IN a rotor, a hollow verticla cylindrical structure rotates about its axis and a person rests asgainst he inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hangs resting against the wall without any floor. If the radius of the roter is 2m and the coefficient of static frictioin between the wall and theperson is 0.2, find the minimum speed at which the floor may be removed Take `g=10 m/s^2`.

Explanation

To solve the problem, we need to analyze the forces acting on the person who is resting against the inner wall of the rotating cylindrical rotor. The key forces to consider are the gravitational force acting downward and the frictional force acting upward.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The gravitational force acting on the person is \( F_g = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.
- The normal force \( N \) exerted by the wall on the person provides the centripetal force necessary for circular motion. This force is given by \( N = \frac{mv^2}{r} \), where \( v \) is the tangential speed and \( r \) is the radius of the rotor.

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