A particle moves in a circle of radius 25 cm at two revolutions per sec. The acceleration of the particle in `m//s^(2)` is:

Options

A. `pi^(2)`
B. `8pi^(2)`
C. `4pi^(2)`
D. `2pi^(2)`

Explanation

To find the acceleration of a particle moving in a circle of radius 25 cm at a frequency of 2 revolutions per second, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the radius from centimeters to meters**:
\[ r = 25 \text{ cm} = 0.25 \text{ m} \]

2. **Identify the frequency of the particle**:
The frequency \( f \) is given as 2 revolutions per second.

3. **Calculate the angular velocity (\( \omega \))**:
The angular velocity can be calculated using the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 2 = 4 \pi \text{ radians per second} \]

4. **Calculate the centripetal acceleration (\( a \))**:
The centripetal acceleration can be calculated using the formula:
\[ a = \omega^2 r \]
First, calculate \( \omega^2 \):
\[ \omega^2 = (4 \pi)^2 = 16 \pi^2 \]
Now substitute \( \omega^2 \) and \( r \) into the acceleration formula:
\[ a = 16 \pi^2 \times 0.25 \]
Simplifying this gives:
\[ a = 4 \pi^2 \text{ m/s}^2 \]

5. **Final Result**:
The acceleration of the particle is:
\[ a \approx 39.478 \text{ m/s}^2 \quad (\text{using } \pi^2 \approx 9.87) \]

### Final Answer:
The acceleration of the particle is approximately \( 39.478 \text{ m/s}^2 \).

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