A block of mass `m` is attached at the end of an inextensible string which is wound over a rough pulley of mass `M` and radius `R`. Assume the string does not slide over the pulley. Find the acceleration of the block when released.

Options

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

Explanation

To find the acceleration of the block when released, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block of mass `m` are:
- The gravitational force acting downward, which is `mg`.
- The tension `T` in the string acting upward.

Using Newton's second law, we can write the equation of motion for the block:
\[ mg - T = ma \]
where `a` is the acceleration of the block.

### Step 2: Analyze the Pulley
The pulley has a mass `M` and a radius `R`. The tension in the string creates a torque on the pulley. The torque `τ` due to the tension is given by:
\[ τ = T \cdot R \]

The moment of inertia `I` of the pulley (assuming it's a solid disk) is:
\[ I = \frac{1}{2} M R^2 \]

Using the relation between torque and angular acceleration, we have:
\[ τ = I \cdot \alpha \]
where `α` is the angular acceleration of the pulley. Since the string does not slide over the pulley, the linear acceleration `a` of the block is related to the angular acceleration `α` of the pulley by:
\[ a = R \cdot \alpha \]
Thus, we can express `α` as:
\[ α = \frac{a}{R} \]

### Step 3: Write the Equation for the Pulley
Substituting the expressions for torque and moment of inertia into the torque equation gives:
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \frac{a}{R} \]
Simplifying this, we get:
\[ T = \frac{1}{2} M a \]

### Step 4: Substitute Tension into the Block's Equation
Now we have two equations:
1. From the block: \( mg - T = ma \)
2. From the pulley: \( T = \frac{1}{2} M a \)

Substituting the expression for `T` from the pulley into the block's equation:
\[ mg - \frac{1}{2} M a = ma \]

### Step 5: Solve for Acceleration
Rearranging the equation gives:
\[ mg = ma + \frac{1}{2} M a \]
\[ mg = a \left( m + \frac{1}{2} M \right) \]
Now, solving for `a`:
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

### Final Result
Thus, the acceleration of the block when released is:
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

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