A simple pendulum of mass m and length L is held in horizontal position. If it is released from this position, then find the angular momentum of the bob about the point of suspension when it is vertically below the point of suspension.

Options

A. `mLsqrt(2gL)`
B. `mLsqrt(gL)`
C. `mLsqrt(3gL)`
D. none

Answer

To find the angular momentum of the bob about the point of suspension when it is vertically below the point of suspension, we can follow these steps:

### Step 1: Understand the Initial and Final States
Initially, the pendulum is held horizontally at a height \( L \) above the lowest point. When released, it swings down to the lowest point, where the height is zero.

### Step 2: Apply Conservation of Mechanical Energy
The mechanical energy conservation principle states that the total mechanical energy (potential + kinetic) remains constant if only conservative forces are acting.

- **Initial Potential Energy (PE_initial)** at height \( L \):
\[ PE_{\text{initial}} = mgh = mgL \]
- **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = 0 \quad (\text{since it is released from rest}) \]

- **Final Potential Energy (PE_final)** at the lowest point:
\[ PE_{\text{final}} = 0 \]

- **Final Kinetic Energy (KE_final)** when the bob is at the lowest point:
\[ KE_{\text{final}} = \frac{1}{2} mv^2 \]

Using conservation of energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]
Substituting the values:
\[ mgL + 0 = 0 + \frac{1}{2} mv^2 \]

### Step 3: Solve for Velocity \( v \)
Rearranging the equation:
\[ mgL = \frac{1}{2} mv^2 \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gL = \frac{1}{2} v^2 \]
Multiplying both sides by 2:
\[ 2gL = v^2 \]
Taking the square root:
\[ v = \sqrt{2gL} \]

### Step 4: Calculate Angular Momentum
The angular momentum \( L \) about the point of suspension is given by:
\[ L = r \cdot p \]
where \( r \) is the distance from the pivot (length of the pendulum \( L \)) and \( p \) is the linear momentum \( p = mv \).

Thus,
\[ L = L \cdot mv = mL \cdot \sqrt{2gL} \]

### Step 5: Final Expression for Angular Momentum
So, the angular momentum \( L \) about the point of suspension is:
\[ L = mL \sqrt{2gL} \]

### Summary
The angular momentum of the bob about the point of suspension when it is vertically below the point of suspension is:
\[ L = mL \sqrt{2gL} \]