Question

A uniform rod is placed on two spinning wheels as shown in the figure. The axes of the wheels are separated by $\mathrm{l}=20\mathrm{cm}$, and the coefficient of friction between the rod and the wheel is $\mathrm{k}=0.18$. Show that the rod performs SHM and find the period of small oscillations.

Answer 1

Step 1: Given information

The axes of the wheels are separated by $\mathrm{l}=20\mathrm{cm}$.

The coefficient of friction between the rod and the wheel is $\mathrm{k}=0.18$.

Step 2: Calculate the period of small oscillations

The rod's C.M. would be between the two revolving wheels somewhere at the equilibrium position. Let's all move the rod horizontally a short distance and afterward remove it. Let us represent the forces operating on the rod at x distances from its equilibrium position. Newton's second law states that there is no net vertical force exerted on the rod:

${\mathrm{N}}_1+{\mathrm{N}}_2=\mathrm{mg}\dots \left(1\right)$

Here,

N₁ and N₂ are normal reaction forces.

m is the mass.

For the translational motion of the rod from the equation.

${\mathrm{kN}}_1-{\mathrm{kN}}_2=\mathrm{mx}...\left(2\right)$

As there is no net torque at an axis perpendicular to the plane of the figure via the rod's C.M.

${\mathrm{N}}_1\left(\frac{\mathrm{l}+\mathrm{x}}{2}\right)={\mathrm{N}}_2\left(\frac{\mathrm{l}-\mathrm{x}}{2}\right)\dots \left(3\right)$

Solve equations $\left(1\right),\left(2\right),$and $\left(3\right)$.

$\mathrm{x}=-\mathrm{k}\left(\frac{2\mathrm{g}}{\mathrm{l}}\right)\mathrm{x}$

Hence, the time period would be,

$$\begin{gathered} \Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{2}\mathrm{kg}} \\ =\mathrm{\pi }\sqrt{\frac{2\mathrm{l}}{\mathrm{kg}}} \\ =1.5\mathrm{s} \end{gathered}$$

Therefore, the period of small oscillations is $1.5\mathrm{s}$.