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A 70kg man stands in contact against the...

A `70kg` man stands in contact against the inner wall of a hollow cylindrical drum of radius `3m` rotating about its verticle axis. 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?

Text Solution

Verified by Experts

Here, `m = 70 kg, r = 3`
`n = 200"rpm", mu = 0.15, omega_("min") = ?`
The necessary centripetal force `(m upsilon^(2)//r)` is provided by horizontal normal reaction `R` of the wall on the man, i.e..,
`R = (m upsilon^(2))/(r ) = m r omega^(2)`
The frictional force `f` acting upwards, balances the weight `(mg)` of the man. The man will remain struck to the wall after the floor is removed, provided.
`f = mu R,i.e., mg le mu (m r omega^(2))`
or `omega^(2) ge (g)/(mu r)`
`:. omega_(min) = sqrt((g)/(mu r)) = sqrt((9.8)/(0.15 xx 3))`
`= sqrt(21.77) = 4.67 rad//s`
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Knowledge Check

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