Laws of Motion
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Question : 39
Total: 40
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 with 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?
Solution:
The cylinder being vertical, the normal reaction of the wall on the man acts horizontally and provides the necessary centripetal force
∴ R = m r ω 2 . . . ( i )
The frictional force f, acting upwards balances his weight i.e.
f = m g . . . ( i i )
The man will remain stuck to the wall after the floor is removed i.e. he will continue to rotate with the cylinder without slipping
if µ ≥
f ≤ µ R
orm g ≤ µ m r ω 2
orω 2 ≥
ω ≥ √
∴ The minimum angular speed of rotation of the cylindrical drum is given by
ω min = √
Givenm = 0.15 , r = 3 m , g = 10 m s – 2
∴ from (iv) and (v), we get= 4.71 s – 1 = 5 s – 1 .
The frictional force f, acting upwards balances his weight i.e.
The man will remain stuck to the wall after the floor is removed i.e. he will continue to rotate with the cylinder without slipping
if
or
or
∴ The minimum angular speed of rotation of the cylindrical drum is given by
Given
∴ from (iv) and (v), we get
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