A uniform rod is placed on two spinning wheels as shown in figure. The axes of the wheels are separated by a distance  $\mathcal{l}$, the coefficient of friction between the rod and the wheels is  $k$. If  the period of small horizontal oscillations of the rod is $\pi \sqrt{\frac{n\mathcal{l}}{kg}}$ Find n is 

In the equilibrium position the centre of mass (C.M) of the rod lies midway between the two rotating wheels. Let us displace the rod horizontally by a small distance and then release it. Let use depict the forces acting on the rod when its C.M. is at distance x from the equilibrium position (see figure). Since there is no net vertical force acting on the rod, Newtons second law gives 
$N_1+N_2=mg\mathrm{..........................}\left(1\right)$ 
For the translational motion of the rod, from the equation $F_x=m{\omega }_{cx}$  we have 
 $k{N}_1-k{N}_2=m\overline{x}\mathrm{..........................}\left(2\right)$
 
As the rod experience no net torque about an axis perpendicular to the plane of the figure through the C.M of the rod
$N_1(\frac{1}{2}\times x)=N_2(\frac{1}{2}-x)\mathrm{....................}\left(3\right)$ 
Solving equations (1), (2) and (3) simultaneously we get $x=-k\frac{2g}{l}x$ 
Hence the sought time period  $T=2\pi \sqrt{\frac{l}{2kg}}=\pi \sqrt{\frac{2l}{kg}}=1.5s$