A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2 rad `s^(-1)`. Its net acceleration in `ms^(-2)` at the end of 2 s is approximately

Options

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

Explanation

To solve the problem step by step, we need to find the net acceleration of a point on the edge of a uniform circular disc after 2 seconds of constant angular acceleration.

### Step 1: Identify Given Values
- Radius of the disc, \( r = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Angular acceleration, \( \alpha = 2 \, \text{rad/s}^2 \)
- Time, \( t = 2 \, \text{s} \)

### Step 2: Calculate Final Angular Velocity
The final angular velocity \( \omega \) after time \( t \) can be calculated using the formula:
\[ \omega = \alpha \cdot t \]
Substituting the known values:
\[ \omega = 2 \, \text{rad/s}^2 \cdot 2 \, \text{s} = 4 \, \text{rad/s} \]

### Step 3: Calculate Centripetal Acceleration
The centripetal acceleration \( a_c \) for a point on the edge of the disc is given by the formula:
\[ a_c = \omega^2 \cdot r \]
Substituting the values we have:
\[ a_c = (4 \, \text{rad/s})^2 \cdot 0.5 \, \text{m} = 16 \cdot 0.5 = 8 \, \text{m/s}^2 \]

### Step 4: Calculate Tangential Acceleration
The tangential acceleration \( a_t \) 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 5: Calculate Net Acceleration
The net acceleration \( a \) is the vector sum of the centripetal and tangential accelerations. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{(8 \, \text{m/s}^2)^2 + (1 \, \text{m/s}^2)^2} = \sqrt{64 + 1} = \sqrt{65} \approx 8.06 \, \text{m/s}^2 \]

### Final Answer
The net acceleration at the end of 2 seconds is approximately \( 8.06 \, \text{m/s}^2 \).

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