A merry go round has a radius 4 m and completes a revolution in 2 seconds. Then acceleration of a point on its rim is

Options

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

Answer

To find the acceleration of a point on the rim of a merry-go-round, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given parameters**:
- Radius of the merry-go-round, \( r = 4 \, \text{m} \)
- Time period for one complete revolution, \( T = 2 \, \text{s} \)

2. **Calculate the linear velocity \( v \)**:
The formula for linear velocity in circular motion is given by:
\[ v = \frac{2 \pi r}{T} \]
Substituting the values:
\[ v = \frac{2 \pi \times 4}{2} = 4 \pi \, \text{m/s} \]

3. **Calculate the centripetal acceleration \( a_c \)**:
The formula for centripetal acceleration is:
\[ a_c = \frac{v^2}{r} \]
First, calculate \( v^2 \):
\[ v^2 = (4 \pi)^2 = 16 \pi^2 \]
Now substitute \( v^2 \) and \( r \) into the centripetal acceleration formula:
\[ a_c = \frac{16 \pi^2}{4} = 4 \pi^2 \, \text{m/s}^2 \]

4. **Final Result**:
The acceleration of a point on the rim of the merry-go-round is:
\[ a_c = 4 \pi^2 \, \text{m/s}^2 \]