To determine the minimum horizontal velocity needed for a charged ball to make a complete revolution while suspended by a string, we can analyze the forces acting on the ball and apply principles from physics, particularly circular motion and electrostatics.
Understanding the Forces Involved
We have two identical balls, each with a charge of 1 µC (microcoulomb) and a mass of 2 × 10^-2 kg (or 0.02 kg). One ball is suspended by a string of length 0.8 m, while the other is positioned at the point of suspension. The electrostatic repulsion between the two balls will play a crucial role in our calculations.
Electrostatic Force Calculation
The electrostatic force \( F_e \) between the two charged balls can be calculated using Coulomb's law:
- Formula: \( F_e = k \frac{|q_1 \cdot q_2|}{r^2} \)
- Where:
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 \) and \( q_2 \) are the charges of the balls (both 1 µC or \( 1 \times 10^{-6} \, \text{C} \))
- \( r \) is the distance between the two charges, which will be equal to the length of the string when the lower ball is at the top of its circular path (0.8 m).
Substituting the values:
\( F_e = 8.99 \times 10^9 \frac{(1 \times 10^{-6})^2}{(0.8)^2} \)
Calculating this gives:
\( F_e = 8.99 \times 10^9 \frac{1 \times 10^{-12}}{0.64} \approx 1.40 \times 10^{-2} \, \text{N} \)
Analyzing Circular Motion
For the lower ball to complete a circular motion, it must have sufficient velocity at the top of the path to maintain tension in the string. At the top of the circular path, the forces acting on the ball are:
- The gravitational force \( F_g = mg \) acting downward
- The electrostatic force \( F_e \) also acting downward
At the top of the circular path, the centripetal force \( F_c \) required to keep the ball in circular motion is provided by the sum of these two forces:
\( F_c = F_g + F_e \)
Substituting the expressions for these forces:
\( \frac{mv^2}{r} = mg + F_e \)
Where \( v \) is the velocity at the top of the circular path and \( r \) is the radius (0.8 m). Rearranging gives:
\( v^2 = r(g + \frac{F_e}{m}) \)
Substituting the known values:
- \( g \approx 9.81 \, \text{m/s}^2 \)
- \( F_e \approx 1.40 \times 10^{-2} \, \text{N} \)
- \( m = 0.02 \, \text{kg} \)
Calculating \( \frac{F_e}{m} \):
\( \frac{1.40 \times 10^{-2}}{0.02} = 0.70 \, \text{m/s}^2 \)
Now substituting back into the equation for \( v^2 \):
\( v^2 = 0.8(9.81 + 0.70) = 0.8 \times 10.51 \approx 8.408 \)
Thus, \( v \approx \sqrt{8.408} \approx 2.9 \, \text{m/s} \).
Final Consideration for Minimum Velocity
To ensure the ball can complete the revolution, it must be given an initial horizontal velocity. The minimum horizontal velocity \( v_0 \) imparted to the lower ball must be at least 2.9 m/s. This velocity ensures that when the ball reaches the top of its path, the forces acting on it will allow it to maintain circular motion without the string going slack.
In summary, the minimum horizontal velocity required for the lower ball to make a complete revolution is approximately 2.9 m/s. This analysis combines concepts from electrostatics and circular motion, illustrating how forces interact in a dynamic system.
