InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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? |
|
Answer» 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` |
|