A disc – like reel with massless thread unwinds itself while falling vertically downwards. The acceleration of its fall is:

Options

A. Zero
B. g
C. `(2g)/(3)`
D. `(g)/(2)`

Explanation

To find the acceleration of a disc-like reel that unwinds itself while falling vertically downwards, we can follow these steps:

### Step 1: Understand the System
We have a disc-like reel that is falling under the influence of gravity. It is unwinding a massless thread as it falls. The forces acting on the disc are the gravitational force (mg) acting downwards and the tension (T) in the thread acting upwards.

### Step 2: Apply Newton's Second Law
For the translational motion of the disc, we can apply Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
Where \( F_{\text{net}} = mg - T \) (the weight of the disc minus the tension in the thread).

Thus, we can write:
\[ mg - T = ma \]
(1)

### Step 3: Analyze the Rotational Motion
The disc is also undergoing rotational motion due to the unwinding of the thread. The torque (\( \tau \)) caused by the tension in the thread is given by:
\[ \tau = T \cdot R \]
Where \( R \) is the radius of the disc.

The relationship between torque and angular acceleration (\( \alpha \)) is given by:
\[ \tau = I \cdot \alpha \]
Where \( I \) is the moment of inertia of the disc. For a disc, the moment of inertia is:
\[ I = \frac{1}{2} m R^2 \]

### Step 4: Relate Angular Acceleration to Linear Acceleration
The angular acceleration (\( \alpha \)) is related to the linear acceleration (\( a \)) of the disc's center of mass by the equation:
\[ \alpha = \frac{a}{R} \]

### Step 5: Substitute into the Torque Equation
Substituting the expression for \( \alpha \) into the torque equation gives:
\[ T \cdot R = \left(\frac{1}{2} m R^2\right) \cdot \left(\frac{a}{R}\right) \]
This simplifies to:
\[ T \cdot R = \frac{1}{2} m R a \]
Dividing both sides by \( R \):
\[ T = \frac{1}{2} m a \]
(2)

### Step 6: Substitute Equation (2) into Equation (1)
Now, we substitute the expression for \( T \) from equation (2) into equation (1):
\[ mg - \frac{1}{2} m a = ma \]
Rearranging gives:
\[ mg = ma + \frac{1}{2} m a \]
Combining the terms on the right:
\[ mg = \frac{3}{2} ma \]

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

### Conclusion
Thus, the acceleration of the fall of the disc-like reel is:
\[ \boxed{\frac{2}{3} g} \]