A human body can safely with stand with an acceleration of `10 g m//s^(2)`. What will be the number of revolution that a space traveller can perform on a rotating platform of radius 10 m ?

Options

A. `4.984 rps`
B. `0.4984 rps`
C. `3.142 rps`
D. `49.84 rps`

Explanation

To solve the problem step by step, we will follow the principles of circular motion and centripetal acceleration.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Maximum safe acceleration for a human body, \( a = 10 \, g \, \text{m/s}^2 \)
- Radius of the rotating platform, \( r = 10 \, \text{m} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate the Maximum Centripetal Acceleration:**
\[ a = 10 \, g = 10 \times 9.8 \, \text{m/s}^2 = 98 \, \text{m/s}^2 \]

3. **Use the Formula for Centripetal Acceleration:**
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the tangential velocity and \( r \) is the radius. We can also express \( a_c \) in terms of angular velocity \( \omega \):
\[ a_c = \omega^2 r \]

4. **Set the Maximum Centripetal Acceleration Equal to the Calculated Value:**
\[ 98 = \omega^2 \times 10 \]

5. **Solve for Angular Velocity \( \omega \):**
\[ \omega^2 = \frac{98}{10} = 9.8 \]
\[ \omega = \sqrt{9.8} \approx 3.13 \, \text{rad/s} \]

6. **Convert Angular Velocity to Revolutions per Second:**
To convert from radians per second to revolutions per second, use the conversion factor \( \frac{1 \, \text{revolution}}{2\pi \, \text{radians}} \):
\[ \text{Revolutions per second} = \frac{\omega}{2\pi} = \frac{3.13}{2\pi} \approx 0.498 \, \text{revolutions/second} \]

7. **Final Answer:**
The number of revolutions that a space traveler can perform on the rotating platform is approximately \( 0.498 \, \text{revolutions/second} \).