A particle moves along a circle if radius `(20 //pi) m` with constant tangential acceleration. If the velocity of the particle is ` 80 m//s` at the end of the second revolution after motion has begun the tangential acceleration is .

Options

A. `160 pi m//s^(2)`
B. `40 pi m//s^(2)`
C. `80 m//s^(2)`
D. `640 pi m//s^(2)`

Answer

To find the tangential acceleration of the particle moving in a circular path, we can follow these steps:

### Step 1: Identify the given values
- Radius of the circle, \( r = \frac{20}{\pi} \) m
- Final velocity after 2 revolutions, \( v_f = 80 \) m/s
- Initial angular velocity, \( \omega_0 = 0 \) rad/s (since the motion has just begun)

### Step 2: Calculate the angular displacement for 2 revolutions
- One revolution corresponds to \( 2\pi \) radians.
- Therefore, for 2 revolutions:
\[ \theta = 2 \times 2\pi = 4\pi \text{ radians} \]

### Step 3: Calculate the final angular velocity
- The final angular velocity \( \omega_f \) can be calculated using the relationship between linear velocity and angular velocity:
\[ \omega_f = \frac{v_f}{r} \]
Substituting the values:
\[ \omega_f = \frac{80}{\frac{20}{\pi}} = 80 \times \frac{\pi}{20} = 4\pi \text{ rad/s} \]

### Step 4: Use the kinematic equation for angular motion
- We can use the equation:
\[ \omega_f^2 = \omega_0^2 + 2\alpha\theta \]
where \( \alpha \) is the angular acceleration. Substituting the known values:
\[ (4\pi)^2 = 0 + 2\alpha(4\pi) \]
Simplifying this gives:
\[ 16\pi^2 = 8\pi\alpha \]

### Step 5: Solve for angular acceleration \( \alpha \)
- Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{16\pi^2}{8\pi} = 2\pi \text{ rad/s}^2 \]

### Step 6: Calculate the tangential acceleration \( a \)
- The tangential acceleration \( a \) is related to angular acceleration by the formula:
\[ a = \alpha r \]
Substituting the values:
\[ a = (2\pi) \left(\frac{20}{\pi}\right) = 2 \times 20 = 40 \text{ m/s}^2 \]

### Final Answer
The tangential acceleration of the particle is \( 40 \text{ m/s}^2 \).

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