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Question -

A70 kg man stands in contact against the inner wall of a hollow cylindrical drumof radius 3 m rotating about its vertical axis with 200 rev/min. Thecoefficient of friction between the wall and his clothing is 0.15. What is theminimum rotational speed of the cylinder to enable the man to remain stuck tothe wall (without falling) when the floor is suddenly removed?



Answer -

Mass of the man,┬аm┬а= 70 kg

Radius of the drum,┬аr┬а= 3 m

Coefficient of friction,┬а╬╝┬а=0.15

Frequencyof rotation, ╬╜ = 200 rev/min┬а

The necessary centripetal force required forthe rotation of the man is provided by the normal force (FN).

When the floor revolves, the man sticks tothe wall of the drum. Hence, the weight of the man (mg) acting downwardis balanced by the frictional force (f┬а=┬а╬╝FN)acting upward.

Hence,the man will not fall until:

mg <┬аf

mg <┬а╬╝FN┬а=┬а╬╝mr╧Й2

g <┬а╬╝r╧Й2

Theminimum angular speed is given as:

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