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. We will use the concepts of angular acceleration, centripetal acceleration, and friction.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Angular acceleration, \( \alpha = 1 \, \text{rad/s}^2 \)
- Radius of the disc, \( r = 0.5 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.255 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Determine the Angular Velocity**:
The angular velocity \( \omega \) at time \( t \) can be expressed as:
\[ \omega = \alpha t = 1 \cdot t = t \, \text{(rad/s)} \]

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

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

5. **Total Acceleration**:
The total acceleration \( a \) acting on the block can be calculated using Pythagoras' theorem:
\[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{(0.5)^2 + (0.5 t^2)^2} = \sqrt{0.25 + 0.25 t^4} \]

6. **Friction Force**:
The maximum friction force that can act on the block is:
\[ F_{\text{friction}} = \mu m g = 0.255 m \cdot 10 = 2.55 m \]

7. **Setting Up the Inequality**:
For the block to not slip, the total acceleration must be less than or equal to the maximum frictional force:
\[ \sqrt{0.25 + 0.25 t^4} \leq 2.55 \]

8. **Square Both Sides**:
Squaring both sides gives:
\[ 0.25 + 0.25 t^4 \leq (2.55)^2 \]
\[ 0.25 + 0.25 t^4 \leq 6.5025 \]

9. **Isolate \( t^4 \)**:
Subtract \( 0.25 \) from both sides:
\[ 0.25 t^4 \leq 6.2525 \]
Divide by \( 0.25 \):
\[ t^4 \leq 25.01 \]

10. **Taking the Fourth Root**:
Taking the fourth root of both sides:
\[ t \leq \sqrt[4]{25.01} \approx 5 \, \text{s} \]

### Conclusion:
The block will start slipping on the disc at approximately \( t \approx \sqrt{5} \, \text{s} \).