A particle is moving in a circle of radius R in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at `t=0` is `v_(0)`, the time taken to complete the first revolution is

Options

A. `(R )/(v_(0))`
B. `(R )/(v_(0))(1-e^(-2pi))`
C. `(R )/(v_(0))e^(-2pi)`
D. `(2piR)/(v_(0))`

Answer

To solve the problem step by step, we will analyze the motion of the particle moving in a circle and relate the tangential retardation and normal acceleration.

### Step 1: Define the Variables
Let:
- \( R \) = radius of the circle
- \( v_0 \) = initial speed of the particle at \( t = 0 \)
- \( a_t \) = tangential acceleration (retardation in this case)
- \( a_n \) = normal (centripetal) acceleration

### Step 2: Set Up the Relationship
According to the problem, we have:
\[ a_t = a_n \]
The normal acceleration for circular motion is given by:
\[ a_n = \frac{v^2}{R} \]
The tangential acceleration (retardation) can be expressed as:
\[ a_t = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = \frac{v^2}{R} \]

### Step 3: Rearranging the Equation
Rearranging the equation gives us:
\[ \frac{dv}{v^2} = \frac{dt}{R} \]

### Step 4: Integrate Both Sides
Now we will integrate both sides. The left side will be integrated from \( v_0 \) to \( v \) and the right side from \( 0 \) to \( t \):
\[ \int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R} \]
This results in:
\[ -\frac{1}{v} \bigg|_{v_0}^{v} = \frac{t}{R} \]
This simplifies to:
\[ -\frac{1}{v} + \frac{1}{v_0} = \frac{t}{R} \]

### Step 5: Solve for \( v \)
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{v_0} - \frac{t}{R} \]
Thus:
\[ v = \frac{1}{\frac{1}{v_0} - \frac{t}{R}} \]

### Step 6: Relate Angular Displacement to Time
The angular velocity \( \omega \) is related to the linear speed \( v \) by:
\[ \omega = \frac{v}{R} \]
We also know that:
\[ \frac{d\theta}{dt} = \omega \]
Substituting for \( v \):
\[ \frac{d\theta}{dt} = \frac{1}{R \left( \frac{1}{v_0} - \frac{t}{R} \right)} \]

### Step 7: Integrate to Find the Total Angle
To find the total angle \( \theta \) for one complete revolution, we need to integrate from \( 0 \) to \( 2\pi \):
\[ \int_{0}^{2\pi} d\theta = \int_{0}^{T} \frac{1}{R \left( \frac{1}{v_0} - \frac{t}{R} \right)} dt \]

### Step 8: Solve the Integral
This integral can be solved, and after some algebra, we find:
\[ 2\pi = \frac{T}{R} \left( \frac{1}{v_0} - \frac{T}{2R} \right) \]

### Step 9: Solve for \( T \)
Rearranging gives us a quadratic equation in \( T \). Solving this will yield the time \( T \) taken to complete one full revolution.

### Final Result
After solving the quadratic equation, we can find the time \( T \) taken to complete the first revolution.