoM is displaced by distance $x$ towards right.


From the FBD,
$N_1+N_2=mg$
Balancing moments about the point of contact of the roller B and the bar, we obtain
$N_1\left(2l\right)=mg(l-x)$
Similarly, about the point of contact of the roller A and the bar
$N_2\left(2l\right)=mg(l+x)$
$\Rightarrow N_1=\frac{mg(l-x)}{2l};N_2=\frac{mg(l+x)}{2l}$
Since $N_2\ne N_1$, the frictional forces $f_1$ and $f_2$ will not be equal.

The net horizontal force acting on the rod is given as
$F=f_1-f_2=\mu (N_1-N_2)$
$\Rightarrow F=\frac{\mu mg}{2l}\left\{\right(l-x)-(l+x\left)\right\}$
$\Rightarrow F=-\left(\frac{\mu mg}{l}\right)x$.
Since $F\propto -x$, the motion of the bar is simple harmonic.
i.e $a=\frac{F}{m}=-{\omega }^{2}x=-\frac{\mu g}{l}x$
$\Rightarrow \omega =\sqrt{\frac{\mu g}{l}}$
$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l}{\mu g}}$