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 step by step, we will analyze the motion of the person on the swing and calculate the work done when the person stands up at the lowest point of the swing.

### Step 1: Understand the system
A person of mass \( M \) is sitting on a swing of length \( L \) and swinging with an angular amplitude \( \theta_0 \). When the swing reaches its lowest point, the person stands up, causing their center of mass to move a distance \( l \) (where \( l < L \)).

### Step 2: Initial and final velocities
At the lowest point of the swing, we can use conservation of angular momentum to relate the initial and final velocities of the person.

- Let the initial velocity of the person at the lowest point be \( v_0 \).
- When the person stands up, their center of mass moves to a new position, and we denote the final velocity as \( v_1 \).

Using conservation of angular momentum:
\[ M v_0 L = M v_1 (L - l) \]
Cancelling \( M \) from both sides, we get:
\[ v_1 = v_0 \frac{L}{L - l} \]

### Step 3: Apply conservation of energy
Next, we apply the conservation of energy to find the work done by the person. The change in kinetic energy will equal the work done by the person and the work done by gravity.

The change in kinetic energy is given by:
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{1}{2} M v_1^2 - \frac{1}{2} M v_0^2 \]

Substituting \( v_1 \):
\[ \Delta KE = \frac{1}{2} M \left( v_0 \frac{L}{L - l} \right)^2 - \frac{1}{2} M v_0^2 \]

### Step 4: Work done by gravity
The work done by gravity when the person stands up is:
\[ W_g = M g l \]

### Step 5: Set up the equation for work done
From conservation of energy:
\[ \Delta KE = W_{person} + W_g \]
Thus,
\[ \frac{1}{2} M \left( v_0 \frac{L}{L - l} \right)^2 - \frac{1}{2} M v_0^2 = W_{person} + M g l \]

### Step 6: Solve for work done by the person
Rearranging gives:
\[ W_{person} = \frac{1}{2} M \left( v_0 \frac{L}{L - l} \right)^2 - \frac{1}{2} M v_0^2 - M g l \]

### Step 7: Substitute \( v_0 \)
To find \( v_0 \), we can use the relationship between the angular amplitude and the velocity:
\[ v_0 = \sqrt{g L} \sin(\theta_0) \]

### Step 8: Final expression for work done
Substituting \( v_0 \) into the equation for \( W_{person} \) gives us the final expression for the work done by the person.

### Final Result
The final expression for the work done by the person when standing up is:
\[ W_{person} = \frac{1}{2} M \left( \frac{g L \sin^2(\theta_0)}{(1 - \frac{l}{L})^2} - g L \sin^2(\theta_0) \right) - M g l \]