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 ure.

When the floor removed, the forces on the person are
a. weight mg downwrd
b. normal force N due to the wall, towards the centre
c. frictioN/Al force `f_s` parallel to teh wll, upward.
The perso is moving in a circle with as uniform speed, so its acceleration is `v^2/r` towards the centre.
Newton's law for the horizontal direction (2nd law) and for the vertical direction (1st law) give
`N=mv^2/r`........i
and `f_s=mg` ...........i
for the minimum speed when the floor may be removed, the friction is limiting one and so equals `mu_sN`.This gives
`mu_sN=mg`
or, ` (mu_smv^2)/r=mg[using i]`
or `v=sqrt((rg)/mu_s)=sqrt((2mxx10m/s^2)/0.2)=10 m/s`