Question

A uniform bar with mass m lies symmetrically across two rapidly rotating fixed rollers, A and B with distance $L=2.0cm$ between the bar's centre of mass and each roller. The rollers, whose directions of rotation are shown in figures slip against the bar with coefficient of kinetic friction ${\mu }_k=0.40$. Suppose the bar is displaced horizontally by a distance x as shown in figure and then released. The angular frequency $\omega$ of the resulting horizontal simple harmonic motion of the bar is(in rad $s^{-1})$

• A. $20$
• B. $15$
• C. $16$
• D. $17$

Answer 1

The correct option is A $20$

Let $N_1$ and $N_2$ be the normal reactions at the rollers, Then

$N_1+N_2=mg$ and $L=2l$

As the bar is in rotational equilibrium

$\sum$ (momentum about G)

$=I_G\left(0\right)=0$

$N_1(l+x)-N_2(l-x)=0$

$\Rightarrow N_1=\frac{mg(l-x)}{2l},N_2=\frac{mg(l+x)}{2l}$

$F=f_1-f_2=\mu [N_1-N_2]=\frac{\mu mg}{2l}\left[\right(l-x)-(l+x\left)\right]=-\left(\frac{\mu mg}{l}\right).x$

$\Rightarrow T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{l}{\mu .g}}=2\pi \sqrt{\frac{L}{2\mu g}}$

$W=\sqrt{\frac{2\mu g}{L}}=20rad/s$