A particle moves in a circle of radius 5 cm with constant speed and time period `0.2pis`. The acceleration of the particle is

Options

A. `25 ms^(-2)`
B. `36 ms^(-2)`
C. `5 ms^(-2)`
D. `15 ms^(-2)`

Explanation

To solve the problem of finding the acceleration of a particle moving in a circle of radius 5 cm with a time period of \(0.2\pi\) seconds, we can follow these steps:

### Step 1: Identify the given values
- Radius \( r = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Time period \( T = 0.2\pi \, \text{s} \)

### Step 2: Calculate the angular velocity (\( \omega \))
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{0.2\pi} = \frac{2}{0.2} = 10 \, \text{rad/s} \]

### Step 3: Calculate the linear velocity (\( v \))
The linear velocity \( v \) can be calculated using the formula:
\[ v = r \cdot \omega \]
Substituting the values of \( r \) and \( \omega \):
\[ v = 0.05 \, \text{m} \cdot 10 \, \text{rad/s} = 0.5 \, \text{m/s} \]

### Step 4: Calculate the centripetal (radial) acceleration (\( a \))
The centripetal acceleration \( a \) is given by the formula:
\[ a = \frac{v^2}{r} \]
Substituting the values of \( v \) and \( r \):
\[ a = \frac{(0.5 \, \text{m/s})^2}{0.05 \, \text{m}} = \frac{0.25 \, \text{m}^2/\text{s}^2}{0.05 \, \text{m}} = 5 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle is \( 5 \, \text{m/s}^2 \).

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