Question

A uniform nonconducting rod of mass m and length $l$, with charge density $\lambda$, as shown in figure, is burged at the midpoint at origin so that it can rotate in a horizontal plane without any friction. A uniform electric field E exists parallel to x-axis in the entire region. Calculate the period of small oscillations of the rod

Answer 1

$\tau ={\int }_0^{//2}d{\tau }_1+{\int }_0^{//2}d{\tau }_2$

$=2{\int }_0^{//2}E\lambda dx\left(xsin\theta \right)=\frac{E\lambda l^2}{4}sin\theta$

As $\theta$ is small, $\tau \approx \frac{E\lambda l^2}{4}\theta$

$\frac{m{l}^2}{12}\alpha =-\frac{E\lambda l^2\theta }{4}$ or $\omega =\sqrt{\frac{3E\lambda }{m}}$

$\therefore T=2\pi \sqrt{\frac{m}{3E\lambda }}$