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 solve the problem, we need to analyze the forces acting on the falling block and the torque acting on the disc. Let's break it down step by step.

### Step 1: Identify the Forces Acting on the Block
The block of mass \( m \) is subject to two forces:
- The gravitational force acting downward: \( F_g = mg \)
- The tension \( T \) in the string acting upward.

According to Newton's second law, the net force acting on the block can be expressed as:
\[ F_{\text{net}} = mg - T = ma \]
where \( a \) is the acceleration of the block.

### Step 2: Express Tension in Terms of Acceleration
From the equation derived in Step 1, we can rearrange it to find the tension \( T \):
\[ T = mg - ma \]
This is our equation (1).

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

### Step 4: Relate Torque to Angular Acceleration
The torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \) of the disc:
\[ \tau = I \alpha \]
For a uniform disc, the moment of inertia \( I \) about its center is given by:
\[ I = \frac{1}{2} M R^2 \]
Thus, we can write:
\[ T \cdot R = \frac{1}{2} M R^2 \alpha \]
Since the linear acceleration \( a \) of the block is related to the angular acceleration \( \alpha \) of the disc by the equation:
\[ \alpha = \frac{a}{R} \]
we can substitute this into the torque equation:
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \frac{a}{R} \]
This simplifies to:
\[ T = \frac{1}{2} M a \]
This is our equation (2).

### Step 5: Substitute Tension into the Block's Equation
Now we have two expressions for tension:
1. From the block: \( T = mg - ma \)
2. From the disc: \( T = \frac{1}{2} M a \)

Setting these equal gives:
\[ mg - ma = \frac{1}{2} M a \]

### Step 6: Rearranging the Equation
Rearranging the above equation to solve for \( a \):
\[ mg = ma + \frac{1}{2} M a \]
Factoring out \( a \):
\[ mg = a \left( m + \frac{1}{2} M \right) \]
Thus, we can solve for \( a \):
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

### Final Step: Simplifying the Expression
To express \( a \) in a simpler form:
\[ a = \frac{2mg}{2m + M} \]

### Conclusion
The magnitude of the acceleration of the falling block \( m \) is given by:
\[ \boxed{a = \frac{2mg}{2m + M}} \]