A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution, after the revolution has started, then the tangential acceleration is

Options

A. `v^2/(8pir)`
B. `v^2/(6pir)`
C. `v^2/(4pir)`
D. `v^2/(10pir)`

Answer

To solve the problem, we need to find the tangential acceleration of a particle moving along a circular path with a constant tangential acceleration. We know that the particle has a velocity \( v \) at the end of the second revolution.

### Step-by-Step Solution:

1. **Understand the Motion**:
- The particle moves along a circular path of radius \( r \) with constant tangential acceleration \( a_t \).
- The angular velocity \( \omega \) is related to the tangential velocity \( v \) by the equation:
\[ \omega = \frac{v}{r} \]

2. **Identify the Angular Displacement**:
- After two complete revolutions, the angular displacement \( \theta \) is given by:
\[ \theta = 2 \times 2\pi = 4\pi \text{ radians} \]

3. **Use the Angular Motion Equation**:
- We can use the third equation of motion for angular motion:
\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
- Here, \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity (which is 0 since the particle starts from rest), \( \alpha \) is the angular acceleration, and \( \theta \) is the angular displacement.

4. **Substituting Values**:
- Since the initial angular velocity \( \omega_i = 0 \), the equation simplifies to:
\[ \omega_f^2 = 2\alpha\theta \]
- Substituting \( \theta = 4\pi \):
\[ \omega_f^2 = 2\alpha(4\pi) = 8\pi\alpha \]

5. **Relate Angular and Tangential Acceleration**:
- The angular acceleration \( \alpha \) is related to the tangential acceleration \( a_t \) by:
\[ \alpha = \frac{a_t}{r} \]
- Substituting this into the equation gives:
\[ \omega_f^2 = 8\pi\left(\frac{a_t}{r}\right) \]

6. **Substituting for Final Angular Velocity**:
- We also know that \( \omega_f = \frac{v}{r} \), so substituting this into the equation gives:
\[ \left(\frac{v}{r}\right)^2 = 8\pi\left(\frac{a_t}{r}\right) \]

7. **Simplifying the Equation**:
- This simplifies to:
\[ \frac{v^2}{r^2} = \frac{8\pi a_t}{r} \]
- Multiplying both sides by \( r^2 \) gives:
\[ v^2 = 8\pi a_t r \]

8. **Solving for Tangential Acceleration**:
- Finally, we can solve for the tangential acceleration \( a_t \):
\[ a_t = \frac{v^2}{8\pi r} \]

### Final Answer:
The tangential acceleration \( a_t \) is:
\[ a_t = \frac{v^2}{8\pi r} \]