We fix a string made of insulating material of length $l$ at point $A$, and attach a small ball of mass $m$ and charge $q$ to its other end. Below point $A$, at a distance of $2l$, we fix a ball of charge $-q$ at point $B$. We launch the ball of mass $m$ such that it moves in a circular path in a horizontal plane. What is the period of the circular motion if the angle between the string and the vertical is $\varphi$? What is the smallest angular velocity with which the ball can still move in a circular path? Given data: $m=0.1 \text{ g}; l=10 \text{ cm}; \varphi=30^{\circ}; q=5 \cdot 10^{-8} \text{ C}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3a86c88d10eea2262dd5g-1.jpg?height=434&width=455&top_left_y=393&top_left_x=824)
The task concerns a conical pendulum (1. Problem 1552, KML 59 [1979] p. 227) on which, in addition to the gravitational force, the force due to the attraction of charges also acts.

![](https://cdn.mathpix.com/cropped/2024_05_02_3b97c82bf096bc84bf4eg-1.jpg?height=341&width=290&top_left_y=285&top_left_x=882)

Based on the second Newton's axiom, we write the equation of motion for a mass \( m \). Using the notation from Figure 1, the components of the equation in \( r \) and \( z \) (with \( a_c \) denoting the centripetal acceleration) are:

![](https://cdn.mathpix.com/cropped/2024_05_02_3b97c82bf096bc84bf4eg-1.jpg?height=418&width=509&top_left_y=821&top_left_x=797)

\[
\begin{gathered}
N \sin \varphi + F \sin \vartheta = m a_c \\
N \cos \varphi - F \cos \vartheta - m g = 0
\end{gathered}
\]

We can express the trigonometric functions of \(\vartheta\) in terms of \(\varphi\):

\[
\cos \vartheta = \frac{2 - \cos \varphi}{\sqrt{5 - 4 \cos \varphi}}, \quad \sin \vartheta = \frac{\sin \varphi}{\sqrt{5 - 4 \cos \varphi}}
\]

According to Coulomb's law,

\[
F = k \frac{q^2}{l^2 (5 - 4 \cos \varphi)}
\]

Using these, we express \( N \) from (1) and substitute it into (2):

\[
m a_c = \tan \varphi \left[ m g + k \frac{q^2 (2 - \cos \varphi)}{l^2 (5 - 4 \cos \varphi)^{3/2}} \right] + k \frac{q^2 \sin \varphi}{l^2 (5 - 4 \cos \varphi)^{3/2}}
\]

We know the centripetal acceleration:

\[
a_c = \omega^2 l \sin \varphi
\]

Substituting this into the equation and rearranging, we get:

\[
\omega^2 = \frac{1}{\cos \varphi} \left[ \frac{g}{l} + \frac{2 k q^2}{(5 - 4 \cos \varphi)^{3/2}} \right]
\]

For the values \( m = 10^{-4} \, \text{kg}, \, l = 0.1 \, \text{m}, \, q = 5 \times 10^{-8} \, \text{C}, \, \varphi = 30^\circ \), and \( k = 9 \times 10^9 \, \text{Vm/C} \):

\[
\omega = 19.65 \, \text{s}^{-1}
\]

The period of revolution is \( T = 2 \pi / \omega = 0.32 \, \text{s} \).

The minimum \(\omega\) value can be found using the condition \(\frac{d(\omega^2)}{d\varphi} = 0\). This results in a higher-order equation, so instead of finding its exact solution, we numerically search for the minimum.

![](https://cdn.mathpix.com/cropped/2024_05_02_3b97c82bf096bc84bf4eg-2.jpg?height=564&width=743&top_left_y=130&top_left_x=623)

Figure 2

The following table records the values of \(\omega(\varphi)\), and the curve is shown in Figure 2.

| \(\varphi (^\circ)\) | 0 | 10 | 20 | 30 | 40 | 45 | 46 | 47 | 48 | 49 | 50 | 60 | 70 | 80 | 90 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| \(\omega (1/\text{s})\) | 23.41 | 22.76 | 21.23 | 19.66 | 18.61 | 18.37 | 18.35 | 18.35 | 18.34 | 18.35 | 18.37 | 19.22 | 21.84 | 29.24 | - |

Thus, we find that the minimum angular velocity \(\omega_{\min} = 18.341 \, \text{s}^{-1}\), and the corresponding angle \(\varphi_0 \approx 48^\circ\).

Gábor Borbély (Kapuvár, Gymnasium, 4th year)

Remarks. 1. In reality, with a fixed \(\varphi\), a circular motion does not always arise from the calculated angular velocity based on the force laws and the equation of motion; the stability of the orbit must be examined. If \( q = 0 \), the stability is clearly evident. A small radial impulse is given to the ball, which, due to the conservation of angular momentum, starts moving on a smaller orbit with a higher angular velocity, which corresponds to a smaller \(\omega\) than the previous one, thus the ball returns to its original orbit. For \( q \neq 0 \), the question must be examined in more detail.

We seek the stability condition against small \(\Delta \varphi\) displacements. The \( z \)-component of the angular momentum is a conserved quantity:

\[
m l^2 \omega(\varphi) \sin^2 \varphi = \text{constant.}
\]

Differentiating this product with respect to \(\varphi\) using the product rule, we get:

\[
m l^2 \left[ \omega \left( \frac{d\omega}{d\varphi} \right) \sin^2 \varphi + \omega \cdot 2 \sin \varphi \cos \varphi \right] = 0
\]

Thus,

\[
\frac{d\omega}{d\varphi} = -2 \omega \cot \varphi
\]

For a small \(\Delta \varphi\) displacement, the change in the angular velocity of the body is approximately:

\[
\Delta \omega \approx - (2 \omega \cot \varphi) \Delta \varphi
\]

The circular motion is stable against \(\Delta \varphi\) if the angular velocity corresponding to the orbit \(\varphi + \Delta \varphi\) (which can be calculated using (3)) is greater than \(\omega + \Delta \omega\) when \(\Delta \varphi > 0\). In this case, the body would start moving on a larger orbit with a smaller angular velocity, and the Coulomb and string forces would pull it back. Similarly, for \(\Delta \varphi < 0\), \(\omega + \Delta \omega\) must be greater than the angular velocity that the centripetal force can provide, so the body would return to its original position. The stability condition is thus that the slope of the \(\omega(\varphi)\) curve must always be greater than \(-2 \omega \cot \varphi\).

The smallest slope of the \(\omega(\varphi)\) curve from (3) is approximately \(\left(1 / 10^\circ\right) \left[ \varphi(40^\circ) - \varphi(30^\circ) \right] = -0.11 /^\circ \text{s}\), which is much greater than the value of \(-2 \omega \cot \varphi\), for example, if \(0 \leq \varphi \leq 89^\circ\). Near \(90^\circ\), the slope of \(\omega(\varphi)\) is positive, while \(-2 \omega \cot \varphi\) is negative.

Thus, the stability condition is satisfied in all cases.

2. The system of equations (1)-(2) only determines the motion if the radiative losses due to the acceleration of the charge \( q \) can be neglected, i.e., as long as the frequency is sufficiently small.