A string is wrapped over the edge a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.

Explanation

To find the downward acceleration of the disc, we can follow these steps:

### Step 1: Identify the Forces Acting on the Disc
The forces acting on the disc are:
- The gravitational force (weight) acting downwards, which is \( mg \).
- The tension force \( T \) in the string acting upwards.

### Step 2: Apply Newton's Second Law
According to Newton's second law, the net force acting on the disc is equal to the mass of the disc multiplied by its acceleration:
\[ \Sigma F = ma \]
The net force can be expressed as:
\[ mg - T = ma \quad \text{(Equation 1)} \]

### Step 3: Relate Linear Acceleration to Angular Acceleration
The disc is rotating as it unwinds the string. The linear acceleration \( a \) of the disc is related to the angular acceleration \( \alpha \) by the equation:
\[ \alpha = \frac{a}{r} \]
where \( r \) is the radius of the disc.

### Step 4: Calculate the Torque Acting on the Disc
The torque \( \tau \) due to the tension \( T \) can be calculated as:
\[ \tau = T \cdot r \]
According to the rotational form of Newton's second law, the torque is also equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]
For a uniform disc, the moment of inertia \( I \) about its center is:
\[ I = \frac{1}{2} m r^2 \]
Thus, we can write:
\[ T \cdot r = \frac{1}{2} m r^2 \cdot \alpha \]

### Step 5: Substitute Angular Acceleration
Substituting \( \alpha = \frac{a}{r} \) into the torque equation gives:
\[ T \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
Simplifying this, we get:
\[ T \cdot r = \frac{1}{2} m r a \]
Dividing both sides by \( r \):
\[ T = \frac{1}{2} m a \quad \text{(Equation 2)} \]

### Step 6: Substitute Tension into the Force Equation
Now, substitute \( T \) from Equation 2 into Equation 1:
\[ mg - \frac{1}{2} ma = ma \]
Rearranging gives:
\[ mg = ma + \frac{1}{2} ma \]
\[ mg = \frac{3}{2} ma \]

### Step 7: Solve for Acceleration
Now, we can solve for \( a \):
\[ a = \frac{2}{3} g \]

### Final Answer
The downward acceleration of the disc is:
\[ a = \frac{2}{3} g \]

---