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 radius of rotor is $2m$ and ${\mu }_s=0.2$ what will be the minimum angular velocity at which floor may be removed :-

• A. $2\text{rad/s}$
• B. $5\text{rad/s}$
• C. $10\text{rad/s}$
• D. None of them

Answer 1

The correct option is B $5\text{rad/s}$

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 $\frac{v^2}{r}$ towards the centre.
Newton's law for the horizontal direction (2nd law) and for the vertical direction (1st law) give
$N=m\frac{v^2}{r}...\left(i\right)$
and $f_s=mg...\left(ii\right)$
For the minimum speed when the floor may be removed, the friction is limiting one and so equals ${\mu }_{s}N$. This gives
${\mu }_{s}N=mg$
$\frac{{\mu }_{s}m{v}^2}{r}=mg$
$v=\sqrt{\frac{rg}{{\mu }_0}}=\sqrt{\frac{2m\times 10{\text{m/s}}^2}{0.2}}=10\text{m/s}$
$\omega =\frac{v}{r}=5rad/s$