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 step by step, we will analyze the motion of the mass attached to the string and determine the minimum distance \( X \) from point \( O \) where the peg \( P \) should be fixed.

### Step 1: Understand the Initial Setup
- A mass \( m \) is attached to a string of length \( l \) and is released from a horizontal position.
- The mass will swing downwards and will reach a vertical position below point \( O \).

### Step 2: Identify the Energy Conservation
- At the initial position (horizontal), the potential energy (PE) is maximum, and kinetic energy (KE) is zero.
- When the mass reaches the vertical position (point B), it has converted some potential energy into kinetic energy.

### Step 3: Calculate Potential and Kinetic Energy
- Initial potential energy at point A (horizontal position):
\[ U_A = mgh = mg \cdot l \]
- Kinetic energy at point B:
\[ K_B = \frac{1}{2} mv^2 \]
- By conservation of mechanical energy:
\[ U_A + K_A = U_B + K_B \]
Since \( K_A = 0 \) (mass is at rest initially):
\[ mg \cdot l = 0 + \frac{1}{2} mv^2 \]
Simplifying gives:
\[ v^2 = 2gl \quad \text{(Equation 1)} \]

### Step 4: Determine Conditions for Completing the Circle
- For the mass to complete a vertical circle around peg \( P \), the speed \( v \) at the top of the circle must satisfy:
\[ v^2 \geq 5gR \]
where \( R = l - X \) (the radius of the circle after the peg).
- Thus, we have:
\[ v^2 \geq 5g(l - X) \quad \text{(Equation 2)} \]

### Step 5: Set Up the Inequality
- From Equations 1 and 2, we equate:
\[ 2gl \geq 5g(l - X) \]
- Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ 2l \geq 5(l - X) \]

### Step 6: Solve for \( X \)
- Expanding the right side:
\[ 2l \geq 5l - 5X \]
- Rearranging gives:
\[ 5X \geq 5l - 2l \]
\[ 5X \geq 3l \]
- Dividing by 5:
\[ X \geq \frac{3l}{5} \]

### Conclusion
The minimum distance \( X \) from point \( O \) where the peg \( P \) should be fixed is:
\[ X = \frac{3l}{5} \]