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. 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?

• A. $2.7rad{s}^{-1}$
• B. $3.2rad{s}^{-1}$
• C. $4.7rad{s}^{-1}$
• D. $5rad{s}^{-1}$

Answer 1

The correct option is C $4.7rad{s}^{-1}$

Given : $r=3$ m $\mu =0.15$ $m=70kg$

In man's reference frame, his FBD is shown above.

For the man to be in equilibrium, the centrifugal force is balanced by the normal force due to inner wall while the friction force balances the gravitational force.

$\therefore$ $N=m{a}_r=mr{w}^2$

Also $f=mg$

OR $\mu N=mg$

OR $\mu \left(mr{w}^2\right)=mg$ $⟹w=\sqrt{\frac{g}{\mu r}}$

$\therefore$ $w=\sqrt{\frac{9.8}{0.15\times 3}}=4.7$ rad/s