A particle starts moving with a constant angular acceleration in a circular path. The time at which the magnitudes of tangential and radial acceleration are equal is

Options

A. `1// alpha`
B. `sqrtalpha`
C. `1// sqrt alpha`
D. `alpha`

Explanation

To solve the problem of finding the time at which the magnitudes of tangential and radial acceleration are equal for a particle moving in a circular path with constant angular acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Tangential acceleration (\(A_t\)) is given by the formula:
\[ A_t = \alpha \cdot r \]
where \(\alpha\) is the angular acceleration and \(r\) is the radius of the circular path.
- Radial (centripetal) acceleration (\(A_r\)) is given by the formula:
\[ A_r = \omega^2 \cdot r \]
where \(\omega\) is the angular velocity.

2. **Set the Accelerations Equal**:
- We need to find the time when the magnitudes of tangential and radial acceleration are equal:
\[ A_t = A_r \]
- Substituting the expressions for \(A_t\) and \(A_r\):
\[ \alpha \cdot r = \omega^2 \cdot r \]
- Since \(r\) is common on both sides, we can divide by \(r\) (assuming \(r \neq 0\)):
\[ \alpha = \omega^2 \]

3. **Relate Angular Velocity to Time**:
- We know that the angular velocity \(\omega\) can be expressed in terms of angular acceleration and time using the equation of motion:
\[ \omega = \omega_0 + \alpha t \]
- Given that the initial angular velocity \(\omega_0 = 0\) (the particle starts from rest):
\[ \omega = \alpha t \]

4. **Substitute \(\omega\) in the Equation**:
- From the previous step, we have \(\omega = \alpha t\). Substitute this into the equation \(\alpha = \omega^2\):
\[ \alpha = (\alpha t)^2 \]
- This simplifies to:
\[ \alpha = \alpha^2 t^2 \]

5. **Solve for Time \(t\)**:
- Dividing both sides by \(\alpha\) (assuming \(\alpha \neq 0\)):
\[ 1 = \alpha t^2 \]
- Rearranging gives:
\[ t^2 = \frac{1}{\alpha} \]
- Taking the square root of both sides:
\[ t = \frac{1}{\sqrt{\alpha}} \]

6. **Conclusion**:
- The time at which the magnitudes of tangential and radial acceleration are equal is:
\[ t = \frac{1}{\sqrt{\alpha}} \]

### Final Answer:
The time at which the magnitudes of tangential and radial acceleration are equal is \( t = \frac{1}{\sqrt{\alpha}} \).