eLECTROSTATICS BYU 3

Solution: Motion of Beads on L-Shaped Rod

Analysis:

Let the position of the bead on the vertical rod be $y$ and the horizontal rod be $x$. The force between them is attractive, $F = \frac{k q_1 q_2}{r^2}$, where $r = \sqrt{x^2+y^2}$.

The equations of motion for the two beads (mass $m$) are:

$$m \ddot{x} = -F \cos \theta = -F \frac{x}{r}$$ $$m \ddot{y} = -F \sin \theta = -F \frac{y}{r}$$

Taking the ratio of the accelerations:

$$\frac{\ddot{x}}{\ddot{y}} = \frac{x}{y} \implies x \ddot{y} = y \ddot{x}$$ $$x \ddot{y} – y \ddot{x} = 0$$

Integrating with respect to time (given initial velocities are zero), we get:

$$x \dot{y} – y \dot{x} = 0 \implies \frac{\dot{y}}{y} = \frac{\dot{x}}{x}$$ $$\ln y = \ln x + C \implies y = A x$$

This implies the beads move such that their coordinates always maintain a constant ratio. From the initial conditions $y_0 = d$ and $x_0 = 2d$, we find the constant $A = 0.5$. So, $y = 0.5x$.