A person stands in contact against a wall of cylindrical drum of radius r. , rotating with an angular velocity $ω$ . If $μ$ is the coefficient of friction between the wall and the person, the minimum rotational speed which enables the person to remain stuck to the wall is

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$\sqrt{\frac{g}{μ r}}$

$\frac{g}{\sqrt{μ r}}$

$\sqrt{\frac{μ r}{g}}$

$\sqrt{\frac{g r}{μ}}$

answer is A.

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Detailed Solution

The person will remain stuck to the wall if the static force of friction μMrω2 is equal to or more than the weight Mg
Hence, μMrω²≥ Mg ⇒ ω² ≥ g/μr
ie ω ≥ $\sqrt{g / μ r }$
so, ω_min = $\sqrt{g / μ r }$

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