__Question 5.39__:

__Solution__:Mass of the man,

*m*= 70 kg

*r*= 3 m

Coefficient of friction,

*μ*= 0.15

Frequency of rotation, ν = 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

(

*f*=

*μ*

*F*

_{N}) acting upward.

Hence, the man will not fall until:

*m*g <

*f*

*m*g <

*μ*

*F*

_{N}=

*μ*

*mrω*

^{2}

g <

*μ*

*rω*

^{2}

*ω > (g /*μr)

^{1/2}

The minimum angular speed is given as:

*ω*

_{min}=*(g /*μr)

^{1/2}

= ( 10 / (0.15 X 3) )

^{1/2}= 4.71 rad s

^{-1}

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