Question

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 2 m and the coefficient of static friction between the wall and the person is 0.2, find the minimum speed of rotor at which the floor may be removed. Take g = 10 m/$s^2$.

• A. 2 m/s
• B. 10 m/s
• C. 20 m/s
• D. None of these

Answer 1

The correct option is B

10 m/s

The situation is shown in below figure

When the floor is removed, the forces on the person are

(a) Weight mg downward

(b) Normal force N due to the wall, towards the centre

(c) Friction force $f_s$ , parallel to the wall, upward.

The person is moving in a circle with a uniform speed, so its acceleration is $\frac{v^2}{r}$ towards the centre.

Newton's law forthe horizontal direction (2nd law) and for the vertical direction (1st law) gives

N = $\frac{m{v}^2}{r}$ ....... (i)

and $f_s$ = mg .........(ii)

for the minimum speed when the floor may be removed, the friction is limiting one and so equals ${\mu }_{s}N$. This gives
$f=mg\mu N=mg\frac{\mu m{v}^2}{r}=mg\frac{\mu v^2}{r}=gv=\sqrt{\frac{rg}{\mu }}=\sqrt{\frac{\left(2\right)\left(10\right)}{0.2}}=10m/s$