A particle begins to move with a tangential acceleration of constant magnitude `0.6m//s^(2)` in a circular path . If it slips when total acceleration becomes `1m//s^(2)`, then the angle through which it would have turned before it stars to slip is `:`

Options

A. `1//3 rad`
B. `2//3 rad`
C. `4//3 rad`
D. `2 rad`

Explanation

To solve the problem, we need to determine the angle through which a particle would have turned before it starts to slip when moving in a circular path with a given tangential acceleration and a condition for slipping based on total acceleration.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Tangential acceleration, \( A_t = 0.6 \, \text{m/s}^2 \)
- Total acceleration at which slipping occurs, \( A_{net} = 1 \, \text{m/s}^2 \)

2. **Understand the Relationship Between Accelerations:**
The total acceleration \( A_{net} \) is the vector sum of the tangential acceleration \( A_t \) and the centripetal acceleration \( A_c \):
\[ A_{net} = \sqrt{A_t^2 + A_c^2} \]

3. **Express Centripetal Acceleration:**
The centripetal acceleration \( A_c \) can be expressed in terms of angular velocity \( \omega \) and radius \( r \):
\[ A_c = \frac{v^2}{r} = \omega^2 r \]
where \( v \) is the linear velocity of the particle.

4. **Relate Angular Velocity to Tangential Acceleration:**
The tangential acceleration is related to the angular acceleration \( \alpha \) by:
\[ A_t = r \alpha \]
Since the particle starts from rest, we can use the kinematic equation:
\[ \omega^2 = \omega_0^2 + 2 \alpha \theta \]
where \( \omega_0 = 0 \) (initial angular velocity).

5. **Substituting Values:**
Since \( \omega_0 = 0 \):
\[ \omega^2 = 2 \alpha \theta \]
And since \( \alpha = \frac{A_t}{r} = \frac{0.6}{r} \):
\[ \omega^2 = 2 \left(\frac{0.6}{r}\right) \theta \]

6. **Substituting into Total Acceleration Equation:**
Now substituting \( A_c \) into the total acceleration equation:
\[ 1 = \sqrt{(0.6)^2 + \left(\frac{0.6 \cdot 2 \theta}{r}\right)^2} \]
This simplifies to:
\[ 1 = \sqrt{0.36 + \left(\frac{1.2 \theta}{r}\right)^2} \]

7. **Squaring Both Sides:**
Squaring both sides gives:
\[ 1 = 0.36 + \frac{1.44 \theta^2}{r^2} \]
Rearranging this gives:
\[ 0.64 = \frac{1.44 \theta^2}{r^2} \]

8. **Solving for \( \theta \):**
Rearranging for \( \theta^2 \):
\[ \theta^2 = \frac{0.64 r^2}{1.44} = \frac{64}{144} r^2 = \frac{4}{9} r^2 \]
Taking the square root:
\[ \theta = \frac{2}{3} r \]

9. **Final Result:**
The angle through which the particle would have turned before it starts to slip is:
\[ \theta = \frac{2}{3} \, \text{radians} \]

### Conclusion:
Thus, the angle through which the particle turns before slipping is \( \frac{2}{3} \) radians.