At time t = 0 , a horizontal disc starts rotating with angular acceleration 1 red s about an axis perpendicular to its plane and passing through its centre. A small block is lying on this disc at a distance 0.5 m from the centre. Coefficient of friction between the surface of block and disc is 0.255. The block will start slipping on the disc at time t . Which is approximately equal to

Options

A. `2sqrt3 s`
B. `2sqrt2 s`
C. 6 s
D. `sqrt5 s `

Answer

To solve the problem, we need to determine the time \( t \) at which the block starts slipping on the rotating disc.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Angular acceleration, \( \alpha = 1 \, \text{rad/s}^2 \)
- Distance from the center, \( r = 0.5 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.255 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

2. **Determine Tangential Acceleration:**
The tangential acceleration \( a_t \) of the block is given by:
\[ a_t = \alpha \cdot r = 1 \cdot 0.5 = 0.5 \, \text{m/s}^2 \]

3. **Determine Angular Velocity:**
The angular velocity \( \omega \) at time \( t \) can be calculated using the equation of motion for angular acceleration:
\[ \omega = \omega_0 + \alpha t \]
Since the disc starts from rest, \( \omega_0 = 0 \):
\[ \omega = 0 + 1 \cdot t = t \, \text{rad/s} \]

4. **Calculate Centripetal Acceleration:**
The centripetal acceleration \( a_c \) is given by:
\[ a_c = \omega^2 \cdot r = (t^2) \cdot 0.5 = 0.5 t^2 \, \text{m/s}^2 \]

5. **Resultant Acceleration:**
The resultant acceleration \( a_r \) can be found using the Pythagorean theorem, as the tangential and centripetal accelerations are perpendicular:
\[ a_r = \sqrt{a_t^2 + a_c^2} = \sqrt{(0.5)^2 + (0.5 t^2)^2} \]
\[ a_r = \sqrt{0.25 + 0.25 t^4} = 0.5 \sqrt{1 + t^4} \]

6. **Maximum Static Friction Force:**
The maximum static friction force \( F_s \) is given by:
\[ F_s = \mu \cdot N = \mu \cdot mg \]
Since the normal force \( N = mg \):
\[ F_s = 0.255 \cdot mg \]

7. **Setting Forces Equal:**
The block will start slipping when the resultant force equals the maximum static friction force:
\[ m a_r = F_s \]
\[ m \cdot 0.5 \sqrt{1 + t^4} = 0.255 \cdot mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ 0.5 \sqrt{1 + t^4} = 0.255 \cdot g \]

8. **Substituting \( g \):**
Substituting \( g = 10 \):
\[ 0.5 \sqrt{1 + t^4} = 0.255 \cdot 10 \]
\[ 0.5 \sqrt{1 + t^4} = 2.55 \]
\[ \sqrt{1 + t^4} = \frac{2.55}{0.5} = 5.1 \]

9. **Squaring Both Sides:**
\[ 1 + t^4 = (5.1)^2 = 26.01 \]
\[ t^4 = 26.01 - 1 = 25.01 \]

10. **Finding \( t \):**
\[ t = \sqrt[4]{25.01} \approx \sqrt[4]{25} = \sqrt{5} \, \text{seconds} \]

### Final Answer:
The block will start slipping on the disc approximately at time \( t \approx \sqrt{5} \) seconds.