A mass m is attached to the end of a rod of length l. The mass goes along a vertical circular path with the other end, hinged at its centre . What should be the minimum speed of mass at the bottom of this circular path

Options

A. `sqrt(4gl)`
B. `sqrt(2gl)`
C. `sqrt(3gl)`
D. None of these

Explanation

To find the minimum speed of a mass \( m \) at the bottom of a vertical circular path, we need to ensure that the mass has enough kinetic energy to reach the top of the circular path. Here’s the step-by-step solution:

### Step 1: Understand the System
We have a mass \( m \) attached to the end of a rod of length \( l \). The rod is hinged at its center, allowing the mass to move in a vertical circular path.

### Step 2: Identify the Conditions for Motion
For the mass to complete the circular motion, it must have enough speed at the bottom to reach the top of the circular path. At the top of the circle, the minimum speed must be sufficient to provide the necessary centripetal force.

### Step 3: Apply Conservation of Energy
We will use the principle of conservation of mechanical energy. The total mechanical energy at the bottom must equal the total mechanical energy at the top.

1. **At the bottom (point A)**:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = 0 (taking the bottom as the reference level)

2. **At the top (point B)**:
- Kinetic Energy (KE) = \( \frac{1}{2} m v'^2 \) (where \( v' \) is the speed at the top)
- Potential Energy (PE) = \( mgh \) (where \( h = 2l \), the height from the bottom to the top)

### Step 4: Set Up the Energy Equation
Using conservation of energy:
\[ \text{KE at A} + \text{PE at A} = \text{KE at B} + \text{PE at B} \]
\[ \frac{1}{2} mv^2 + 0 = \frac{1}{2} mv'^2 + mg(2l) \]

### Step 5: Determine the Minimum Speed at the Top
At the top of the circular path, the minimum speed \( v' \) must be such that the gravitational force provides the necessary centripetal force. Thus:
\[ mg = \frac{mv'^2}{l} \]
This simplifies to:
\[ v' = \sqrt{gl} \]

### Step 6: Substitute \( v' \) Back into the Energy Equation
Substituting \( v' \) into the energy equation:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m (\sqrt{gl})^2 + mg(2l) \]
This simplifies to:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mgl + 2mgl \]
\[ \frac{1}{2} mv^2 = \frac{1}{2} mgl + 4mgl = \frac{9}{2} mgl \]

### Step 7: Solve for \( v \)
Cancelling \( m \) from both sides and solving for \( v \):
\[ \frac{1}{2} v^2 = 9gl \]
\[ v^2 = 18gl \]
\[ v = \sqrt{18gl} = 3\sqrt{2gl} \]

### Final Answer
The minimum speed of the mass at the bottom of the circular path is:
\[ v = 3\sqrt{2gl} \]