A flywheel of diameter 1 m is rotating at 600 r.p.m., the acceleration of a point on the rim of the fly wheel is

Options

A. `100 pi ^(2) m//s^(2)`
B. `150 pi ^(2) m//s^(2)`
C. `200 pi ^(2)m//s^(2)`
D. `300 pi ^(2) m//s^(2)`

Explanation

To solve the problem of finding the acceleration of a point on the rim of a flywheel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the flywheel, \( D = 1 \, \text{m} \)
- Rotational speed, \( N = 600 \, \text{r.p.m.} \)

2. **Convert Rotational Speed to SI Units:**
- We need to convert the rotational speed from rotations per minute (r.p.m.) to hertz (Hz), which is rotations per second.
- \( N = \frac{600 \, \text{r.p.m.}}{60} = 10 \, \text{Hz} \)

3. **Calculate the Radius:**
- The radius \( r \) of the flywheel is half of the diameter.
- \( r = \frac{D}{2} = \frac{1 \, \text{m}}{2} = 0.5 \, \text{m} \)

4. **Determine Angular Frequency (\( \omega \)):**
- The angular frequency \( \omega \) in radians per second can be calculated using the formula:
\[ \omega = 2\pi N \]
- Substituting the value of \( N \):
\[ \omega = 2\pi \times 10 = 20\pi \, \text{rad/s} \]

5. **Calculate the Centripetal Acceleration:**
- The formula for centripetal acceleration \( a \) at a point on the rim of the flywheel is given by:
\[ a = r \omega^2 \]
- Substituting the values of \( r \) and \( \omega \):
\[ a = 0.5 \times (20\pi)^2 \]
- Calculating \( (20\pi)^2 \):
\[ (20\pi)^2 = 400\pi^2 \]
- Therefore,
\[ a = 0.5 \times 400\pi^2 = 200\pi^2 \, \text{m/s}^2 \]

6. **Final Result:**
- The acceleration of a point on the rim of the flywheel is:
\[ a = 200\pi^2 \, \text{m/s}^2 \]