A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :

Options

A. `1.6`
B. `4`
C. `15.6`
D. `13.2`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations of motion for circular motion.

### Step 1: Understand the problem
We have a particle moving in a circle of radius \( r = \frac{20}{\pi} \) meters with constant tangential acceleration. The velocity at the end of the second revolution is \( v = 50 \, \text{m/s} \).

### Step 2: Determine the angular displacement
The angular displacement for two revolutions is:
\[ \theta = 2 \times 2\pi = 4\pi \, \text{radians} \]

### Step 3: Relate linear velocity to angular velocity
Using the relationship between linear velocity \( v \) and angular velocity \( \omega \):
\[ v = r \omega \implies \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{50}{\frac{20}{\pi}} = 50 \times \frac{\pi}{20} = \frac{50\pi}{20} = \frac{5\pi}{2} \, \text{rad/s} \]

### Step 4: Use the angular motion equation
We use the equation of motion for angular displacement:
\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]
Here, \( \omega_f = \frac{5\pi}{2} \, \text{rad/s} \), \( \omega_i = 0 \) (since the particle starts from rest), and \( \theta = 4\pi \).

Substituting these values:
\[ \left(\frac{5\pi}{2}\right)^2 = 0 + 2\alpha (4\pi) \]
Calculating \( \left(\frac{5\pi}{2}\right)^2 \):
\[ \frac{25\pi^2}{4} = 8\pi \alpha \]

### Step 5: Solve for angular acceleration \( \alpha \)
Rearranging the equation:
\[ \alpha = \frac{\frac{25\pi^2}{4}}{8\pi} = \frac{25\pi}{32} \]

### Step 6: Convert angular acceleration to linear acceleration
The linear acceleration \( a \) is given by:
\[ a = \alpha \cdot r \]
Substituting \( \alpha \) and \( r \):
\[ a = \left(\frac{25\pi}{32}\right) \cdot \left(\frac{20}{\pi}\right) \]
The \( \pi \) cancels out:
\[ a = \frac{25 \cdot 20}{32} = \frac{500}{32} = 15.625 \, \text{m/s}^2 \]

### Step 7: Final answer
Thus, the tangential acceleration is approximately:
\[ a \approx 15.6 \, \text{m/s}^2 \]

### Summary of Steps
1. Determine the angular displacement for two revolutions.
2. Relate linear velocity to angular velocity.
3. Use the angular motion equation to find angular acceleration.
4. Convert angular acceleration to linear acceleration using the radius.