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, we need to find the time at which the magnitudes of tangential acceleration (\(A_T\)) and radial acceleration (\(A_R\)) are equal for a particle moving in a circular path with constant angular acceleration.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The particle starts from rest, which means the initial angular velocity (\(\omega_0\)) is 0.
- The angular acceleration (\(\alpha\)) is constant.
- We need to find the time (\(t\)) when the magnitudes of tangential and radial accelerations are equal.

2. **Define Tangential Acceleration**:
- The tangential acceleration (\(A_T\)) is given by the formula:
\[ A_T = \alpha \cdot r \]
where \(r\) is the radius of the circular path.

3. **Define Radial Acceleration**:
- The radial (or centripetal) acceleration (\(A_R\)) can be expressed in terms of linear velocity (\(v\)) or angular velocity (\(\omega\)):
\[ A_R = \frac{v^2}{r} = \omega^2 \cdot r \]

4. **Set the Accelerations Equal**:
- We need to find the time when \(A_T = A_R\):
\[ \alpha \cdot r = \omega^2 \cdot r \]
- We can cancel \(r\) from both sides (assuming \(r \neq 0\)):
\[ \alpha = \omega^2 \]

5. **Relate Angular Velocity to Time**:
- Using the equation of motion for angular velocity:
\[ \omega = \omega_0 + \alpha \cdot t \]
- Since \(\omega_0 = 0\), this simplifies to:
\[ \omega = \alpha \cdot t \]

6. **Substitute \(\omega\) into the Equation**:
- From the equality \(\alpha = \omega^2\), substitute \(\omega\):
\[ \alpha = (\alpha \cdot t)^2 \]
- This simplifies to:
\[ \alpha = \alpha^2 \cdot t^2 \]
- Dividing both sides by \(\alpha\) (assuming \(\alpha \neq 0\)):
\[ 1 = \alpha \cdot t^2 \]

7. **Solve for Time \(t\)**:
- Rearranging gives:
\[ t^2 = \frac{1}{\alpha} \]
- Taking the square root:
\[ t = \frac{1}{\sqrt{\alpha}} \]

### Final Answer:
The time at which the magnitudes of tangential and radial accelerations are equal is:
\[ t = \frac{1}{\sqrt{\alpha}} \]