A particle is moving in circular path with constant acceleration. In time t after the beginning of motion the direction of net acceleration is at `30^(@)` to the radius vector at that instant. The angular acceleration of the particle at that time t is

Options

A. `(3)/(t^(2))`
B. `(1)/(t^(2))`
C. `(sqrt3)/(t)`
D. `(sqrt3)/(t^(2))`

Explanation

To solve the problem, we need to find the angular acceleration (α) of a particle moving in a circular path with constant acceleration, given that the net acceleration makes an angle of 30 degrees with the radius vector at time t.

### Step-by-Step Solution:

1. **Understanding the Components of Acceleration**:
- The net acceleration (A) of the particle can be broken down into two components:
- Tangential acceleration (At), which is directed along the tangent to the circular path.
- Centripetal acceleration (Ac), which is directed towards the center of the circular path.
- Given that the angle between the net acceleration and the radius vector is 30 degrees, we can use trigonometric relationships to express these components.

2. **Using Trigonometry**:
- From the angle given (30 degrees), we can write:
\[ \tan(30^\circ) = \frac{A_t}{A_c} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\). Therefore, we can express this relationship as:
\[ \frac{A_t}{A_c} = \frac{1}{\sqrt{3}} \]

3. **Expressing Accelerations in Terms of Angular Acceleration**:
- The tangential acceleration \(A_t\) can be expressed as:
\[ A_t = \alpha r \]
- The centripetal acceleration \(A_c\) can be expressed as:
\[ A_c = \omega^2 r \]
- Substituting these into our earlier equation gives:
\[ \frac{\alpha r}{\omega^2 r} = \frac{1}{\sqrt{3}} \]
- The radius \(r\) cancels out, leading to:
\[ \frac{\alpha}{\omega^2} = \frac{1}{\sqrt{3}} \]

4. **Relating Angular Velocity to Angular Acceleration**:
- We know that the angular velocity \(\omega\) can be expressed in terms of angular acceleration \(\alpha\) and time \(t\):
\[ \omega = \alpha t \]
- Substituting this into the equation \(\frac{\alpha}{\omega^2} = \frac{1}{\sqrt{3}}\) gives:
\[ \frac{\alpha}{(\alpha t)^2} = \frac{1}{\sqrt{3}} \]
- This simplifies to:
\[ \frac{\alpha}{\alpha^2 t^2} = \frac{1}{\sqrt{3}} \]
- Cancelling one \(\alpha\) from the numerator and denominator, we have:
\[ \frac{1}{\alpha t^2} = \frac{1}{\sqrt{3}} \]

5. **Solving for Angular Acceleration**:
- Rearranging the equation gives:
\[ \alpha t^2 = \sqrt{3} \]
- Therefore, the angular acceleration \(\alpha\) is:
\[ \alpha = \frac{\sqrt{3}}{t^2} \]

### Final Answer:
The angular acceleration of the particle at time \(t\) is:
\[ \alpha = \frac{\sqrt{3}}{t^2} \]