A non-conducting ring of mass m and radius R is charged as shown in figure and placed on a rough horizontal nonconducting plane. The charge per unit length on the charged quadrants of ring is $\mathrm{\lambda }$. At time t = 0, a uniform electric field $\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\hat{\mathrm{i}}$ is switched on and the ring starts rolling without sliding. If ${\mathrm{E}}_0=15\mathrm{N}/\mathrm{C}, \mathrm{R}=0.50\mathrm{m}\text{ and }\mathrm{\lambda }=2\mathrm{C}/\mathrm{m}$, then find the magnitude of friction force (in N) acting on the ring when it starts rolling.

 

We consider an angular element as shown in figure. Force on element is

     $\mathrm{dF}=\mathrm{\lambda }(\mathrm{R}\cdot \mathrm{d\theta })\cdot {\mathrm{E}}_0$

Perpendicular distance between two equal and opposite force pairs of dF will be

       $\mathrm{r}=2\mathrm{Rsin}\mathrm{\theta }$

torque on ring is

      $\mathrm{d\tau }=\mathrm{dF}.\mathrm{r}=2{\mathrm{\lambda R}}^2{\mathrm{E}}_0\sin \mathrm{\theta }.\mathrm{d\theta }$

$\Rightarrow \mathrm{\tau }={\int }_0^{\mathrm{\pi }/2} \mathrm{d\tau }=2{\mathrm{\lambda R}}^2{\mathrm{E}}_0$

These pair of forces will not provide net force but due to rotation tendency force of friction on ring is/in forward direction as shown.

For pure rolling to take place, we use

$\begin{array}{l} \mathrm{a}=\mathrm{R\alpha }\\ \Rightarrow \frac{\mathrm{f}}{\mathrm{m}}=\mathrm{R}\left[\frac{\mathrm{\tau }-\mathrm{fR}}{{\mathrm{mR}}^2}\right]\\ \Rightarrow \mathrm{f}=\frac{\mathrm{\tau }}{\mathrm{R}}-\mathrm{f}\\ \Rightarrow \mathrm{f}=\frac{\mathrm{\tau }}{2\mathrm{R}}={\mathrm{\lambda RE}}_0=2\times 0.5\times 15=15\mathrm{N}\end{array}$