a 70 kg man stands in contact of against the inner wall of the hollow cylindrical drum of radius 3 m rotating about vertical axis with 200 rev/hr the coefficient of friction between the wall and his clothing is 0.15 what is the minimum rotational speed of cylinder to enable the man to remain stuck the ground when the floor is suddenly removed

Asked by  | 14th Sep, 2012, 09:47: PM

Expert Answer:

the centrifugal force on the man at the time of rotation will be:
F= m?2r
this will be the normal force betwen the cylinder's wall and the man's clothing.
thus the friction will be: 0.15F
this should be equal to the man's weight to prevent him from falling
thus mg = m?2r
?=(g/r)1/2= (10/3) = 1.83 rad/s

Answered by  | 15th Sep, 2012, 02:03: PM

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