Question

A person of mass M stands in contact against the wall of a cylindrical drum of radius r rotating with an angular velocity w. If the coefficient of friction between the wall and the man is u. the minimum rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor is removed is ??​

Answer 1

Answer:

$\omega_{min}= \frac{g}{\mu\times r}$

Step-by-step explanation:

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 = μF_N) acting upward.

Hence, the man will not fall until:

mg < f

mg < μF_N = μ×m×r×ω^2

g < μrω^2

ω > (g / μr)^1/2

The minimum angular speed is given as:

ω_min = (g / μr)^1/2

$\omega_{min}= \frac{g}{\mu\times r}$