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 (in $rad/s$) of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
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 (in $rad/s$) of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Mass of the man, $m=70 \,kg$
Radius of the drum, $r=3 \,m$
Coefficient of friction, $\mu=0.15$
Frequency of rotation, $v=200$ $rev/min$ $=200 / 60=10 / 3$ $rev/s$
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 ( $m g$ ) acting downward is balanced by the frictional force $\left(f=\mu F_{N}\right)$ acting upward.
Hence, the man will not fall until:
$m g \,<\, f m g \,<\, \mu F_{N}$
$=\mu m r \omega^{2} g \,<\, \mu r \omega^{2}$
$\omega\,>\,\sqrt{\frac{ g }{\mu r}}$
The minimum angular speed is given as:
$\omega_{\min }=\sqrt{\frac{g}{\mu r}}$
$=\sqrt{\frac{10}{0.15 \times 3}}=4.71\, rad \,s ^{-1}$
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