A string of negligible mass is wrapped on a cylindrical pulley of mass M and radius R. The other end of string is tied to a bucket of mass m. If the pulley rotates about a horizontal axis then the tension in the string is :

Options

A. mg
B. `(M+m)g`
C. `(Mmg)/((M+2m))`
D. `(2mg)/((M+2m))`

Answer

To find the tension in the string wrapped around a cylindrical pulley with a bucket hanging from it, we can follow these steps:

### Step 1: Identify the forces acting on the bucket
The forces acting on the bucket of mass \( m \) are:
- The gravitational force downward, which is \( mg \) (where \( g \) is the acceleration due to gravity).
- The tension \( T \) in the string acting upward.

### Step 2: Write the equation of motion for the bucket
Since the bucket is accelerating downwards, we can write the equation of motion as:
\[ mg - T = ma \]
where \( a \) is the acceleration of the bucket.

### Step 3: Identify the torque acting on the pulley
The tension \( T \) in the string creates a torque on the pulley. The torque \( \tau \) can be expressed as:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the pulley.

### Step 4: Relate torque to angular acceleration
The torque is also related to the angular acceleration \( \alpha \) of the pulley by the equation:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the pulley. For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} M R^2 \]
Thus, we can write:
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \alpha \]

### Step 5: Relate linear acceleration to angular acceleration
The linear acceleration \( a \) of the bucket is related to the angular acceleration \( \alpha \) of the pulley by:
\[ a = R \alpha \]
Substituting this into the torque equation gives:
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \frac{a}{R} \]
Simplifying this, we find:
\[ T = \frac{1}{2} M a \]

### Step 6: Substitute \( a \) back into the equation of motion
Now we have two equations:
1. \( mg - T = ma \)
2. \( T = \frac{1}{2} M a \)

Substituting \( T \) from the second equation into the first gives:
\[ mg - \frac{1}{2} M a = ma \]
Rearranging this leads to:
\[ mg = ma + \frac{1}{2} M a \]
Factoring out \( a \):
\[ mg = a \left( m + \frac{1}{2} M \right) \]
Solving for \( a \):
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

### Step 7: Substitute \( a \) back to find \( T \)
Now substitute \( a \) back into the equation for \( T \):
\[ T = \frac{1}{2} M a = \frac{1}{2} M \left( \frac{mg}{m + \frac{1}{2} M} \right) \]
This simplifies to:
\[ T = \frac{Mmg}{2(m + \frac{1}{2} M)} = \frac{Mmg}{2m + M} \]

### Final Answer
Thus, the tension in the string is:
\[ T = \frac{Mmg}{2m + M} \]