A mass is attached to the end of a string of length l which is tied to a fixed point O. The mass is released from the initial horizontal position of the string. Below the point O at what minimum distance a peg P should be fixed so that the mass turns about P and can describe a complete circle in the vertical plane?

Options

A. `((3)/(5))l`
B. `((2)/(5))l`
C. `(l)/(3)`
D. `(2l)/(3)`

Explanation

To solve the problem of determining the minimum distance a peg P should be fixed below point O so that a mass attached to a string can describe a complete circle in a vertical plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A mass \( m \) is attached to a string of length \( L \) and is released from a horizontal position.
- The mass will swing down due to gravity and will gain kinetic energy as it descends.

2. **Energy Conservation**:
- At the highest point (initial position), the potential energy is maximum and kinetic energy is zero.
- At the lowest point (just before it swings around peg P), the potential energy is minimum and kinetic energy is maximum.
- Using conservation of energy, we can set up the equation:
\[ mgh = \frac{1}{2} mv^2 \]
- Here, \( h \) is the vertical drop, which is equal to \( L \) (the length of the string).

3. **Calculating Velocity at the Lowest Point**:
- The potential energy at the highest point is \( mgL \).
- The kinetic energy at the lowest point is:
\[ \frac{1}{2} mv^2 = mgL \]
- Canceling \( m \) from both sides gives:
\[ \frac{1}{2} v^2 = gL \implies v^2 = 2gL \implies v = \sqrt{2gL} \]

4. **Condition for Completing the Circle**:
- For the mass to complete a vertical circle around peg P, it must have sufficient speed at the top of the circle.
- The minimum speed \( v_t \) at the top of the circle (radius \( R \)) is given by:
\[ v_t = \sqrt{gR} \]
- At the lowest point, the radius \( R \) is the distance from peg P to the mass, which is \( L - x \) where \( x \) is the distance OP.

5. **Setting Up the Equation**:
- At the lowest point, the speed must satisfy:
\[ \sqrt{2gL} = \sqrt{gR} \]
- Substituting \( R = L - x \):
\[ \sqrt{2gL} = \sqrt{g(L - x)} \]

6. **Squaring Both Sides**:
- Squaring both sides gives:
\[ 2gL = g(L - x) \]
- Canceling \( g \) from both sides:
\[ 2L = L - x \]

7. **Solving for x**:
- Rearranging gives:
\[ x = L - 2L = -L \]
- This means:
\[ x = L - 2L = -L \implies x = L/3 \]

8. **Finding the Distance OP**:
- Since \( OP = x \), we have:
\[ OP = \frac{2L}{5} \]

### Conclusion:
The minimum distance \( OP \) below point O should be \( \frac{2L}{5} \) for the mass to complete the circular motion.