A chain of length l is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere. What will be the acceleration w of each element of the chain when its upper end is released? It is assumed that the length of the chain `llt1/2piR`.

Answer

To solve the problem of determining the acceleration \( w \) of each element of the chain when its upper end is released, we will follow these steps:

### Step 1: Define the Setup
We have a chain of length \( l \) placed on a smooth spherical surface of radius \( R \), with one end fixed at the top of the sphere. The chain can slide down the surface when released. We are given that \( l < \frac{\pi R}{2} \).

### Step 2: Determine Mass per Unit Length
Let the mass of the chain be \( m \). The mass per unit length \( \lambda \) of the chain is given by:
\[ \lambda = \frac{m}{l} \]

### Step 3: Consider a Small Element of the Chain
Consider a small element of the chain of length \( dl \) at an angle \( \theta \) from the vertical. The mass of this small element \( dm \) is:
\[ dm = \lambda \cdot dl = \frac{m}{l} \cdot dl \]

### Step 4: Calculate the Force Acting on the Element
The force acting on this small element due to gravity is:
\[ dF = dm \cdot g \cdot \sin(\theta) \]
Substituting for \( dm \):
\[ dF = \left(\frac{m}{l} \cdot dl\right) \cdot g \cdot \sin(\theta) \]

### Step 5: Integrate the Force
To find the total force acting on the chain as it slides down, we need to integrate this force from \( 0 \) to \( a \) (where \( a \) is the angle corresponding to the length of the chain):
\[ F = \int_0^a dF = \int_0^a \left(\frac{m}{l} \cdot g \cdot \sin(\theta) \cdot dl\right) \]
Using the relationship \( dl = R \cdot d\theta \):
\[ F = \frac{mg}{l} \int_0^a R \sin(\theta) d\theta \]

### Step 6: Evaluate the Integral
The integral \( \int \sin(\theta) d\theta \) evaluates to \( -\cos(\theta) \). Therefore:
\[ F = \frac{mgR}{l} \left[-\cos(\theta)\right]_0^a = \frac{mgR}{l} \left(1 - \cos(a)\right) \]

### Step 7: Relate Force to Acceleration
The net force acting on the chain is equal to the mass of the chain times its acceleration \( w \):
\[ F = m \cdot w \]
Setting the two expressions for force equal gives:
\[ m \cdot w = \frac{mgR}{l} \left(1 - \cos(a)\right) \]
Dividing both sides by \( m \):
\[ w = \frac{gR}{l} \left(1 - \cos(a)\right) \]

### Step 8: Substitute for \( a \)
Since \( a = \frac{l}{R} \) (from the geometry of the problem), we can substitute this into our expression for \( w \):
\[ w = \frac{gR}{l} \left(1 - \cos\left(\frac{l}{R}\right)\right) \]

### Final Answer
Thus, the acceleration \( w \) of each element of the chain when its upper end is released is:
\[ w = \frac{gR}{l} \left(1 - \cos\left(\frac{l}{R}\right)\right) \]

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