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A $70 \mathrm{~kg}$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3 \mathrm{~m}$ rotating about its vertical axis with $200 \mathrm{riv} / \mathrm{min}$. 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?
minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
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Verified Answer
Given, $m=70 \mathrm{~kg}, r=3 \mathrm{~m}, n=200 \mathrm{rpm}=200 / 60 \mathrm{rps}$, and $\mu=0.15$
Horizontal forceprovided bythe wall on the man $\mathrm{N}=$ centripetal force $=m r \omega^2$. Frictional force, acting vertically upwards, opposes the weight of the man.
If floor is removed, man would stuck the wall, when $\mathrm{mg}=f$ $ < \mu \mathrm{N}$ i.e. $g < \mu r \omega^2$.
$\therefore$ Minimum angular speed of rotation of the cylinder is $\omega=\sqrt{ }(g / \mu r)=\sqrt{[10 /(0.15 \times 3)}]=4.7 \mathrm{rad} / \mathrm{s}$.
Horizontal forceprovided bythe wall on the man $\mathrm{N}=$ centripetal force $=m r \omega^2$. Frictional force, acting vertically upwards, opposes the weight of the man.
If floor is removed, man would stuck the wall, when $\mathrm{mg}=f$ $ < \mu \mathrm{N}$ i.e. $g < \mu r \omega^2$.
$\therefore$ Minimum angular speed of rotation of the cylinder is $\omega=\sqrt{ }(g / \mu r)=\sqrt{[10 /(0.15 \times 3)}]=4.7 \mathrm{rad} / \mathrm{s}$.
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