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InterviewSolution
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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 200 rev/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 ? |
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Answer» Solution :The frictional force f (vertically upwards) opposes the weight MG. The man remains stuck to the wall after the floor is removed if `mg le f_(L) i.e., mg LT MU m R omega^(2)`. The minimum ANGULAR speed of rotation of the cylinder is `omega_(MIN)=sqrt((g)/(mu R))=sqrt((9.8)/(0.15xx3))=4.67 s^(-1)`. |
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