Question

A uniform rod is placed on two spinning wheels as shown in figure. The axes of the wheels ore separated by a distance i = 20 cm, the coefficient of friction between the rod and the wheels is k = 0.18. Demonstrate that in this case the rod performs harmonic oscillations. Find the period of these oscillations.

Answer 1

In the equilibrium position the C.M of the rod lies way between the two rotating wheels.Let us displaced the rod horizontally by some small distance and then release it. Let us depit the forces acting on the rod when its C.M. is at distance $x$ from its equilibrium position (Fig). Since there is no net vertical force acting on the rod, Newton's law gives :
$N_1+N_2=mg\left(1\right)$
For the translational motion of the rod from the Eqsn : $F_x=m{w}_{\alpha }$
$k{N}_1-k{N}_2=m\ddot{x}$
As the rod experiences no net torque about an axis perpendicular to the plane of the Fig. through the C.M of the rod
$N_1\left(\frac{l+x}{2}\right)=N_2\left(\frac{l-x}{2}\right)\left(3\right)$
Solving Wqns $\left(1\right),\left(2\right)$ and$\left(3\right)$ simultaneously we get
$\ddot{x}=-k\frac{2g}{l}x$
Hence th2 sough time period
$T=2\pi \sqrt{\frac{l}{2kg}}=\pi \sqrt{\frac{2l}{kg}}=1.5s$