Solution

According to Problem 3.1.13, in relation to Problem 3.1.12, we find the following configuration: a small charged body of mass $m$ can slide freely on a vertical spoke, at the lower point of which there is a charge of the same sign as the charge of the body. The equilibrium position of the body is at a distance $R$ from this charge. Then a small mass $\Delta m$ is added jointly to the charged body of mass $m$.

Evidently, this will alter the value of $R$ in the interaction of forces, first established to be:

$$K\frac{Qq}{r^{2}} = mg \qquad \Rightarrow \qquad r = \sqrt{\frac{KQq}{mg}} .$$

After adding the mass:

$$K\frac{Qq}{r'^2} = (m+\Delta m)g \qquad \Rightarrow \qquad r' = \sqrt{\frac{KQq}{(m+\Delta m)g}} .$$

The simplest way to solve this problem is with the Universal Energetic Conservation Law, by which $\Delta \Sigma m = 0$, allowing:

$$E_{el,pot}(R) + E_{gr,pot}(R) = E_{el,pot}(R') + E_{gr,pot}(R') + E_{kin}$$

Expanding:

$$(m+\Delta m) g r + \frac{KQq}{r} = (m+\Delta m) g r' + \frac{KQq}{r'} + \frac{(m+\Delta m)v^{2}}{2}$$

Since

$$r' = \sqrt{\frac{KQq}{(m+\Delta m)g}},$$

the equation becomes:

$$(m+\Delta m) g r + \frac{KQq}{r} = (m+\Delta m) g \sqrt{\frac{KQq}{(m+\Delta m)g}} + \frac{KQq}{\sqrt{\frac{KQq}{(m+\Delta m)g}}} + \frac{(m+\Delta m)v^{2}}{2}$$

Solving for $v$, the maximum velocity is:

$$v_{max} = \sqrt{ 2gR \left( 1 + \frac{m}{m+\Delta m} - 2\sqrt{\frac{m}{m+\Delta m}} \right) }$$

This can be proved by examining the motion it will produce. As no air drag or other forces act, we are working with a conservative force field, implying the body will describe an unrestrained Simple Harmonic Motion (SHM). According to the SHM velocity formula:

$$v = \omega \sqrt{A^{2} - x^{2}},$$

the maximum speed occurs at $x=0$, so:

$$v_{\max} = A\omega .$$

The amplitude is the difference between the original equilibrium position and the new one:

$$A = r - r' = \sqrt{\frac{KQq}{mg}} - \sqrt{\frac{KQq}{(m+\Delta m)g}}$$

Substituting this into the SHM formulas yields an equivalent expression to the one obtained using the Universal Energy Conservation Law, although doing so directly would require a Taylor expansion, complicating the problem.