A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 `m//s`, the time taken to complete the first revolution will be `frac{1}{alpha}[1-e^(-2pi)]`s, where `alpha` = ______.

Answer

To solve the problem step by step, we need to analyze the motion of the particle and the conditions given in the question.

### Step 1: Understand the Components of Acceleration
The particle is moving in a circle of radius \( r = 50 \) cm, which we convert to meters:
\[ r = \frac{50}{100} = 0.5 \text{ m} \]
We know that in circular motion, the normal (centripetal) acceleration \( a_n \) and the tangential acceleration \( a_t \) are given by:
\[ a_n = \frac{v^2}{r} \]
\[ a_t = \frac{dv}{dt} \]
According to the problem, at any instant, these two components are equal:
\[ \frac{v^2}{r} = \frac{dv}{dt} \]

### Step 2: Set Up the Equation
Substituting the value of \( r \):
\[ \frac{v^2}{0.5} = \frac{dv}{dt} \]
This simplifies to:
\[ 2v^2 = \frac{dv}{dt} \]

### Step 3: Separate Variables and Integrate
We can rearrange the equation to separate variables:
\[ \frac{1}{v^2} dv = 2 dt \]
Now, we integrate both sides. The limits for \( v \) will be from the initial speed \( v_0 = 4 \text{ m/s} \) to \( v \) at time \( t \), and for \( t \) from \( 0 \) to \( t \):
\[ \int_{4}^{v} \frac{1}{v^2} dv = \int_{0}^{t} 2 dt \]
Calculating the left side:
\[ \left[-\frac{1}{v}\right]_{4}^{v} = -\frac{1}{v} + \frac{1}{4} \]
Calculating the right side:
\[ \left[2t\right]_{0}^{t} = 2t \]
Thus, we have:
\[ -\frac{1}{v} + \frac{1}{4} = 2t \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{4} - 2t \]

### Step 4: Find the Distance for One Revolution
The distance \( s \) covered in one complete revolution is the circumference of the circle:
\[ s = 2\pi r = 2\pi \times 0.5 = \pi \text{ m} \]

### Step 5: Relate Speed and Distance
We know that speed \( v \) can also be expressed in terms of distance \( s \) and time \( t \):
\[ v = \frac{ds}{dt} \]
Substituting \( ds = s = \pi \):
\[ \int_{0}^{t} v dt = \int_{0}^{\pi} ds \]
This gives:
\[ \int_{0}^{t} \left( \frac{1}{\frac{1}{4} - 2t} \right) dt = \pi \]

### Step 6: Solve for Time \( t \)
To solve for \( t \), we need to integrate:
\[ \int_{0}^{t} \frac{1}{\frac{1}{4} - 2t} dt \]
This integral can be solved using logarithmic integration. The result will yield:
\[ \log\left(\frac{\frac{1}{4} - 2t}{\frac{1}{4}}\right) = -\pi \]
Exponentiating both sides gives:
\[ \frac{\frac{1}{4} - 2t}{\frac{1}{4}} = e^{-\pi} \]
From this, we can solve for \( t \):
\[ \frac{1}{4} - 2t = \frac{1}{4} e^{-\pi} \]
\[ 2t = \frac{1}{4} - \frac{1}{4} e^{-\pi} \]
\[ t = \frac{1}{8}(1 - e^{-\pi}) \]

### Step 7: Compare with Given Expression
The problem states that the time taken to complete the first revolution is given by:
\[ t = \frac{1}{\alpha}(1 - e^{-2\pi}) \]
Setting the two expressions for \( t \) equal:
\[ \frac{1}{8}(1 - e^{-\pi}) = \frac{1}{\alpha}(1 - e^{-2\pi}) \]
To find \( \alpha \), we can manipulate this equation:
\[ \alpha = 8 \cdot \frac{(1 - e^{-2\pi})}{(1 - e^{-\pi})} \]
Using the identity \( 1 - e^{-2\pi} = (1 - e^{-\pi})(1 + e^{-\pi}) \):
\[ \alpha = 8(1 + e^{-\pi}) \]
Since \( e^{-\pi} \) is a small number, we approximate:
\[ \alpha \approx 8 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 8 \]