A person of mass M is setting on a swing of length L and swinging with an angular amplitude `theta_(0)` if the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance `l(lt lt L)`

Options

A. `Mgl(1-theta_(0))^(2)`
B. `Mgl(1+heta_(0)^(2))`
C. `Mgl`
D. `Mgl(1+(theta_(0)^(2))/(2))`

Answer

To solve the problem, we need to analyze the situation step by step. The person is sitting on a swing of length \( L \) and swings with an angular amplitude \( \theta_0 \). When the swing passes through its lowest point, the person stands up, and we need to find the work done by the person, assuming that his center of mass moves a distance \( l \) (where \( l \ll L \)).

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The person has a mass \( M \).
- The swing has a length \( L \).
- The angular amplitude is \( \theta_0 \).
- The swing passes through its lowest point with maximum speed \( v_0 \).

2. **Determine the Initial Angular Momentum**:
- At the lowest point, the angular momentum \( L_i \) about the point of suspension is given by:
\[ L_i = M \cdot v_0 \cdot L \]

3. **Determine the Final Angular Momentum**:
- After the person stands up, the new speed \( v_1 \) can be expressed in terms of the distance moved by the center of mass:
\[ L_f = M \cdot v_1 \cdot (L - l) \]
- By conservation of angular momentum, we have:
\[ M \cdot v_0 \cdot L = M \cdot v_1 \cdot (L - l) \]
- Simplifying gives:
\[ v_1 = \frac{v_0 \cdot L}{L - l} \]

4. **Apply the Work-Energy Theorem**:
- The change in kinetic energy is equal to the work done on the person:
\[ \Delta KE = KE_f - KE_i \]
- This can be expressed as:
\[ \Delta KE = \frac{1}{2} M v_1^2 - \frac{1}{2} M v_0^2 \]

5. **Substituting for \( v_1 \)**:
- Substitute \( v_1 \) into the kinetic energy equation:
\[ \Delta KE = \frac{1}{2} M \left(\frac{v_0 L}{L - l}\right)^2 - \frac{1}{2} M v_0^2 \]

6. **Calculate the Work Done by Gravity**:
- The work done by gravity when the center of mass moves a distance \( l \) is:
\[ W_g = M g l \]

7. **Set Up the Equation**:
- According to the work-energy theorem:
\[ \Delta KE = W_{person} + W_g \]
- Rearranging gives:
\[ W_{person} = \Delta KE - W_g \]

8. **Substituting Values**:
- Substitute the expressions for \( \Delta KE \) and \( W_g \) into the equation to find \( W_{person} \).

9. **Final Expression**:
- After simplification and applying the binomial approximation for \( l \ll L \), we arrive at:
\[ W_{person} = M g l (1 + \theta_0^2) \]

### Final Answer:
The work done by the person when he stands up is:
\[ W_{person} = M g l (1 + \theta_0^2) \]