A uniform disc of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block (m) is

Options

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

Answer

To find the magnitude of the acceleration of the falling block (m), we can follow these steps:

### Step 1: Analyze the Forces on the Block
The forces acting on the block of mass \( m \) are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T \) in the string acting upwards.

Using Newton's second law, we can write the equation for the block:
\[ mg - T = ma \quad \text{(1)} \]
where \( a \) is the acceleration of the block.

### Step 2: Analyze the Torque on the Disc
The tension in the string also creates a torque on the disc. The torque \( \tau \) is given by:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the disc.

The moment of inertia \( I \) of a uniform disc about its axis is:
\[ I = \frac{1}{2} M R^2 \]

Using the relation between torque and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
we can substitute the values:
\[ T \cdot R = \frac{1}{2} M R^2 \alpha \quad \text{(2)} \]

### Step 3: Relate Linear and Angular Acceleration
Since the string does not slip on the disc, the linear acceleration \( a \) of the block is related to the angular acceleration \( \alpha \) of the disc by:
\[ a = R \alpha \quad \text{(3)} \]
Substituting \( \alpha \) from equation (3) into equation (2):
\[ T \cdot R = \frac{1}{2} M R^2 \left(\frac{a}{R}\right) \]
This simplifies to:
\[ T = \frac{1}{2} M a \quad \text{(4)} \]

### Step 4: Substitute Tension in the Block's Equation
Now, substitute equation (4) into equation (1):
\[ mg - \frac{1}{2} M a = ma \]
Rearranging gives:
\[ mg = ma + \frac{1}{2} M a \]
Factoring out \( a \) from the right side:
\[ mg = a \left(m + \frac{1}{2} M\right) \]

### Step 5: Solve for Acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

### Final Expression for Acceleration
Thus, the magnitude of the acceleration of the falling block is:
\[ a = \frac{2mg}{2m + M} \]