A particle is rotated in a vertical circle by connecting it to a string of length `l` and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

Options

A. `sqrt(gl)`
B. `sqrt(2gl)`
C. `sqrt(3gl)`
D. `sqrt(5gl)`

Answer

To find the minimum speed of a particle when the string is horizontal for it to complete a vertical circle, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Problem
A particle is attached to a string of length \( l \) and is rotated in a vertical circle. We need to determine the minimum speed of the particle at the horizontal position (when the string is horizontal) so that it can complete the full circular motion.

### Step 2: Set Up the Energy Conservation Equation
We will use the conservation of mechanical energy. The total mechanical energy at the horizontal position must equal the total mechanical energy at the lowest point of the circle.

### Step 3: Define the Energies
1. **At the horizontal position** (let's denote this position as point A):
- Height from the reference point (lowest point) = \( l \)
- Potential Energy (PE) at A = \( mgh = mg \cdot l \)
- Kinetic Energy (KE) at A = \( \frac{1}{2} mv^2 \) (where \( v \) is the speed at the horizontal position)

2. **At the lowest point** (let's denote this position as point B):
- Height from the reference point = 0
- Potential Energy (PE) at B = \( 0 \)
- Kinetic Energy (KE) at B = \( \frac{1}{2} m u^2 \) (where \( u \) is the speed at the lowest point)

### Step 4: Write the Conservation of Energy Equation
Using the conservation of energy:
\[ \text{Total Energy at A} = \text{Total Energy at B} \]
\[ PE_A + KE_A = PE_B + KE_B \]
Substituting the expressions we defined:
\[ mg \cdot l + \frac{1}{2} mv^2 = 0 + \frac{1}{2} mu^2 \]

### Step 5: Substitute the Minimum Speed at the Lowest Point
From physics, we know that the minimum speed \( u \) required at the lowest point to maintain circular motion is given by:
\[ u = \sqrt{5gl} \]
Substituting \( u \) into the energy equation:
\[ mg \cdot l + \frac{1}{2} mv^2 = \frac{1}{2} m(5gl) \]

### Step 6: Simplify the Equation
Cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ gl + \frac{1}{2} v^2 = \frac{5}{2} gl \]
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{5}{2} gl - gl \]
\[ \frac{1}{2} v^2 = \frac{3}{2} gl \]

### Step 7: Solve for \( v \)
Multiply both sides by 2:
\[ v^2 = 3gl \]
Taking the square root:
\[ v = \sqrt{3gl} \]

### Final Answer
The minimum speed of the particle when the string is horizontal for it to complete the circle is:
\[ v = \sqrt{3gl} \]