A string is wrapped around a cylinder of mass `M` and radius `R`. The string is pulled vertically upwards to prevent the centre of mass from falling as the cylinder unwinds the string. The tension in the string is

Options

A. `(Mg)/6`
B. `(Mg)/3`
C. `(Mg)/2`
D. `(2Mg)/3`

Explanation

To find the tension in the string wrapped around a cylinder of mass \( M \) and radius \( R \), we can follow these steps:

### Step 1: Identify the forces acting on the cylinder
The forces acting on the cylinder are:
- The weight of the cylinder, \( W = Mg \) (acting downwards)
- The tension in the string, \( T \) (acting upwards)

### Step 2: Write the equation of motion for the cylinder
Since the center of mass of the cylinder is not falling, we can write the net force equation as:
\[ Mg - T = Ma \]
where \( a \) is the downward acceleration of the cylinder.

### Step 3: Relate linear acceleration to angular acceleration
The cylinder will also have an angular acceleration \( \alpha \) due to the unwinding of the string. The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by:
\[ a = \alpha R \]

### Step 4: Write the torque equation
The torque \( \tau \) about the center of the cylinder due to the tension \( T \) is given by:
\[ \tau = T \cdot R \]
This torque is also related to the moment of inertia \( I \) of the cylinder and its angular acceleration \( \alpha \):
\[ \tau = I \alpha \]

### Step 5: Moment of inertia of the cylinder
For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} M R^2 \]

### Step 6: Substitute the torque equation
Substituting the expressions for torque, we have:
\[ T \cdot R = \frac{1}{2} M R^2 \alpha \]
Now, substituting \( \alpha = \frac{a}{R} \):
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \frac{a}{R} \]
This simplifies to:
\[ T \cdot R = \frac{1}{2} M R a \]
Dividing both sides by \( R \):
\[ T = \frac{1}{2} M a \]

### Step 7: Substitute \( a \) in terms of \( T \)
From the net force equation \( Mg - T = Ma \), we can express \( a \):
\[ a = \frac{Mg - T}{M} \]
Substituting this expression for \( a \) into the tension equation:
\[ T = \frac{1}{2} M \left(\frac{Mg - T}{M}\right) \]
This simplifies to:
\[ T = \frac{1}{2} (Mg - T) \]
Multiplying through by 2:
\[ 2T = Mg - T \]
Adding \( T \) to both sides:
\[ 3T = Mg \]
Thus, we find:
\[ T = \frac{Mg}{3} \]

### Final Answer
The tension in the string is:
\[ T = \frac{Mg}{3} \]

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