A 70 kg man stands in contact against the inner wall of a hollow
Question.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/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?

solution:

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction, $\mu=0.15$

Frequency of rotation, $v=200 \mathrm{rev} / \mathrm{min}=\frac{200}{60}=\frac{10}{3} \mathrm{rev} / \mathrm{s}$

The necessary centripetal force required for the rotation of the man is provided by the normal force $\left(F_{N}\right)$.

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man ( $m g$ ) acting downward is balanced by the frictional force $\left(f=\mu F_{N}\right)$ acting upward.

Hence, the man will not fall until:

$m g
$m g<\mu F_{N}=\mu m r \omega^{2}$

Frequency of rotation, $v=200 \mathrm{rev} / \mathrm{min}=\frac{200}{60}=\frac{10}{3} \mathrm{rev} / \mathrm{s}$

The necessary centripetal force required for the rotation of the man is provided by the normal force $\left(F_{N}\right)$.

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man $(m g)$ acting downward is balanced by the frictional force $\left(f=\mu F_{N}\right)$ acting upward.

Hence, the man will not fall until:

$m g
$m g<\mu F_{N}=\mu m r \omega^{2}$

$g<\mu r \omega^{2}$

$\omega>\sqrt{\frac{\mathrm{g}}{\mu r}}$

The minimum angular speed is given as:

$\omega_{\min }=\sqrt{\frac{\mathrm{g}}{\mu r}}$

$=\sqrt{\frac{10}{0.15 \times 3}}=4.71 \mathrm{rad} \mathrm{s}^{-1}$
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