A disc of mass 2 kg and diameter 40 cm is free to rotate about an axis passing through its centre and perpendicular to its plane. If a force of 50 N is applied to the disc tangentially Its angular acceleration will be

Options

A. `100 rad//s^(2)`
B. `25 rad//s^(2)`
C. `250 rad//s^(2)`
D. `500 rad//s^(2)`

Explanation

To find the angular acceleration of the disc when a tangential force is applied, we can follow these steps:

### Step 1: Identify the given values
- Mass of the disc (m) = 2 kg
- Diameter of the disc = 40 cm, hence the radius (r) = 20 cm = 0.2 m
- Tangential force (F) = 50 N

### Step 2: Calculate the moment of inertia (I) of the disc
The moment of inertia (I) for a solid disc rotating about an axis through its center is given by the formula:
\[ I = \frac{1}{2} m r^2 \]

Substituting the values:
\[ I = \frac{1}{2} \times 2 \, \text{kg} \times (0.2 \, \text{m})^2 \]
\[ I = \frac{1}{2} \times 2 \times 0.04 \]
\[ I = \frac{1}{2} \times 0.08 \]
\[ I = 0.04 \, \text{kg m}^2 \]

### Step 3: Calculate the torque (τ) applied to the disc
The torque (τ) due to the tangential force is given by:
\[ \tau = F \times r \]

Substituting the values:
\[ \tau = 50 \, \text{N} \times 0.2 \, \text{m} \]
\[ \tau = 10 \, \text{N m} \]

### Step 4: Use the relation between torque and angular acceleration
The relationship between torque, moment of inertia, and angular acceleration (α) is given by:
\[ \tau = I \alpha \]

Rearranging this to find α:
\[ \alpha = \frac{\tau}{I} \]

### Step 5: Substitute the values to find angular acceleration
Substituting the values of τ and I:
\[ \alpha = \frac{10 \, \text{N m}}{0.04 \, \text{kg m}^2} \]
\[ \alpha = 250 \, \text{rad/s}^2 \]

### Conclusion
The angular acceleration of the disc is \( \alpha = 250 \, \text{rad/s}^2 \).

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