Figure shows a smooth non-conducting rod of radius r charged with uniform linear charge density $\lambda$, fixed horizontally. A neutral and smooth ring Q of mass M can slide freely on the rod which happens to just fit in it, and P is a non-conducting particle having a mass m and charge q, attached to the ring Q by means of a non-conducting and inextensible string of length R. P is released from the position shown in figure, when the string becomes vertical

The particle P executes its motion within the influence of two conservative fields, one the gravitational field of earth and the other the electrostatic field of the charged horizontal rod.
The loss in gravitational and electrostatic potential energy of the system appears as gain in its kinetic energy.
As the work done by the tension in the string is zero, we have.
 $LossinG.PE+LossinE.PE=GaininKE\mathrm{...}\left(i\right)$
Now, loss in $G.PE=mgR$
And loss in $E.PE=q{\int }_r^{r+R}\left(\frac{\lambda dx}{2\pi {\epsilon }_{0}x}\right)$
$=\frac{q\lambda }{2\pi {\epsilon }_0}\ln (1+\frac{R}{r}) or, v^2(1+\frac{m}{M})=2gR+\frac{q\lambda }{\pi {\epsilon }_{0}m}\ln (1+\frac{R}{r})$

If v’ and v be the velocities attained by the ring Q and particle P, when the string becomes vertical then gain in

$KE=\frac{1}{2}M{\text{'}}^2+\frac{1}{2}m{v}^2 or v={\left[{(1+\frac{m}{M})}^{-1}(2gR+\frac{q\lambda }{\pi {\epsilon }_{0}m}\ln (1+\frac{R}{r}))\right]}^{1/2}$

Since, there is no force (external) acting on the system horizontal, so applying, the law of conservation of momentum

$$\begin{gathered} Mv\text{'}=mv or v\text{'}=\frac{mv}{M} \\ \therefore GaininKE=\frac{m{v}^2}{2}(1+\frac{m}{M}) \\ \therefore Fromeqn.\left(i\right), \\ mgR+\frac{q\lambda }{2\pi {\epsilon }_0}\ln (1+\frac{R}{r})=\frac{1}{2}\times m{v}^2(1+\frac{m}{M}) \end{gathered}$$

If T be the required tension in the string, then, the net force toward the centre of the circular path, acting on the charged particle (vertically up) will be
$$\begin{gathered} T=mg-qE \\ Where E=\frac{\lambda }{2\pi {\epsilon }_0(r+R)}=\frac{m{v}_{net}^2}{R} \end{gathered}$$
From, mechanics of circular motion,
Or  $T=mg+\frac{q\lambda }{2\pi {\epsilon }_0(r+R)}+\frac{m{v}^2}{R}{(1+\frac{m}{M})}^2$