In a rotor, a hollow vertical cylindrical structure rotates about its axis and a person rests against the 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 rotor is 2m and the coefficient of static friction between the wall and the person 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 downward
b. normal force N due to the wall, towards the centre
c. Frictional force `f_s` parallel to the wall, upward.
The person 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` ...........ii
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`