A uniform circular disc of radius `50 cm` at rest is free to turn about an axis, which is perpendicular to the plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of `2.0 rad//s^(2)`. Its net acceleration in `m//s^(2)` at the end of `2.0 s` is approximately

Options

A. 7
B. 6
C. 3
D. 8

Explanation

To solve the problem step by step, we will calculate the net acceleration of the uniform circular disc at the end of 2 seconds, given the angular acceleration and radius.

### Step 1: Identify the given values
- Radius of the disc, \( r = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
- Time, \( t = 2 \, \text{s} \)
- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \)

### Step 2: Calculate the angular velocity at \( t = 2 \, \text{s} \)
Using the formula for angular velocity:
\[ \omega = \omega_0 + \alpha t \]
Substituting the known values:
\[ \omega = 0 + (2 \, \text{rad/s}^2)(2 \, \text{s}) = 4 \, \text{rad/s} \]

### Step 3: Calculate the tangential acceleration \( A_t \)
The tangential acceleration is given by:
\[ A_t = \alpha \cdot r \]
Substituting the values:
\[ A_t = 2 \, \text{rad/s}^2 \cdot 0.5 \, \text{m} = 1 \, \text{m/s}^2 \]

### Step 4: Calculate the centripetal acceleration \( A_c \)
The centripetal acceleration is given by:
\[ A_c = \frac{\omega^2 \cdot r}{1} \]
Substituting the values:
\[ A_c = (4 \, \text{rad/s})^2 \cdot 0.5 \, \text{m} = 16 \cdot 0.5 = 8 \, \text{m/s}^2 \]

### Step 5: Calculate the net acceleration \( A \)
The net acceleration is the resultant of the tangential and centripetal accelerations:
\[ A = \sqrt{A_t^2 + A_c^2} \]
Substituting the values:
\[ A = \sqrt{(1 \, \text{m/s}^2)^2 + (8 \, \text{m/s}^2)^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8 \, \text{m/s}^2 \]

### Conclusion
The net acceleration at the end of 2 seconds is approximately \( 8 \, \text{m/s}^2 \).