a particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at `t=0` is `v_(0)` then time it takes to complete the first revolution is `R/(alphav_(0))(1-e^(-betapi))`. Find the value of `(alpha+beta)`.

Explanation

To solve the problem, we need to analyze the motion of a particle moving in a circle with a radius \( R \) where the normal (centripetal) and tangential components of acceleration are equal. Let's break down the solution step by step.

### Step 1: Understanding the Acceleration Components
The normal component of acceleration \( a_n \) is given by:
\[ a_n = \frac{v^2}{R} \]
where \( v \) is the speed of the particle.

The tangential component of acceleration \( a_t \) is given by:
\[ a_t = \frac{dv}{dt} \]
According to the problem, these two components are equal:
\[ \frac{v^2}{R} = \frac{dv}{dt} \]

### Step 2: Rearranging the Equation
We can rearrange the equation to separate variables:
\[ \frac{dv}{v^2} = \frac{dt}{R} \]

### Step 3: Integrating Both Sides
Now, we integrate both sides. The left side integrates from \( v_0 \) to \( v \) and the right side from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R} \]
The integral of \( \frac{1}{v^2} \) is \( -\frac{1}{v} \), so we have:
\[ -\frac{1}{v} + \frac{1}{v_0} = \frac{t}{R} \]
Rearranging gives:
\[ t = R \left( \frac{1}{v_0} - \frac{1}{v} \right) \]

### Step 4: Finding the Final Velocity After One Revolution
Next, we analyze the motion over one complete revolution. The distance traveled in one revolution is \( 2\pi R \). The tangential acceleration is also equal to the normal acceleration, which gives us:
\[ a_t = \frac{dv}{dt} = \frac{v^2}{R} \]
Using the relationship \( ds = v dt \), we can express the tangential acceleration as:
\[ a_t = v \frac{dv}{ds} \]
Setting these equal gives:
\[ \frac{v^2}{R} = v \frac{dv}{ds} \]
This simplifies to:
\[ \frac{dv}{v} = \frac{ds}{R} \]

### Step 5: Integrating for One Revolution
Integrating from \( 0 \) to \( 2\pi R \) gives:
\[ \int_{v_0}^{v} \frac{dv}{v} = \int_{0}^{2\pi R} \frac{ds}{R} \]
This results in:
\[ \ln\left(\frac{v}{v_0}\right) = \frac{2\pi R}{R} = 2\pi \]
Exponentiating both sides gives:
\[ \frac{v}{v_0} = e^{2\pi} \implies v = v_0 e^{2\pi} \]

### Step 6: Substituting Back to Find Time
Substituting \( v \) back into the time equation:
\[ t = R \left( \frac{1}{v_0} - \frac{1}{v_0 e^{2\pi}} \right) = R \left( \frac{1}{v_0} - \frac{e^{-2\pi}}{v_0} \right) \]
This simplifies to:
\[ t = \frac{R}{v_0} (1 - e^{-2\pi}) \]

### Step 7: Comparing with Given Expression
The problem states that the time taken to complete the first revolution is given by:
\[ t = \frac{R}{\alpha v_0} (1 - e^{-\beta \pi}) \]
Comparing both expressions, we find:
\[ \alpha = 1 \quad \text{and} \quad \beta = 2 \]

### Final Step: Calculate \( \alpha + \beta \)
Thus, we find:
\[ \alpha + \beta = 1 + 2 = 3 \]

### Conclusion
The final answer is:
\[ \boxed{3} \]