A small ball of mass 2 * 10^-3 kg having a charge of 1 muC is suspended by a string of length 0.8 m.

Q: Why is tension zero at the highest point?

For the minimum velocity to complete the circle, the string just barely stays taut at the highest point. Any slower, and it would slacken, causing the ball to fall out of the circular path.

Q: Why doesn't electrostatic potential energy change?

The electrostatic potential energy depends on the distance between the two charges. Since the ball moves in a circle of radius $l$ around the fixed charge, this distance is constant, so the potential energy remains constant.

Q: What if the charges were opposite?

If the charges were opposite, the electrostatic force would be attractive (inwards). This would add to the required centripetal force, meaning you would need a much higher initial velocity to complete the circle.

Q: Why do we use $2l$ for the height difference?

The ball starts at the lowest point (distance $l$ below the center) and goes to the highest point (distance $l$ above the center). The total vertical displacement is $l + l = 2l$.

Q: Can we use kinematics equations here?

No, because the net force (and therefore acceleration) is not constant. The direction of gravity and tension changes continuously as the ball moves along the circular path. We must use energy conservation.

Question

A small ball of mass $2 	imes 10^{-3}	ext{ kg}$ having a charge of $1\mu	ext{C}$ is suspended by a string of length $0.8	ext{ m}$. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball, so that it can make complete revolution.

Sigma Percentile · Accepted Answer

### The Setup
Imagine a small charged ball suspended by a string of length $l = 0.8	ext{ m}$. Right at the point of suspension, another identical charge is fixed. We need to find the minimum horizontal velocity $u$ to give the lower ball so it completes a full vertical circle.

This is a classic vertical circular motion problem, but with an electrifying twist! Instead of just gravity and tension, we now have a constant radial electrostatic force pushing the ball outwards.

### Conquering the Highest Point
To ensure the ball completes the circle, we must analyze the forces at the highest point. Gravity pulls it down with a force $mg$, the string's tension $T$ also pulls it down, but the electrostatic repulsion $F_e$ pushes it radially outward, which is upwards in this position.

For the minimum velocity, the string must just barely stay taut at the top. This means the tension $T$ becomes zero. The net inward force providing the centripetal acceleration is simply the weight minus the electrostatic repulsion:

$$ \frac{mv^2}{l} = mg - F_e $$

Using Coulomb's law, the electrostatic force is $F_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l^2}$. Substituting this into our equation gives:

$$ v^2 = \frac{l}{m} \left( mg - \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l^2} ight) $$

Plugging in the given values ($m = 2 	imes 10^{-3}	ext{ kg}$, $l = 0.8	ext{ m}$, $q = 10^{-6}	ext{ C}$), we calculate the square of the velocity at the top:

$$ v^2 = \frac{0.8}{2 	imes 10^{-3}} \left( 2 	imes 10^{-3} 	imes 10 - \frac{9 	imes 10^9 	imes (10^{-6})^2}{(0.8)^2} ight) = 2.4 	ext{ m}^2/	ext{s}^2 $$

### The Energy Bridge
Now, how do we relate this top velocity $v$ to the initial velocity $u$ at the bottom? We use the conservation of mechanical energy. Notice that the electrostatic potential energy $U_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{l}$ is exactly the same at the top and bottom because the distance from the fixed charge is always $l$!

Since the electrostatic potential energy doesn't change, the loss in kinetic energy simply equals the gain in gravitational potential energy:

$$ \frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mg(2l) $$

### The Final Calculation
We can cancel the mass $m$ and rearrange to solve for $u^2$:

$$ u^2 = v^2 + 4gl $$

Substituting $v^2 = 2.4$, $g = 10$, and $l = 0.8$, we get:

$$ u^2 = 2.4 + 4(10)(0.8) = 34.4 	ext{ m}^2/	ext{s}^2 $$

Taking the square root, we find the minimum initial velocity $u$ to be approximately $5.86	ext{ m/s}$. This is the exact speed needed to conquer the circle!

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Forces at the Highest Point$

$Critical Condition for Loop$

$Electrostatic Force$

$Calculating v^{2}$

$Conservation of Mechanical Energy$

$Calculating Initial Velocity u$

$Final Answer$

The Sigma Insight: Coulomb's Law

The Setup Imagine a small charged ball suspended by a string of length $l = 0.8 m$

Conquering the Highest Point To ensure the ball completes the circle, we must analyze the forces at the highest point

The Energy Bridge Now, how do we relate this top velocity $v$ to the initial velocity $u$ at the bottom? We use the conservation of mechanical energy

The Final Calculation

Similar Questions

Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$ . What is the equilibrium separation when each thread makes a small angle $θ$ with the vertical ?

Two charges each equal to $q$ , are kept at $x = - a$ and $x = a$ on the x-axis. A particle of mass $m$ and charge $q_{0} = q /2$ is placed at the origin. If charge $q_{0}$ is given, a small displacement ( $y << a$ ) along the y-axis, the net force acting on the particle is proportional to

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces at the Highest Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Critical Condition for Loop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrostatic Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating v2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Initial Velocity u

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer