Question

A $70kg$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3m$ rotating about its vertical axis with $200rev/min$. The coefficient of friction between the wall and his clothing is $0.15$. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Answer 1

Mass of the man, $m=70kg$

Radius of the drum, $r=3m$

Coefficient of friction, $\mu$ = $0.15$

Frequency of rotation, $\omega$= $200rev/min=200/60=10/3rev/s$

The necessary centripetal force required for the rotation of the man is provided by the normal force ($F_N$).

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force

($f=$ $F_N$) acting upward.

Hence, the man will not fall until:

$mg<$$F_N=\mu mr{\omega }^2$

$\omega >\sqrt{g/r\mu }$

= ${\left(\frac{10}{(0.15\times 3)}\right)}^{1/2}=4.71rad/s$