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

The situation is shown in figure below

When the floor is removed, the forces on the person are
(i) weight mg downward.
(ii) normal force N due to the wall towards the centre.
(iii) frictional force `f_(s)` parallel to the wall, upwards.
The person is moving in a circle with a uniform speed, so its acceleration is `v^(2)//r` towards the centre.
For the minimum speed when the floor may be removed, the friction is limiting one and so equals `mu_(s)N`.
Theis gives
`mu_(s)N=mg" or "(mu_(s)mv^(2))/(r)=mg`
or `v=sqrt((rg)/(mu_(s)))=sqrt((2xx10)/(0.2))=10ms^(-1)`