A unifrom flexible chain of length `3/2` m rests on a fixed smooth sphere of radius `R= 2/pi` m such that one end A of chain is on the top of the sphere while the other end B is hanging freely. Chain is helpd stationary by a horizontal thread PA as shown . Calculate the acceleration of chain when the horizontal string PA is burnt ( g = 10 `m//s^(2))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a uniform flexible chain of length \( L = \frac{3}{2} \) m resting on a smooth sphere of radius \( R = \frac{2}{\pi} \) m. One end of the chain is at the top of the sphere (point A), and the other end (point B) is hanging freely. The chain is held stationary by a horizontal thread PA. We need to calculate the acceleration of the chain when the thread PA is burnt.

### Step 2: Analyze the Forces Acting on the Chain
When the thread is burnt, the forces acting on the chain will be:
- The weight of the hanging portion of the chain, which we denote as \( F_1 \).
- The tension in the chain due to the portion resting on the sphere, which we denote as \( F_2 \).

### Step 3: Calculate the Weight of the Hanging Portion
The length of the hanging portion of the chain is \( \frac{L}{2} = \frac{3/2}{2} = \frac{3}{4} \) m. The weight of this portion is given by:
\[ F_1 = \text{mass} \times g = \left(\frac{3}{4} \cdot \frac{m}{L}\right) \cdot g = \frac{3}{4} \cdot \frac{m}{\frac{3}{2}} \cdot 10 = \frac{3}{4} \cdot \frac{2m}{3} \cdot 10 = \frac{5m}{2} \text{ N} \]

### Step 4: Calculate the Tension in the Chain Resting on the Sphere
The tension \( F_2 \) can be calculated by integrating the forces acting on the chain resting on the sphere. The total force supporting the motion can be expressed as:
\[ F_2 = \int_0^{\frac{\pi}{2}} mg \cos(\theta) d\theta \]
where \( \theta \) is the angle subtended by the chain at the center of the sphere. The limits of integration are from \( 0 \) to \( \frac{\pi}{2} \).

Calculating this integral gives:
\[ F_2 = mg \left[ \sin(\theta) \right]_0^{\frac{\pi}{2}} = mg \cdot 1 = mg \]

### Step 5: Total Force Acting on the Chain
The total force acting on the chain when the thread is burnt is:
\[ F_{\text{net}} = F_1 + F_2 = \frac{5m}{2} + mg \]

### Step 6: Calculate the Net Acceleration
Using Newton's second law, the net acceleration \( a \) of the chain can be calculated as:
\[ a = \frac{F_{\text{net}}}{m} = \frac{\frac{5m}{2} + mg}{m} = \frac{5}{2} + g \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ a = \frac{5}{2} + 10 = 2.5 + 10 = 12.5 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the chain when the horizontal string PA is burnt is \( 12.5 \, \text{m/s}^2 \).

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